3.131 \(\int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=216 \[ \frac {a \left (a^2-3 b^2\right ) \sin (c+d x)}{d \left (a^2+b^2\right )^3}+\frac {b \left (3 a^2-b^2\right ) \cos (c+d x)}{d \left (a^2+b^2\right )^3}-\frac {3 b^2 \left (4 a^2-b^2\right ) \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{7/2}}+\frac {b^4 \sin (c+d x)}{2 a d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^2}-\frac {b^3 \left (8 a^2+b^2\right )}{2 a d \left (a^2+b^2\right )^3 (a \cos (c+d x)+b \sin (c+d x))} \]

[Out]

-3*b^2*(4*a^2-b^2)*arctanh((b-a*tan(1/2*d*x+1/2*c))/(a^2+b^2)^(1/2))/(a^2+b^2)^(7/2)/d+b*(3*a^2-b^2)*cos(d*x+c
)/(a^2+b^2)^3/d+a*(a^2-3*b^2)*sin(d*x+c)/(a^2+b^2)^3/d+1/2*b^4*sin(d*x+c)/a/(a^2+b^2)^2/d/(a*cos(d*x+c)+b*sin(
d*x+c))^2-1/2*b^3*(8*a^2+b^2)/a/(a^2+b^2)^3/d/(a*cos(d*x+c)+b*sin(d*x+c))

________________________________________________________________________________________

Rubi [B]  time = 1.74, antiderivative size = 492, normalized size of antiderivative = 2.28, number of steps used = 15, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6742, 639, 203, 638, 618, 206, 614} \[ -\frac {3 b^4 \left (a^2+2 b^2\right ) \left (b-a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 d \left (a^2+b^2\right )^3 \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 b^4 \left (\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )+a b\right )}{a^3 d \left (a^2+b^2\right )^2 \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 b^3 \left (a b \left (3 a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )+2 a^4-b^4\right )}{a^3 d \left (a^2+b^2\right )^3 \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 \left (a \left (a^2-3 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )+b \left (3 a^2-b^2\right )\right )}{d \left (a^2+b^2\right )^3 \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )}-\frac {3 b^4 \left (a^2+2 b^2\right ) \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^2 d \left (a^2+b^2\right )^{7/2}}+\frac {4 b^4 \left (3 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^2 d \left (a^2+b^2\right )^{7/2}}-\frac {2 b^2 \left (3 a^2 b^2+6 a^4+b^4\right ) \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^2 d \left (a^2+b^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

(-3*b^4*(a^2 + 2*b^2)*ArcTanh[(b - a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2*(a^2 + b^2)^(7/2)*d) + (4*b^4*(3
*a^2 + 2*b^2)*ArcTanh[(b - a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2*(a^2 + b^2)^(7/2)*d) - (2*b^2*(6*a^4 + 3
*a^2*b^2 + b^4)*ArcTanh[(b - a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2*(a^2 + b^2)^(7/2)*d) + (2*(b*(3*a^2 -
b^2) + a*(a^2 - 3*b^2)*Tan[(c + d*x)/2]))/((a^2 + b^2)^3*d*(1 + Tan[(c + d*x)/2]^2)) + (2*b^4*(a*b + (a^2 + 2*
b^2)*Tan[(c + d*x)/2]))/(a^3*(a^2 + b^2)^2*d*(a + 2*b*Tan[(c + d*x)/2] - a*Tan[(c + d*x)/2]^2)^2) - (3*b^4*(a^
2 + 2*b^2)*(b - a*Tan[(c + d*x)/2]))/(a^3*(a^2 + b^2)^3*d*(a + 2*b*Tan[(c + d*x)/2] - a*Tan[(c + d*x)/2]^2)) -
(4*b^3*(2*a^4 - b^4 + a*b*(3*a^2 + 2*b^2)*Tan[(c + d*x)/2]))/(a^3*(a^2 + b^2)^3*d*(a + 2*b*Tan[(c + d*x)/2] -
a*Tan[(c + d*x)/2]^2))

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^4}{\left (1+x^2\right )^2 \left (a+2 b x-a x^2\right )^3} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{d}\\ &=\frac {2 \operatorname {Subst}\left (\int \left (\frac {2 \left (a \left (a^2-3 b^2\right )-b \left (3 a^2-b^2\right ) x\right )}{\left (a^2+b^2\right )^3 \left (1+x^2\right )^2}-\frac {a \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^3 \left (1+x^2\right )}+\frac {4 b^3 \left (-b \left (a^2+b^2\right )-a \left (2 a^2+b^2\right ) x\right )}{a^3 \left (a^2+b^2\right )^2 \left (a+2 b x-a x^2\right )^2}-\frac {4 b^4 (a+2 b x)}{a^3 \left (a^2+b^2\right ) \left (-a-2 b x+a x^2\right )^3}-\frac {b^2 \left (6 a^4+3 a^2 b^2+b^4\right )}{a^2 \left (a^2+b^2\right )^3 \left (-a-2 b x+a x^2\right )}\right ) \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{d}\\ &=\frac {4 \operatorname {Subst}\left (\int \frac {a \left (a^2-3 b^2\right )-b \left (3 a^2-b^2\right ) x}{\left (1+x^2\right )^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^3 d}-\frac {\left (2 a \left (a^2-3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^3 d}+\frac {\left (8 b^3\right ) \operatorname {Subst}\left (\int \frac {-b \left (a^2+b^2\right )-a \left (2 a^2+b^2\right ) x}{\left (a+2 b x-a x^2\right )^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^2 d}-\frac {\left (8 b^4\right ) \operatorname {Subst}\left (\int \frac {a+2 b x}{\left (-a-2 b x+a x^2\right )^3} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right ) d}-\frac {\left (2 b^2 \left (6 a^4+3 a^2 b^2+b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-a-2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 \left (a^2+b^2\right )^3 d}\\ &=-\frac {a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^3}+\frac {2 \left (b \left (3 a^2-b^2\right )+a \left (a^2-3 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^3 d \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 b^4 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 b^3 \left (2 a^4-b^4+a b \left (3 a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^3 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\left (2 a \left (a^2-3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^3 d}+\frac {\left (6 b^4 \left (a^2+2 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-a-2 b x+a x^2\right )^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^2 d}-\frac {\left (4 b^4 \left (3 a^2+2 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x-a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 \left (a^2+b^2\right )^3 d}+\frac {\left (4 b^2 \left (6 a^4+3 a^2 b^2+b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,-2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 \left (a^2+b^2\right )^3 d}\\ &=-\frac {2 b^2 \left (6 a^4+3 a^2 b^2+b^4\right ) \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^2 \left (a^2+b^2\right )^{7/2} d}+\frac {2 \left (b \left (3 a^2-b^2\right )+a \left (a^2-3 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^3 d \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 b^4 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {3 b^4 \left (a^2+2 b^2\right ) \left (b-a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^3 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}-\frac {4 b^3 \left (2 a^4-b^4+a b \left (3 a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^3 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}-\frac {\left (3 b^4 \left (a^2+2 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-a-2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 \left (a^2+b^2\right )^3 d}+\frac {\left (8 b^4 \left (3 a^2+2 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 \left (a^2+b^2\right )^3 d}\\ &=\frac {4 b^4 \left (3 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^2 \left (a^2+b^2\right )^{7/2} d}-\frac {2 b^2 \left (6 a^4+3 a^2 b^2+b^4\right ) \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^2 \left (a^2+b^2\right )^{7/2} d}+\frac {2 \left (b \left (3 a^2-b^2\right )+a \left (a^2-3 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^3 d \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 b^4 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {3 b^4 \left (a^2+2 b^2\right ) \left (b-a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^3 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}-\frac {4 b^3 \left (2 a^4-b^4+a b \left (3 a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^3 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\left (6 b^4 \left (a^2+2 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,-2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 \left (a^2+b^2\right )^3 d}\\ &=-\frac {3 b^4 \left (a^2+2 b^2\right ) \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^2 \left (a^2+b^2\right )^{7/2} d}+\frac {4 b^4 \left (3 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^2 \left (a^2+b^2\right )^{7/2} d}-\frac {2 b^2 \left (6 a^4+3 a^2 b^2+b^4\right ) \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^2 \left (a^2+b^2\right )^{7/2} d}+\frac {2 \left (b \left (3 a^2-b^2\right )+a \left (a^2-3 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^3 d \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 b^4 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {3 b^4 \left (a^2+2 b^2\right ) \left (b-a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^3 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}-\frac {4 b^3 \left (2 a^4-b^4+a b \left (3 a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^3 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 1.12, size = 211, normalized size = 0.98 \[ \frac {\frac {2 a \left (a^2-3 b^2\right ) \sin (c+d x)}{\left (a^2+b^2\right )^3}-\frac {2 b \left (b^2-3 a^2\right ) \cos (c+d x)}{\left (a^2+b^2\right )^3}-\frac {6 b^2 \left (b^2-4 a^2\right ) \tanh ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}}-\frac {b^3 \left (8 a^2+b^2\right )}{a \left (a^2+b^2\right )^3 (a \cos (c+d x)+b \sin (c+d x))}+\frac {b^4 \sin (c+d x)}{a (a-i b)^2 (a+i b)^2 (a \cos (c+d x)+b \sin (c+d x))^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

((-6*b^2*(-4*a^2 + b^2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(7/2) - (2*b*(-3*a^2 +
b^2)*Cos[c + d*x])/(a^2 + b^2)^3 + (2*a*(a^2 - 3*b^2)*Sin[c + d*x])/(a^2 + b^2)^3 + (b^4*Sin[c + d*x])/(a*(a
- I*b)^2*(a + I*b)^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2) - (b^3*(8*a^2 + b^2))/(a*(a^2 + b^2)^3*(a*Cos[c + d*
x] + b*Sin[c + d*x])))/(2*d)

________________________________________________________________________________________

fricas [B]  time = 0.57, size = 480, normalized size = 2.22 \[ \frac {4 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, a^{2} b^{4} - b^{6} + {\left (4 \, a^{4} b^{2} - 5 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (4 \, a^{3} b^{3} - a b^{5}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 2 \, {\left (4 \, a^{6} b - 10 \, a^{4} b^{3} - 17 \, a^{2} b^{5} - 3 \, b^{7}\right )} \cos \left (d x + c\right ) + 2 \, {\left (2 \, a^{5} b^{2} - 11 \, a^{3} b^{4} - 13 \, a b^{6} + 2 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{10} + 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} - 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} - b^{10}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{9} b + 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} + 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{8} b^{2} + 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} + 4 \, a^{2} b^{8} + b^{10}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(4*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cos(d*x + c)^3 - 3*(4*a^2*b^4 - b^6 + (4*a^4*b^2 - 5*a^2*b^4 + b^
6)*cos(d*x + c)^2 + 2*(4*a^3*b^3 - a*b^5)*cos(d*x + c)*sin(d*x + c))*sqrt(a^2 + b^2)*log((2*a*b*cos(d*x + c)*s
in(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 - 2*a^2 - b^2 - 2*sqrt(a^2 + b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/
(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2)) + 2*(4*a^6*b - 10*a^4*b^3 - 17*a^2*b^5 -
3*b^7)*cos(d*x + c) + 2*(2*a^5*b^2 - 11*a^3*b^4 - 13*a*b^6 + 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cos(d*x
+ c)^2)*sin(d*x + c))/((a^10 + 3*a^8*b^2 + 2*a^6*b^4 - 2*a^4*b^6 - 3*a^2*b^8 - b^10)*d*cos(d*x + c)^2 + 2*(a^9
*b + 4*a^7*b^3 + 6*a^5*b^5 + 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)*sin(d*x + c) + (a^8*b^2 + 4*a^6*b^4 + 6*a^4*b^6
+ 4*a^2*b^8 + b^10)*d)

________________________________________________________________________________________

giac [A]  time = 0.99, size = 399, normalized size = 1.85 \[ -\frac {\frac {3 \, {\left (4 \, a^{2} b^{2} - b^{4}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} - \frac {4 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2} b - b^{3}\right )}}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}} - \frac {2 \, {\left (9 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 23 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, a^{4} b^{3} - a^{2} b^{5}\right )}}{{\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(3*(4*a^2*b^2 - b^4)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/
2*c) - 2*b + 2*sqrt(a^2 + b^2)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sqrt(a^2 + b^2)) - 4*(a^3*tan(1/2*d*x +
1/2*c) - 3*a*b^2*tan(1/2*d*x + 1/2*c) + 3*a^2*b - b^3)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*(tan(1/2*d*x + 1/2
*c)^2 + 1)) - 2*(9*a^3*b^4*tan(1/2*d*x + 1/2*c)^3 + 2*a*b^6*tan(1/2*d*x + 1/2*c)^3 + 8*a^4*b^3*tan(1/2*d*x + 1
/2*c)^2 - 15*a^2*b^5*tan(1/2*d*x + 1/2*c)^2 - 2*b^7*tan(1/2*d*x + 1/2*c)^2 - 23*a^3*b^4*tan(1/2*d*x + 1/2*c) -
2*a*b^6*tan(1/2*d*x + 1/2*c) - 8*a^4*b^3 - a^2*b^5)/((a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6)*(a*tan(1/2*d*x +
1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)^2))/d

________________________________________________________________________________________

maple [A]  time = 0.27, size = 283, normalized size = 1.31 \[ \frac {-\frac {2 \left (\left (-a^{3}+3 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3 a^{2} b +b^{3}\right )}{\left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 b^{2} \left (\frac {-\frac {b^{2} \left (9 a^{2}+2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {b \left (8 a^{4}-15 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2}}+\frac {b^{2} \left (23 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}+4 a^{2} b +\frac {b^{3}}{2}}{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )^{2}}-\frac {3 \left (4 a^{2}-b^{2}\right ) \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

1/d*(-2/(a^6+3*a^4*b^2+3*a^2*b^4+b^6)*((-a^3+3*a*b^2)*tan(1/2*d*x+1/2*c)-3*a^2*b+b^3)/(tan(1/2*d*x+1/2*c)^2+1)
-2*b^2/(a^2+b^2)^3*((-1/2*b^2*(9*a^2+2*b^2)/a*tan(1/2*d*x+1/2*c)^3-1/2*b*(8*a^4-15*a^2*b^2-2*b^4)/a^2*tan(1/2*
d*x+1/2*c)^2+1/2*b^2*(23*a^2+2*b^2)/a*tan(1/2*d*x+1/2*c)+4*a^2*b+1/2*b^3)/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*
d*x+1/2*c)-a)^2-3/2*(4*a^2-b^2)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))))

________________________________________________________________________________________

maxima [B]  time = 0.44, size = 658, normalized size = 3.05 \[ -\frac {\frac {3 \, {\left (4 \, a^{2} b^{2} - b^{4}\right )} \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (6 \, a^{6} b - 10 \, a^{4} b^{3} - a^{2} b^{5} + \frac {{\left (2 \, a^{7} + 18 \, a^{5} b^{2} - 31 \, a^{3} b^{4} - 2 \, a b^{6}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, {\left (2 \, a^{6} b - 2 \, a^{4} b^{3} + 12 \, a^{2} b^{5} + b^{7}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, {\left (2 \, a^{7} + 2 \, a^{5} b^{2} + 15 \, a^{3} b^{4}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {{\left (2 \, a^{6} b - 30 \, a^{4} b^{3} + 15 \, a^{2} b^{5} + 2 \, b^{7}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {{\left (2 \, a^{7} - 6 \, a^{5} b^{2} + 9 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{10} + 3 \, a^{8} b^{2} + 3 \, a^{6} b^{4} + a^{4} b^{6} + \frac {4 \, {\left (a^{9} b + 3 \, a^{7} b^{3} + 3 \, a^{5} b^{5} + a^{3} b^{7}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {{\left (a^{10} - a^{8} b^{2} - 9 \, a^{6} b^{4} - 11 \, a^{4} b^{6} - 4 \, a^{2} b^{8}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {{\left (a^{10} - a^{8} b^{2} - 9 \, a^{6} b^{4} - 11 \, a^{4} b^{6} - 4 \, a^{2} b^{8}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 \, {\left (a^{9} b + 3 \, a^{7} b^{3} + 3 \, a^{5} b^{5} + a^{3} b^{7}\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {{\left (a^{10} + 3 \, a^{8} b^{2} + 3 \, a^{6} b^{4} + a^{4} b^{6}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*(3*(4*a^2*b^2 - b^4)*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + sqrt(a^2 + b^2))/(b - a*sin(d*x + c)/(c
os(d*x + c) + 1) - sqrt(a^2 + b^2)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sqrt(a^2 + b^2)) - 2*(6*a^6*b - 10*a
^4*b^3 - a^2*b^5 + (2*a^7 + 18*a^5*b^2 - 31*a^3*b^4 - 2*a*b^6)*sin(d*x + c)/(cos(d*x + c) + 1) - 2*(2*a^6*b -
2*a^4*b^3 + 12*a^2*b^5 + b^7)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 2*(2*a^7 + 2*a^5*b^2 + 15*a^3*b^4)*sin(d*x
+ c)^3/(cos(d*x + c) + 1)^3 - (2*a^6*b - 30*a^4*b^3 + 15*a^2*b^5 + 2*b^7)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
+ (2*a^7 - 6*a^5*b^2 + 9*a^3*b^4 + 2*a*b^6)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a^10 + 3*a^8*b^2 + 3*a^6*b^
4 + a^4*b^6 + 4*(a^9*b + 3*a^7*b^3 + 3*a^5*b^5 + a^3*b^7)*sin(d*x + c)/(cos(d*x + c) + 1) - (a^10 - a^8*b^2 -
9*a^6*b^4 - 11*a^4*b^6 - 4*a^2*b^8)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - (a^10 - a^8*b^2 - 9*a^6*b^4 - 11*a^4
*b^6 - 4*a^2*b^8)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*(a^9*b + 3*a^7*b^3 + 3*a^5*b^5 + a^3*b^7)*sin(d*x +
c)^5/(cos(d*x + c) + 1)^5 + (a^10 + 3*a^8*b^2 + 3*a^6*b^4 + a^4*b^6)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6))/d

________________________________________________________________________________________

mupad [B]  time = 4.25, size = 610, normalized size = 2.82 \[ -\frac {\frac {-6\,a^4\,b+10\,a^2\,b^3+b^5}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^5+2\,a^3\,b^2+15\,a\,b^4\right )}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^6\,b-30\,a^4\,b^3+15\,a^2\,b^5+2\,b^7\right )}{a^2\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^6+18\,a^4\,b^2-31\,a^2\,b^4-2\,b^6\right )}{a\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a^6-6\,a^4\,b^2+9\,a^2\,b^4+2\,b^6\right )}{a\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^6\,b-2\,a^4\,b^3+12\,a^2\,b^5+b^7\right )}{a^2\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+a^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2-4\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-4\,b^2\right )-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {\mathrm {atan}\left (\frac {-1{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^7+a^6\,b\,1{}\mathrm {i}-3{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^5\,b^2+a^4\,b^3\,3{}\mathrm {i}-3{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b^4+a^2\,b^5\,3{}\mathrm {i}-1{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^6+b^7\,1{}\mathrm {i}}{{\left (a^2+b^2\right )}^{7/2}}\right )\,\left (3\,b^4-12\,a^2\,b^2\right )\,1{}\mathrm {i}}{d\,{\left (a^2+b^2\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(a*cos(c + d*x) + b*sin(c + d*x))^3,x)

[Out]

- ((b^5 - 6*a^4*b + 10*a^2*b^3)/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) + (2*tan(c/2 + (d*x)/2)^3*(15*a*b^4 + 2*a^
5 + 2*a^3*b^2))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) + (tan(c/2 + (d*x)/2)^4*(2*a^6*b + 2*b^7 + 15*a^2*b^5 - 30
*a^4*b^3))/(a^2*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) - (tan(c/2 + (d*x)/2)*(2*a^6 - 2*b^6 - 31*a^2*b^4 + 18*a^
4*b^2))/(a*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) - (tan(c/2 + (d*x)/2)^5*(2*a^6 + 2*b^6 + 9*a^2*b^4 - 6*a^4*b^2
))/(a*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)^2*(2*a^6*b + b^7 + 12*a^2*b^5 - 2*a^4*b^3))
/(a^2*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)))/(d*(a^2*tan(c/2 + (d*x)/2)^6 + a^2 - tan(c/2 + (d*x)/2)^2*(a^2 - 4
*b^2) - tan(c/2 + (d*x)/2)^4*(a^2 - 4*b^2) - 4*a*b*tan(c/2 + (d*x)/2)^5 + 4*a*b*tan(c/2 + (d*x)/2))) - (atan((
a^6*b*1i + b^7*1i + a^2*b^5*3i + a^4*b^3*3i - a^7*tan(c/2 + (d*x)/2)*1i - a*b^6*tan(c/2 + (d*x)/2)*1i - a^3*b^
4*tan(c/2 + (d*x)/2)*3i - a^5*b^2*tan(c/2 + (d*x)/2)*3i)/(a^2 + b^2)^(7/2))*(3*b^4 - 12*a^2*b^2)*1i)/(d*(a^2 +
b^2)^(7/2))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4/(a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________