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

[Out]

(-3*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)*d) + (b*(3*a^2 - b
^2)*Cos[c + d*x])/((a^2 + b^2)^3*d) + (a*(a^2 - 3*b^2)*Sin[c + d*x])/((a^2 + b^2)^3*d) + (b^4*Sin[c + d*x])/(2
*a*(a^2 + b^2)^2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^2) - (b^3*(8*a^2 + b^2))/(2*a*(a^2 + b^2)^3*d*(a*Cos[c +
d*x] + b*Sin[c + d*x]))

________________________________________________________________________________________

Rubi [B]  time = 1.74227, antiderivative size = 492, normalized size of antiderivative = 2.28, number of steps used = 15, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 6742

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

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 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 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 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 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]

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 = 1.03486, size = 211, normalized size = 0.98 \[ \frac{\frac{2 a \left (a^2-3 b^2\right ) \sin (c+d x)}{\left (a^2+b^2\right )^3}-\frac{2 b \left (b^2-3 a^2\right ) \cos (c+d x)}{\left (a^2+b^2\right )^3}-\frac{b^3 \left (8 a^2+b^2\right )}{a \left (a^2+b^2\right )^3 (a \cos (c+d x)+b \sin (c+d x))}-\frac{6 b^2 \left (b^2-4 a^2\right ) \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}}+\frac{b^4 \sin (c+d x)}{a (a-i b)^2 (a+i b)^2 (a \cos (c+d x)+b \sin (c+d x))^2}}{2 d} \]

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)

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Maple [A]  time = 0.213, size = 283, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{{b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{3}} \left ({\frac{1}{ \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) ^{2}} \left ( -1/2\,{\frac{{b}^{2} \left ( 9\,{a}^{2}+2\,{b}^{2} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{a}}-1/2\,{\frac{b \left ( 8\,{a}^{4}-15\,{a}^{2}{b}^{2}-2\,{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{{a}^{2}}}+1/2\,{\frac{{b}^{2} \left ( 23\,{a}^{2}+2\,{b}^{2} \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{a}}+4\,{a}^{2}b+1/2\,{b}^{3} \right ) }-3/2\,{\frac{4\,{a}^{2}-{b}^{2}}{\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \right ) }-2\,{\frac{ \left ( -{a}^{3}+3\,a{b}^{2} \right ) \tan \left ( 1/2\,dx+c/2 \right ) -3\,{a}^{2}b+{b}^{3}}{ \left ({a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6} \right ) \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }} \right ) } \end{align*}

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*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*tan(1
/2*d*x+1/2*c)*b-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))
)-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)/(1+tan(1/2*d*x+1/2*c)^2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 0.609099, size = 1073, normalized size = 4.97 \begin{align*} \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 )}} \end{align*}

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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

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Giac [A]  time = 1.37751, size = 539, normalized size = 2.5 \begin{align*} -\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} \end{align*}

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