3.123 \(\int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=138 \[ \frac {\left (a^2-b^2\right ) \sin (c+d x)}{d \left (a^2+b^2\right )^2}+\frac {2 a b \cos (c+d x)}{d \left (a^2+b^2\right )^2}-\frac {3 a b^2 \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{5/2}}-\frac {b^3}{d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))} \]

[Out]

-3*a*b^2*arctanh((b*cos(d*x+c)-a*sin(d*x+c))/(a^2+b^2)^(1/2))/(a^2+b^2)^(5/2)/d+2*a*b*cos(d*x+c)/(a^2+b^2)^2/d
+(a^2-b^2)*sin(d*x+c)/(a^2+b^2)^2/d-b^3/(a^2+b^2)^2/d/(a*cos(d*x+c)+b*sin(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 1.05, antiderivative size = 231, normalized size of antiderivative = 1.67, number of steps used = 11, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6742, 639, 203, 638, 618, 206} \[ -\frac {2 b^3 \left (a+b \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a d \left (a^2+b^2\right )^2 \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 \left (\left (a^2-b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )+2 a b\right )}{d \left (a^2+b^2\right )^2 \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )}+\frac {2 b^4 \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a d \left (a^2+b^2\right )^{5/2}}-\frac {2 b^2 \left (3 a^2+b^2\right ) \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a d \left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

(2*b^4*ArcTanh[(b - a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a*(a^2 + b^2)^(5/2)*d) - (2*b^2*(3*a^2 + b^2)*ArcTa
nh[(b - a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a*(a^2 + b^2)^(5/2)*d) + (2*(2*a*b + (a^2 - b^2)*Tan[(c + d*x)/
2]))/((a^2 + b^2)^2*d*(1 + Tan[(c + d*x)/2]^2)) - (2*b^3*(a + b*Tan[(c + d*x)/2]))/(a*(a^2 + b^2)^2*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 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 ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^3}{\left (1+x^2\right )^2 \left (a+2 b x-a x^2\right )^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{d}\\ &=\frac {2 \operatorname {Subst}\left (\int \left (\frac {2 \left (a^2-b^2-2 a b x\right )}{\left (a^2+b^2\right )^2 \left (1+x^2\right )^2}+\frac {-a^2+b^2}{\left (a^2+b^2\right )^2 \left (1+x^2\right )}-\frac {2 b^3 x}{a \left (a^2+b^2\right ) \left (-a-2 b x+a x^2\right )^2}-\frac {b^2 \left (3 a^2+b^2\right )}{a \left (a^2+b^2\right )^2 \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^2-b^2-2 a b 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 )^2 d}-\frac {\left (2 \left (a^2-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 )^2 d}-\frac {\left (4 b^3\right ) \operatorname {Subst}\left (\int \frac {x}{\left (-a-2 b x+a x^2\right )^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a \left (a^2+b^2\right ) d}-\frac {\left (2 b^2 \left (3 a^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 \left (a^2+b^2\right )^2 d}\\ &=-\frac {\left (a^2-b^2\right ) x}{\left (a^2+b^2\right )^2}+\frac {2 \left (2 a b+\left (a^2-b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^2 d \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}-\frac {2 b^3 \left (a+b \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a \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 )}+\frac {\left (2 b^4\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 \left (a^2+b^2\right )^2 d}+\frac {\left (2 \left (a^2-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 )^2 d}+\frac {\left (4 b^2 \left (3 a^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 \left (a^2+b^2\right )^2 d}\\ &=-\frac {2 b^2 \left (3 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 )}{a \left (a^2+b^2\right )^{5/2} d}+\frac {2 \left (2 a b+\left (a^2-b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^2 d \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}-\frac {2 b^3 \left (a+b \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a \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 )}-\frac {\left (4 b^4\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 \left (a^2+b^2\right )^2 d}\\ &=\frac {2 b^4 \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right )^{5/2} d}-\frac {2 b^2 \left (3 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 )}{a \left (a^2+b^2\right )^{5/2} d}+\frac {2 \left (2 a b+\left (a^2-b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^2 d \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}-\frac {2 b^3 \left (a+b \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a \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 )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.76, size = 130, normalized size = 0.94 \[ \frac {\frac {a \left (a^2+b^2\right ) \sin (2 (c+d x))+b \left (a^2+b^2\right ) \cos (2 (c+d x))+3 b \left (a^2-b^2\right )}{\left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))}+\frac {12 a b^2 \tanh ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

((12*a*b^2*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) + (3*b*(a^2 - b^2) + b*(a^2 +
b^2)*Cos[2*(c + d*x)] + a*(a^2 + b^2)*Sin[2*(c + d*x)])/((a^2 + b^2)^2*(a*Cos[c + d*x] + b*Sin[c + d*x])))/(2
*d)

________________________________________________________________________________________

fricas [B]  time = 0.57, size = 302, normalized size = 2.19 \[ \frac {2 \, a^{4} b - 2 \, a^{2} b^{3} - 4 \, b^{5} + 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, {\left (a^{2} b^{2} \cos \left (d x + c\right ) + a b^{3} \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (-\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right )}{2 \, {\left ({\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} d \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/2*(2*a^4*b - 2*a^2*b^3 - 4*b^5 + 2*(a^4*b + 2*a^2*b^3 + b^5)*cos(d*x + c)^2 + 2*(a^5 + 2*a^3*b^2 + a*b^4)*co
s(d*x + c)*sin(d*x + c) + 3*(a^2*b^2*cos(d*x + c) + a*b^3*sin(d*x + c))*sqrt(a^2 + b^2)*log(-(2*a*b*cos(d*x +
c)*sin(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)))/((a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b
^6)*d*cos(d*x + c) + (a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*d*sin(d*x + c))

________________________________________________________________________________________

giac [B]  time = 0.32, size = 286, normalized size = 2.07 \[ -\frac {\frac {3 \, a b^{2} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a^{3} b + a b^{3}\right )}}{{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-(3*a*b^2*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*s
qrt(a^2 + b^2)))/((a^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) - 2*(a^4*tan(1/2*d*x + 1/2*c)^3 - a^2*b^2*tan(1/2*d
*x + 1/2*c)^3 + b^4*tan(1/2*d*x + 1/2*c)^3 + 3*a*b^3*tan(1/2*d*x + 1/2*c)^2 - a^4*tan(1/2*d*x + 1/2*c) - 3*a^2
*b^2*tan(1/2*d*x + 1/2*c) + b^4*tan(1/2*d*x + 1/2*c) - 2*a^3*b + a*b^3)/((a^5 + 2*a^3*b^2 + a*b^4)*(a*tan(1/2*
d*x + 1/2*c)^4 - 2*b*tan(1/2*d*x + 1/2*c)^3 - 2*b*tan(1/2*d*x + 1/2*c) - a)))/d

________________________________________________________________________________________

maple [A]  time = 0.24, size = 172, normalized size = 1.25 \[ \frac {-\frac {2 \left (\left (-a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 a b \right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 b^{2} \left (\frac {-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-b}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a}-\frac {3 a \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^2,x)

[Out]

1/d*(-2/(a^4+2*a^2*b^2+b^4)*((-a^2+b^2)*tan(1/2*d*x+1/2*c)-2*a*b)/(tan(1/2*d*x+1/2*c)^2+1)-2*b^2/(a^2+b^2)^2*(
(-b^2/a*tan(1/2*d*x+1/2*c)-b)/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)-3*a/(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 = 2.55, size = 348, normalized size = 2.52 \[ -\frac {\frac {3 \, a b^{2} \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (2 \, a^{3} b - a b^{3} - \frac {3 \, a b^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {{\left (a^{4} + 3 \, a^{2} b^{2} - b^{4}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {{\left (a^{4} - a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4} + \frac {2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {{\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-(3*a*b^2*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + sqrt(a^2 + b^2))/(b - a*sin(d*x + c)/(cos(d*x + c) + 1)
- sqrt(a^2 + b^2)))/((a^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) - 2*(2*a^3*b - a*b^3 - 3*a*b^3*sin(d*x + c)^2/(
cos(d*x + c) + 1)^2 + (a^4 + 3*a^2*b^2 - b^4)*sin(d*x + c)/(cos(d*x + c) + 1) - (a^4 - a^2*b^2 + b^4)*sin(d*x
+ c)^3/(cos(d*x + c) + 1)^3)/(a^6 + 2*a^4*b^2 + a^2*b^4 + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*sin(d*x + c)/(cos(d*x
+ c) + 1) + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - (a^6 + 2*a^4*b^2 + a^2*b^4)*si
n(d*x + c)^4/(cos(d*x + c) + 1)^4))/d

________________________________________________________________________________________

mupad [B]  time = 2.81, size = 286, normalized size = 2.07 \[ \frac {\frac {4\,a^2\,b-2\,b^3}{a^4+2\,a^2\,b^2+b^4}-\frac {6\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4+3\,a^2\,b^2-b^4\right )}{a\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^4-2\,a^2\,b^2+2\,b^4\right )}{a\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}-\frac {6\,a\,b^2\,\mathrm {atanh}\left (\frac {a^4\,b+b^5+2\,a^2\,b^3-a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{{\left (a^2+b^2\right )}^{5/2}}\right )}{d\,{\left (a^2+b^2\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/(a*cos(c + d*x) + b*sin(c + d*x))^2,x)

[Out]

((4*a^2*b - 2*b^3)/(a^4 + b^4 + 2*a^2*b^2) - (6*b^3*tan(c/2 + (d*x)/2)^2)/(a^4 + b^4 + 2*a^2*b^2) + (2*tan(c/2
+ (d*x)/2)*(a^4 - b^4 + 3*a^2*b^2))/(a*(a^4 + b^4 + 2*a^2*b^2)) - (tan(c/2 + (d*x)/2)^3*(2*a^4 + 2*b^4 - 2*a^
2*b^2))/(a*(a^4 + b^4 + 2*a^2*b^2)))/(d*(a + 2*b*tan(c/2 + (d*x)/2) - a*tan(c/2 + (d*x)/2)^4 + 2*b*tan(c/2 + (
d*x)/2)^3)) - (6*a*b^2*atanh((a^4*b + b^5 + 2*a^2*b^3 - a*tan(c/2 + (d*x)/2)*(a^4 + b^4 + 2*a^2*b^2))/(a^2 + b
^2)^(5/2)))/(d*(a^2 + b^2)^(5/2))

________________________________________________________________________________________

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)**3/(a*cos(d*x+c)+b*sin(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________