3.51 \(\int \frac{\sin ^5(c+d x)}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=274 \[ \frac{a^2 b \sin ^3(c+d x)}{3 d \left (a^2+b^2\right )^2}+\frac{b \sin ^5(c+d x)}{5 d \left (a^2+b^2\right )}+\frac{a^4 b \sin (c+d x)}{d \left (a^2+b^2\right )^3}-\frac{a \cos ^5(c+d x)}{5 d \left (a^2+b^2\right )}+\frac{2 a \cos ^3(c+d x)}{3 d \left (a^2+b^2\right )}-\frac{a b^2 \cos ^3(c+d x)}{3 d \left (a^2+b^2\right )^2}+\frac{a^3 b^2 \cos (c+d x)}{d \left (a^2+b^2\right )^3}-\frac{a \cos (c+d x)}{d \left (a^2+b^2\right )}+\frac{a b^2 \cos (c+d x)}{d \left (a^2+b^2\right )^2}+\frac{a^5 b \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{7/2}} \]

[Out]

(a^5*b*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(7/2)*d) + (a^3*b^2*Cos[c + d*
x])/((a^2 + b^2)^3*d) + (a*b^2*Cos[c + d*x])/((a^2 + b^2)^2*d) - (a*Cos[c + d*x])/((a^2 + b^2)*d) - (a*b^2*Cos
[c + d*x]^3)/(3*(a^2 + b^2)^2*d) + (2*a*Cos[c + d*x]^3)/(3*(a^2 + b^2)*d) - (a*Cos[c + d*x]^5)/(5*(a^2 + b^2)*
d) + (a^4*b*Sin[c + d*x])/((a^2 + b^2)^3*d) + (a^2*b*Sin[c + d*x]^3)/(3*(a^2 + b^2)^2*d) + (b*Sin[c + d*x]^5)/
(5*(a^2 + b^2)*d)

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Rubi [A]  time = 0.353902, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3518, 3109, 2564, 30, 2633, 3099, 3074, 206, 2638} \[ \frac{a^2 b \sin ^3(c+d x)}{3 d \left (a^2+b^2\right )^2}+\frac{b \sin ^5(c+d x)}{5 d \left (a^2+b^2\right )}+\frac{a^4 b \sin (c+d x)}{d \left (a^2+b^2\right )^3}-\frac{a \cos ^5(c+d x)}{5 d \left (a^2+b^2\right )}+\frac{2 a \cos ^3(c+d x)}{3 d \left (a^2+b^2\right )}-\frac{a b^2 \cos ^3(c+d x)}{3 d \left (a^2+b^2\right )^2}+\frac{a^3 b^2 \cos (c+d x)}{d \left (a^2+b^2\right )^3}-\frac{a \cos (c+d x)}{d \left (a^2+b^2\right )}+\frac{a b^2 \cos (c+d x)}{d \left (a^2+b^2\right )^2}+\frac{a^5 b \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^5/(a + b*Tan[c + d*x]),x]

[Out]

(a^5*b*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(7/2)*d) + (a^3*b^2*Cos[c + d*
x])/((a^2 + b^2)^3*d) + (a*b^2*Cos[c + d*x])/((a^2 + b^2)^2*d) - (a*Cos[c + d*x])/((a^2 + b^2)*d) - (a*b^2*Cos
[c + d*x]^3)/(3*(a^2 + b^2)^2*d) + (2*a*Cos[c + d*x]^3)/(3*(a^2 + b^2)*d) - (a*Cos[c + d*x]^5)/(5*(a^2 + b^2)*
d) + (a^4*b*Sin[c + d*x])/((a^2 + b^2)^3*d) + (a^2*b*Sin[c + d*x]^3)/(3*(a^2 + b^2)^2*d) + (b*Sin[c + d*x]^5)/
(5*(a^2 + b^2)*d)

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3109

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[b/(a^2 + b^2), Int[Cos[c + d*x]^m*Sin[c + d*x]^(n - 1), x], x] + (Dist[
a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^n, x], x] - Dist[(a*b)/(a^2 + b^2), Int[(Cos[c + d*x]^(m
- 1)*Sin[c + d*x]^(n - 1))/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b
^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 3099

Int[sin[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(a*Sin[c + d*x]^(m - 1))/(d*(a^2 + b^2)*(m - 1)), x] + (Dist[a^2/(a^2 + b^2), Int[Sin[c + d*x]^(m - 2)/
(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x] + Dist[b/(a^2 + b^2), Int[Sin[c + d*x]^(m - 1), x], x]) /; FreeQ[{a,
b, c, d}, x] && NeQ[a^2 + b^2, 0] && GtQ[m, 1]

Rule 3074

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sin ^5(c+d x)}{a+b \tan (c+d x)} \, dx &=\int \frac{\cos (c+d x) \sin ^5(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx\\ &=\frac{a \int \sin ^5(c+d x) \, dx}{a^2+b^2}+\frac{b \int \cos (c+d x) \sin ^4(c+d x) \, dx}{a^2+b^2}-\frac{(a b) \int \frac{\sin ^4(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{a^2+b^2}\\ &=\frac{a^2 b \sin ^3(c+d x)}{3 \left (a^2+b^2\right )^2 d}-\frac{\left (a^3 b\right ) \int \frac{\sin ^2(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{\left (a^2+b^2\right )^2}-\frac{\left (a b^2\right ) \int \sin ^3(c+d x) \, dx}{\left (a^2+b^2\right )^2}-\frac{a \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{\left (a^2+b^2\right ) d}+\frac{b \operatorname{Subst}\left (\int x^4 \, dx,x,\sin (c+d x)\right )}{\left (a^2+b^2\right ) d}\\ &=-\frac{a \cos (c+d x)}{\left (a^2+b^2\right ) d}+\frac{2 a \cos ^3(c+d x)}{3 \left (a^2+b^2\right ) d}-\frac{a \cos ^5(c+d x)}{5 \left (a^2+b^2\right ) d}+\frac{a^4 b \sin (c+d x)}{\left (a^2+b^2\right )^3 d}+\frac{a^2 b \sin ^3(c+d x)}{3 \left (a^2+b^2\right )^2 d}+\frac{b \sin ^5(c+d x)}{5 \left (a^2+b^2\right ) d}-\frac{\left (a^5 b\right ) \int \frac{1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{\left (a^2+b^2\right )^3}-\frac{\left (a^3 b^2\right ) \int \sin (c+d x) \, dx}{\left (a^2+b^2\right )^3}+\frac{\left (a b^2\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{\left (a^2+b^2\right )^2 d}\\ &=\frac{a^3 b^2 \cos (c+d x)}{\left (a^2+b^2\right )^3 d}+\frac{a b^2 \cos (c+d x)}{\left (a^2+b^2\right )^2 d}-\frac{a \cos (c+d x)}{\left (a^2+b^2\right ) d}-\frac{a b^2 \cos ^3(c+d x)}{3 \left (a^2+b^2\right )^2 d}+\frac{2 a \cos ^3(c+d x)}{3 \left (a^2+b^2\right ) d}-\frac{a \cos ^5(c+d x)}{5 \left (a^2+b^2\right ) d}+\frac{a^4 b \sin (c+d x)}{\left (a^2+b^2\right )^3 d}+\frac{a^2 b \sin ^3(c+d x)}{3 \left (a^2+b^2\right )^2 d}+\frac{b \sin ^5(c+d x)}{5 \left (a^2+b^2\right ) d}+\frac{\left (a^5 b\right ) \operatorname{Subst}\left (\int \frac{1}{a^2+b^2-x^2} \, dx,x,b \cos (c+d x)-a \sin (c+d x)\right )}{\left (a^2+b^2\right )^3 d}\\ &=\frac{a^5 b \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2} d}+\frac{a^3 b^2 \cos (c+d x)}{\left (a^2+b^2\right )^3 d}+\frac{a b^2 \cos (c+d x)}{\left (a^2+b^2\right )^2 d}-\frac{a \cos (c+d x)}{\left (a^2+b^2\right ) d}-\frac{a b^2 \cos ^3(c+d x)}{3 \left (a^2+b^2\right )^2 d}+\frac{2 a \cos ^3(c+d x)}{3 \left (a^2+b^2\right ) d}-\frac{a \cos ^5(c+d x)}{5 \left (a^2+b^2\right ) d}+\frac{a^4 b \sin (c+d x)}{\left (a^2+b^2\right )^3 d}+\frac{a^2 b \sin ^3(c+d x)}{3 \left (a^2+b^2\right )^2 d}+\frac{b \sin ^5(c+d x)}{5 \left (a^2+b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 3.12434, size = 289, normalized size = 1.05 \[ \frac{\sqrt{a^2+b^2} \left (120 a^2 b^3 \sin (c+d x)-50 a^2 b^3 \sin (3 (c+d x))+6 a^2 b^3 \sin (5 (c+d x))-6 a^3 b^2 \cos (5 (c+d x))-30 a \left (-4 a^2 b^2+5 a^4-b^4\right ) \cos (c+d x)+5 a \left (6 a^2 b^2+5 a^4+b^4\right ) \cos (3 (c+d x))+330 a^4 b \sin (c+d x)-35 a^4 b \sin (3 (c+d x))+3 a^4 b \sin (5 (c+d x))-3 a^5 \cos (5 (c+d x))-3 a b^4 \cos (5 (c+d x))+30 b^5 \sin (c+d x)-15 b^5 \sin (3 (c+d x))+3 b^5 \sin (5 (c+d x))\right )-480 a^5 b \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{240 d \left (a^2+b^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^5/(a + b*Tan[c + d*x]),x]

[Out]

(-480*a^5*b*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]] + Sqrt[a^2 + b^2]*(-30*a*(5*a^4 - 4*a^2*b^2 - b
^4)*Cos[c + d*x] + 5*a*(5*a^4 + 6*a^2*b^2 + b^4)*Cos[3*(c + d*x)] - 3*a^5*Cos[5*(c + d*x)] - 6*a^3*b^2*Cos[5*(
c + d*x)] - 3*a*b^4*Cos[5*(c + d*x)] + 330*a^4*b*Sin[c + d*x] + 120*a^2*b^3*Sin[c + d*x] + 30*b^5*Sin[c + d*x]
- 35*a^4*b*Sin[3*(c + d*x)] - 50*a^2*b^3*Sin[3*(c + d*x)] - 15*b^5*Sin[3*(c + d*x)] + 3*a^4*b*Sin[5*(c + d*x)
] + 6*a^2*b^3*Sin[5*(c + d*x)] + 3*b^5*Sin[5*(c + d*x)]))/(240*(a^2 + b^2)^(7/2)*d)

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Maple [A]  time = 0.079, size = 361, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{1}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ({a}^{2}+{b}^{2} \right ) \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{5}} \left ( -{a}^{4}b \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{9}-{a}^{3}{b}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+ \left ( -16/3\,{a}^{4}b-4/3\,{a}^{2}{b}^{3} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}+ \left ( -6\,{a}^{3}{b}^{2}-2\,a{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+ \left ( -{\frac{178\,{a}^{4}b}{15}}-{\frac{136\,{a}^{2}{b}^{3}}{15}}-{\frac{16\,{b}^{5}}{5}} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+ \left ( 16/3\,{a}^{5}+2/3\,a{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( -16/3\,{a}^{4}b-4/3\,{a}^{2}{b}^{3} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+ \left ( -2\,{a}^{3}{b}^{2}+8/3\,{a}^{5}-2/3\,a{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-{a}^{4}b\tan \left ( 1/2\,dx+c/2 \right ) +{\frac{8\,{a}^{5}}{15}}-3/5\,{a}^{3}{b}^{2}-2/15\,a{b}^{4} \right ) }-64\,{\frac{{a}^{5}b}{ \left ( 32\,{a}^{6}+96\,{a}^{4}{b}^{2}+96\,{a}^{2}{b}^{4}+32\,{b}^{6} \right ) \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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^5/(a+b*tan(d*x+c)),x)

[Out]

1/d*(-2/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)*(-a^4*b*tan(1/2*d*x+1/2*c)^9-a^3*b^2*tan(1/2*d*x+1/2*c)^8+(-16/3*a^4*b-4
/3*a^2*b^3)*tan(1/2*d*x+1/2*c)^7+(-6*a^3*b^2-2*a*b^4)*tan(1/2*d*x+1/2*c)^6+(-178/15*a^4*b-136/15*a^2*b^3-16/5*
b^5)*tan(1/2*d*x+1/2*c)^5+(16/3*a^5+2/3*a*b^4)*tan(1/2*d*x+1/2*c)^4+(-16/3*a^4*b-4/3*a^2*b^3)*tan(1/2*d*x+1/2*
c)^3+(-2*a^3*b^2+8/3*a^5-2/3*a*b^4)*tan(1/2*d*x+1/2*c)^2-a^4*b*tan(1/2*d*x+1/2*c)+8/15*a^5-3/5*a^3*b^2-2/15*a*
b^4)/(1+tan(1/2*d*x+1/2*c)^2)^5-64*a^5*b/(32*a^6+96*a^4*b^2+96*a^2*b^4+32*b^6)/(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)))

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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(sin(d*x+c)^5/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.3272, size = 840, normalized size = 3.07 \begin{align*} \frac{15 \, \sqrt{a^{2} + b^{2}} a^{5} b \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 ) - 6 \,{\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{5} + 10 \,{\left (2 \, a^{7} + 5 \, a^{5} b^{2} + 4 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} - 30 \,{\left (a^{7} + a^{5} b^{2}\right )} \cos \left (d x + c\right ) + 2 \,{\left (23 \, a^{6} b + 34 \, a^{4} b^{3} + 14 \, a^{2} b^{5} + 3 \, b^{7} + 3 \,{\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{4} -{\left (11 \, a^{6} b + 28 \, a^{4} b^{3} + 23 \, a^{2} b^{5} + 6 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{30 \,{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^5/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/30*(15*sqrt(a^2 + b^2)*a^5*b*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)) - 6*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cos(d*x + c)^5 + 10*(2*a^7 + 5*a^5*b^2 + 4*a^3*b^4 +
a*b^6)*cos(d*x + c)^3 - 30*(a^7 + a^5*b^2)*cos(d*x + c) + 2*(23*a^6*b + 34*a^4*b^3 + 14*a^2*b^5 + 3*b^7 + 3*(
a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cos(d*x + c)^4 - (11*a^6*b + 28*a^4*b^3 + 23*a^2*b^5 + 6*b^7)*cos(d*x + c
)^2)*sin(d*x + c))/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*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(sin(d*x+c)**5/(a+b*tan(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.34169, size = 626, normalized size = 2.28 \begin{align*} \frac{\frac{15 \, a^{5} b \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{2 \,{\left (15 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 15 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 80 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 20 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 90 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 30 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 178 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 136 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 48 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 80 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 10 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 80 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 20 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 30 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 10 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 15 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 8 \, a^{5} + 9 \, a^{3} b^{2} + 2 \, a b^{4}\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 )}^{5}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^5/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

1/15*(15*a^5*b*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)) + 2*(15*a^4*b*tan(1/2*d*x + 1/2*c)
^9 + 15*a^3*b^2*tan(1/2*d*x + 1/2*c)^8 + 80*a^4*b*tan(1/2*d*x + 1/2*c)^7 + 20*a^2*b^3*tan(1/2*d*x + 1/2*c)^7 +
90*a^3*b^2*tan(1/2*d*x + 1/2*c)^6 + 30*a*b^4*tan(1/2*d*x + 1/2*c)^6 + 178*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 136*
a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 48*b^5*tan(1/2*d*x + 1/2*c)^5 - 80*a^5*tan(1/2*d*x + 1/2*c)^4 - 10*a*b^4*tan(
1/2*d*x + 1/2*c)^4 + 80*a^4*b*tan(1/2*d*x + 1/2*c)^3 + 20*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 40*a^5*tan(1/2*d*x
+ 1/2*c)^2 + 30*a^3*b^2*tan(1/2*d*x + 1/2*c)^2 + 10*a*b^4*tan(1/2*d*x + 1/2*c)^2 + 15*a^4*b*tan(1/2*d*x + 1/2*
c) - 8*a^5 + 9*a^3*b^2 + 2*a*b^4)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*(tan(1/2*d*x + 1/2*c)^2 + 1)^5))/d