3.1359 \(\int \frac{\sin ^3(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=240 \[ -\frac{a^8 \log (a+b \sin (c+d x))}{b^3 d \left (a^2-b^2\right )^3}-\frac{\left (35 a^2+57 a b+24 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac{\left (35 a^2-57 a b+24 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right )}+\frac{\sec ^2(c+d x) \left (4 b \left (4 a^2-3 b^2\right )-a \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac{a \sin (c+d x)}{b^2 d}-\frac{\sin ^2(c+d x)}{2 b d} \]

[Out]

-((35*a^2 + 57*a*b + 24*b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^3*d) + ((35*a^2 - 57*a*b + 24*b^2)*Log[1 + Sin
[c + d*x]])/(16*(a - b)^3*d) - (a^8*Log[a + b*Sin[c + d*x]])/(b^3*(a^2 - b^2)^3*d) + (a*Sin[c + d*x])/(b^2*d)
- Sin[c + d*x]^2/(2*b*d) - (Sec[c + d*x]^4*(b - a*Sin[c + d*x]))/(4*(a^2 - b^2)*d) + (Sec[c + d*x]^2*(4*b*(4*a
^2 - 3*b^2) - a*(13*a^2 - 9*b^2)*Sin[c + d*x]))/(8*(a^2 - b^2)^2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.617745, antiderivative size = 240, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2837, 12, 1647, 1629} \[ -\frac{a^8 \log (a+b \sin (c+d x))}{b^3 d \left (a^2-b^2\right )^3}-\frac{\left (35 a^2+57 a b+24 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac{\left (35 a^2-57 a b+24 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right )}+\frac{\sec ^2(c+d x) \left (4 b \left (4 a^2-3 b^2\right )-a \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac{a \sin (c+d x)}{b^2 d}-\frac{\sin ^2(c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[c + d*x]^3*Tan[c + d*x]^5)/(a + b*Sin[c + d*x]),x]

[Out]

-((35*a^2 + 57*a*b + 24*b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^3*d) + ((35*a^2 - 57*a*b + 24*b^2)*Log[1 + Sin
[c + d*x]])/(16*(a - b)^3*d) - (a^8*Log[a + b*Sin[c + d*x]])/(b^3*(a^2 - b^2)^3*d) + (a*Sin[c + d*x])/(b^2*d)
- Sin[c + d*x]^2/(2*b*d) - (Sec[c + d*x]^4*(b - a*Sin[c + d*x]))/(4*(a^2 - b^2)*d) + (Sec[c + d*x]^2*(4*b*(4*a
^2 - 3*b^2) - a*(13*a^2 - 9*b^2)*Sin[c + d*x]))/(8*(a^2 - b^2)^2*d)

Rule 2837

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin{align*} \int \frac{\sin ^3(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{x^8}{b^8 (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^8}{(a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=-\frac{\sec ^4(c+d x) \left (\frac{b}{a^2-b^2}-\frac{a \sin (c+d x)}{a^2-b^2}\right )}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{a^2 b^8}{a^2-b^2}+\frac{3 a b^8 x}{a^2-b^2}-4 b^6 x^2-4 b^4 x^4-4 b^2 x^6}{(a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 b^5 d}\\ &=\frac{\sec ^2(c+d x) \left (4 b \left (4 a^2-3 b^2\right )-a \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}-\frac{\sec ^4(c+d x) \left (\frac{b}{a^2-b^2}-\frac{a \sin (c+d x)}{a^2-b^2}\right )}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{\frac{a^2 b^8 \left (11 a^2-7 b^2\right )}{\left (a^2-b^2\right )^2}-\frac{a b^8 \left (13 a^2-9 b^2\right ) x}{\left (a^2-b^2\right )^2}+16 b^6 x^2+8 b^4 x^4}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 b^7 d}\\ &=\frac{\sec ^2(c+d x) \left (4 b \left (4 a^2-3 b^2\right )-a \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}-\frac{\sec ^4(c+d x) \left (\frac{b}{a^2-b^2}-\frac{a \sin (c+d x)}{a^2-b^2}\right )}{4 d}+\frac{\operatorname{Subst}\left (\int \left (8 a b^4+\frac{b^7 \left (35 a^2+57 a b+24 b^2\right )}{2 (a+b)^3 (b-x)}-8 b^4 x-\frac{8 a^8 b^4}{(a-b)^3 (a+b)^3 (a+x)}+\frac{b^7 \left (35 a^2-57 a b+24 b^2\right )}{2 (a-b)^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^7 d}\\ &=-\frac{\left (35 a^2+57 a b+24 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac{\left (35 a^2-57 a b+24 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}-\frac{a^8 \log (a+b \sin (c+d x))}{b^3 \left (a^2-b^2\right )^3 d}+\frac{a \sin (c+d x)}{b^2 d}-\frac{\sin ^2(c+d x)}{2 b d}+\frac{\sec ^2(c+d x) \left (4 b \left (4 a^2-3 b^2\right )-a \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}-\frac{\sec ^4(c+d x) \left (\frac{b}{a^2-b^2}-\frac{a \sin (c+d x)}{a^2-b^2}\right )}{4 d}\\ \end{align*}

Mathematica [A]  time = 3.03721, size = 212, normalized size = 0.88 \[ \frac{-\frac{16 a^8 \log (a+b \sin (c+d x))}{b^3 (a-b)^3 (a+b)^3}-\frac{\left (35 a^2+57 a b+24 b^2\right ) \log (1-\sin (c+d x))}{(a+b)^3}+\frac{\left (35 a^2-57 a b+24 b^2\right ) \log (\sin (c+d x)+1)}{(a-b)^3}+\frac{16 a \sin (c+d x)}{b^2}+\frac{13 a+11 b}{(a+b)^2 (\sin (c+d x)-1)}+\frac{13 a-11 b}{(a-b)^2 (\sin (c+d x)+1)}+\frac{1}{(a+b) (\sin (c+d x)-1)^2}-\frac{1}{(a-b) (\sin (c+d x)+1)^2}-\frac{8 \sin ^2(c+d x)}{b}}{16 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[c + d*x]^3*Tan[c + d*x]^5)/(a + b*Sin[c + d*x]),x]

[Out]

(-(((35*a^2 + 57*a*b + 24*b^2)*Log[1 - Sin[c + d*x]])/(a + b)^3) + ((35*a^2 - 57*a*b + 24*b^2)*Log[1 + Sin[c +
d*x]])/(a - b)^3 - (16*a^8*Log[a + b*Sin[c + d*x]])/((a - b)^3*b^3*(a + b)^3) + 1/((a + b)*(-1 + Sin[c + d*x]
)^2) + (13*a + 11*b)/((a + b)^2*(-1 + Sin[c + d*x])) + (16*a*Sin[c + d*x])/b^2 - (8*Sin[c + d*x]^2)/b - 1/((a
- b)*(1 + Sin[c + d*x])^2) + (13*a - 11*b)/((a - b)^2*(1 + Sin[c + d*x])))/(16*d)

________________________________________________________________________________________

Maple [A]  time = 0.095, size = 338, normalized size = 1.4 \begin{align*} -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,bd}}+{\frac{a\sin \left ( dx+c \right ) }{{b}^{2}d}}-{\frac{{a}^{8}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{b}^{3} \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}}}+{\frac{1}{2\,d \left ( 8\,a+8\,b \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}+{\frac{13\,a}{16\,d \left ( a+b \right ) ^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{11\,b}{16\,d \left ( a+b \right ) ^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{35\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ){a}^{2}}{16\,d \left ( a+b \right ) ^{3}}}-{\frac{57\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) ab}{16\,d \left ( a+b \right ) ^{3}}}-{\frac{3\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ){b}^{2}}{2\,d \left ( a+b \right ) ^{3}}}-{\frac{1}{2\,d \left ( 8\,a-8\,b \right ) \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{13\,a}{16\,d \left ( a-b \right ) ^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{11\,b}{16\,d \left ( a-b \right ) ^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{35\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ){a}^{2}}{16\,d \left ( a-b \right ) ^{3}}}-{\frac{57\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) ab}{16\,d \left ( a-b \right ) ^{3}}}+{\frac{3\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ){b}^{2}}{2\,d \left ( a-b \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^8/(a+b*sin(d*x+c)),x)

[Out]

-1/2*sin(d*x+c)^2/b/d+a*sin(d*x+c)/b^2/d-1/d/b^3*a^8/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c))+1/2/d/(8*a+8*b)/(sin(d
*x+c)-1)^2+13/16/d/(a+b)^2/(sin(d*x+c)-1)*a+11/16/d/(a+b)^2/(sin(d*x+c)-1)*b-35/16/d/(a+b)^3*ln(sin(d*x+c)-1)*
a^2-57/16/d/(a+b)^3*ln(sin(d*x+c)-1)*a*b-3/2/d/(a+b)^3*ln(sin(d*x+c)-1)*b^2-1/2/d/(8*a-8*b)/(1+sin(d*x+c))^2+1
3/16/d/(a-b)^2/(1+sin(d*x+c))*a-11/16/d/(a-b)^2/(1+sin(d*x+c))*b+35/16/d/(a-b)^3*ln(1+sin(d*x+c))*a^2-57/16/d/
(a-b)^3*ln(1+sin(d*x+c))*a*b+3/2/d/(a-b)^3*ln(1+sin(d*x+c))*b^2

________________________________________________________________________________________

Maxima [A]  time = 1.03944, size = 427, normalized size = 1.78 \begin{align*} -\frac{\frac{16 \, a^{8} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} b^{3} - 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} - b^{9}} - \frac{{\left (35 \, a^{2} - 57 \, a b + 24 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{{\left (35 \, a^{2} + 57 \, a b + 24 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac{2 \,{\left ({\left (13 \, a^{3} - 9 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} + 14 \, a^{2} b - 10 \, b^{3} - 4 \,{\left (4 \, a^{2} b - 3 \, b^{3}\right )} \sin \left (d x + c\right )^{2} -{\left (11 \, a^{3} - 7 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}} + \frac{8 \,{\left (b \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )\right )}}{b^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^8/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/16*(16*a^8*log(b*sin(d*x + c) + a)/(a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9) - (35*a^2 - 57*a*b + 24*b^2)*log
(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (35*a^2 + 57*a*b + 24*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3
*a^2*b + 3*a*b^2 + b^3) - 2*((13*a^3 - 9*a*b^2)*sin(d*x + c)^3 + 14*a^2*b - 10*b^3 - 4*(4*a^2*b - 3*b^3)*sin(d
*x + c)^2 - (11*a^3 - 7*a*b^2)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 -
2*(a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^2) + 8*(b*sin(d*x + c)^2 - 2*a*sin(d*x + c))/b^2)/d

________________________________________________________________________________________

Fricas [A]  time = 3.75418, size = 946, normalized size = 3.94 \begin{align*} -\frac{16 \, a^{8} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, a^{4} b^{4} - 8 \, a^{2} b^{6} + 4 \, b^{8} - 8 \,{\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (d x + c\right )^{6} -{\left (35 \, a^{5} b^{3} + 48 \, a^{4} b^{4} - 42 \, a^{3} b^{5} - 64 \, a^{2} b^{6} + 15 \, a b^{7} + 24 \, b^{8}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (35 \, a^{5} b^{3} - 48 \, a^{4} b^{4} - 42 \, a^{3} b^{5} + 64 \, a^{2} b^{6} + 15 \, a b^{7} - 24 \, b^{8}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \,{\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (d x + c\right )^{4} - 8 \,{\left (4 \, a^{4} b^{4} - 7 \, a^{2} b^{6} + 3 \, b^{8}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (2 \, a^{5} b^{3} - 4 \, a^{3} b^{5} + 2 \, a b^{7} + 8 \,{\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \cos \left (d x + c\right )^{4} -{\left (13 \, a^{5} b^{3} - 22 \, a^{3} b^{5} + 9 \, a b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \,{\left (a^{6} b^{3} - 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} - b^{9}\right )} d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^8/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/16*(16*a^8*cos(d*x + c)^4*log(b*sin(d*x + c) + a) + 4*a^4*b^4 - 8*a^2*b^6 + 4*b^8 - 8*(a^6*b^2 - 3*a^4*b^4
+ 3*a^2*b^6 - b^8)*cos(d*x + c)^6 - (35*a^5*b^3 + 48*a^4*b^4 - 42*a^3*b^5 - 64*a^2*b^6 + 15*a*b^7 + 24*b^8)*co
s(d*x + c)^4*log(sin(d*x + c) + 1) + (35*a^5*b^3 - 48*a^4*b^4 - 42*a^3*b^5 + 64*a^2*b^6 + 15*a*b^7 - 24*b^8)*c
os(d*x + c)^4*log(-sin(d*x + c) + 1) + 4*(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(d*x + c)^4 - 8*(4*a^4*b^4
- 7*a^2*b^6 + 3*b^8)*cos(d*x + c)^2 - 2*(2*a^5*b^3 - 4*a^3*b^5 + 2*a*b^7 + 8*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 -
a*b^7)*cos(d*x + c)^4 - (13*a^5*b^3 - 22*a^3*b^5 + 9*a*b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^6*b^3 - 3*a^4*b
^5 + 3*a^2*b^7 - b^9)*d*cos(d*x + c)^4)

________________________________________________________________________________________

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(sec(d*x+c)**5*sin(d*x+c)**8/(a+b*sin(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.26324, size = 544, normalized size = 2.27 \begin{align*} -\frac{\frac{16 \, a^{8} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b^{3} - 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} - b^{9}} - \frac{{\left (35 \, a^{2} - 57 \, a b + 24 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{{\left (35 \, a^{2} + 57 \, a b + 24 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac{8 \,{\left (b \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )\right )}}{b^{2}} + \frac{2 \,{\left (36 \, a^{4} b \sin \left (d x + c\right )^{4} - 48 \, a^{2} b^{3} \sin \left (d x + c\right )^{4} + 18 \, b^{5} \sin \left (d x + c\right )^{4} - 13 \, a^{5} \sin \left (d x + c\right )^{3} + 22 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} - 9 \, a b^{4} \sin \left (d x + c\right )^{3} - 56 \, a^{4} b \sin \left (d x + c\right )^{2} + 68 \, a^{2} b^{3} \sin \left (d x + c\right )^{2} - 24 \, b^{5} \sin \left (d x + c\right )^{2} + 11 \, a^{5} \sin \left (d x + c\right ) - 18 \, a^{3} b^{2} \sin \left (d x + c\right ) + 7 \, a b^{4} \sin \left (d x + c\right ) + 22 \, a^{4} b - 24 \, a^{2} b^{3} + 8 \, b^{5}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^8/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/16*(16*a^8*log(abs(b*sin(d*x + c) + a))/(a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9) - (35*a^2 - 57*a*b + 24*b^2
)*log(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (35*a^2 + 57*a*b + 24*b^2)*log(abs(sin(d*x + c)
- 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 8*(b*sin(d*x + c)^2 - 2*a*sin(d*x + c))/b^2 + 2*(36*a^4*b*sin(d*x + c
)^4 - 48*a^2*b^3*sin(d*x + c)^4 + 18*b^5*sin(d*x + c)^4 - 13*a^5*sin(d*x + c)^3 + 22*a^3*b^2*sin(d*x + c)^3 -
9*a*b^4*sin(d*x + c)^3 - 56*a^4*b*sin(d*x + c)^2 + 68*a^2*b^3*sin(d*x + c)^2 - 24*b^5*sin(d*x + c)^2 + 11*a^5*
sin(d*x + c) - 18*a^3*b^2*sin(d*x + c) + 7*a*b^4*sin(d*x + c) + 22*a^4*b - 24*a^2*b^3 + 8*b^5)/((a^6 - 3*a^4*b
^2 + 3*a^2*b^4 - b^6)*(sin(d*x + c)^2 - 1)^2))/d