\(\int \frac {\sin ^8(c+d x)}{(a-b \sin ^4(c+d x))^3} \, dx\) [179]

Optimal result

Integrand size = 24, antiderivative size = 306 \[ \int \frac {\sin ^8(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=-\frac {\left (2 \sqrt {a}-5 \sqrt {b}\right ) \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^{5/2} b^{3/2} d}+\frac {\left (2 \sqrt {a}+5 \sqrt {b}\right ) \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^{5/2} b^{3/2} d}-\frac {a \tan (c+d x) \left (3 a+b+4 (a+b) \tan ^2(c+d x)\right )}{8 (a-b)^3 d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )^2}-\frac {\tan (c+d x) \left (\frac {(a-9 b) (a+b)}{(a-b)^3}+\frac {(a+19 b) \tan ^2(c+d x)}{(a-b)^2}\right )}{32 b d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )} \] Output:

-1/64*(2*a^(1/2)-5*b^(1/2))*arctan((a^(1/2)-b^(1/2))^(1/2)*tan(d*x+c)/a^(1 
/4))/a^(3/4)/(a^(1/2)-b^(1/2))^(5/2)/b^(3/2)/d+1/64*(2*a^(1/2)+5*b^(1/2))* 
arctan((a^(1/2)+b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))/a^(3/4)/(a^(1/2)+b^(1/2 
))^(5/2)/b^(3/2)/d-1/8*a*tan(d*x+c)*(3*a+b+4*(a+b)*tan(d*x+c)^2)/(a-b)^3/d 
/(a+2*a*tan(d*x+c)^2+(a-b)*tan(d*x+c)^4)^2-1/32*tan(d*x+c)*((a-9*b)*(a+b)/ 
(a-b)^3+(a+19*b)*tan(d*x+c)^2/(a-b)^2)/b/d/(a+2*a*tan(d*x+c)^2+(a-b)*tan(d 
*x+c)^4)
 

Mathematica [A] (verified)

Time = 11.74 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.08 \[ \int \frac {\sin ^8(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\frac {\frac {\left (2 a^{3/2} \sqrt {b}+a b-8 \sqrt {a} b^{3/2}+5 b^2\right ) \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {a+\sqrt {a} \sqrt {b}}}+\frac {\left (2 \sqrt {a}-5 \sqrt {b}\right ) \left (\sqrt {a}+\sqrt {b}\right )^2 \sqrt {b} \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {-a+\sqrt {a} \sqrt {b}}}+\frac {8 b (5 a-14 b+(-2 a+5 b) \cos (2 (c+d x))) \sin (2 (c+d x))}{8 a-3 b+4 b \cos (2 (c+d x))-b \cos (4 (c+d x))}+\frac {64 a (a-b) b (-6 \sin (2 (c+d x))+\sin (4 (c+d x)))}{(-8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x)))^2}}{64 (a-b)^2 b^2 d} \] Input:

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

Output:

(((2*a^(3/2)*Sqrt[b] + a*b - 8*Sqrt[a]*b^(3/2) + 5*b^2)*ArcTan[((Sqrt[a] + 
 Sqrt[b])*Tan[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[a + Sqrt 
[a]*Sqrt[b]]) + ((2*Sqrt[a] - 5*Sqrt[b])*(Sqrt[a] + Sqrt[b])^2*Sqrt[b]*Arc 
Tanh[((Sqrt[a] - Sqrt[b])*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]])/(Sqrt 
[a]*Sqrt[-a + Sqrt[a]*Sqrt[b]]) + (8*b*(5*a - 14*b + (-2*a + 5*b)*Cos[2*(c 
 + d*x)])*Sin[2*(c + d*x)])/(8*a - 3*b + 4*b*Cos[2*(c + d*x)] - b*Cos[4*(c 
 + d*x)]) + (64*a*(a - b)*b*(-6*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(-8* 
a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)])^2)/(64*(a - b)^2*b^2* 
d)
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.15, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3696, 1598, 27, 1440, 27, 1602, 27, 1602, 27, 1480, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sin ^8(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^8}{\left (a-b \sin (c+d x)^4\right )^3}dx\)

\(\Big \downarrow \) 3696

\(\displaystyle \frac {\int \frac {\tan ^8(c+d x) \left (\tan ^2(c+d x)+1\right )}{\left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^3}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 1598

\(\displaystyle \frac {\frac {\int -\frac {2 b \tan ^8(c+d x)}{\left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{16 a b}+\frac {\tan ^9(c+d x)}{8 a \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\tan ^9(c+d x)}{8 a \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}-\frac {\int \frac {\tan ^8(c+d x)}{\left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{8 a}}{d}\)

\(\Big \downarrow \) 1440

\(\displaystyle \frac {\frac {\tan ^9(c+d x)}{8 a \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}-\frac {\frac {\tan ^5(c+d x) \left (\tan ^2(c+d x)+1\right )}{4 b \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}-\frac {\int \frac {2 a \tan ^4(c+d x) \left (3 \tan ^2(c+d x)+5\right )}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{8 a b}}{8 a}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\tan ^9(c+d x)}{8 a \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}-\frac {\frac {\tan ^5(c+d x) \left (\tan ^2(c+d x)+1\right )}{4 b \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}-\frac {\int \frac {\tan ^4(c+d x) \left (3 \tan ^2(c+d x)+5\right )}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{4 b}}{8 a}}{d}\)

\(\Big \downarrow \) 1602

\(\displaystyle \frac {\frac {\tan ^9(c+d x)}{8 a \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}-\frac {\frac {\tan ^5(c+d x) \left (\tan ^2(c+d x)+1\right )}{4 b \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}-\frac {\frac {\tan ^3(c+d x)}{a-b}-\frac {\int \frac {3 \tan ^2(c+d x) \left ((a+5 b) \tan ^2(c+d x)+3 a\right )}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{3 (a-b)}}{4 b}}{8 a}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\tan ^9(c+d x)}{8 a \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}-\frac {\frac {\tan ^5(c+d x) \left (\tan ^2(c+d x)+1\right )}{4 b \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}-\frac {\frac {\tan ^3(c+d x)}{a-b}-\frac {\int \frac {\tan ^2(c+d x) \left ((a+5 b) \tan ^2(c+d x)+3 a\right )}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}}{4 b}}{8 a}}{d}\)

\(\Big \downarrow \) 1602

\(\displaystyle \frac {\frac {\tan ^9(c+d x)}{8 a \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}-\frac {\frac {\tan ^5(c+d x) \left (\tan ^2(c+d x)+1\right )}{4 b \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}-\frac {\frac {\tan ^3(c+d x)}{a-b}-\frac {\frac {(a+5 b) \tan (c+d x)}{a-b}-\frac {\int \frac {a \left (-\left ((a-13 b) \tan ^2(c+d x)\right )+a+5 b\right )}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}}{a-b}}{4 b}}{8 a}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\tan ^9(c+d x)}{8 a \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}-\frac {\frac {\tan ^5(c+d x) \left (\tan ^2(c+d x)+1\right )}{4 b \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}-\frac {\frac {\tan ^3(c+d x)}{a-b}-\frac {\frac {(a+5 b) \tan (c+d x)}{a-b}-\frac {a \int \frac {-\left ((a-13 b) \tan ^2(c+d x)\right )+a+5 b}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}}{a-b}}{4 b}}{8 a}}{d}\)

\(\Big \downarrow \) 1480

\(\displaystyle \frac {\frac {\tan ^9(c+d x)}{8 a \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}-\frac {\frac {\tan ^5(c+d x) \left (\tan ^2(c+d x)+1\right )}{4 b \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}-\frac {\frac {\tan ^3(c+d x)}{a-b}-\frac {\frac {(a+5 b) \tan (c+d x)}{a-b}-\frac {a \left (\frac {\left (\sqrt {a}-\sqrt {b}\right )^3 \left (2 \sqrt {a}+5 \sqrt {b}\right ) \int \frac {1}{(a-b) \tan ^2(c+d x)+\sqrt {a} \left (\sqrt {a}-\sqrt {b}\right )}d\tan (c+d x)}{2 \sqrt {a} \sqrt {b}}-\frac {\left (2 \sqrt {a}-5 \sqrt {b}\right ) \left (\sqrt {a}+\sqrt {b}\right )^3 \int \frac {1}{(a-b) \tan ^2(c+d x)+\sqrt {a} \left (\sqrt {a}+\sqrt {b}\right )}d\tan (c+d x)}{2 \sqrt {a} \sqrt {b}}\right )}{a-b}}{a-b}}{4 b}}{8 a}}{d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\tan ^9(c+d x)}{8 a \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}-\frac {\frac {\tan ^5(c+d x) \left (\tan ^2(c+d x)+1\right )}{4 b \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}-\frac {\frac {\tan ^3(c+d x)}{a-b}-\frac {\frac {(a+5 b) \tan (c+d x)}{a-b}-\frac {a \left (\frac {\left (\sqrt {a}-\sqrt {b}\right )^2 \left (2 \sqrt {a}+5 \sqrt {b}\right ) \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {b} \sqrt {\sqrt {a}+\sqrt {b}}}-\frac {\left (2 \sqrt {a}-5 \sqrt {b}\right ) \left (\sqrt {a}+\sqrt {b}\right )^2 \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {b} \sqrt {\sqrt {a}-\sqrt {b}}}\right )}{a-b}}{a-b}}{4 b}}{8 a}}{d}\)

Input:

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

Output:

(Tan[c + d*x]^9/(8*a*(a + 2*a*Tan[c + d*x]^2 + (a - b)*Tan[c + d*x]^4)^2) 
- ((Tan[c + d*x]^5*(1 + Tan[c + d*x]^2))/(4*b*(a + 2*a*Tan[c + d*x]^2 + (a 
 - b)*Tan[c + d*x]^4)) - (Tan[c + d*x]^3/(a - b) - (-((a*(-1/2*((2*Sqrt[a] 
 - 5*Sqrt[b])*(Sqrt[a] + Sqrt[b])^2*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c 
+ d*x])/a^(1/4)])/(a^(3/4)*Sqrt[Sqrt[a] - Sqrt[b]]*Sqrt[b]) + ((Sqrt[a] - 
Sqrt[b])^2*(2*Sqrt[a] + 5*Sqrt[b])*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + 
 d*x])/a^(1/4)])/(2*a^(3/4)*Sqrt[Sqrt[a] + Sqrt[b]]*Sqrt[b])))/(a - b)) + 
((a + 5*b)*Tan[c + d*x])/(a - b))/(a - b))/(4*b))/(8*a))/d
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

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

rule 1440

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

rule 1480

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 

rule 1598

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

rule 1602

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3696

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^( 
p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1 
)/f   Subst[Int[x^m*((a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2) 
^(m/2 + 2*p + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] & 
& IntegerQ[m/2] && IntegerQ[p]
 

Maple [A] (verified)

Time = 5.84 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.22

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\left (a +19 b \right ) \tan \left (d x +c \right )^{7}}{32 \left (a -b \right ) b}-\frac {3 \left (a^{2}+10 a b -3 b^{2}\right ) \tan \left (d x +c \right )^{5}}{32 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 \left (a +7 b \right ) a \tan \left (d x +c \right )^{3}}{32 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {a \left (a +5 b \right ) \tan \left (d x +c \right )}{32 b \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (d x +c \right )^{4} a -\tan \left (d x +c \right )^{4} b +2 a \tan \left (d x +c \right )^{2}+a \right )^{2}}+\frac {\left (a -b \right ) \left (\frac {\left (-a \sqrt {a b}+13 \sqrt {a b}\, b +2 a^{2}-9 a b -5 b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (-a \sqrt {a b}+13 \sqrt {a b}\, b -2 a^{2}+9 a b +5 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{32 b \left (a^{2}-2 a b +b^{2}\right )}}{d}\)	\(374\)
default	\(\frac {\frac {-\frac {\left (a +19 b \right ) \tan \left (d x +c \right )^{7}}{32 \left (a -b \right ) b}-\frac {3 \left (a^{2}+10 a b -3 b^{2}\right ) \tan \left (d x +c \right )^{5}}{32 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 \left (a +7 b \right ) a \tan \left (d x +c \right )^{3}}{32 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {a \left (a +5 b \right ) \tan \left (d x +c \right )}{32 b \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (d x +c \right )^{4} a -\tan \left (d x +c \right )^{4} b +2 a \tan \left (d x +c \right )^{2}+a \right )^{2}}+\frac {\left (a -b \right ) \left (\frac {\left (-a \sqrt {a b}+13 \sqrt {a b}\, b +2 a^{2}-9 a b -5 b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (-a \sqrt {a b}+13 \sqrt {a b}\, b -2 a^{2}+9 a b +5 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{32 b \left (a^{2}-2 a b +b^{2}\right )}}{d}\)	\(374\)
risch	\(\text {Expression too large to display}\)	\(1589\)

Input:

int(sin(d*x+c)^8/(a-b*sin(d*x+c)^4)^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*((-1/32*(a+19*b)/(a-b)/b*tan(d*x+c)^7-3/32*(a^2+10*a*b-3*b^2)/b/(a^2-2 
*a*b+b^2)*tan(d*x+c)^5-3/32*(a+7*b)*a/b/(a^2-2*a*b+b^2)*tan(d*x+c)^3-1/32* 
a*(a+5*b)/b/(a^2-2*a*b+b^2)*tan(d*x+c))/(tan(d*x+c)^4*a-tan(d*x+c)^4*b+2*a 
*tan(d*x+c)^2+a)^2+1/32/b/(a^2-2*a*b+b^2)*(a-b)*(1/2*(-a*(a*b)^(1/2)+13*(a 
*b)^(1/2)*b+2*a^2-9*a*b-5*b^2)/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)-a)*(a-b))^( 
1/2)*arctanh((-a+b)*tan(d*x+c)/(((a*b)^(1/2)-a)*(a-b))^(1/2))+1/2*(-a*(a*b 
)^(1/2)+13*(a*b)^(1/2)*b-2*a^2+9*a*b+5*b^2)/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2 
)+a)*(a-b))^(1/2)*arctan((a-b)*tan(d*x+c)/(((a*b)^(1/2)+a)*(a-b))^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5219 vs. \(2 (254) = 508\).

Time = 1.11 (sec) , antiderivative size = 5219, normalized size of antiderivative = 17.06 \[ \int \frac {\sin ^8(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(sin(d*x+c)^8/(a-b*sin(d*x+c)^4)^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^8(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Timed out} \] Input:

integrate(sin(d*x+c)**8/(a-b*sin(d*x+c)**4)**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sin ^8(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\int { -\frac {\sin \left (d x + c\right )^{8}}{{\left (b \sin \left (d x + c\right )^{4} - a\right )}^{3}} \,d x } \] Input:

integrate(sin(d*x+c)^8/(a-b*sin(d*x+c)^4)^3,x, algorithm="maxima")
 

Output:

-1/8*(4*(72*a^2*b^2 - 155*a*b^3 + 26*b^4)*cos(4*d*x + 4*c)*sin(2*d*x + 2*c 
) + ((a*b^3 - 4*b^4)*sin(14*d*x + 14*c) - (32*a^2*b^2 - 58*a*b^3 - b^4)*si 
n(12*d*x + 12*c) + 3*(48*a^2*b^2 - 73*a*b^3 + 20*b^4)*sin(10*d*x + 10*c) + 
 (256*a^3*b - 832*a^2*b^2 + 550*a*b^3 - 175*b^4)*sin(8*d*x + 8*c) + (112*a 
^2*b^2 - 533*a*b^3 + 220*b^4)*sin(6*d*x + 6*c) - (32*a^2*b^2 - 158*a*b^3 + 
 141*b^4)*sin(4*d*x + 4*c) - (17*a*b^3 - 44*b^4)*sin(2*d*x + 2*c))*cos(16* 
d*x + 16*c) + 2*(2*(72*a^2*b^2 - 155*a*b^3 + 26*b^4)*sin(12*d*x + 12*c) - 
8*(80*a^2*b^2 - 145*a*b^3 + 44*b^4)*sin(10*d*x + 10*c) - 3*(384*a^3*b - 13 
12*a^2*b^2 + 873*a*b^3 - 280*b^4)*sin(8*d*x + 8*c) - 16*(32*a^2*b^2 - 151* 
a*b^3 + 62*b^4)*sin(6*d*x + 6*c) + 2*(72*a^2*b^2 - 355*a*b^3 + 310*b^4)*si 
n(4*d*x + 4*c) + 24*(3*a*b^3 - 8*b^4)*sin(2*d*x + 2*c))*cos(14*d*x + 14*c) 
 - 2*(2*(128*a^3*b - 456*a^2*b^2 + 1233*a*b^3 - 434*b^4)*sin(10*d*x + 10*c 
) - (6400*a^3*b - 13888*a^2*b^2 + 8566*a*b^3 - 2485*b^4)*sin(8*d*x + 8*c) 
- 2*(128*a^3*b + 2744*a^2*b^2 - 4711*a*b^3 + 1554*b^4)*sin(6*d*x + 6*c) + 
4*(400*a^2*b^2 - 918*a*b^3 + 497*b^4)*sin(4*d*x + 4*c) - 2*(72*a^2*b^2 - 3 
55*a*b^3 + 310*b^4)*sin(2*d*x + 2*c))*cos(12*d*x + 12*c) - 2*((2048*a^4 + 
18560*a^3*b - 24752*a^2*b^2 + 13175*a*b^3 - 2800*b^4)*sin(8*d*x + 8*c) + 8 
*(256*a^3*b + 2400*a^2*b^2 - 2379*a*b^3 + 560*b^4)*sin(6*d*x + 6*c) - 2*(1 
28*a^3*b + 2744*a^2*b^2 - 4711*a*b^3 + 1554*b^4)*sin(4*d*x + 4*c) + 16*(32 
*a^2*b^2 - 151*a*b^3 + 62*b^4)*sin(2*d*x + 2*c))*cos(10*d*x + 10*c) - 2...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1989 vs. \(2 (254) = 508\).

Time = 1.42 (sec) , antiderivative size = 1989, normalized size of antiderivative = 6.50 \[ \int \frac {\sin ^8(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(sin(d*x+c)^8/(a-b*sin(d*x+c)^4)^3,x, algorithm="giac")
 

Output:

1/64*(((3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^3 - 45*sqrt(a^2 
- a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b + 77*sqrt(a^2 - a*b + sqrt(a*b) 
*(a - b))*sqrt(a*b)*a*b^2 + 13*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a* 
b)*b^3)*(a^2*b - 2*a*b^2 + b^3)^2*abs(-a + b) + (3*sqrt(a^2 - a*b + sqrt(a 
*b)*(a - b))*a^6*b - 49*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^4*b^3 + 112* 
sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^3*b^4 - 87*sqrt(a^2 - a*b + sqrt(a*b 
)*(a - b))*a^2*b^5 + 16*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a*b^6 + 5*sqrt 
(a^2 - a*b + sqrt(a*b)*(a - b))*b^7)*abs(a^2*b - 2*a*b^2 + b^3)*abs(-a + b 
) - (6*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^8*b - 63*sqrt(a^2 - 
 a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^7*b^2 + 229*sqrt(a^2 - a*b + sqrt(a* 
b)*(a - b))*sqrt(a*b)*a^6*b^3 - 367*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sq 
rt(a*b)*a^5*b^4 + 233*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^4*b^ 
5 + 27*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^3*b^6 - 89*sqrt(a^2 
 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b^7 + 19*sqrt(a^2 - a*b + sqrt(a 
*b)*(a - b))*sqrt(a*b)*a*b^8 + 5*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt( 
a*b)*b^9)*abs(-a + b))*(pi*floor((d*x + c)/pi + 1/2) + arctan(tan(d*x + c) 
/sqrt((a^3*b - 2*a^2*b^2 + a*b^3 + sqrt((a^3*b - 2*a^2*b^2 + a*b^3)^2 - (a 
^3*b - 2*a^2*b^2 + a*b^3)*(a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)))/(a^3*b - 3 
*a^2*b^2 + 3*a*b^3 - b^4))))/((3*a^10*b^2 - 27*a^9*b^3 + 104*a^8*b^4 - 224 
*a^7*b^5 + 294*a^6*b^6 - 238*a^5*b^7 + 112*a^4*b^8 - 24*a^3*b^9 - a^2*b...
 

Mupad [B] (verification not implemented)

Time = 40.83 (sec) , antiderivative size = 5508, normalized size of antiderivative = 18.00 \[ \int \frac {\sin ^8(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:

int(sin(c + d*x)^8/(a - b*sin(c + d*x)^4)^3,x)
 

Output:

(atan(((((229376*a^2*b^6 - 81920*a*b^7 - 196608*a^3*b^5 + 32768*a^4*b^4 + 
16384*a^5*b^3)/(32768*(3*a*b^5 - b^6 - 3*a^2*b^4 + a^3*b^3)) - (tan(c + d* 
x)*((25*b^2*(a^3*b^9)^(1/2) - 35*a^2*(a^3*b^9)^(1/2) + 105*a^2*b^6 + 70*a^ 
3*b^5 - 35*a^4*b^4 + 4*a^5*b^3 + 154*a*b*(a^3*b^9)^(1/2))/(16384*(a^3*b^11 
 - 5*a^4*b^10 + 10*a^5*b^9 - 10*a^6*b^8 + 5*a^7*b^7 - a^8*b^6)))^(1/2)*(16 
384*a^2*b^8 - 81920*a^3*b^7 + 163840*a^4*b^6 - 163840*a^5*b^5 + 81920*a^6* 
b^4 - 16384*a^7*b^3))/(256*(3*a*b^4 - b^5 - 3*a^2*b^3 + a^3*b^2)))*((25*b^ 
2*(a^3*b^9)^(1/2) - 35*a^2*(a^3*b^9)^(1/2) + 105*a^2*b^6 + 70*a^3*b^5 - 35 
*a^4*b^4 + 4*a^5*b^3 + 154*a*b*(a^3*b^9)^(1/2))/(16384*(a^3*b^11 - 5*a^4*b 
^10 + 10*a^5*b^9 - 10*a^6*b^8 + 5*a^7*b^7 - a^8*b^6)))^(1/2) + (tan(c + d* 
x)*(259*a*b^3 - 35*a^3*b + 4*a^4 + 25*b^4 + 35*a^2*b^2))/(256*(3*a*b^4 - b 
^5 - 3*a^2*b^3 + a^3*b^2)))*((25*b^2*(a^3*b^9)^(1/2) - 35*a^2*(a^3*b^9)^(1 
/2) + 105*a^2*b^6 + 70*a^3*b^5 - 35*a^4*b^4 + 4*a^5*b^3 + 154*a*b*(a^3*b^9 
)^(1/2))/(16384*(a^3*b^11 - 5*a^4*b^10 + 10*a^5*b^9 - 10*a^6*b^8 + 5*a^7*b 
^7 - a^8*b^6)))^(1/2)*1i - (((229376*a^2*b^6 - 81920*a*b^7 - 196608*a^3*b^ 
5 + 32768*a^4*b^4 + 16384*a^5*b^3)/(32768*(3*a*b^5 - b^6 - 3*a^2*b^4 + a^3 
*b^3)) + (tan(c + d*x)*((25*b^2*(a^3*b^9)^(1/2) - 35*a^2*(a^3*b^9)^(1/2) + 
 105*a^2*b^6 + 70*a^3*b^5 - 35*a^4*b^4 + 4*a^5*b^3 + 154*a*b*(a^3*b^9)^(1/ 
2))/(16384*(a^3*b^11 - 5*a^4*b^10 + 10*a^5*b^9 - 10*a^6*b^8 + 5*a^7*b^7 - 
a^8*b^6)))^(1/2)*(16384*a^2*b^8 - 81920*a^3*b^7 + 163840*a^4*b^6 - 1638...
 

Reduce [F]

\[ \int \frac {\sin ^8(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=-\left (\int \frac {\sin \left (d x +c \right )^{8}}{\sin \left (d x +c \right )^{12} b^{3}-3 \sin \left (d x +c \right )^{8} a \,b^{2}+3 \sin \left (d x +c \right )^{4} a^{2} b -a^{3}}d x \right ) \] Input:

int(sin(d*x+c)^8/(a-b*sin(d*x+c)^4)^3,x)
 

Output:

 - int(sin(c + d*x)**8/(sin(c + d*x)**12*b**3 - 3*sin(c + d*x)**8*a*b**2 + 
 3*sin(c + d*x)**4*a**2*b - a**3),x)