\(\int \frac {\sin ^3(c+d x)}{(a-b \sin ^4(c+d x))^2} \, dx\) [164]

Optimal result

Integrand size = 24, antiderivative size = 186 \[ \int \frac {\sin ^3(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{8 \sqrt {a} \left (\sqrt {a}-\sqrt {b}\right )^{3/2} b^{3/4} d}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{8 \sqrt {a} \left (\sqrt {a}+\sqrt {b}\right )^{3/2} b^{3/4} d}-\frac {\cos (c+d x) \left (2-\cos ^2(c+d x)\right )}{4 (a-b) d \left (a-b+2 b \cos ^2(c+d x)-b \cos ^4(c+d x)\right )} \] Output:

-1/8*arctan(b^(1/4)*cos(d*x+c)/(a^(1/2)-b^(1/2))^(1/2))/a^(1/2)/(a^(1/2)-b 
^(1/2))^(3/2)/b^(3/4)/d+1/8*arctanh(b^(1/4)*cos(d*x+c)/(a^(1/2)+b^(1/2))^( 
1/2))/a^(1/2)/(a^(1/2)+b^(1/2))^(3/2)/b^(3/4)/d-1/4*cos(d*x+c)*(2-cos(d*x+ 
c)^2)/(a-b)/d/(a-b+2*b*cos(d*x+c)^2-b*cos(d*x+c)^4)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 1.10 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.85 \[ \int \frac {\sin ^3(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\frac {\frac {16 (-5 \cos (c+d x)+\cos (3 (c+d x)))}{8 a-3 b+4 b \cos (2 (c+d x))-b \cos (4 (c+d x))}-i \text {RootSum}\left [b-4 b \text {$\#$1}^2-16 a \text {$\#$1}^4+6 b \text {$\#$1}^4-4 b \text {$\#$1}^6+b \text {$\#$1}^8\&,\frac {-2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )+14 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2-7 i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2-14 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4+7 i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4+2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^6-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^6}{-b \text {$\#$1}-8 a \text {$\#$1}^3+3 b \text {$\#$1}^3-3 b \text {$\#$1}^5+b \text {$\#$1}^7}\&\right ]}{32 (a-b) d} \] Input:

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

Output:

((16*(-5*Cos[c + d*x] + Cos[3*(c + d*x)]))/(8*a - 3*b + 4*b*Cos[2*(c + d*x 
)] - b*Cos[4*(c + d*x)]) - I*RootSum[b - 4*b*#1^2 - 16*a*#1^4 + 6*b*#1^4 - 
 4*b*#1^6 + b*#1^8 & , (-2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] + I*Lo 
g[1 - 2*Cos[c + d*x]*#1 + #1^2] + 14*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - # 
1)]*#1^2 - (7*I)*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 - 14*ArcTan[Sin[c 
+ d*x]/(Cos[c + d*x] - #1)]*#1^4 + (7*I)*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] 
*#1^4 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^6 - I*Log[1 - 2*Cos[ 
c + d*x]*#1 + #1^2]*#1^6)/(-(b*#1) - 8*a*#1^3 + 3*b*#1^3 - 3*b*#1^5 + b*#1 
^7) & ])/(32*(a - b)*d)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3042, 3694, 1492, 27, 1480, 218, 221}

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.

\(\displaystyle \int \frac {\sin ^3(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3694

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

\(\Big \downarrow \) 1492

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

\(\displaystyle -\frac {\frac {\frac {\left (\frac {\sqrt {b}}{\sqrt {a}}+1\right ) \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 b^{3/4} \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {\left (1-\frac {\sqrt {b}}{\sqrt {a}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 b^{3/4} \sqrt {\sqrt {a}+\sqrt {b}}}}{4 (a-b)}+\frac {\cos (c+d x) \left (2-\cos ^2(c+d x)\right )}{4 (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )}}{d}\)

Input:

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

Output:

-(((((1 + Sqrt[b]/Sqrt[a])*ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] - Sq 
rt[b]]])/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^(3/4)) - ((1 - Sqrt[b]/Sqrt[a])*ArcT 
anh[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]])/(2*Sqrt[Sqrt[a] + Sqr 
t[b]]*b^(3/4)))/(4*(a - b)) + (Cos[c + d*x]*(2 - Cos[c + d*x]^2))/(4*(a - 
b)*(a - b + 2*b*Cos[c + d*x]^2 - b*Cos[c + d*x]^4)))/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 221

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

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 1492

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symb 
ol] :> Simp[x*(a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2)*((a + b*x^2 + 
 c*x^4)^(p + 1)/(2*a*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 
 - 4*a*c))   Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 
7)*(d*b - 2*a*e)*c*x^2, x]*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, 
 b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && 
 LtQ[p, -1] && IntegerQ[2*p]
 

rule 3042

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

rule 3694

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

Maple [A] (verified)

Time = 2.00 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.10

method	result	size
derivativedivides	\(\frac {b^{2} \left (\frac {\sqrt {a b}\, \left (\frac {\cos \left (d x +c \right )}{2 \left (\sqrt {a b}+b \right ) \left (-b \cos \left (d x +c \right )^{2}+\sqrt {a b}+b \right )}+\frac {\operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 \left (\sqrt {a b}+b \right ) \sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{4 b^{2} a}-\frac {\sqrt {a b}\, \left (\frac {\cos \left (d x +c \right )}{2 \left (\sqrt {a b}-b \right ) \left (b \cos \left (d x +c \right )^{2}+\sqrt {a b}-b \right )}+\frac {\arctan \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 \left (\sqrt {a b}-b \right ) \sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{4 b^{2} a}\right )}{d}\)	\(204\)
default	\(\frac {b^{2} \left (\frac {\sqrt {a b}\, \left (\frac {\cos \left (d x +c \right )}{2 \left (\sqrt {a b}+b \right ) \left (-b \cos \left (d x +c \right )^{2}+\sqrt {a b}+b \right )}+\frac {\operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 \left (\sqrt {a b}+b \right ) \sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{4 b^{2} a}-\frac {\sqrt {a b}\, \left (\frac {\cos \left (d x +c \right )}{2 \left (\sqrt {a b}-b \right ) \left (b \cos \left (d x +c \right )^{2}+\sqrt {a b}-b \right )}+\frac {\arctan \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 \left (\sqrt {a b}-b \right ) \sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{4 b^{2} a}\right )}{d}\)	\(204\)
risch	\(-\frac {{\mathrm e}^{7 i \left (d x +c \right )}-5 \,{\mathrm e}^{5 i \left (d x +c \right )}-5 \,{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}}{2 \left (a -b \right ) d \left ({\mathrm e}^{8 i \left (d x +c \right )} b -4 \,{\mathrm e}^{6 i \left (d x +c \right )} b -16 \,{\mathrm e}^{4 i \left (d x +c \right )} a +6 b \,{\mathrm e}^{4 i \left (d x +c \right )}-4 b \,{\mathrm e}^{2 i \left (d x +c \right )}+b \right )}+\frac {i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (-1+\left (16 a^{5} b^{3} d^{4}-48 a^{4} b^{4} d^{4}+48 a^{3} b^{5} d^{4}-16 a^{2} b^{6} d^{4}\right ) \textit {\_Z}^{4}+\left (-24 a^{2} b^{2} d^{2}-8 a \,b^{3} d^{2}\right ) \textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\left (-\frac {16 i a^{5} b^{2} d^{3}}{a +3 b}+\frac {32 i a^{4} b^{3} d^{3}}{a +3 b}-\frac {32 i a^{2} b^{5} d^{3}}{a +3 b}+\frac {16 i a \,b^{6} d^{3}}{a +3 b}\right ) \textit {\_R}^{3}+\left (\frac {20 i b d \,a^{2}}{a +3 b}+\frac {40 i b^{2} d a}{a +3 b}+\frac {4 i b^{3} d}{a +3 b}\right ) \textit {\_R} \right ) {\mathrm e}^{i \left (d x +c \right )}+1\right )\right )}{8}\)	\(352\)

Input:

int(sin(d*x+c)^3/(a-b*sin(d*x+c)^4)^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*b^2*(1/4*(a*b)^(1/2)/b^2/a*(1/2*cos(d*x+c)/((a*b)^(1/2)+b)/(-b*cos(d*x 
+c)^2+(a*b)^(1/2)+b)+1/2/((a*b)^(1/2)+b)/(((a*b)^(1/2)+b)*b)^(1/2)*arctanh 
(b*cos(d*x+c)/(((a*b)^(1/2)+b)*b)^(1/2)))-1/4*(a*b)^(1/2)/b^2/a*(1/2*cos(d 
*x+c)/((a*b)^(1/2)-b)/(b*cos(d*x+c)^2+(a*b)^(1/2)-b)+1/2/((a*b)^(1/2)-b)/( 
((a*b)^(1/2)-b)*b)^(1/2)*arctan(b*cos(d*x+c)/(((a*b)^(1/2)-b)*b)^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2049 vs. \(2 (141) = 282\).

Time = 0.23 (sec) , antiderivative size = 2049, normalized size of antiderivative = 11.02 \[ \int \frac {\sin ^3(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(sin(d*x+c)^3/(a-b*sin(d*x+c)^4)^2,x, algorithm="fricas")
 

Output:

-1/16*(4*cos(d*x + c)^3 - ((a*b - b^2)*d*cos(d*x + c)^4 - 2*(a*b - b^2)*d* 
cos(d*x + c)^2 - (a^2 - 2*a*b + b^2)*d)*sqrt(-((a^4*b - 3*a^3*b^2 + 3*a^2* 
b^3 - a*b^4)*d^2*sqrt((a^2 + 6*a*b + 9*b^2)/((a^7*b^3 - 6*a^6*b^4 + 15*a^5 
*b^5 - 20*a^4*b^6 + 15*a^3*b^7 - 6*a^2*b^8 + a*b^9)*d^4)) + 3*a + b)/((a^4 
*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*d^2))*log((a + 3*b)*cos(d*x + c) - ((a 
^5*b^2 - 2*a^4*b^3 + 2*a^2*b^5 - a*b^6)*d^3*sqrt((a^2 + 6*a*b + 9*b^2)/((a 
^7*b^3 - 6*a^6*b^4 + 15*a^5*b^5 - 20*a^4*b^6 + 15*a^3*b^7 - 6*a^2*b^8 + a* 
b^9)*d^4)) - 2*(a^2*b + 3*a*b^2)*d)*sqrt(-((a^4*b - 3*a^3*b^2 + 3*a^2*b^3 
- a*b^4)*d^2*sqrt((a^2 + 6*a*b + 9*b^2)/((a^7*b^3 - 6*a^6*b^4 + 15*a^5*b^5 
 - 20*a^4*b^6 + 15*a^3*b^7 - 6*a^2*b^8 + a*b^9)*d^4)) + 3*a + b)/((a^4*b - 
 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*d^2))) + ((a*b - b^2)*d*cos(d*x + c)^4 - 2 
*(a*b - b^2)*d*cos(d*x + c)^2 - (a^2 - 2*a*b + b^2)*d)*sqrt(((a^4*b - 3*a^ 
3*b^2 + 3*a^2*b^3 - a*b^4)*d^2*sqrt((a^2 + 6*a*b + 9*b^2)/((a^7*b^3 - 6*a^ 
6*b^4 + 15*a^5*b^5 - 20*a^4*b^6 + 15*a^3*b^7 - 6*a^2*b^8 + a*b^9)*d^4)) - 
3*a - b)/((a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*d^2))*log((a + 3*b)*cos( 
d*x + c) - ((a^5*b^2 - 2*a^4*b^3 + 2*a^2*b^5 - a*b^6)*d^3*sqrt((a^2 + 6*a* 
b + 9*b^2)/((a^7*b^3 - 6*a^6*b^4 + 15*a^5*b^5 - 20*a^4*b^6 + 15*a^3*b^7 - 
6*a^2*b^8 + a*b^9)*d^4)) + 2*(a^2*b + 3*a*b^2)*d)*sqrt(((a^4*b - 3*a^3*b^2 
 + 3*a^2*b^3 - a*b^4)*d^2*sqrt((a^2 + 6*a*b + 9*b^2)/((a^7*b^3 - 6*a^6*b^4 
 + 15*a^5*b^5 - 20*a^4*b^6 + 15*a^3*b^7 - 6*a^2*b^8 + a*b^9)*d^4)) - 3*...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(sin(d*x+c)^3/(a-b*sin(d*x+c)^4)^2,x, algorithm="maxima")
 

Output:

1/2*(4*b*cos(2*d*x + 2*c)*cos(d*x + c) - 20*b*sin(3*d*x + 3*c)*sin(2*d*x + 
 2*c) + 4*b*sin(2*d*x + 2*c)*sin(d*x + c) - (b*cos(7*d*x + 7*c) - 5*b*cos( 
5*d*x + 5*c) - 5*b*cos(3*d*x + 3*c) + b*cos(d*x + c))*cos(8*d*x + 8*c) + ( 
4*b*cos(6*d*x + 6*c) + 2*(8*a - 3*b)*cos(4*d*x + 4*c) + 4*b*cos(2*d*x + 2* 
c) - b)*cos(7*d*x + 7*c) - 4*(5*b*cos(5*d*x + 5*c) + 5*b*cos(3*d*x + 3*c) 
- b*cos(d*x + c))*cos(6*d*x + 6*c) - 5*(2*(8*a - 3*b)*cos(4*d*x + 4*c) + 4 
*b*cos(2*d*x + 2*c) - b)*cos(5*d*x + 5*c) - 2*(5*(8*a - 3*b)*cos(3*d*x + 3 
*c) - (8*a - 3*b)*cos(d*x + c))*cos(4*d*x + 4*c) - 5*(4*b*cos(2*d*x + 2*c) 
 - b)*cos(3*d*x + 3*c) - b*cos(d*x + c) + 2*((a*b^2 - b^3)*d*cos(8*d*x + 8 
*c)^2 + 16*(a*b^2 - b^3)*d*cos(6*d*x + 6*c)^2 + 4*(64*a^3 - 112*a^2*b + 57 
*a*b^2 - 9*b^3)*d*cos(4*d*x + 4*c)^2 + 16*(a*b^2 - b^3)*d*cos(2*d*x + 2*c) 
^2 + (a*b^2 - b^3)*d*sin(8*d*x + 8*c)^2 + 16*(a*b^2 - b^3)*d*sin(6*d*x + 6 
*c)^2 + 4*(64*a^3 - 112*a^2*b + 57*a*b^2 - 9*b^3)*d*sin(4*d*x + 4*c)^2 + 1 
6*(8*a^2*b - 11*a*b^2 + 3*b^3)*d*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*(a 
*b^2 - b^3)*d*sin(2*d*x + 2*c)^2 - 8*(a*b^2 - b^3)*d*cos(2*d*x + 2*c) + (a 
*b^2 - b^3)*d - 2*(4*(a*b^2 - b^3)*d*cos(6*d*x + 6*c) + 2*(8*a^2*b - 11*a* 
b^2 + 3*b^3)*d*cos(4*d*x + 4*c) + 4*(a*b^2 - b^3)*d*cos(2*d*x + 2*c) - (a* 
b^2 - b^3)*d)*cos(8*d*x + 8*c) + 8*(2*(8*a^2*b - 11*a*b^2 + 3*b^3)*d*cos(4 
*d*x + 4*c) + 4*(a*b^2 - b^3)*d*cos(2*d*x + 2*c) - (a*b^2 - b^3)*d)*cos(6* 
d*x + 6*c) + 4*(4*(8*a^2*b - 11*a*b^2 + 3*b^3)*d*cos(2*d*x + 2*c) - (8*...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 577 vs. \(2 (141) = 282\).

Time = 0.90 (sec) , antiderivative size = 577, normalized size of antiderivative = 3.10 \[ \int \frac {\sin ^3(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=-\frac {\frac {\cos \left (d x + c\right )^{3}}{d} - \frac {2 \, \cos \left (d x + c\right )}{d}}{4 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} - a + b\right )} {\left (a - b\right )}} + \frac {{\left ({\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \sqrt {-b^{2} + \sqrt {a b} b} d^{4} - 2 \, \sqrt {a b} \sqrt {-b^{2} + \sqrt {a b} b} {\left (a - b\right )} d^{2} {\left | -a d^{2} + b d^{2} \right |} + {\left (a d^{2} - b d^{2}\right )}^{2} \sqrt {-b^{2} + \sqrt {a b} b} a\right )} \arctan \left (\frac {\cos \left (d x + c\right )}{d \sqrt {-\frac {a b d^{2} - b^{2} d^{2} + \sqrt {{\left (a b d^{2} - b^{2} d^{2}\right )}^{2} + {\left (a b d^{4} - b^{2} d^{4}\right )} {\left (a^{2} - 2 \, a b + b^{2}\right )}}}{a b d^{4} - b^{2} d^{4}}}}\right )}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b} d^{3} {\left | -a d^{2} + b d^{2} \right |} {\left | b \right |}} - \frac {{\left ({\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \sqrt {-b^{2} - \sqrt {a b} b} d^{4} + 2 \, \sqrt {a b} \sqrt {-b^{2} - \sqrt {a b} b} {\left (a - b\right )} d^{2} {\left | -a d^{2} + b d^{2} \right |} + {\left (a d^{2} - b d^{2}\right )}^{2} \sqrt {-b^{2} - \sqrt {a b} b} a\right )} \arctan \left (\frac {\cos \left (d x + c\right )}{d \sqrt {-\frac {a b d^{2} - b^{2} d^{2} - \sqrt {{\left (a b d^{2} - b^{2} d^{2}\right )}^{2} + {\left (a b d^{4} - b^{2} d^{4}\right )} {\left (a^{2} - 2 \, a b + b^{2}\right )}}}{a b d^{4} - b^{2} d^{4}}}}\right )}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b} d^{3} {\left | -a d^{2} + b d^{2} \right |} {\left | b \right |}} \] Input:

integrate(sin(d*x+c)^3/(a-b*sin(d*x+c)^4)^2,x, algorithm="giac")
 

Output:

-1/4*(cos(d*x + c)^3/d - 2*cos(d*x + c)/d)/((b*cos(d*x + c)^4 - 2*b*cos(d* 
x + c)^2 - a + b)*(a - b)) + 1/8*((a^2*b - 2*a*b^2 + b^3)*sqrt(-b^2 + sqrt 
(a*b)*b)*d^4 - 2*sqrt(a*b)*sqrt(-b^2 + sqrt(a*b)*b)*(a - b)*d^2*abs(-a*d^2 
 + b*d^2) + (a*d^2 - b*d^2)^2*sqrt(-b^2 + sqrt(a*b)*b)*a)*arctan(cos(d*x + 
 c)/(d*sqrt(-(a*b*d^2 - b^2*d^2 + sqrt((a*b*d^2 - b^2*d^2)^2 + (a*b*d^4 - 
b^2*d^4)*(a^2 - 2*a*b + b^2)))/(a*b*d^4 - b^2*d^4))))/((a^3 - 3*a^2*b + 3* 
a*b^2 - b^3)*sqrt(a*b)*d^3*abs(-a*d^2 + b*d^2)*abs(b)) - 1/8*((a^2*b - 2*a 
*b^2 + b^3)*sqrt(-b^2 - sqrt(a*b)*b)*d^4 + 2*sqrt(a*b)*sqrt(-b^2 - sqrt(a* 
b)*b)*(a - b)*d^2*abs(-a*d^2 + b*d^2) + (a*d^2 - b*d^2)^2*sqrt(-b^2 - sqrt 
(a*b)*b)*a)*arctan(cos(d*x + c)/(d*sqrt(-(a*b*d^2 - b^2*d^2 - sqrt((a*b*d^ 
2 - b^2*d^2)^2 + (a*b*d^4 - b^2*d^4)*(a^2 - 2*a*b + b^2)))/(a*b*d^4 - b^2* 
d^4))))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqrt(a*b)*d^3*abs(-a*d^2 + b*d^2) 
*abs(b))
 

Mupad [B] (verification not implemented)

Time = 38.53 (sec) , antiderivative size = 3060, normalized size of antiderivative = 16.45 \[ \int \frac {\sin ^3(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(cos(c + d*x)^3/(4*(a - b)) - cos(c + d*x)/(2*(a - b)))/(d*(a - b + 2*b*co 
s(c + d*x)^2 - b*cos(c + d*x)^4)) - (atan(((((512*a*b^4 - 512*a^2*b^3)/(64 
*(a^2 - 2*a*b + b^2)) - (cos(c + d*x)*((a*(a^3*b^3)^(1/2) + 3*b*(a^3*b^3)^ 
(1/2) + a*b^3 + 3*a^2*b^2)/(256*(a^2*b^6 - 3*a^3*b^5 + 3*a^4*b^4 - a^5*b^3 
)))^(1/2)*(256*a*b^6 - 512*a^2*b^5 + 256*a^3*b^4))/(4*(a^2 - 2*a*b + b^2)) 
)*((a*(a^3*b^3)^(1/2) + 3*b*(a^3*b^3)^(1/2) + a*b^3 + 3*a^2*b^2)/(256*(a^2 
*b^6 - 3*a^3*b^5 + 3*a^4*b^4 - a^5*b^3)))^(1/2) + (cos(c + d*x)*(a*b^2 + b 
^3))/(4*(a^2 - 2*a*b + b^2)))*((a*(a^3*b^3)^(1/2) + 3*b*(a^3*b^3)^(1/2) + 
a*b^3 + 3*a^2*b^2)/(256*(a^2*b^6 - 3*a^3*b^5 + 3*a^4*b^4 - a^5*b^3)))^(1/2 
)*1i - (((512*a*b^4 - 512*a^2*b^3)/(64*(a^2 - 2*a*b + b^2)) + (cos(c + d*x 
)*((a*(a^3*b^3)^(1/2) + 3*b*(a^3*b^3)^(1/2) + a*b^3 + 3*a^2*b^2)/(256*(a^2 
*b^6 - 3*a^3*b^5 + 3*a^4*b^4 - a^5*b^3)))^(1/2)*(256*a*b^6 - 512*a^2*b^5 + 
 256*a^3*b^4))/(4*(a^2 - 2*a*b + b^2)))*((a*(a^3*b^3)^(1/2) + 3*b*(a^3*b^3 
)^(1/2) + a*b^3 + 3*a^2*b^2)/(256*(a^2*b^6 - 3*a^3*b^5 + 3*a^4*b^4 - a^5*b 
^3)))^(1/2) - (cos(c + d*x)*(a*b^2 + b^3))/(4*(a^2 - 2*a*b + b^2)))*((a*(a 
^3*b^3)^(1/2) + 3*b*(a^3*b^3)^(1/2) + a*b^3 + 3*a^2*b^2)/(256*(a^2*b^6 - 3 
*a^3*b^5 + 3*a^4*b^4 - a^5*b^3)))^(1/2)*1i)/(b/(32*(a^2 - 2*a*b + b^2)) + 
(((512*a*b^4 - 512*a^2*b^3)/(64*(a^2 - 2*a*b + b^2)) - (cos(c + d*x)*((a*( 
a^3*b^3)^(1/2) + 3*b*(a^3*b^3)^(1/2) + a*b^3 + 3*a^2*b^2)/(256*(a^2*b^6 - 
3*a^3*b^5 + 3*a^4*b^4 - a^5*b^3)))^(1/2)*(256*a*b^6 - 512*a^2*b^5 + 256...
 

Reduce [F]

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

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

Output:

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