\(\int \frac {\sin ^4(c+d x)}{(a-b \sin ^4(c+d x))^2} \, dx\) [169]

Optimal result

Integrand size = 24, antiderivative size = 186 \[ \int \frac {\sin ^4(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\frac {\arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^{3/2} \sqrt {b} d}-\frac {\arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^{3/2} \sqrt {b} d}-\frac {\tan (c+d x) \left (1+2 \tan ^2(c+d x)\right )}{4 (a-b) d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )} \] Output:

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

Mathematica [A] (verified)

Time = 11.36 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.21 \[ \int \frac {\sin ^4(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=-\frac {\frac {\left (\sqrt {a}-\sqrt {b}\right ) \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {a+\sqrt {a} \sqrt {b}} \sqrt {b}}+\frac {\left (\sqrt {a}+\sqrt {b}\right ) \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {-a+\sqrt {a} \sqrt {b}} \sqrt {b}}-\frac {2 (-6 \sin (2 (c+d x))+\sin (4 (c+d x)))}{8 a-3 b+4 b \cos (2 (c+d x))-b \cos (4 (c+d x))}}{8 (a-b) d} \] Input:

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

Output:

-1/8*(((Sqrt[a] - Sqrt[b])*ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x])/Sqrt[ 
a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[a + Sqrt[a]*Sqrt[b]]*Sqrt[b]) + ((Sqr 
t[a] + Sqrt[b])*ArcTanh[((Sqrt[a] - Sqrt[b])*Tan[c + d*x])/Sqrt[-a + Sqrt[ 
a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[-a + Sqrt[a]*Sqrt[b]]*Sqrt[b]) - (2*(-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)]))/((a - b)*d)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.28, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3696, 1598, 27, 1442, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3696

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

\(\Big \downarrow \) 1598

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1442

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(-1/4*(-((a*(((Sqrt[a] + 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]) + ((2 - (a + b 
)/(Sqrt[a]*Sqrt[b]))*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4) 
])/(2*a^(1/4)*(Sqrt[a] - Sqrt[b])*Sqrt[Sqrt[a] + Sqrt[b]])))/(a - b)) + Ta 
n[c + d*x]/(a - b))/a + Tan[c + d*x]^5/(4*a*(a + 2*a*Tan[c + d*x]^2 + (a - 
 b)*Tan[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 1442

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> Simp[d^3*(d*x)^(m - 3)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), 
x] - Simp[d^4/(c*(m + 4*p + 1))   Int[(d*x)^(m - 4)*Simp[a*(m - 3) + b*(m + 
 2*p - 1)*x^2, x]*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x 
] && NeQ[b^2 - 4*a*c, 0] && GtQ[m, 3] && NeQ[m + 4*p + 1, 0] && 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 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 = 1.91 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\tan \left (d x +c \right )^{3}}{2 \left (a -b \right )}-\frac {\tan \left (d x +c \right )}{4 \left (a -b \right )}}{\tan \left (d x +c \right )^{4} a -\tan \left (d x +c \right )^{4} b +2 a \tan \left (d x +c \right )^{2}+a}+\frac {\left (-a -b +2 \sqrt {a b}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{8 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (a +b +2 \sqrt {a b}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{8 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}}{d}\)	\(218\)
default	\(\frac {\frac {-\frac {\tan \left (d x +c \right )^{3}}{2 \left (a -b \right )}-\frac {\tan \left (d x +c \right )}{4 \left (a -b \right )}}{\tan \left (d x +c \right )^{4} a -\tan \left (d x +c \right )^{4} b +2 a \tan \left (d x +c \right )^{2}+a}+\frac {\left (-a -b +2 \sqrt {a b}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{8 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (a +b +2 \sqrt {a b}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{8 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}}{d}\)	\(218\)
risch	\(-\frac {i \left ({\mathrm e}^{6 i \left (d x +c \right )} b -8 \,{\mathrm e}^{4 i \left (d x +c \right )} a +3 b \,{\mathrm e}^{4 i \left (d x +c \right )}-5 b \,{\mathrm e}^{2 i \left (d x +c \right )}+b \right )}{2 b \left (a -b \right ) d \left ({\mathrm e}^{8 i \left (d x +c \right )} b -4 \,{\mathrm e}^{6 i \left (d x +c \right )} b -16 \,{\mathrm e}^{4 i \left (d x +c \right )} a +6 b \,{\mathrm e}^{4 i \left (d x +c \right )}-4 b \,{\mathrm e}^{2 i \left (d x +c \right )}+b \right )}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (1+\left (a^{6} b^{2} d^{4}-3 a^{5} b^{3} d^{4}+3 a^{4} b^{4} d^{4}-a^{3} b^{5} d^{4}\right ) \textit {\_Z}^{4}+\left (2 a^{3} b \,d^{2}+6 a^{2} b^{2} d^{2}\right ) \textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (-\frac {4 i d^{3} a^{6} b^{2}}{3 a b +b^{2}}+\frac {12 i d^{3} a^{5} b^{3}}{3 a b +b^{2}}-\frac {12 i d^{3} a^{4} b^{4}}{3 a b +b^{2}}+\frac {4 i d^{3} a^{3} b^{5}}{3 a b +b^{2}}\right ) \textit {\_R}^{3}+\left (-\frac {2 d^{2} a^{5} b}{3 a b +b^{2}}+\frac {6 d^{2} b^{2} a^{4}}{3 a b +b^{2}}-\frac {6 d^{2} b^{3} a^{3}}{3 a b +b^{2}}+\frac {2 d^{2} b^{4} a^{2}}{3 a b +b^{2}}\right ) \textit {\_R}^{2}+\left (-\frac {10 i d \,a^{3} b}{3 a b +b^{2}}-\frac {20 i d \,a^{2} b^{2}}{3 a b +b^{2}}-\frac {2 i d \,b^{3} a}{3 a b +b^{2}}\right ) \textit {\_R} -\frac {2 a^{2}}{3 a b +b^{2}}-\frac {9 a b}{3 a b +b^{2}}-\frac {b^{2}}{3 a b +b^{2}}\right )\right )}{16}\)	\(507\)

Input:

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

Output:

1/d*((-1/2/(a-b)*tan(d*x+c)^3-1/4/(a-b)*tan(d*x+c))/(tan(d*x+c)^4*a-tan(d* 
x+c)^4*b+2*a*tan(d*x+c)^2+a)+1/8*(-a-b+2*(a*b)^(1/2))/(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))+1/8*(a+b+2*(a*b)^(1/2))/(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. 2796 vs. \(2 (144) = 288\).

Time = 0.39 (sec) , antiderivative size = 2796, normalized size of antiderivative = 15.03 \[ \int \frac {\sin ^4(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/32*(((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*sq 
rt((9*a^2 + 6*a*b + b^2)/((a^9*b - 6*a^8*b^2 + 15*a^7*b^3 - 20*a^6*b^4 + 1 
5*a^5*b^5 - 6*a^4*b^6 + a^3*b^7)*d^4)) + a + 3*b)/((a^4*b - 3*a^3*b^2 + 3* 
a^2*b^3 - a*b^4)*d^2))*log(1/4*(3*a + b)*cos(d*x + c)^2 + 1/2*(2*(a^6*b - 
3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*d^3*sqrt((9*a^2 + 6*a*b + b^2)/((a^9*b - 
6*a^8*b^2 + 15*a^7*b^3 - 20*a^6*b^4 + 15*a^5*b^5 - 6*a^4*b^6 + a^3*b^7)*d^ 
4))*cos(d*x + c)*sin(d*x + c) - (3*a^3 + 4*a^2*b + a*b^2)*d*cos(d*x + c)*s 
in(d*x + c))*sqrt(-((a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*d^2*sqrt((9*a^ 
2 + 6*a*b + b^2)/((a^9*b - 6*a^8*b^2 + 15*a^7*b^3 - 20*a^6*b^4 + 15*a^5*b^ 
5 - 6*a^4*b^6 + a^3*b^7)*d^4)) + a + 3*b)/((a^4*b - 3*a^3*b^2 + 3*a^2*b^3 
- a*b^4)*d^2)) - 1/4*(2*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*d^2*cos(d*x 
+ c)^2 - (a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*d^2)*sqrt((9*a^2 + 6*a*b + 
b^2)/((a^9*b - 6*a^8*b^2 + 15*a^7*b^3 - 20*a^6*b^4 + 15*a^5*b^5 - 6*a^4*b^ 
6 + a^3*b^7)*d^4)) - 3/4*a - 1/4*b) - ((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((9*a^2 + 6*a*b + b^2)/((a^9*b - 6*a^8*b^ 
2 + 15*a^7*b^3 - 20*a^6*b^4 + 15*a^5*b^5 - 6*a^4*b^6 + a^3*b^7)*d^4)) + a 
+ 3*b)/((a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*d^2))*log(1/4*(3*a + b)*co 
s(d*x + c)^2 - 1/2*(2*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*d^3*sqr...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

1/2*(6*(8*a - 3*b)*cos(4*d*x + 4*c)*sin(2*d*x + 2*c) + (b*sin(6*d*x + 6*c) 
 - (8*a - 3*b)*sin(4*d*x + 4*c) - 5*b*sin(2*d*x + 2*c))*cos(8*d*x + 8*c) + 
 6*((8*a - 3*b)*sin(4*d*x + 4*c) + 4*b*sin(2*d*x + 2*c))*cos(6*d*x + 6*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 + 1 
6*(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 + 16*(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*a^2*b - 11*a*b^2 + 3*b^3)*d)*cos(4*d*x + 4* 
c) - 4*(2*(a*b^2 - b^3)*d*sin(6*d*x + 6*c) + (8*a^2*b - 11*a*b^2 + 3*b^3)* 
d*sin(4*d*x + 4*c) + 2*(a*b^2 - b^3)*d*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) 
+ 16*((8*a^2*b - 11*a*b^2 + 3*b^3)*d*sin(4*d*x + 4*c) + 2*(a*b^2 - b^3)*d* 
sin(2*d*x + 2*c))*sin(6*d*x + 6*c))*integrate((4*b*cos(6*d*x + 6*c)^2 - 12 
*(8*a - 3*b)*cos(4*d*x + 4*c)^2 + 4*b*cos(2*d*x + 2*c)^2 + 4*b*sin(6*d*...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1264 vs. \(2 (144) = 288\).

Time = 0.87 (sec) , antiderivative size = 1264, normalized size of antiderivative = 6.80 \[ \int \frac {\sin ^4(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/8*((3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^5 - 9*sqrt(a^2 - a 
*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^4*b + 2*sqrt(a^2 - a*b - sqrt(a*b)*(a 
- b))*sqrt(a*b)*a^3*b^2 + 10*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b) 
*a^2*b^3 - 5*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a*b^4 - sqrt(a^ 
2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*b^5 - 2*(3*sqrt(a^2 - a*b - sqrt(a* 
b)*(a - b))*sqrt(a*b)*a^2*b - 6*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a 
*b)*a*b^2 - sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*b^3)*(a - b)^2 + 
 (3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^4*b - 12*sqrt(a^2 - a*b - sqrt(a 
*b)*(a - b))*a^3*b^2 + 14*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^2*b^3 - 4* 
sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a*b^4 - sqrt(a^2 - a*b - sqrt(a*b)*(a 
- b))*b^5)*abs(-a + b))*(pi*floor((d*x + c)/pi + 1/2) + arctan(tan(d*x + c 
)/sqrt((a^2 - a*b + sqrt((a^2 - a*b)^2 - (a^2 - a*b)*(a^2 - 2*a*b + b^2))) 
/(a^2 - 2*a*b + b^2))))/(3*a^8*b - 21*a^7*b^2 + 59*a^6*b^3 - 85*a^5*b^4 + 
65*a^4*b^5 - 23*a^3*b^6 + a^2*b^7 + a*b^8) - (3*sqrt(a^2 - a*b + sqrt(a*b) 
*(a - b))*sqrt(a*b)*a^5 - 9*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)* 
a^4*b + 2*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^3*b^2 + 10*sqrt( 
a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b^3 - 5*sqrt(a^2 - a*b + sqrt 
(a*b)*(a - b))*sqrt(a*b)*a*b^4 - sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt( 
a*b)*b^5 - 2*(3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b - 6*sq 
rt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a*b^2 - sqrt(a^2 - a*b + sq...
 

Mupad [B] (verification not implemented)

Time = 38.83 (sec) , antiderivative size = 2980, normalized size of antiderivative = 16.02 \[ \int \frac {\sin ^4(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

- (atan(((((128*a*b^3 + 128*a^3*b - 256*a^2*b^2)/(32*(a - b)) - (tan(c + d 
*x)*((3*a*(a^3*b^3)^(1/2) + b*(a^3*b^3)^(1/2) + a^3*b + 3*a^2*b^2)/(256*(a 
^3*b^5 - 3*a^4*b^4 + 3*a^5*b^3 - a^6*b^2)))^(1/2)*(256*a^5*b - 256*a^2*b^4 
 + 768*a^3*b^3 - 768*a^4*b^2))/(4*(a - b)))*((3*a*(a^3*b^3)^(1/2) + b*(a^3 
*b^3)^(1/2) + a^3*b + 3*a^2*b^2)/(256*(a^3*b^5 - 3*a^4*b^4 + 3*a^5*b^3 - a 
^6*b^2)))^(1/2) - (tan(c + d*x)*(6*a*b + a^2 + b^2))/(4*(a - b)))*((3*a*(a 
^3*b^3)^(1/2) + b*(a^3*b^3)^(1/2) + a^3*b + 3*a^2*b^2)/(256*(a^3*b^5 - 3*a 
^4*b^4 + 3*a^5*b^3 - a^6*b^2)))^(1/2)*1i - (((128*a*b^3 + 128*a^3*b - 256* 
a^2*b^2)/(32*(a - b)) + (tan(c + d*x)*((3*a*(a^3*b^3)^(1/2) + b*(a^3*b^3)^ 
(1/2) + a^3*b + 3*a^2*b^2)/(256*(a^3*b^5 - 3*a^4*b^4 + 3*a^5*b^3 - a^6*b^2 
)))^(1/2)*(256*a^5*b - 256*a^2*b^4 + 768*a^3*b^3 - 768*a^4*b^2))/(4*(a - b 
)))*((3*a*(a^3*b^3)^(1/2) + b*(a^3*b^3)^(1/2) + a^3*b + 3*a^2*b^2)/(256*(a 
^3*b^5 - 3*a^4*b^4 + 3*a^5*b^3 - a^6*b^2)))^(1/2) + (tan(c + d*x)*(6*a*b + 
 a^2 + b^2))/(4*(a - b)))*((3*a*(a^3*b^3)^(1/2) + b*(a^3*b^3)^(1/2) + a^3* 
b + 3*a^2*b^2)/(256*(a^3*b^5 - 3*a^4*b^4 + 3*a^5*b^3 - a^6*b^2)))^(1/2)*1i 
)/((((128*a*b^3 + 128*a^3*b - 256*a^2*b^2)/(32*(a - b)) - (tan(c + d*x)*(( 
3*a*(a^3*b^3)^(1/2) + b*(a^3*b^3)^(1/2) + a^3*b + 3*a^2*b^2)/(256*(a^3*b^5 
 - 3*a^4*b^4 + 3*a^5*b^3 - a^6*b^2)))^(1/2)*(256*a^5*b - 256*a^2*b^4 + 768 
*a^3*b^3 - 768*a^4*b^2))/(4*(a - b)))*((3*a*(a^3*b^3)^(1/2) + b*(a^3*b^3)^ 
(1/2) + a^3*b + 3*a^2*b^2)/(256*(a^3*b^5 - 3*a^4*b^4 + 3*a^5*b^3 - a^6*...
 

Reduce [F]

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

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

Output:

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