\(\int \frac {\sin ^6(c+d x)}{(a-b \sin ^4(c+d x))^2} \, dx\) [168]

Optimal result

Integrand size = 24, antiderivative size = 221 \[ \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=-\frac {\left (2 \sqrt {a}-3 \sqrt {b}\right ) \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 \sqrt [4]{a} \left (\sqrt {a}-\sqrt {b}\right )^{3/2} b^{3/2} d}+\frac {\left (2 \sqrt {a}+3 \sqrt {b}\right ) \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 \sqrt [4]{a} \left (\sqrt {a}+\sqrt {b}\right )^{3/2} b^{3/2} d}-\frac {\tan (c+d x) \left (a+(a+b) \tan ^2(c+d x)\right )}{4 (a-b) b d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )} \] Output:

-1/8*(2*a^(1/2)-3*b^(1/2))*arctan((a^(1/2)-b^(1/2))^(1/2)*tan(d*x+c)/a^(1/ 
4))/a^(1/4)/(a^(1/2)-b^(1/2))^(3/2)/b^(3/2)/d+1/8*(2*a^(1/2)+3*b^(1/2))*ar 
ctan((a^(1/2)+b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))/a^(1/4)/(a^(1/2)+b^(1/2)) 
^(3/2)/b^(3/2)/d-1/4*tan(d*x+c)*(a+(a+b)*tan(d*x+c)^2)/(a-b)/b/d/(a+2*a*ta 
n(d*x+c)^2+(a-b)*tan(d*x+c)^4)
 

Mathematica [A] (verified)

Time = 5.61 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.08 \[ \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\frac {\frac {\left (2 a+\sqrt {a} \sqrt {b}-3 b\right ) \sqrt {b} \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a+\sqrt {a} \sqrt {b}}}-\frac {\sqrt {b} \left (-2 a+\sqrt {a} \sqrt {b}+3 b\right ) \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {-a+\sqrt {a} \sqrt {b}}}+\frac {4 b (-2 a-b+b \cos (2 (c+d x))) \sin (2 (c+d x))}{8 a-3 b+4 b \cos (2 (c+d x))-b \cos (4 (c+d x))}}{8 (a-b) b^2 d} \] Input:

Integrate[Sin[c + d*x]^6/(a - b*Sin[c + d*x]^4)^2,x]
 

Output:

(((2*a + Sqrt[a]*Sqrt[b] - 3*b)*Sqrt[b]*ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c 
+ d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/Sqrt[a + Sqrt[a]*Sqrt[b]] - (Sqrt[b]*( 
-2*a + Sqrt[a]*Sqrt[b] + 3*b)*ArcTanh[((Sqrt[a] - Sqrt[b])*Tan[c + d*x])/S 
qrt[-a + Sqrt[a]*Sqrt[b]]])/Sqrt[-a + Sqrt[a]*Sqrt[b]] + (4*b*(-2*a - b + 
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)]))/(8*(a - b)*b^2*d)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.23, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3042, 3696, 1440, 27, 1602, 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.

\(\displaystyle \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^6}{\left (a-b \sin (c+d x)^4\right )^2}dx\)

\(\Big \downarrow \) 3696

\(\displaystyle \frac {\int \frac {\tan ^6(c+d x)}{\left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 1440

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1602

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 218

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

Input:

Int[Sin[c + d*x]^6/(a - b*Sin[c + d*x]^4)^2,x]
 

Output:

(-1/4*(-((-1/2*((a + (2*Sqrt[a]*(a - 2*b))/Sqrt[b] - 3*b)*ArcTan[(Sqrt[Sqr 
t[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(a^(1/4)*Sqrt[Sqrt[a] - Sqrt[b]]*( 
Sqrt[a] + Sqrt[b])) - ((a - (2*Sqrt[a]*(a - 2*b))/Sqrt[b] - 3*b)*ArcTan[(S 
qrt[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)) + Tan[c + d*x]/(a - b))/b + (Tan[c 
+ d*x]^3*(1 + Tan[c + d*x]^2))/(4*b*(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 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 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 = 2.41 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.19

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\left (a +b \right ) \tan \left (d x +c \right )^{3}}{4 b \left (a -b \right )}-\frac {a \tan \left (d x +c \right )}{4 b \left (a -b \right )}}{\tan \left (d x +c \right )^{4} a -\tan \left (d x +c \right )^{4} b +2 a \tan \left (d x +c \right )^{2}+a}+\frac {\frac {\left (-a \sqrt {a b}+3 \sqrt {a b}\, b +2 a^{2}-4 a b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (-a \sqrt {a b}+3 \sqrt {a b}\, b -2 a^{2}+4 a b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}}{4 b}}{d}\)	\(262\)
default	\(\frac {\frac {-\frac {\left (a +b \right ) \tan \left (d x +c \right )^{3}}{4 b \left (a -b \right )}-\frac {a \tan \left (d x +c \right )}{4 b \left (a -b \right )}}{\tan \left (d x +c \right )^{4} a -\tan \left (d x +c \right )^{4} b +2 a \tan \left (d x +c \right )^{2}+a}+\frac {\frac {\left (-a \sqrt {a b}+3 \sqrt {a b}\, b +2 a^{2}-4 a b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (-a \sqrt {a b}+3 \sqrt {a b}\, b -2 a^{2}+4 a b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}}{4 b}}{d}\)	\(262\)
risch	\(-\frac {i \left (2 \,{\mathrm e}^{6 i \left (d x +c \right )} a -{\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 )}-2 a \,{\mathrm e}^{2 i \left (d x +c \right )}-3 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 (\left (a^{4} b^{6} d^{4}-3 a^{3} b^{7} d^{4}+3 a^{2} b^{8} d^{4}-a \,b^{9} d^{4}\right ) \textit {\_Z}^{4}+\left (128 a^{3} b^{3} d^{2}-480 a^{2} b^{4} d^{2}+480 a \,b^{5} d^{2}\right ) \textit {\_Z}^{2}+4096 a^{2}-18432 a b +20736 b^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\frac {i a^{5} b^{5} d^{3}}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}-\frac {6 i a^{4} b^{6} d^{3}}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}+\frac {12 i a^{3} b^{7} d^{3}}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}-\frac {10 i a^{2} b^{8} d^{3}}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}+\frac {3 i a \,b^{9} d^{3}}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}\right ) \textit {\_R}^{3}+\left (\frac {16 d^{2} b^{3} a^{5}}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}-\frac {84 d^{2} b^{4} a^{4}}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}+\frac {156 d^{2} b^{5} a^{3}}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}-\frac {124 d^{2} b^{6} a^{2}}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}+\frac {36 d^{2} b^{7} a}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}\right ) \textit {\_R}^{2}+\left (\frac {64 i a^{4} b^{2} d}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}-\frac {592 i a^{3} b^{3} d}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}+\frac {1568 i a^{2} b^{4} d}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}-\frac {1296 i a \,b^{5} d}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}\right ) \textit {\_R} +\frac {1024 a^{4}}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}-\frac {6144 a^{3} b}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}+\frac {11840 a^{2} b^{2}}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}-\frac {6048 a \,b^{3}}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}-\frac {2592 b^{4}}{640 a^{2} b^{2}-2592 a \,b^{3}+2592 b^{4}}\right )\right )}{64}\)	\(881\)

Input:

int(sin(d*x+c)^6/(a-b*sin(d*x+c)^4)^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*((-1/4*(a+b)/b/(a-b)*tan(d*x+c)^3-1/4/b*a/(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)+1/4/b*(1/2*(-a*(a*b)^(1/2)+3*(a*b 
)^(1/2)*b+2*a^2-4*a*b)/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)-a)*(a-b))^(1/2)*arc 
tanh((-a+b)*tan(d*x+c)/(((a*b)^(1/2)-a)*(a-b))^(1/2))+1/2*(-a*(a*b)^(1/2)+ 
3*(a*b)^(1/2)*b-2*a^2+4*a*b)/(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. 3135 vs. \(2 (171) = 342\).

Time = 0.69 (sec) , antiderivative size = 3135, normalized size of antiderivative = 14.19 \[ \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(sin(d*x+c)^6/(a-b*sin(d*x+c)^4)^2,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Timed out} \] Input:

integrate(sin(d*x+c)**6/(a-b*sin(d*x+c)**4)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{6}}{{\left (b \sin \left (d x + c\right )^{4} - a\right )}^{2}} \,d x } \] Input:

integrate(sin(d*x+c)^6/(a-b*sin(d*x+c)^4)^2,x, algorithm="maxima")
 

Output:

1/2*(2*(16*a^2 + 2*a*b - 3*b^2)*cos(4*d*x + 4*c)*sin(2*d*x + 2*c) + ((2*a* 
b - b^2)*sin(6*d*x + 6*c) - (8*a*b - 3*b^2)*sin(4*d*x + 4*c) - (2*a*b + 3* 
b^2)*sin(2*d*x + 2*c))*cos(8*d*x + 8*c) + 2*((16*a^2 + 2*a*b - 3*b^2)*sin( 
4*d*x + 4*c) + 4*(2*a*b + b^2)*sin(2*d*x + 2*c))*cos(6*d*x + 6*c) - 2*((a* 
b^3 - b^4)*d*cos(8*d*x + 8*c)^2 + 16*(a*b^3 - b^4)*d*cos(6*d*x + 6*c)^2 + 
4*(64*a^3*b - 112*a^2*b^2 + 57*a*b^3 - 9*b^4)*d*cos(4*d*x + 4*c)^2 + 16*(a 
*b^3 - b^4)*d*cos(2*d*x + 2*c)^2 + (a*b^3 - b^4)*d*sin(8*d*x + 8*c)^2 + 16 
*(a*b^3 - b^4)*d*sin(6*d*x + 6*c)^2 + 4*(64*a^3*b - 112*a^2*b^2 + 57*a*b^3 
 - 9*b^4)*d*sin(4*d*x + 4*c)^2 + 16*(8*a^2*b^2 - 11*a*b^3 + 3*b^4)*d*sin(4 
*d*x + 4*c)*sin(2*d*x + 2*c) + 16*(a*b^3 - b^4)*d*sin(2*d*x + 2*c)^2 - 8*( 
a*b^3 - b^4)*d*cos(2*d*x + 2*c) + (a*b^3 - b^4)*d - 2*(4*(a*b^3 - b^4)*d*c 
os(6*d*x + 6*c) + 2*(8*a^2*b^2 - 11*a*b^3 + 3*b^4)*d*cos(4*d*x + 4*c) + 4* 
(a*b^3 - b^4)*d*cos(2*d*x + 2*c) - (a*b^3 - b^4)*d)*cos(8*d*x + 8*c) + 8*( 
2*(8*a^2*b^2 - 11*a*b^3 + 3*b^4)*d*cos(4*d*x + 4*c) + 4*(a*b^3 - b^4)*d*co 
s(2*d*x + 2*c) - (a*b^3 - b^4)*d)*cos(6*d*x + 6*c) + 4*(4*(8*a^2*b^2 - 11* 
a*b^3 + 3*b^4)*d*cos(2*d*x + 2*c) - (8*a^2*b^2 - 11*a*b^3 + 3*b^4)*d)*cos( 
4*d*x + 4*c) - 4*(2*(a*b^3 - b^4)*d*sin(6*d*x + 6*c) + (8*a^2*b^2 - 11*a*b 
^3 + 3*b^4)*d*sin(4*d*x + 4*c) + 2*(a*b^3 - b^4)*d*sin(2*d*x + 2*c))*sin(8 
*d*x + 8*c) + 16*((8*a^2*b^2 - 11*a*b^3 + 3*b^4)*d*sin(4*d*x + 4*c) + 2*(a 
*b^3 - b^4)*d*sin(2*d*x + 2*c))*sin(6*d*x + 6*c))*integrate(-(4*(2*a*b ...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1481 vs. \(2 (171) = 342\).

Time = 0.86 (sec) , antiderivative size = 1481, normalized size of antiderivative = 6.70 \[ \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(sin(d*x+c)^6/(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^3 - 15*sqrt(a^2 - 
 a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b + 17*sqrt(a^2 - a*b - sqrt(a*b)* 
(a - b))*sqrt(a*b)*a*b^2 + 3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b) 
*b^3)*(a*b - b^2)^2*abs(-a + b) + (3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a 
^5*b - 12*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^4*b^2 + 14*sqrt(a^2 - a*b 
- sqrt(a*b)*(a - b))*a^3*b^3 - 4*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^2*b 
^4 - sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a*b^5)*abs(-a*b + b^2)*abs(-a + b 
) - 2*(3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^6*b - 18*sqrt(a^2 
 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^5*b^2 + 38*sqrt(a^2 - a*b - sqrt(a 
*b)*(a - b))*sqrt(a*b)*a^4*b^3 - 32*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sq 
rt(a*b)*a^3*b^4 + 7*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b^5 
+ 2*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a*b^6)*abs(-a + b))*(pi* 
floor((d*x + c)/pi + 1/2) + arctan(tan(d*x + c)/sqrt((a^2*b - a*b^2 + sqrt 
((a^2*b - a*b^2)^2 - (a^2*b - a*b^2)*(a^2*b - 2*a*b^2 + b^3)))/(a^2*b - 2* 
a*b^2 + b^3))))/((3*a^8*b^2 - 21*a^7*b^3 + 59*a^6*b^4 - 85*a^5*b^5 + 65*a^ 
4*b^6 - 23*a^3*b^7 + a^2*b^8 + a*b^9)*abs(-a*b + b^2)) - ((3*sqrt(a^2 - a* 
b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^3 - 15*sqrt(a^2 - a*b + sqrt(a*b)*(a - 
b))*sqrt(a*b)*a^2*b + 17*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a*b 
^2 + 3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*b^3)*(a*b - b^2)^2*ab 
s(-a + b) - (3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^5*b - 12*sqrt(a^2 ...
 

Mupad [B] (verification not implemented)

Time = 39.03 (sec) , antiderivative size = 3400, normalized size of antiderivative = 15.38 \[ \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

int(sin(c + d*x)^6/(a - b*sin(c + d*x)^4)^2,x)
 

Output:

(atan(((((256*a^2*b^5 - 512*a^3*b^4 + 256*a^4*b^3)/(64*(a*b^3 - b^4)) - (t 
an(c + d*x)*((15*a*b^5 - 5*a*(a*b^9)^(1/2) + 9*b*(a*b^9)^(1/2) - 15*a^2*b^ 
4 + 4*a^3*b^3)/(256*(a*b^9 - 3*a^2*b^8 + 3*a^3*b^7 - a^4*b^6)))^(1/2)*(256 
*a^2*b^6 - 768*a^3*b^5 + 768*a^4*b^4 - 256*a^5*b^3))/(4*(a*b^2 - b^3)))*(( 
15*a*b^5 - 5*a*(a*b^9)^(1/2) + 9*b*(a*b^9)^(1/2) - 15*a^2*b^4 + 4*a^3*b^3) 
/(256*(a*b^9 - 3*a^2*b^8 + 3*a^3*b^7 - a^4*b^6)))^(1/2) + (tan(c + d*x)*(9 
*a*b^3 - 15*a^3*b + 4*a^4 + 10*a^2*b^2))/(4*(a*b^2 - b^3)))*((15*a*b^5 - 5 
*a*(a*b^9)^(1/2) + 9*b*(a*b^9)^(1/2) - 15*a^2*b^4 + 4*a^3*b^3)/(256*(a*b^9 
 - 3*a^2*b^8 + 3*a^3*b^7 - a^4*b^6)))^(1/2)*1i - (((256*a^2*b^5 - 512*a^3* 
b^4 + 256*a^4*b^3)/(64*(a*b^3 - b^4)) + (tan(c + d*x)*((15*a*b^5 - 5*a*(a* 
b^9)^(1/2) + 9*b*(a*b^9)^(1/2) - 15*a^2*b^4 + 4*a^3*b^3)/(256*(a*b^9 - 3*a 
^2*b^8 + 3*a^3*b^7 - a^4*b^6)))^(1/2)*(256*a^2*b^6 - 768*a^3*b^5 + 768*a^4 
*b^4 - 256*a^5*b^3))/(4*(a*b^2 - b^3)))*((15*a*b^5 - 5*a*(a*b^9)^(1/2) + 9 
*b*(a*b^9)^(1/2) - 15*a^2*b^4 + 4*a^3*b^3)/(256*(a*b^9 - 3*a^2*b^8 + 3*a^3 
*b^7 - a^4*b^6)))^(1/2) - (tan(c + d*x)*(9*a*b^3 - 15*a^3*b + 4*a^4 + 10*a 
^2*b^2))/(4*(a*b^2 - b^3)))*((15*a*b^5 - 5*a*(a*b^9)^(1/2) + 9*b*(a*b^9)^( 
1/2) - 15*a^2*b^4 + 4*a^3*b^3)/(256*(a*b^9 - 3*a^2*b^8 + 3*a^3*b^7 - a^4*b 
^6)))^(1/2)*1i)/((27*a*b^2 - 21*a^2*b + 4*a^3)/(32*(a*b^3 - b^4)) + (((256 
*a^2*b^5 - 512*a^3*b^4 + 256*a^4*b^3)/(64*(a*b^3 - b^4)) - (tan(c + d*x)*( 
(15*a*b^5 - 5*a*(a*b^9)^(1/2) + 9*b*(a*b^9)^(1/2) - 15*a^2*b^4 + 4*a^3*...
 

Reduce [F]

\[ \int \frac {\sin ^6(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\int \frac {\sin \left (d x +c \right )^{6}}{\sin \left (d x +c \right )^{8} b^{2}-2 \sin \left (d x +c \right )^{4} a b +a^{2}}d x \] Input:

int(sin(d*x+c)^6/(a-b*sin(d*x+c)^4)^2,x)
 

Output:

int(sin(c + d*x)**6/(sin(c + d*x)**8*b**2 - 2*sin(c + d*x)**4*a*b + a**2), 
x)