\(\int \frac {\sin (c+d x)}{(a-b \sin ^4(c+d x))^2} \, dx\) [165]

Optimal result

Integrand size = 22, antiderivative size = 221 \[ \int \frac {\sin (c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=-\frac {\left (3 \sqrt {a}-2 \sqrt {b}\right ) \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{8 a^{3/2} \left (\sqrt {a}-\sqrt {b}\right )^{3/2} \sqrt [4]{b} d}-\frac {\left (3 \sqrt {a}+2 \sqrt {b}\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{8 a^{3/2} \left (\sqrt {a}+\sqrt {b}\right )^{3/2} \sqrt [4]{b} d}-\frac {\cos (c+d x) \left (a+b-b \cos ^2(c+d x)\right )}{4 a (a-b) d \left (a-b+2 b \cos ^2(c+d x)-b \cos ^4(c+d x)\right )} \] Output:

-1/8*(3*a^(1/2)-2*b^(1/2))*arctan(b^(1/4)*cos(d*x+c)/(a^(1/2)-b^(1/2))^(1/ 
2))/a^(3/2)/(a^(1/2)-b^(1/2))^(3/2)/b^(1/4)/d-1/8*(3*a^(1/2)+2*b^(1/2))*ar 
ctanh(b^(1/4)*cos(d*x+c)/(a^(1/2)+b^(1/2))^(1/2))/a^(3/2)/(a^(1/2)+b^(1/2) 
)^(3/2)/b^(1/4)/d-1/4*cos(d*x+c)*(a+b-b*cos(d*x+c)^2)/a/(a-b)/d/(a-b+2*b*c 
os(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 = 1.20 (sec) , antiderivative size = 469, normalized size of antiderivative = 2.12 \[ \int \frac {\sin (c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=-\frac {\frac {32 \cos (c+d x) (2 a+b-b \cos (2 (c+d x)))}{8 a-3 b+4 b \cos (2 (c+d x))-b \cos (4 (c+d x))}+i \text {RootSum}\left [b-4 b \text {$\#$1}^2-16 a \text {$\#$1}^4+6 b \text {$\#$1}^4-4 b \text {$\#$1}^6+b \text {$\#$1}^8\&,\frac {-2 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+i b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )+24 a \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2-10 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2-12 i a \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2+5 i b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2-24 a \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4+10 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4+12 i a \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4-5 i b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4+2 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^6-i b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^6}{-b \text {$\#$1}-8 a \text {$\#$1}^3+3 b \text {$\#$1}^3-3 b \text {$\#$1}^5+b \text {$\#$1}^7}\&\right ]}{32 a (a-b) d} \] Input:

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

Output:

-1/32*((32*Cos[c + d*x]*(2*a + b - b*Cos[2*(c + d*x)]))/(8*a - 3*b + 4*b*C 
os[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*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x 
] - #1)] + I*b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] + 24*a*ArcTan[Sin[c + d*x 
]/(Cos[c + d*x] - #1)]*#1^2 - 10*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1) 
]*#1^2 - (12*I)*a*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 + (5*I)*b*Log[1 - 
 2*Cos[c + d*x]*#1 + #1^2]*#1^2 - 24*a*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - 
 #1)]*#1^4 + 10*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 + (12*I)*a 
*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4 - (5*I)*b*Log[1 - 2*Cos[c + d*x]*# 
1 + #1^2]*#1^4 + 2*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^6 - I*b*L 
og[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) & ])/(a*(a - b)*d)
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {3042, 3694, 1405, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3694

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

\(\Big \downarrow \) 1405

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-((((((3*a - 2*b)/Sqrt[a] + Sqrt[b])*ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sq 
rt[a] - Sqrt[b]]])/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^(1/4)) + (((3*a - 2*b)/Sqr 
t[a] - Sqrt[b])*ArcTanh[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]])/( 
2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(1/4)))/(4*a*(a - b)) + (Cos[c + d*x]*(a + b - 
 b*Cos[c + d*x]^2))/(4*a*(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 1405

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(-x)*(b^2 
- 2*a*c + b*c*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[(b^2 - 2*a*c + 2*(p + 1)*( 
b^2 - 4*a*c) + b*c*(4*p + 7)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; Fr 
eeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IntegerQ[2*p]
 

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 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 = 2.39 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(-\frac {b^{2} \left (\frac {\frac {\left (\sqrt {a b}+a \right ) \cos \left (d x +c \right )}{2 b \left (a -b \right ) \left (\cos \left (d x +c \right )^{2}+\frac {\sqrt {a b}}{b}-1\right )}+\frac {\left (\sqrt {a b}+3 a -2 b \right ) \arctan \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 \left (a -b \right ) \sqrt {\left (\sqrt {a b}-b \right ) b}}}{4 \sqrt {a b}\, a b}+\frac {-\frac {\left (-\sqrt {a b}+a \right ) \cos \left (d x +c \right )}{2 b \left (a -b \right ) \left (\cos \left (d x +c \right )^{2}-1-\frac {\sqrt {a b}}{b}\right )}-\frac {\left (\sqrt {a b}-3 a +2 b \right ) \operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 \left (a -b \right ) \sqrt {\left (\sqrt {a b}+b \right ) b}}}{4 \sqrt {a b}\, a b}\right )}{d}\)	\(241\)
default	\(-\frac {b^{2} \left (\frac {\frac {\left (\sqrt {a b}+a \right ) \cos \left (d x +c \right )}{2 b \left (a -b \right ) \left (\cos \left (d x +c \right )^{2}+\frac {\sqrt {a b}}{b}-1\right )}+\frac {\left (\sqrt {a b}+3 a -2 b \right ) \arctan \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 \left (a -b \right ) \sqrt {\left (\sqrt {a b}-b \right ) b}}}{4 \sqrt {a b}\, a b}+\frac {-\frac {\left (-\sqrt {a b}+a \right ) \cos \left (d x +c \right )}{2 b \left (a -b \right ) \left (\cos \left (d x +c \right )^{2}-1-\frac {\sqrt {a b}}{b}\right )}-\frac {\left (\sqrt {a b}-3 a +2 b \right ) \operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 \left (a -b \right ) \sqrt {\left (\sqrt {a b}+b \right ) b}}}{4 \sqrt {a b}\, a b}\right )}{d}\)	\(241\)
risch	\(-\frac {b \,{\mathrm e}^{7 i \left (d x +c \right )}-4 a \,{\mathrm e}^{5 i \left (d x +c \right )}-b \,{\mathrm e}^{5 i \left (d x +c \right )}-4 a \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{3 i \left (d x +c \right )}+b \,{\mathrm e}^{i \left (d x +c \right )}}{2 a \left (a -b \right ) d \left ({\mathrm e}^{8 i \left (d x +c \right )} b -4 \,{\mathrm e}^{6 i \left (d x +c \right )} b -16 \,{\mathrm e}^{4 i \left (d x +c \right )} a +6 b \,{\mathrm e}^{4 i \left (d x +c \right )}-4 b \,{\mathrm e}^{2 i \left (d x +c \right )}+b \right )}-\frac {i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4096 a^{9} b \,d^{4}-12288 a^{8} b^{2} d^{4}+12288 a^{7} b^{3} d^{4}-4096 a^{6} b^{4} d^{4}\right ) \textit {\_Z}^{4}+\left (-1920 a^{5} b \,d^{2}+1920 a^{4} b^{2} d^{2}-512 a^{3} b^{3} d^{2}\right ) \textit {\_Z}^{2}-81 a^{2}+72 a b -16 b^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\left (-\frac {4096 i a^{7} d^{3} b}{81 a^{2}-81 a b +20 b^{2}}+\frac {14336 i a^{6} d^{3} b^{2}}{81 a^{2}-81 a b +20 b^{2}}-\frac {18432 i a^{5} d^{3} b^{3}}{81 a^{2}-81 a b +20 b^{2}}+\frac {10240 i a^{4} b^{4} d^{3}}{81 a^{2}-81 a b +20 b^{2}}-\frac {2048 i a^{3} b^{5} d^{3}}{81 a^{2}-81 a b +20 b^{2}}\right ) \textit {\_R}^{3}+\left (\frac {432 i a^{4} d}{81 a^{2}-81 a b +20 b^{2}}+\frac {576 i a^{3} d b}{81 a^{2}-81 a b +20 b^{2}}-\frac {1360 i a^{2} d \,b^{2}}{81 a^{2}-81 a b +20 b^{2}}+\frac {736 i a \,b^{3} d}{81 a^{2}-81 a b +20 b^{2}}-\frac {128 i b^{4} d}{81 a^{2}-81 a b +20 b^{2}}\right ) \textit {\_R} \right ) {\mathrm e}^{i \left (d x +c \right )}+1\right )\right )}{2}\)	\(557\)

Input:

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

Output:

-1/d*b^2*(1/4/(a*b)^(1/2)/a/b*(1/2*((a*b)^(1/2)+a)/b/(a-b)*cos(d*x+c)/(cos 
(d*x+c)^2+(a*b)^(1/2)/b-1)+1/2*((a*b)^(1/2)+3*a-2*b)/(a-b)/(((a*b)^(1/2)-b 
)*b)^(1/2)*arctan(b*cos(d*x+c)/(((a*b)^(1/2)-b)*b)^(1/2)))+1/4/(a*b)^(1/2) 
/a/b*(-1/2*(-(a*b)^(1/2)+a)/b/(a-b)*cos(d*x+c)/(cos(d*x+c)^2-1-(a*b)^(1/2) 
/b)-1/2*((a*b)^(1/2)-3*a+2*b)/(a-b)/(((a*b)^(1/2)+b)*b)^(1/2)*arctanh(b*co 
s(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. 2269 vs. \(2 (173) = 346\).

Time = 0.32 (sec) , antiderivative size = 2269, normalized size of antiderivative = 10.27 \[ \int \frac {\sin (c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/16*(4*b*cos(d*x + c)^3 - ((a^2*b - a*b^2)*d*cos(d*x + c)^4 - 2*(a^2*b - 
 a*b^2)*d*cos(d*x + c)^2 - (a^3 - 2*a^2*b + a*b^2)*d)*sqrt(-((a^6 - 3*a^5* 
b + 3*a^4*b^2 - a^3*b^3)*d^2*sqrt((81*a^2 - 90*a*b + 25*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)) 
 + 15*a^2 - 15*a*b + 4*b^2)/((a^6 - 3*a^5*b + 3*a^4*b^2 - a^3*b^3)*d^2))*l 
og((81*a^2 - 81*a*b + 20*b^2)*cos(d*x + c) + (2*(2*a^7*b - 7*a^6*b^2 + 9*a 
^5*b^3 - 5*a^4*b^4 + a^3*b^5)*d^3*sqrt((81*a^2 - 90*a*b + 25*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)) - (27*a^4 - 24*a^3*b + 5*a^2*b^2)*d)*sqrt(-((a^6 - 3*a^5*b + 3*a^4*b 
^2 - a^3*b^3)*d^2*sqrt((81*a^2 - 90*a*b + 25*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)) + 15*a^2 - 
 15*a*b + 4*b^2)/((a^6 - 3*a^5*b + 3*a^4*b^2 - a^3*b^3)*d^2))) + ((a^2*b - 
 a*b^2)*d*cos(d*x + c)^4 - 2*(a^2*b - a*b^2)*d*cos(d*x + c)^2 - (a^3 - 2*a 
^2*b + a*b^2)*d)*sqrt(((a^6 - 3*a^5*b + 3*a^4*b^2 - a^3*b^3)*d^2*sqrt((81* 
a^2 - 90*a*b + 25*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)) - 15*a^2 + 15*a*b - 4*b^2)/((a^6 - 3* 
a^5*b + 3*a^4*b^2 - a^3*b^3)*d^2))*log((81*a^2 - 81*a*b + 20*b^2)*cos(d*x 
+ c) + (2*(2*a^7*b - 7*a^6*b^2 + 9*a^5*b^3 - 5*a^4*b^4 + a^3*b^5)*d^3*sqrt 
((81*a^2 - 90*a*b + 25*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)) + (27*a^4 - 24*a^3*b + 5*a^2*...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

1/2*(4*b^2*cos(2*d*x + 2*c)*cos(d*x + c) + 4*b^2*sin(2*d*x + 2*c)*sin(d*x 
+ c) - b^2*cos(d*x + c) - 4*(4*a*b + b^2)*sin(3*d*x + 3*c)*sin(2*d*x + 2*c 
) - (b^2*cos(7*d*x + 7*c) + b^2*cos(d*x + c) - (4*a*b + b^2)*cos(5*d*x + 5 
*c) - (4*a*b + b^2)*cos(3*d*x + 3*c))*cos(8*d*x + 8*c) + (4*b^2*cos(6*d*x 
+ 6*c) + 4*b^2*cos(2*d*x + 2*c) - b^2 + 2*(8*a*b - 3*b^2)*cos(4*d*x + 4*c) 
)*cos(7*d*x + 7*c) + 4*(b^2*cos(d*x + c) - (4*a*b + b^2)*cos(5*d*x + 5*c) 
- (4*a*b + b^2)*cos(3*d*x + 3*c))*cos(6*d*x + 6*c) + (4*a*b + b^2 - 2*(32* 
a^2 - 4*a*b - 3*b^2)*cos(4*d*x + 4*c) - 4*(4*a*b + b^2)*cos(2*d*x + 2*c))* 
cos(5*d*x + 5*c) - 2*((32*a^2 - 4*a*b - 3*b^2)*cos(3*d*x + 3*c) - (8*a*b - 
 3*b^2)*cos(d*x + c))*cos(4*d*x + 4*c) + (4*a*b + b^2 - 4*(4*a*b + b^2)*co 
s(2*d*x + 2*c))*cos(3*d*x + 3*c) + 2*((a^2*b^2 - a*b^3)*d*cos(8*d*x + 8*c) 
^2 + 16*(a^2*b^2 - a*b^3)*d*cos(6*d*x + 6*c)^2 + 4*(64*a^4 - 112*a^3*b + 5 
7*a^2*b^2 - 9*a*b^3)*d*cos(4*d*x + 4*c)^2 + 16*(a^2*b^2 - a*b^3)*d*cos(2*d 
*x + 2*c)^2 + (a^2*b^2 - a*b^3)*d*sin(8*d*x + 8*c)^2 + 16*(a^2*b^2 - a*b^3 
)*d*sin(6*d*x + 6*c)^2 + 4*(64*a^4 - 112*a^3*b + 57*a^2*b^2 - 9*a*b^3)*d*s 
in(4*d*x + 4*c)^2 + 16*(8*a^3*b - 11*a^2*b^2 + 3*a*b^3)*d*sin(4*d*x + 4*c) 
*sin(2*d*x + 2*c) + 16*(a^2*b^2 - a*b^3)*d*sin(2*d*x + 2*c)^2 - 8*(a^2*b^2 
 - a*b^3)*d*cos(2*d*x + 2*c) + (a^2*b^2 - a*b^3)*d - 2*(4*(a^2*b^2 - a*b^3 
)*d*cos(6*d*x + 6*c) + 2*(8*a^3*b - 11*a^2*b^2 + 3*a*b^3)*d*cos(4*d*x + 4* 
c) + 4*(a^2*b^2 - a*b^3)*d*cos(2*d*x + 2*c) - (a^2*b^2 - a*b^3)*d)*cos(...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 693 vs. \(2 (173) = 346\).

Time = 0.75 (sec) , antiderivative size = 693, normalized size of antiderivative = 3.14 \[ \int \frac {\sin (c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=-\frac {\frac {b \cos \left (d x + c\right )^{3}}{d} - \frac {a \cos \left (d x + c\right )}{d} - \frac {b \cos \left (d x + c\right )}{d}}{4 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} - a + b\right )} {\left (a^{2} - a b\right )}} + \frac {{\left ({\left (3 \, a^{4} b - 8 \, a^{3} b^{2} + 7 \, a^{2} b^{3} - 2 \, a b^{4}\right )} \sqrt {-b^{2} + \sqrt {a b} b} d^{4} - {\left (3 \, a^{2} - 4 \, a b + b^{2}\right )} \sqrt {a b} \sqrt {-b^{2} + \sqrt {a b} b} d^{2} {\left | -a^{2} d^{2} + a b d^{2} \right |} + {\left (a^{2} d^{2} - a b d^{2}\right )}^{2} \sqrt {-b^{2} + \sqrt {a b} b} b\right )} \arctan \left (\frac {\cos \left (d x + c\right )}{d \sqrt {-\frac {a^{2} b d^{2} - a b^{2} d^{2} + \sqrt {{\left (a^{2} b d^{2} - a b^{2} d^{2}\right )}^{2} + {\left (a^{2} b d^{4} - a b^{2} d^{4}\right )} {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )}}}{a^{2} b d^{4} - a b^{2} d^{4}}}}\right )}{8 \, {\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \sqrt {a b} d^{3} {\left | -a^{2} d^{2} + a b d^{2} \right |} {\left | b \right |}} - \frac {{\left ({\left (3 \, a^{4} b - 8 \, a^{3} b^{2} + 7 \, a^{2} b^{3} - 2 \, a b^{4}\right )} \sqrt {-b^{2} - \sqrt {a b} b} d^{4} + {\left (3 \, a^{2} - 4 \, a b + b^{2}\right )} \sqrt {a b} \sqrt {-b^{2} - \sqrt {a b} b} d^{2} {\left | -a^{2} d^{2} + a b d^{2} \right |} + {\left (a^{2} d^{2} - a b d^{2}\right )}^{2} \sqrt {-b^{2} - \sqrt {a b} b} b\right )} \arctan \left (\frac {\cos \left (d x + c\right )}{d \sqrt {-\frac {a^{2} b d^{2} - a b^{2} d^{2} - \sqrt {{\left (a^{2} b d^{2} - a b^{2} d^{2}\right )}^{2} + {\left (a^{2} b d^{4} - a b^{2} d^{4}\right )} {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )}}}{a^{2} b d^{4} - a b^{2} d^{4}}}}\right )}{8 \, {\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \sqrt {a b} d^{3} {\left | -a^{2} d^{2} + a b d^{2} \right |} {\left | b \right |}} \] Input:

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

Output:

-1/4*(b*cos(d*x + c)^3/d - a*cos(d*x + c)/d - b*cos(d*x + c)/d)/((b*cos(d* 
x + c)^4 - 2*b*cos(d*x + c)^2 - a + b)*(a^2 - a*b)) + 1/8*((3*a^4*b - 8*a^ 
3*b^2 + 7*a^2*b^3 - 2*a*b^4)*sqrt(-b^2 + sqrt(a*b)*b)*d^4 - (3*a^2 - 4*a*b 
 + b^2)*sqrt(a*b)*sqrt(-b^2 + sqrt(a*b)*b)*d^2*abs(-a^2*d^2 + a*b*d^2) + ( 
a^2*d^2 - a*b*d^2)^2*sqrt(-b^2 + sqrt(a*b)*b)*b)*arctan(cos(d*x + c)/(d*sq 
rt(-(a^2*b*d^2 - a*b^2*d^2 + sqrt((a^2*b*d^2 - a*b^2*d^2)^2 + (a^2*b*d^4 - 
 a*b^2*d^4)*(a^3 - 2*a^2*b + a*b^2)))/(a^2*b*d^4 - a*b^2*d^4))))/((a^4 - 3 
*a^3*b + 3*a^2*b^2 - a*b^3)*sqrt(a*b)*d^3*abs(-a^2*d^2 + a*b*d^2)*abs(b)) 
- 1/8*((3*a^4*b - 8*a^3*b^2 + 7*a^2*b^3 - 2*a*b^4)*sqrt(-b^2 - sqrt(a*b)*b 
)*d^4 + (3*a^2 - 4*a*b + b^2)*sqrt(a*b)*sqrt(-b^2 - sqrt(a*b)*b)*d^2*abs(- 
a^2*d^2 + a*b*d^2) + (a^2*d^2 - a*b*d^2)^2*sqrt(-b^2 - sqrt(a*b)*b)*b)*arc 
tan(cos(d*x + c)/(d*sqrt(-(a^2*b*d^2 - a*b^2*d^2 - sqrt((a^2*b*d^2 - a*b^2 
*d^2)^2 + (a^2*b*d^4 - a*b^2*d^4)*(a^3 - 2*a^2*b + a*b^2)))/(a^2*b*d^4 - a 
*b^2*d^4))))/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*sqrt(a*b)*d^3*abs(-a^2*d 
^2 + a*b*d^2)*abs(b))
 

Mupad [B] (verification not implemented)

Time = 38.91 (sec) , antiderivative size = 3507, normalized size of antiderivative = 15.87 \[ \int \frac {\sin (c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((b*cos(c + d*x)^3)/(4*a*(a - b)) - (cos(c + d*x)*(a + b))/(4*a*(a - b)))/ 
(d*(a - b + 2*b*cos(c + d*x)^2 - b*cos(c + d*x)^4)) + (atan(((((256*a^3*b^ 
5 - 1024*a^4*b^4 + 768*a^5*b^3)/(64*(a^5 - 2*a^4*b + a^3*b^2)) - (cos(c + 
d*x)*(256*a^3*b^6 - 512*a^4*b^5 + 256*a^5*b^4)*(-(15*a^5*b - 9*a*(a^9*b)^( 
1/2) + 5*b*(a^9*b)^(1/2) + 4*a^3*b^3 - 15*a^4*b^2)/(256*(a^9*b - a^6*b^4 + 
 3*a^7*b^3 - 3*a^8*b^2)))^(1/2))/(4*(a^4 - 2*a^3*b + a^2*b^2)))*(-(15*a^5* 
b - 9*a*(a^9*b)^(1/2) + 5*b*(a^9*b)^(1/2) + 4*a^3*b^3 - 15*a^4*b^2)/(256*( 
a^9*b - a^6*b^4 + 3*a^7*b^3 - 3*a^8*b^2)))^(1/2) + (cos(c + d*x)*(4*b^5 - 
11*a*b^4 + 9*a^2*b^3))/(4*(a^4 - 2*a^3*b + a^2*b^2)))*(-(15*a^5*b - 9*a*(a 
^9*b)^(1/2) + 5*b*(a^9*b)^(1/2) + 4*a^3*b^3 - 15*a^4*b^2)/(256*(a^9*b - a^ 
6*b^4 + 3*a^7*b^3 - 3*a^8*b^2)))^(1/2)*1i - (((256*a^3*b^5 - 1024*a^4*b^4 
+ 768*a^5*b^3)/(64*(a^5 - 2*a^4*b + a^3*b^2)) + (cos(c + d*x)*(256*a^3*b^6 
 - 512*a^4*b^5 + 256*a^5*b^4)*(-(15*a^5*b - 9*a*(a^9*b)^(1/2) + 5*b*(a^9*b 
)^(1/2) + 4*a^3*b^3 - 15*a^4*b^2)/(256*(a^9*b - a^6*b^4 + 3*a^7*b^3 - 3*a^ 
8*b^2)))^(1/2))/(4*(a^4 - 2*a^3*b + a^2*b^2)))*(-(15*a^5*b - 9*a*(a^9*b)^( 
1/2) + 5*b*(a^9*b)^(1/2) + 4*a^3*b^3 - 15*a^4*b^2)/(256*(a^9*b - a^6*b^4 + 
 3*a^7*b^3 - 3*a^8*b^2)))^(1/2) - (cos(c + d*x)*(4*b^5 - 11*a*b^4 + 9*a^2* 
b^3))/(4*(a^4 - 2*a^3*b + a^2*b^2)))*(-(15*a^5*b - 9*a*(a^9*b)^(1/2) + 5*b 
*(a^9*b)^(1/2) + 4*a^3*b^3 - 15*a^4*b^2)/(256*(a^9*b - a^6*b^4 + 3*a^7*b^3 
 - 3*a^8*b^2)))^(1/2)*1i)/((9*a*b^3 - 4*b^4)/(32*(a^5 - 2*a^4*b + a^3*b...
 

Reduce [F]

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

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

Output:

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