\(\int \frac {\sin ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) [155]

Optimal result

Integrand size = 24, antiderivative size = 125 \[ \int \frac {\sin ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt {\sqrt {a}-\sqrt {b}} \sqrt {b} d}-\frac {\arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt {\sqrt {a}+\sqrt {b}} \sqrt {b} d} \] Output:

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))^(1/2)/b^(1/2)/d-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))^(1/2)/b^(1/2)/d
 

Mathematica [A] (verified)

Time = 3.12 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.10 \[ \int \frac {\sin ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {\arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{2 \sqrt {a+\sqrt {a} \sqrt {b}} \sqrt {b} d}-\frac {\text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{2 \sqrt {-a+\sqrt {a} \sqrt {b}} \sqrt {b} d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.34, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3696, 1450, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3696

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

\(\Big \downarrow \) 1450

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Defintions of rubi rules used

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 1450

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

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 [C] (verified)

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

Time = 0.66 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.91

method	result	size
risch	\(-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (1+\left (a^{2} b^{2} d^{4}-a \,b^{3} d^{4}\right ) \textit {\_Z}^{4}+2 a \,d^{2} \textit {\_Z}^{2} b \right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (2 i a^{2} b \,d^{3}-2 i a \,b^{2} d^{3}\right ) \textit {\_R}^{3}+\left (-2 a^{2} d^{2}+2 b \,d^{2} a \right ) \textit {\_R}^{2}+4 i a d \textit {\_R} -\frac {2 a}{b}-1\right )\right )}{4}\)	\(114\)
derivativedivides	\(\frac {\left (a -b \right ) \left (\frac {\left (\sqrt {a b}+a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \left (a -b \right ) \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (\sqrt {a b}-a \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{d}\)	\(145\)
default	\(\frac {\left (a -b \right ) \left (\frac {\left (\sqrt {a b}+a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \left (a -b \right ) \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (\sqrt {a b}-a \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{d}\)	\(145\)

Input:

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

Output:

-1/4*sum(_R*ln(exp(2*I*(d*x+c))+(2*I*a^2*b*d^3-2*I*a*b^2*d^3)*_R^3+(-2*a^2 
*d^2+2*a*b*d^2)*_R^2+4*I*a*d*_R-2/b*a-1),_R=RootOf(1+(a^2*b^2*d^4-a*b^3*d^ 
4)*_Z^4+2*a*d^2*_Z^2*b))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1087 vs. \(2 (85) = 170\).

Time = 0.19 (sec) , antiderivative size = 1087, normalized size of antiderivative = 8.70 \[ \int \frac {\sin ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/8*sqrt(-((a*b - b^2)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + 1) 
/((a*b - b^2)*d^2))*log(1/4*cos(d*x + c)^2 + 1/2*((a^2*b - a*b^2)*d^3*sqrt 
(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4))*cos(d*x + c)*sin(d*x + c) - a*d*cos( 
d*x + c)*sin(d*x + c))*sqrt(-((a*b - b^2)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + 
 a*b^3)*d^4)) + 1)/((a*b - b^2)*d^2)) - 1/4*(2*(a^2 - a*b)*d^2*cos(d*x + c 
)^2 - (a^2 - a*b)*d^2)*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) - 1/4) + 
1/8*sqrt(-((a*b - b^2)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + 1)/ 
((a*b - b^2)*d^2))*log(1/4*cos(d*x + c)^2 - 1/2*((a^2*b - a*b^2)*d^3*sqrt( 
1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4))*cos(d*x + c)*sin(d*x + c) - a*d*cos(d 
*x + c)*sin(d*x + c))*sqrt(-((a*b - b^2)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + 
a*b^3)*d^4)) + 1)/((a*b - b^2)*d^2)) - 1/4*(2*(a^2 - a*b)*d^2*cos(d*x + c) 
^2 - (a^2 - a*b)*d^2)*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) - 1/4) - 1 
/8*sqrt(((a*b - b^2)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) - 1)/(( 
a*b - b^2)*d^2))*log(-1/4*cos(d*x + c)^2 + 1/2*((a^2*b - a*b^2)*d^3*sqrt(1 
/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4))*cos(d*x + c)*sin(d*x + c) + a*d*cos(d* 
x + c)*sin(d*x + c))*sqrt(((a*b - b^2)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a* 
b^3)*d^4)) - 1)/((a*b - b^2)*d^2)) - 1/4*(2*(a^2 - a*b)*d^2*cos(d*x + c)^2 
 - (a^2 - a*b)*d^2)*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + 1/4) + 1/8 
*sqrt(((a*b - b^2)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) - 1)/((a* 
b - b^2)*d^2))*log(-1/4*cos(d*x + c)^2 - 1/2*((a^2*b - a*b^2)*d^3*sqrt(...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 397 vs. \(2 (85) = 170\).

Time = 0.74 (sec) , antiderivative size = 397, normalized size of antiderivative = 3.18 \[ \int \frac {\sin ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {\frac {{\left (3 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a^{2} - 6 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a b - \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (d x + c\right )}{\sqrt {\frac {4 \, a + \sqrt {-16 \, {\left (a - b\right )} a + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left | a - b \right |}}{3 \, a^{5} b - 12 \, a^{4} b^{2} + 14 \, a^{3} b^{3} - 4 \, a^{2} b^{4} - a b^{5}} + \frac {{\left (3 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a^{2} - 6 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a b - \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (d x + c\right )}{\sqrt {\frac {4 \, a - \sqrt {-16 \, {\left (a - b\right )} a + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left | a - b \right |}}{3 \, a^{5} b - 12 \, a^{4} b^{2} + 14 \, a^{3} b^{3} - 4 \, a^{2} b^{4} - a b^{5}}}{2 \, d} \] Input:

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

Output:

-1/2*((3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2 - 6*sqrt(a^2 - 
a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a*b - sqrt(a^2 - a*b + sqrt(a*b)*(a - b 
))*sqrt(a*b)*b^2)*(pi*floor((d*x + c)/pi + 1/2) + arctan(2*tan(d*x + c)/sq 
rt((4*a + sqrt(-16*(a - b)*a + 16*a^2))/(a - b))))*abs(a - b)/(3*a^5*b - 1 
2*a^4*b^2 + 14*a^3*b^3 - 4*a^2*b^4 - a*b^5) + (3*sqrt(a^2 - a*b - sqrt(a*b 
)*(a - b))*sqrt(a*b)*a^2 - 6*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b) 
*a*b - sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*b^2)*(pi*floor((d*x + 
 c)/pi + 1/2) + arctan(2*tan(d*x + c)/sqrt((4*a - sqrt(-16*(a - b)*a + 16* 
a^2))/(a - b))))*abs(a - b)/(3*a^5*b - 12*a^4*b^2 + 14*a^3*b^3 - 4*a^2*b^4 
 - a*b^5))/d
 

Mupad [B] (verification not implemented)

Time = 38.37 (sec) , antiderivative size = 443, normalized size of antiderivative = 3.54 \[ \int \frac {\sin ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\ln \left (a\,b-a^2-\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\left (a-b\right )\,\sqrt {-\frac {1}{a\,b+\sqrt {a\,b^3}}}\,\left (2\,a\,b^2+a\,\sqrt {a\,b^3}+b\,\sqrt {a\,b^3}\right )}{a\,b+\sqrt {a\,b^3}}\right )\,\sqrt {-\frac {1}{a\,b+\sqrt {a\,b^3}}}}{4\,d}-\frac {\ln \left (a\,b-a^2-\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{a\,b-\sqrt {a\,b^3}}}\,\left (a-b\right )\,\left (a\,\sqrt {a\,b^3}-2\,a\,b^2+b\,\sqrt {a\,b^3}\right )}{a\,b-\sqrt {a\,b^3}}\right )\,\sqrt {\frac {a\,b+\sqrt {a\,b^3}}{16\,\left (a\,b^3-a^2\,b^2\right )}}}{d}+\frac {\ln \left (a\,b-a^2+\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{a\,b-\sqrt {a\,b^3}}}\,\left (a-b\right )\,\left (a\,\sqrt {a\,b^3}-2\,a\,b^2+b\,\sqrt {a\,b^3}\right )}{a\,b-\sqrt {a\,b^3}}\right )\,\sqrt {-\frac {1}{a\,b-\sqrt {a\,b^3}}}}{4\,d}-\frac {\ln \left (a\,b-a^2+\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\left (a-b\right )\,\sqrt {-\frac {1}{a\,b+\sqrt {a\,b^3}}}\,\left (2\,a\,b^2+a\,\sqrt {a\,b^3}+b\,\sqrt {a\,b^3}\right )}{a\,b+\sqrt {a\,b^3}}\right )\,\sqrt {\frac {a\,b-\sqrt {a\,b^3}}{16\,\left (a\,b^3-a^2\,b^2\right )}}}{d} \] Input:

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

Output:

(log(a*b - a^2 - (a*tan(c + d*x)*(a - b)*(-1/(a*b + (a*b^3)^(1/2)))^(1/2)* 
(2*a*b^2 + a*(a*b^3)^(1/2) + b*(a*b^3)^(1/2)))/(a*b + (a*b^3)^(1/2)))*(-1/ 
(a*b + (a*b^3)^(1/2)))^(1/2))/(4*d) - (log(a*b - a^2 - (a*tan(c + d*x)*(-1 
/(a*b - (a*b^3)^(1/2)))^(1/2)*(a - b)*(a*(a*b^3)^(1/2) - 2*a*b^2 + b*(a*b^ 
3)^(1/2)))/(a*b - (a*b^3)^(1/2)))*((a*b + (a*b^3)^(1/2))/(16*(a*b^3 - a^2* 
b^2)))^(1/2))/d + (log(a*b - a^2 + (a*tan(c + d*x)*(-1/(a*b - (a*b^3)^(1/2 
)))^(1/2)*(a - b)*(a*(a*b^3)^(1/2) - 2*a*b^2 + b*(a*b^3)^(1/2)))/(a*b - (a 
*b^3)^(1/2)))*(-1/(a*b - (a*b^3)^(1/2)))^(1/2))/(4*d) - (log(a*b - a^2 + ( 
a*tan(c + d*x)*(a - b)*(-1/(a*b + (a*b^3)^(1/2)))^(1/2)*(2*a*b^2 + a*(a*b^ 
3)^(1/2) + b*(a*b^3)^(1/2)))/(a*b + (a*b^3)^(1/2)))*((a*b - (a*b^3)^(1/2)) 
/(16*(a*b^3 - a^2*b^2)))^(1/2))/d
 

Reduce [F]

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

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

Output:

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