\(\int \frac {\sin ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) [206]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3296, 1144, 211} \[ \int \frac {\sin ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt {b} d \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {\arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt {b} d \sqrt {\sqrt {a}+\sqrt {b}}} \]

[In]

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

[Out]

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

Rule 211

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

Rule 1144

Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(
d^2/2)*(b/q + 1), Int[(d*x)^(m - 2)/(b/2 + q/2 + c*x^2), x], x] - Dist[(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 3296

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^2}{a+2 a x^2+(a-b) x^4} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\left (1-\frac {\sqrt {a}}{\sqrt {b}}\right ) \text {Subst}\left (\int \frac {1}{a-\sqrt {a} \sqrt {b}+(a-b) x^2} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (1+\frac {\sqrt {a}}{\sqrt {b}}\right ) \text {Subst}\left (\int \frac {1}{a+\sqrt {a} \sqrt {b}+(a-b) x^2} \, dx,x,\tan (c+d x)\right )}{2 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}-\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.54 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.10 \[ \int \frac {\sin ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {\arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{2 \sqrt {a+\sqrt {a} \sqrt {b}} \sqrt {b} d}-\frac {\text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{2 \sqrt {-a+\sqrt {a} \sqrt {b}} \sqrt {b} d} \]

[In]

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

[Out]

-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)

Maple [C] (verified)

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

Time = 0.69 (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	\(\frac {\left (a -b \right ) \left (\frac {\left (\sqrt {a b}+a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \left (a -b \right ) \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (\sqrt {a b}-a \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{d}\)	\(145\)

[In]

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

[Out]

-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 = 0.41 (sec) , antiderivative size = 1087, normalized size of antiderivative = 8.70 \[ \int \frac {\sin ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-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*co
s(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*co
s(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)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

-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 = 0.79 (sec) , antiderivative size = 397, normalized size of antiderivative = 3.18 \[ \int \frac {\sin ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {\frac {{\left (3 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a^{2} - 6 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a b - \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (d x + c\right )}{\sqrt {\frac {4 \, a + \sqrt {-16 \, {\left (a - b\right )} a + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left | a - b \right |}}{3 \, a^{5} b - 12 \, a^{4} b^{2} + 14 \, a^{3} b^{3} - 4 \, a^{2} b^{4} - a b^{5}} + \frac {{\left (3 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a^{2} - 6 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a b - \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (d x + c\right )}{\sqrt {\frac {4 \, a - \sqrt {-16 \, {\left (a - b\right )} a + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left | a - b \right |}}{3 \, a^{5} b - 12 \, a^{4} b^{2} + 14 \, a^{3} b^{3} - 4 \, a^{2} b^{4} - a b^{5}}}{2 \, d} \]

[In]

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

[Out]

-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)/
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) + (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 = 16.78 (sec) , antiderivative size = 443, normalized size of antiderivative = 3.54 \[ \int \frac {\sin ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\ln \left (a\,b-a^2-\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\left (a-b\right )\,\sqrt {-\frac {1}{a\,b+\sqrt {a\,b^3}}}\,\left (2\,a\,b^2+a\,\sqrt {a\,b^3}+b\,\sqrt {a\,b^3}\right )}{a\,b+\sqrt {a\,b^3}}\right )\,\sqrt {-\frac {1}{a\,b+\sqrt {a\,b^3}}}}{4\,d}-\frac {\ln \left (a\,b-a^2-\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{a\,b-\sqrt {a\,b^3}}}\,\left (a-b\right )\,\left (a\,\sqrt {a\,b^3}-2\,a\,b^2+b\,\sqrt {a\,b^3}\right )}{a\,b-\sqrt {a\,b^3}}\right )\,\sqrt {\frac {a\,b+\sqrt {a\,b^3}}{16\,\left (a\,b^3-a^2\,b^2\right )}}}{d}+\frac {\ln \left (a\,b-a^2+\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{a\,b-\sqrt {a\,b^3}}}\,\left (a-b\right )\,\left (a\,\sqrt {a\,b^3}-2\,a\,b^2+b\,\sqrt {a\,b^3}\right )}{a\,b-\sqrt {a\,b^3}}\right )\,\sqrt {-\frac {1}{a\,b-\sqrt {a\,b^3}}}}{4\,d}-\frac {\ln \left (a\,b-a^2+\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\left (a-b\right )\,\sqrt {-\frac {1}{a\,b+\sqrt {a\,b^3}}}\,\left (2\,a\,b^2+a\,\sqrt {a\,b^3}+b\,\sqrt {a\,b^3}\right )}{a\,b+\sqrt {a\,b^3}}\right )\,\sqrt {\frac {a\,b-\sqrt {a\,b^3}}{16\,\left (a\,b^3-a^2\,b^2\right )}}}{d} \]

[In]

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

[Out]

(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