3.206 \(\int \frac{\sin ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx\)
Optimal. Leaf size=125 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt{b} d \sqrt{\sqrt{a}-\sqrt{b}}}-\frac{\tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt{b} d \sqrt{\sqrt{a}+\sqrt{b}}} \]
[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)
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Rubi [A] time = 0.112376, antiderivative size = 125, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used =
{3217, 1130, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt{b} d \sqrt{\sqrt{a}-\sqrt{b}}}-\frac{\tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt{b} d \sqrt{\sqrt{a}+\sqrt{b}}} \]
Antiderivative was successfully verified.
[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 3217
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]
Rule 1130
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*(b/q + 1))/2, Int[(d*x)^(m - 2)/(b/2 + q/2 + c*x^2), x], x] - Dist[(d^2*(b/q - 1))/2, 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 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{\sin ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx &=\frac{\operatorname{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 ) \operatorname{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 ) \operatorname{Subst}\left (\int \frac{1}{a+\sqrt{a} \sqrt{b}+(a-b) x^2} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{\tan ^{-1}\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{\tan ^{-1}\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] time = 0.34862, size = 137, normalized size = 1.1 \[ -\frac{\tan ^{-1}\left (\frac{\left (\sqrt{a}+\sqrt{b}\right ) \tan (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}+a}}\right )}{2 \sqrt{b} d \sqrt{\sqrt{a} \sqrt{b}+a}}-\frac{\tanh ^{-1}\left (\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tan (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}-a}}\right )}{2 \sqrt{b} d \sqrt{\sqrt{a} \sqrt{b}-a}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sin[c + d*x]^2/(a - b*Sin[c + d*x]^4),x]
[Out]
-ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]]/(2*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)
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Maple [B] time = 0.096, size = 492, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(d*x+c)^2/(a-b*sin(d*x+c)^4),x)
[Out]
1/2/d*a/(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))+1/2/d*a^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))+1/2/d*a/
(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))-1/2/d*a^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))-1/2/d*b/(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))-1/2/d*a/(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))*b-1/2/d*b/(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))+1/2/d*a/(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))*b
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{\sin \left (d x + c\right )^{2}}{b \sin \left (d x + c\right )^{4} - a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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Fricas [B] time = 3.27401, size = 2390, normalized size = 19.12 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(d*x+c)**2/(a-b*sin(d*x+c)**4),x)
[Out]
Timed out
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(d*x+c)^2/(a-b*sin(d*x+c)^4),x, algorithm="giac")
[Out]
Exception raised: NotImplementedError