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

Optimal. Leaf size=499 \[ -\frac{\cos ^2(c+d x) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}\right ) \sqrt{(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}{2 \sqrt{b} d \sqrt{a+b \sin ^4(c+d x)}}-\frac{\sqrt [4]{a} \left (\sqrt{a+b}+\sqrt{a}\right ) \cos ^2(c+d x) \left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right ) \sqrt{\frac{(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2} \left (1-\frac{\sqrt{a}}{\sqrt{a+b}}\right )\right )}{2 b d \sqrt [4]{a+b} \sqrt{a+b \sin ^4(c+d x)}}+\frac{\left (\sqrt{a+b}+\sqrt{a}\right )^2 \cos ^2(c+d x) \left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right ) \sqrt{\frac{(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right )^2}} \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{a+b}\right )^2}{4 \sqrt{a} \sqrt{a+b}};2 \tan ^{-1}\left (\frac{\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2} \left (1-\frac{\sqrt{a}}{\sqrt{a+b}}\right )\right )}{4 \sqrt [4]{a} b d \sqrt [4]{a+b} \sqrt{a+b \sin ^4(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.664941, antiderivative size = 499, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3219, 1319, 1103, 1706} \[ -\frac{\cos ^2(c+d x) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}\right ) \sqrt{(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}{2 \sqrt{b} d \sqrt{a+b \sin ^4(c+d x)}}-\frac{\sqrt [4]{a} \left (\sqrt{a+b}+\sqrt{a}\right ) \cos ^2(c+d x) \left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right ) \sqrt{\frac{(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2} \left (1-\frac{\sqrt{a}}{\sqrt{a+b}}\right )\right )}{2 b d \sqrt [4]{a+b} \sqrt{a+b \sin ^4(c+d x)}}+\frac{\left (\sqrt{a+b}+\sqrt{a}\right )^2 \cos ^2(c+d x) \left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right ) \sqrt{\frac{(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right )^2}} \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{a+b}\right )^2}{4 \sqrt{a} \sqrt{a+b}};2 \tan ^{-1}\left (\frac{\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2} \left (1-\frac{\sqrt{a}}{\sqrt{a+b}}\right )\right )}{4 \sqrt [4]{a} b d \sqrt [4]{a+b} \sqrt{a+b \sin ^4(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3219

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_), x_Symbol] :> With[{ff = FreeFa
ctors[Tan[e + f*x], x]}, Dist[(ff^(m + 1)*(a + b*Sin[e + f*x]^4)^p*(Sec[e + f*x]^2)^(2*p))/(f*Apart[a*(1 + Tan
[e + f*x]^2)^2 + b*Tan[e + f*x]^4]^p), Subst[Int[(x^m*ExpandToSum[a*(1 + ff^2*x^2)^2 + b*ff^4*x^4, x]^p)/(1 +
ff^2*x^2)^(m/2 + 2*p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integ
erQ[p - 1/2]

Rule 1319

Int[(x_)^2/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]
}, -Dist[(a*(e + d*q))/(c*d^2 - a*e^2), Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + Dist[(a*d*(e + d*q))/(c*d^2 -
a*e^2), Int[(1 + q*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2
- 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && PosQ[c/a] && NeQ[c*d^2 - a*e^2, 0]

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1706

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[
{q = Rt[B/A, 2]}, -Simp[((B*d - A*e)*ArcTan[(Rt[-b + (c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + b*x^2 + c*x^4]])/(2*d*e
*Rt[-b + (c*d)/e + (a*e)/d, 2]), x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + b*x^2 + c*x^4))/(a*(A + B*x
^2)^2)]*EllipticPi[Cancel[-((B*d - A*e)^2/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2 - (b*A)/(4*a*B)])/(4*d*e*A*q*Sqrt[
a + b*x^2 + c*x^4]), x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^
2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rubi steps

\begin{align*} \int \frac{\sin ^2(c+d x)}{\sqrt{a+b \sin ^4(c+d x)}} \, dx &=\frac{\left (\cos ^2(c+d x) \sqrt{a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right ) \sqrt{a+2 a x^2+(a+b) x^4}} \, dx,x,\tan (c+d x)\right )}{d \sqrt{a+b \sin ^4(c+d x)}}\\ &=-\frac{\left (a \left (1+\frac{\sqrt{a+b}}{\sqrt{a}}\right ) \cos ^2(c+d x) \sqrt{a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+2 a x^2+(a+b) x^4}} \, dx,x,\tan (c+d x)\right )}{b d \sqrt{a+b \sin ^4(c+d x)}}+\frac{\left (a \left (1+\frac{\sqrt{a+b}}{\sqrt{a}}\right ) \cos ^2(c+d x) \sqrt{a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1+\frac{\sqrt{a+b} x^2}{\sqrt{a}}}{\left (1+x^2\right ) \sqrt{a+2 a x^2+(a+b) x^4}} \, dx,x,\tan (c+d x)\right )}{b d \sqrt{a+b \sin ^4(c+d x)}}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}}\right ) \cos ^2(c+d x) \sqrt{a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}}{2 \sqrt{b} d \sqrt{a+b \sin ^4(c+d x)}}-\frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{a+b}\right ) \cos ^2(c+d x) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2} \left (1-\frac{\sqrt{a}}{\sqrt{a+b}}\right )\right ) \left (\sqrt{a}+\sqrt{a+b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{a+b} \tan ^2(c+d x)\right )^2}}}{2 b \sqrt [4]{a+b} d \sqrt{a+b \sin ^4(c+d x)}}+\frac{\left (\sqrt{a}+\sqrt{a+b}\right )^2 \cos ^2(c+d x) \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{a+b}\right )^2}{4 \sqrt{a} \sqrt{a+b}};2 \tan ^{-1}\left (\frac{\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2} \left (1-\frac{\sqrt{a}}{\sqrt{a+b}}\right )\right ) \left (\sqrt{a}+\sqrt{a+b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{a+b} \tan ^2(c+d x)\right )^2}}}{4 \sqrt [4]{a} b \sqrt [4]{a+b} d \sqrt{a+b \sin ^4(c+d x)}}\\ \end{align*}

Mathematica [C]  time = 2.77753, size = 287, normalized size = 0.58 \[ -\frac{2 i \cos ^2(c+d x) \sqrt{1+\left (1+\frac{i \sqrt{b}}{\sqrt{a}}\right ) \tan ^2(c+d x)} \sqrt{2+\left (2-\frac{2 i \sqrt{b}}{\sqrt{a}}\right ) \tan ^2(c+d x)} \left (F\left (i \sinh ^{-1}\left (\sqrt{1-\frac{i \sqrt{b}}{\sqrt{a}}} \tan (c+d x)\right )|\frac{\sqrt{a}+i \sqrt{b}}{\sqrt{a}-i \sqrt{b}}\right )-\Pi \left (\frac{\sqrt{a}}{\sqrt{a}-i \sqrt{b}};i \sinh ^{-1}\left (\sqrt{1-\frac{i \sqrt{b}}{\sqrt{a}}} \tan (c+d x)\right )|\frac{\sqrt{a}+i \sqrt{b}}{\sqrt{a}-i \sqrt{b}}\right )\right )}{d \sqrt{1-\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{8 a-4 b \cos (2 (c+d x))+b \cos (4 (c+d x))+3 b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*Cos[c + d*x]^2*(EllipticF[I*ArcSinh[Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]*Tan[c + d*x]], (Sqrt[a] + I*Sqrt[b])
/(Sqrt[a] - I*Sqrt[b])] - EllipticPi[Sqrt[a]/(Sqrt[a] - I*Sqrt[b]), I*ArcSinh[Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]*Ta
n[c + d*x]], (Sqrt[a] + I*Sqrt[b])/(Sqrt[a] - I*Sqrt[b])])*Sqrt[1 + (1 + (I*Sqrt[b])/Sqrt[a])*Tan[c + d*x]^2]*
Sqrt[2 + (2 - ((2*I)*Sqrt[b])/Sqrt[a])*Tan[c + d*x]^2])/(Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]*d*Sqrt[8*a + 3*b - 4*b*
Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)]])

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Maple [A]  time = 4.555, size = 881, normalized size = 1.8 \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)^(1/2),x)

[Out]

-1/2*((4*a+cos(2*d*x+2*c)^2*b+b-2*b*cos(2*d*x+2*c))*sin(2*d*x+2*c)^2)^(1/2)*(-a*b)^(1/2)*((-b+(-a*b)^(1/2))*(-
1+cos(2*d*x+2*c))/(-a*b)^(1/2)/(cos(2*d*x+2*c)+1))^(1/2)*(cos(2*d*x+2*c)+1)^2*((-b*cos(2*d*x+2*c)+2*(-a*b)^(1/
2)+b)/(-a*b)^(1/2)/(cos(2*d*x+2*c)+1))^(1/2)*((b*cos(2*d*x+2*c)+2*(-a*b)^(1/2)-b)/(-a*b)^(1/2)/(cos(2*d*x+2*c)
+1))^(1/2)*(EllipticF(((-b+(-a*b)^(1/2))*(-1+cos(2*d*x+2*c))/(-a*b)^(1/2)/(cos(2*d*x+2*c)+1))^(1/2),((b+(-a*b)
^(1/2))/(-b+(-a*b)^(1/2)))^(1/2))-2*EllipticPi(((-b+(-a*b)^(1/2))*(-1+cos(2*d*x+2*c))/(-a*b)^(1/2)/(cos(2*d*x+
2*c)+1))^(1/2),(-a*b)^(1/2)/(-b+(-a*b)^(1/2)),((b+(-a*b)^(1/2))/(-b+(-a*b)^(1/2)))^(1/2)))/(-b+(-a*b)^(1/2))/(
1/b*(-1+cos(2*d*x+2*c))*(cos(2*d*x+2*c)+1)*(-b*cos(2*d*x+2*c)+2*(-a*b)^(1/2)+b)*(b*cos(2*d*x+2*c)+2*(-a*b)^(1/
2)-b))^(1/2)/sin(2*d*x+2*c)/(4*a+cos(2*d*x+2*c)^2*b+b-2*b*cos(2*d*x+2*c))^(1/2)/d-1/2*((4*a+cos(2*d*x+2*c)^2*b
+b-2*b*cos(2*d*x+2*c))*sin(2*d*x+2*c)^2)^(1/2)*(-a*b)^(1/2)*((-b+(-a*b)^(1/2))*(-1+cos(2*d*x+2*c))/(-a*b)^(1/2
)/(cos(2*d*x+2*c)+1))^(1/2)*(cos(2*d*x+2*c)+1)^2*((-b*cos(2*d*x+2*c)+2*(-a*b)^(1/2)+b)/(-a*b)^(1/2)/(cos(2*d*x
+2*c)+1))^(1/2)*((b*cos(2*d*x+2*c)+2*(-a*b)^(1/2)-b)/(-a*b)^(1/2)/(cos(2*d*x+2*c)+1))^(1/2)*EllipticF(((-b+(-a
*b)^(1/2))*(-1+cos(2*d*x+2*c))/(-a*b)^(1/2)/(cos(2*d*x+2*c)+1))^(1/2),((b+(-a*b)^(1/2))/(-b+(-a*b)^(1/2)))^(1/
2))/(-b+(-a*b)^(1/2))/(1/b*(-1+cos(2*d*x+2*c))*(cos(2*d*x+2*c)+1)*(-b*cos(2*d*x+2*c)+2*(-a*b)^(1/2)+b)*(b*cos(
2*d*x+2*c)+2*(-a*b)^(1/2)-b))^(1/2)/sin(2*d*x+2*c)/(4*a+cos(2*d*x+2*c)^2*b+b-2*b*cos(2*d*x+2*c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (d x + c\right )^{2}}{\sqrt{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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\cos \left (d x + c\right )^{2} - 1}{\sqrt{b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b}}, x\right ) \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)^(1/2),x, algorithm="fricas")

[Out]

integral(-(cos(d*x + c)^2 - 1)/sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b), x)

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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)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (d x + c\right )^{2}}{\sqrt{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)^(1/2),x, algorithm="giac")

[Out]

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