3.326 \(\int \frac{1}{(a+b x^2)^{3/4} (c+d x^2)} \, dx\)

Optimal. Leaf size=152 \[ \frac{\sqrt [4]{a} \sqrt{-\frac{b x^2}{a}} \Pi \left (-\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{x (b c-a d)}+\frac{\sqrt [4]{a} \sqrt{-\frac{b x^2}{a}} \Pi \left (\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{x (b c-a d)} \]

[Out]

(a^(1/4)*Sqrt[-((b*x^2)/a)]*EllipticPi[-((Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d]), ArcSin[(a + b*x^2)^(1/4)/a^(1/
4)], -1])/((b*c - a*d)*x) + (a^(1/4)*Sqrt[-((b*x^2)/a)]*EllipticPi[(Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d], ArcSi
n[(a + b*x^2)^(1/4)/a^(1/4)], -1])/((b*c - a*d)*x)

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Rubi [A]  time = 0.117927, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {401, 108, 409, 1218} \[ \frac{\sqrt [4]{a} \sqrt{-\frac{b x^2}{a}} \Pi \left (-\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{x (b c-a d)}+\frac{\sqrt [4]{a} \sqrt{-\frac{b x^2}{a}} \Pi \left (\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{x (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^2)^(3/4)*(c + d*x^2)),x]

[Out]

(a^(1/4)*Sqrt[-((b*x^2)/a)]*EllipticPi[-((Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d]), ArcSin[(a + b*x^2)^(1/4)/a^(1/
4)], -1])/((b*c - a*d)*x) + (a^(1/4)*Sqrt[-((b*x^2)/a)]*EllipticPi[(Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d], ArcSi
n[(a + b*x^2)^(1/4)/a^(1/4)], -1])/((b*c - a*d)*x)

Rule 401

Int[1/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[Sqrt[-((b*x^2)/a)]/(2*x), Subst[I
nt[1/(Sqrt[-((b*x)/a)]*(a + b*x)^(3/4)*(c + d*x)), x], x, x^2], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0]

Rule 108

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(3/4)), x_Symbol] :> Dist[-4, Subst[
Int[1/((b*e - a*f - b*x^4)*Sqrt[c - (d*e)/f + (d*x^4)/f]), x], x, (e + f*x)^(1/4)], x] /; FreeQ[{a, b, c, d, e
, f}, x] && GtQ[-(f/(d*e - c*f)), 0]

Rule 409

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

Rule 1218

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(c/a), 4]}, Simp[(1*Ellipt
icPi[-(e/(d*q^2)), ArcSin[q*x], -1])/(d*Sqrt[a]*q), x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx &=\frac{\sqrt{-\frac{b x^2}{a}} \operatorname{Subst}\left (\int \frac{1}{\sqrt{-\frac{b x}{a}} (a+b x)^{3/4} (c+d x)} \, dx,x,x^2\right )}{2 x}\\ &=-\frac{\left (2 \sqrt{-\frac{b x^2}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^4}{a}} \left (-b c+a d-d x^4\right )} \, dx,x,\sqrt [4]{a+b x^2}\right )}{x}\\ &=\frac{\sqrt{-\frac{b x^2}{a}} \operatorname{Subst}\left (\int \frac{1}{\left (1-\frac{\sqrt{d} x^2}{\sqrt{-b c+a d}}\right ) \sqrt{1-\frac{x^4}{a}}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{(b c-a d) x}+\frac{\sqrt{-\frac{b x^2}{a}} \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{\sqrt{d} x^2}{\sqrt{-b c+a d}}\right ) \sqrt{1-\frac{x^4}{a}}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{(b c-a d) x}\\ &=\frac{\sqrt [4]{a} \sqrt{-\frac{b x^2}{a}} \Pi \left (-\frac{\sqrt{a} \sqrt{d}}{\sqrt{-b c+a d}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{(b c-a d) x}+\frac{\sqrt [4]{a} \sqrt{-\frac{b x^2}{a}} \Pi \left (\frac{\sqrt{a} \sqrt{d}}{\sqrt{-b c+a d}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{(b c-a d) x}\\ \end{align*}

Mathematica [A]  time = 0.0374438, size = 123, normalized size = 0.81 \[ \frac{b x \left (\Pi \left (-\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .-\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )+\Pi \left (\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .-\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )\right )}{a^{3/4} \sqrt{-\frac{b x^2}{a}} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x^2)^(3/4)*(c + d*x^2)),x]

[Out]

(b*x*(EllipticPi[-((Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d]), -ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1] + EllipticPi
[(Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d], -ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1]))/(a^(3/4)*(b*c - a*d)*Sqrt[-((
b*x^2)/a)])

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Maple [F]  time = 0.043, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{d{x}^{2}+c} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a)^(3/4)/(d*x^2+c),x)

[Out]

int(1/(b*x^2+a)^(3/4)/(d*x^2+c),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{3}{4}}{\left (d x^{2} + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(3/4)/(d*x^2+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(3/4)*(d*x^2 + c)), x)

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Fricas [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(1/(b*x^2+a)^(3/4)/(d*x^2+c),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x^{2}\right )^{\frac{3}{4}} \left (c + d x^{2}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+a)**(3/4)/(d*x**2+c),x)

[Out]

Integral(1/((a + b*x**2)**(3/4)*(c + d*x**2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{3}{4}}{\left (d x^{2} + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(3/4)/(d*x^2+c),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(3/4)*(d*x^2 + c)), x)