3.4.26 \(\int \frac {1}{(a+b x^2)^{3/4} (c+d x^2)} \, dx\) [326]

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

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {410, 109, 418, 1232} \begin {gather*} \frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\text {ArcSin}\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 .\text {ArcSin}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{x (b c-a d)} \end {gather*}

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 109

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 410

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 418

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 1232

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-c/a, 4]}, Simp[(1/(d*Sqrt[
a]*q))*EllipticPi[-e/(d*q^2), ArcSin[q*x], -1], 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}} \text {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 ) \text {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}} \text {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}} \text {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*}

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Mathematica [A]
time = 7.60, size = 120, normalized size = 0.79 \begin {gather*} \frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} \left (\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 )+\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 )\right )}{(b c-a d) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[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] + 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)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (d \,x^{2}+c \right )}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x^{2}\right )^{\frac {3}{4}} \left (c + d x^{2}\right )}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (b\,x^2+a\right )}^{3/4}\,\left (d\,x^2+c\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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