3.1152 \(\int \frac{1}{x^7 (a+b x^4)^{5/4}} \, dx\)

Optimal. Leaf size=104 \[ \frac{7 b^{3/2} \sqrt [4]{\frac{b x^4}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{4 a^{5/2} \sqrt [4]{a+b x^4}}+\frac{7 b}{12 a^2 x^2 \sqrt [4]{a+b x^4}}-\frac{1}{6 a x^6 \sqrt [4]{a+b x^4}} \]

[Out]

-1/(6*a*x^6*(a + b*x^4)^(1/4)) + (7*b)/(12*a^2*x^2*(a + b*x^4)^(1/4)) + (7*b^(3/2)*(1 + (b*x^4)/a)^(1/4)*Ellip
ticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(4*a^(5/2)*(a + b*x^4)^(1/4))

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Rubi [A]  time = 0.0639458, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {275, 286, 197, 196} \[ \frac{7 b^{3/2} \sqrt [4]{\frac{b x^4}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{4 a^{5/2} \sqrt [4]{a+b x^4}}+\frac{7 b}{12 a^2 x^2 \sqrt [4]{a+b x^4}}-\frac{1}{6 a x^6 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(a + b*x^4)^(5/4)),x]

[Out]

-1/(6*a*x^6*(a + b*x^4)^(1/4)) + (7*b)/(12*a^2*x^2*(a + b*x^4)^(1/4)) + (7*b^(3/2)*(1 + (b*x^4)/a)^(1/4)*Ellip
ticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(4*a^(5/2)*(a + b*x^4)^(1/4))

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 286

Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[(c*x)^(m + 1)/(a*c*(m + 1)*(a + b*x^2)^(1
/4)), x] - Dist[(b*(2*m + 1))/(2*a*c^2*(m + 1)), Int[(c*x)^(m + 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, c
}, x] && PosQ[b/a] && IntegerQ[2*m] && LtQ[m, -1]

Rule 197

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(1/4)/(a*(a + b*x^2)^(1/4)), Int[1/(1 + (b
*x^2)/a)^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a] && PosQ[b/a]

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rubi steps

\begin{align*} \int \frac{1}{x^7 \left (a+b x^4\right )^{5/4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^4 \left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right )\\ &=-\frac{1}{6 a x^6 \sqrt [4]{a+b x^4}}-\frac{(7 b) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right )}{12 a}\\ &=-\frac{1}{6 a x^6 \sqrt [4]{a+b x^4}}+\frac{7 b}{12 a^2 x^2 \sqrt [4]{a+b x^4}}+\frac{\left (7 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right )}{8 a^2}\\ &=-\frac{1}{6 a x^6 \sqrt [4]{a+b x^4}}+\frac{7 b}{12 a^2 x^2 \sqrt [4]{a+b x^4}}+\frac{\left (7 b^2 \sqrt [4]{1+\frac{b x^4}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{5/4}} \, dx,x,x^2\right )}{8 a^3 \sqrt [4]{a+b x^4}}\\ &=-\frac{1}{6 a x^6 \sqrt [4]{a+b x^4}}+\frac{7 b}{12 a^2 x^2 \sqrt [4]{a+b x^4}}+\frac{7 b^{3/2} \sqrt [4]{1+\frac{b x^4}{a}} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{4 a^{5/2} \sqrt [4]{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0114801, size = 54, normalized size = 0.52 \[ -\frac{\sqrt [4]{\frac{b x^4}{a}+1} \, _2F_1\left (-\frac{3}{2},\frac{5}{4};-\frac{1}{2};-\frac{b x^4}{a}\right )}{6 a x^6 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(a + b*x^4)^(5/4)),x]

[Out]

-((1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[-3/2, 5/4, -1/2, -((b*x^4)/a)])/(6*a*x^6*(a + b*x^4)^(1/4))

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Maple [F]  time = 0.053, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{7}} \left ( b{x}^{4}+a \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(b*x^4+a)^(5/4),x)

[Out]

int(1/x^7/(b*x^4+a)^(5/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{4} + a\right )}^{\frac{5}{4}} x^{7}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(5/4)*x^7), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{4} + a\right )}^{\frac{3}{4}}}{b^{2} x^{15} + 2 \, a b x^{11} + a^{2} x^{7}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)/(b^2*x^15 + 2*a*b*x^11 + a^2*x^7), x)

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Sympy [C]  time = 2.45471, size = 32, normalized size = 0.31 \begin{align*} - \frac{{{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{2}, \frac{5}{4} \\ - \frac{1}{2} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{6 a^{\frac{5}{4}} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(b*x**4+a)**(5/4),x)

[Out]

-hyper((-3/2, 5/4), (-1/2,), b*x**4*exp_polar(I*pi)/a)/(6*a**(5/4)*x**6)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{4} + a\right )}^{\frac{5}{4}} x^{7}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(5/4)*x^7), x)