3.1123 \(\int \frac{1}{x^7 (a+b x^4)^{3/4}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0628616, 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, 325, 233, 231} \[ \frac{5 b^{3/2} \left (\frac{b x^4}{a}+1\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{12 a^{3/2} \left (a+b x^4\right )^{3/4}}+\frac{5 b \sqrt [4]{a+b x^4}}{12 a^2 x^2}-\frac{\sqrt [4]{a+b x^4}}{6 a x^6} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(a + b*x^4)^(3/4)),x]

[Out]

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

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 233

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(3/4)/(a + b*x^2)^(3/4), Int[1/(1 + (b*x^2
)/a)^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*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 )^{3/4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^4 \left (a+b x^2\right )^{3/4}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt [4]{a+b x^4}}{6 a x^6}-\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )^{3/4}} \, dx,x,x^2\right )}{12 a}\\ &=-\frac{\sqrt [4]{a+b x^4}}{6 a x^6}+\frac{5 b \sqrt [4]{a+b x^4}}{12 a^2 x^2}+\frac{\left (5 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{3/4}} \, dx,x,x^2\right )}{24 a^2}\\ &=-\frac{\sqrt [4]{a+b x^4}}{6 a x^6}+\frac{5 b \sqrt [4]{a+b x^4}}{12 a^2 x^2}+\frac{\left (5 b^2 \left (1+\frac{b x^4}{a}\right )^{3/4}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{3/4}} \, dx,x,x^2\right )}{24 a^2 \left (a+b x^4\right )^{3/4}}\\ &=-\frac{\sqrt [4]{a+b x^4}}{6 a x^6}+\frac{5 b \sqrt [4]{a+b x^4}}{12 a^2 x^2}+\frac{5 b^{3/2} \left (1+\frac{b x^4}{a}\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{12 a^{3/2} \left (a+b x^4\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0108236, size = 51, normalized size = 0.49 \[ -\frac{\left (\frac{b x^4}{a}+1\right )^{3/4} \, _2F_1\left (-\frac{3}{2},\frac{3}{4};-\frac{1}{2};-\frac{b x^4}{a}\right )}{6 x^6 \left (a+b x^4\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(a + b*x^4)^(3/4)),x]

[Out]

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

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Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{7}} \left ( b{x}^{4}+a \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(b*x^4+a)^(3/4),x)

[Out]

int(1/x^7/(b*x^4+a)^(3/4),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{3}{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)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(3/4)*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{1}{4}}}{b x^{11} + a 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)^(3/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(1/4)/(b*x^11 + a*x^7), x)

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Sympy [C]  time = 1.4409, size = 32, normalized size = 0.31 \begin{align*} - \frac{{{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{2}, \frac{3}{4} \\ - \frac{1}{2} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{6 a^{\frac{3}{4}} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(b*x**4+a)**(3/4),x)

[Out]

-hyper((-3/2, 3/4), (-1/2,), b*x**4*exp_polar(I*pi)/a)/(6*a**(3/4)*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{3}{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)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(3/4)*x^7), x)