3.1061 \(\int \frac{(a+b x^4)^{5/4}}{x^{15}} \, dx\)

Optimal. Leaf size=146 \[ \frac{5 b^{7/2} \left (\frac{b x^4}{a}+1\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right ),2\right )}{336 a^{3/2} \left (a+b x^4\right )^{3/4}}+\frac{5 b^3 \sqrt [4]{a+b x^4}}{336 a^2 x^2}-\frac{b^2 \sqrt [4]{a+b x^4}}{168 a x^6}-\frac{b \sqrt [4]{a+b x^4}}{28 x^{10}}-\frac{\left (a+b x^4\right )^{5/4}}{14 x^{14}} \]

[Out]

-(b*(a + b*x^4)^(1/4))/(28*x^10) - (b^2*(a + b*x^4)^(1/4))/(168*a*x^6) + (5*b^3*(a + b*x^4)^(1/4))/(336*a^2*x^
2) - (a + b*x^4)^(5/4)/(14*x^14) + (5*b^(7/2)*(1 + (b*x^4)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2,
2])/(336*a^(3/2)*(a + b*x^4)^(3/4))

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Rubi [A]  time = 0.100437, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {275, 277, 325, 233, 231} \[ \frac{5 b^3 \sqrt [4]{a+b x^4}}{336 a^2 x^2}+\frac{5 b^{7/2} \left (\frac{b x^4}{a}+1\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{336 a^{3/2} \left (a+b x^4\right )^{3/4}}-\frac{b^2 \sqrt [4]{a+b x^4}}{168 a x^6}-\frac{b \sqrt [4]{a+b x^4}}{28 x^{10}}-\frac{\left (a+b x^4\right )^{5/4}}{14 x^{14}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(5/4)/x^15,x]

[Out]

-(b*(a + b*x^4)^(1/4))/(28*x^10) - (b^2*(a + b*x^4)^(1/4))/(168*a*x^6) + (5*b^3*(a + b*x^4)^(1/4))/(336*a^2*x^
2) - (a + b*x^4)^(5/4)/(14*x^14) + (5*b^(7/2)*(1 + (b*x^4)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2,
2])/(336*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 277

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

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{\left (a+b x^4\right )^{5/4}}{x^{15}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^{5/4}}{x^8} \, dx,x,x^2\right )\\ &=-\frac{\left (a+b x^4\right )^{5/4}}{14 x^{14}}+\frac{1}{28} (5 b) \operatorname{Subst}\left (\int \frac{\sqrt [4]{a+b x^2}}{x^6} \, dx,x,x^2\right )\\ &=-\frac{b \sqrt [4]{a+b x^4}}{28 x^{10}}-\frac{\left (a+b x^4\right )^{5/4}}{14 x^{14}}+\frac{1}{56} b^2 \operatorname{Subst}\left (\int \frac{1}{x^4 \left (a+b x^2\right )^{3/4}} \, dx,x,x^2\right )\\ &=-\frac{b \sqrt [4]{a+b x^4}}{28 x^{10}}-\frac{b^2 \sqrt [4]{a+b x^4}}{168 a x^6}-\frac{\left (a+b x^4\right )^{5/4}}{14 x^{14}}-\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )^{3/4}} \, dx,x,x^2\right )}{336 a}\\ &=-\frac{b \sqrt [4]{a+b x^4}}{28 x^{10}}-\frac{b^2 \sqrt [4]{a+b x^4}}{168 a x^6}+\frac{5 b^3 \sqrt [4]{a+b x^4}}{336 a^2 x^2}-\frac{\left (a+b x^4\right )^{5/4}}{14 x^{14}}+\frac{\left (5 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{3/4}} \, dx,x,x^2\right )}{672 a^2}\\ &=-\frac{b \sqrt [4]{a+b x^4}}{28 x^{10}}-\frac{b^2 \sqrt [4]{a+b x^4}}{168 a x^6}+\frac{5 b^3 \sqrt [4]{a+b x^4}}{336 a^2 x^2}-\frac{\left (a+b x^4\right )^{5/4}}{14 x^{14}}+\frac{\left (5 b^4 \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 )}{672 a^2 \left (a+b x^4\right )^{3/4}}\\ &=-\frac{b \sqrt [4]{a+b x^4}}{28 x^{10}}-\frac{b^2 \sqrt [4]{a+b x^4}}{168 a x^6}+\frac{5 b^3 \sqrt [4]{a+b x^4}}{336 a^2 x^2}-\frac{\left (a+b x^4\right )^{5/4}}{14 x^{14}}+\frac{5 b^{7/2} \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 )}{336 a^{3/2} \left (a+b x^4\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0123068, size = 52, normalized size = 0.36 \[ -\frac{a \sqrt [4]{a+b x^4} \, _2F_1\left (-\frac{7}{2},-\frac{5}{4};-\frac{5}{2};-\frac{b x^4}{a}\right )}{14 x^{14} \sqrt [4]{\frac{b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(5/4)/x^15,x]

[Out]

-(a*(a + b*x^4)^(1/4)*Hypergeometric2F1[-7/2, -5/4, -5/2, -((b*x^4)/a)])/(14*x^14*(1 + (b*x^4)/a)^(1/4))

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Maple [F]  time = 0.04, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{15}} \left ( b{x}^{4}+a \right ) ^{{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^(5/4)/x^15,x)

[Out]

int((b*x^4+a)^(5/4)/x^15,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{4} + a\right )}^{\frac{5}{4}}}{x^{15}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(5/4)/x^15,x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(5/4)/x^15, 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{5}{4}}}{x^{15}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(5/4)/x^15,x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(5/4)/x^15, x)

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Sympy [C]  time = 10.0357, size = 34, normalized size = 0.23 \begin{align*} - \frac{a^{\frac{5}{4}}{{}_{2}F_{1}\left (\begin{matrix} - \frac{7}{2}, - \frac{5}{4} \\ - \frac{5}{2} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{14 x^{14}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**(5/4)/x**15,x)

[Out]

-a**(5/4)*hyper((-7/2, -5/4), (-5/2,), b*x**4*exp_polar(I*pi)/a)/(14*x**14)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{4} + a\right )}^{\frac{5}{4}}}{x^{15}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(5/4)/x^15,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(5/4)/x^15, x)