3.11.61 \(\int \frac {(a+b x^4)^{5/4}}{x^{15}} \, dx\) [1061]

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

[Out]

-1/28*b*(b*x^4+a)^(1/4)/x^10-1/168*b^2*(b*x^4+a)^(1/4)/a/x^6+5/336*b^3*(b*x^4+a)^(1/4)/a^2/x^2-1/14*(b*x^4+a)^
(5/4)/x^14+5/336*b^(7/2)*(1+b*x^4/a)^(3/4)*(cos(1/2*arctan(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x^2*b
^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arctan(x^2*b^(1/2)/a^(1/2))),2^(1/2))/a^(3/2)/(b*x^4+a)^(3/4)

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Rubi [A]
time = 0.07, antiderivative size = 146, normalized size of antiderivative = 1.00, 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 = {281, 283, 331, 239, 237} \begin {gather*} \frac {5 b^{7/2} \left (\frac {b x^4}{a}+1\right )^{3/4} F\left (\left .\frac {1}{2} \text {ArcTan}\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 {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 {\left (a+b x^4\right )^{5/4}}{14 x^{14}}-\frac {b \sqrt [4]{a+b x^4}}{28 x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(5/4)/x^15,x]

[Out]

-1/28*(b*(a + b*x^4)^(1/4))/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 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 239

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 281

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 283

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 331

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]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^4\right )^{5/4}}{x^{15}} \, dx &=\frac {1}{2} \text {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) \text {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 \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.01, size = 52, normalized size = 0.36 \begin {gather*} -\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]{1+\frac {b x^4}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(5/4)/x^15,x]

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{4}+a \right )^{\frac {5}{4}}}{x^{15}}\, dx\]

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 = 0.00, 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((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 = 0.08, size = 15, normalized size = 0.10 \begin {gather*} {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{x^{15}}, x\right ) \end {gather*}

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] Result contains complex when optimal does not.
time = 1.52, size = 34, normalized size = 0.23 \begin {gather*} - \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 {gather*}

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 = 0.00, 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((b*x^4+a)^(5/4)/x^15,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(5/4)/x^15, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^4+a\right )}^{5/4}}{x^{15}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(5/4)/x^15,x)

[Out]

int((a + b*x^4)^(5/4)/x^15, x)

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