3.12.51 \(\int \frac {1}{x^3 (a+b x^4)^{5/4}} \, dx\) [1151]

Optimal. Leaf size=82 \[ -\frac {1}{2 a x^2 \sqrt [4]{a+b x^4}}-\frac {3 \sqrt {b} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 a^{3/2} \sqrt [4]{a+b x^4}} \]

[Out]

-1/2/a/x^2/(b*x^4+a)^(1/4)-3/2*(1+b*x^4/a)^(1/4)*(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)))*EllipticE(sin(1/2*arctan(x^2*b^(1/2)/a^(1/2))),2^(1/2))*b^(1/2)/a^(3/2)/(b*x^4+a)^(1/4)

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Rubi [A]
time = 0.03, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {281, 292, 203, 202} \begin {gather*} -\frac {3 \sqrt {b} \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 a^{3/2} \sqrt [4]{a+b x^4}}-\frac {1}{2 a x^2 \sqrt [4]{a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^4)^(5/4)),x]

[Out]

-1/2*1/(a*x^2*(a + b*x^4)^(1/4)) - (3*Sqrt[b]*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2,
2])/(2*a^(3/2)*(a + b*x^4)^(1/4))

Rule 202

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

Rule 203

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 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 292

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]

Rubi steps

\begin {align*} \int \frac {1}{x^3 \left (a+b x^4\right )^{5/4}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right )\\ &=-\frac {1}{2 a x^2 \sqrt [4]{a+b x^4}}-\frac {(3 b) \text {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right )}{4 a}\\ &=-\frac {1}{2 a x^2 \sqrt [4]{a+b x^4}}-\frac {\left (3 b \sqrt [4]{1+\frac {b x^4}{a}}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx,x,x^2\right )}{4 a^2 \sqrt [4]{a+b x^4}}\\ &=-\frac {1}{2 a x^2 \sqrt [4]{a+b x^4}}-\frac {3 \sqrt {b} \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 )}{2 a^{3/2} \sqrt [4]{a+b x^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 = 10.02, size = 54, normalized size = 0.66 \begin {gather*} -\frac {\sqrt [4]{1+\frac {b x^4}{a}} \, _2F_1\left (-\frac {1}{2},\frac {5}{4};\frac {1}{2};-\frac {b x^4}{a}\right )}{2 a x^2 \sqrt [4]{a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*x^4)^(5/4)),x]

[Out]

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

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{3} \left (b \,x^{4}+a \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x^4+a)^(5/4),x)

[Out]

int(1/x^3/(b*x^4+a)^(5/4),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/x^3/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(5/4)*x^3), x)

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Fricas [F]
time = 0.07, size = 36, normalized size = 0.44 \begin {gather*} {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{b^{2} x^{11} + 2 \, a b x^{7} + a^{2} x^{3}}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 0.54, size = 31, normalized size = 0.38 \begin {gather*} - \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x**4+a)**(5/4),x)

[Out]

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

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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/x^3/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(5/4)*x^3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^3\,{\left (b\,x^4+a\right )}^{5/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x^4)^(5/4)),x)

[Out]

int(1/(x^3*(a + b*x^4)^(5/4)), x)

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