3.9.4 \(\int \frac {(a+b x^2)^{3/4}}{x^6} \, dx\) [804]

Optimal. Leaf size=145 \[ -\frac {3 b^3 x}{20 a^2 \sqrt [4]{a+b x^2}}-\frac {\left (a+b x^2\right )^{3/4}}{5 x^5}-\frac {b \left (a+b x^2\right )^{3/4}}{10 a x^3}+\frac {3 b^2 \left (a+b x^2\right )^{3/4}}{20 a^2 x}+\frac {3 b^{5/2} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 a^{3/2} \sqrt [4]{a+b x^2}} \]

[Out]

-3/20*b^3*x/a^2/(b*x^2+a)^(1/4)-1/5*(b*x^2+a)^(3/4)/x^5-1/10*b*(b*x^2+a)^(3/4)/a/x^3+3/20*b^2*(b*x^2+a)^(3/4)/
a^2/x+3/20*b^(5/2)*(1+b*x^2/a)^(1/4)*(cos(1/2*arctan(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x*b^(1/2)/a^(
1/2)))*EllipticE(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))/a^(3/2)/(b*x^2+a)^(1/4)

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Rubi [A]
time = 0.04, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {283, 331, 235, 233, 202} \begin {gather*} \frac {3 b^{5/2} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 a^{3/2} \sqrt [4]{a+b x^2}}-\frac {3 b^3 x}{20 a^2 \sqrt [4]{a+b x^2}}+\frac {3 b^2 \left (a+b x^2\right )^{3/4}}{20 a^2 x}-\frac {\left (a+b x^2\right )^{3/4}}{5 x^5}-\frac {b \left (a+b x^2\right )^{3/4}}{10 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/4)/x^6,x]

[Out]

(-3*b^3*x)/(20*a^2*(a + b*x^2)^(1/4)) - (a + b*x^2)^(3/4)/(5*x^5) - (b*(a + b*x^2)^(3/4))/(10*a*x^3) + (3*b^2*
(a + b*x^2)^(3/4))/(20*a^2*x) + (3*b^(5/2)*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/
(20*a^(3/2)*(a + b*x^2)^(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 233

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

Rule 235

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

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^2\right )^{3/4}}{x^6} \, dx &=-\frac {\left (a+b x^2\right )^{3/4}}{5 x^5}+\frac {1}{10} (3 b) \int \frac {1}{x^4 \sqrt [4]{a+b x^2}} \, dx\\ &=-\frac {\left (a+b x^2\right )^{3/4}}{5 x^5}-\frac {b \left (a+b x^2\right )^{3/4}}{10 a x^3}-\frac {\left (3 b^2\right ) \int \frac {1}{x^2 \sqrt [4]{a+b x^2}} \, dx}{20 a}\\ &=-\frac {\left (a+b x^2\right )^{3/4}}{5 x^5}-\frac {b \left (a+b x^2\right )^{3/4}}{10 a x^3}+\frac {3 b^2 \left (a+b x^2\right )^{3/4}}{20 a^2 x}-\frac {\left (3 b^3\right ) \int \frac {1}{\sqrt [4]{a+b x^2}} \, dx}{40 a^2}\\ &=-\frac {\left (a+b x^2\right )^{3/4}}{5 x^5}-\frac {b \left (a+b x^2\right )^{3/4}}{10 a x^3}+\frac {3 b^2 \left (a+b x^2\right )^{3/4}}{20 a^2 x}-\frac {\left (3 b^3 \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx}{40 a^2 \sqrt [4]{a+b x^2}}\\ &=-\frac {3 b^3 x}{20 a^2 \sqrt [4]{a+b x^2}}-\frac {\left (a+b x^2\right )^{3/4}}{5 x^5}-\frac {b \left (a+b x^2\right )^{3/4}}{10 a x^3}+\frac {3 b^2 \left (a+b x^2\right )^{3/4}}{20 a^2 x}+\frac {\left (3 b^3 \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx}{40 a^2 \sqrt [4]{a+b x^2}}\\ &=-\frac {3 b^3 x}{20 a^2 \sqrt [4]{a+b x^2}}-\frac {\left (a+b x^2\right )^{3/4}}{5 x^5}-\frac {b \left (a+b x^2\right )^{3/4}}{10 a x^3}+\frac {3 b^2 \left (a+b x^2\right )^{3/4}}{20 a^2 x}+\frac {3 b^{5/2} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 a^{3/2} \sqrt [4]{a+b x^2}}\\ \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 = 51, normalized size = 0.35 \begin {gather*} -\frac {\left (a+b x^2\right )^{3/4} \, _2F_1\left (-\frac {5}{2},-\frac {3}{4};-\frac {3}{2};-\frac {b x^2}{a}\right )}{5 x^5 \left (1+\frac {b x^2}{a}\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/4)/x^6,x]

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{2}+a \right )^{\frac {3}{4}}}{x^{6}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/4)/x^6,x)

[Out]

int((b*x^2+a)^(3/4)/x^6,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^2+a)^(3/4)/x^6,x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(3/4)/x^6, x)

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Fricas [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^2+a)^(3/4)/x^6,x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(3/4)/x^6, x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.68, size = 34, normalized size = 0.23 \begin {gather*} - \frac {a^{\frac {3}{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, - \frac {3}{4} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/4)/x**6,x)

[Out]

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

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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^2+a)^(3/4)/x^6,x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(3/4)/x^6, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^{3/4}}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(3/4)/x^6,x)

[Out]

int((a + b*x^2)^(3/4)/x^6, x)

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