3.803 \(\int \frac{(a+b x^2)^{3/4}}{x^4} \, dx\)

Optimal. Leaf size=121 \[ \frac{b^2 x}{2 a \sqrt [4]{a+b x^2}}-\frac{b^{3/2} \sqrt [4]{\frac{b x^2}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{2 \sqrt{a} \sqrt [4]{a+b x^2}}-\frac{b \left (a+b x^2\right )^{3/4}}{2 a x}-\frac{\left (a+b x^2\right )^{3/4}}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.0378355, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {277, 325, 229, 227, 196} \[ \frac{b^2 x}{2 a \sqrt [4]{a+b x^2}}-\frac{b^{3/2} \sqrt [4]{\frac{b x^2}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{2 \sqrt{a} \sqrt [4]{a+b x^2}}-\frac{b \left (a+b x^2\right )^{3/4}}{2 a x}-\frac{\left (a+b x^2\right )^{3/4}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/4)/x^4,x]

[Out]

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

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 229

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 227

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 196

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

Mathematica [C]  time = 0.0103978, size = 51, normalized size = 0.42 \[ -\frac{\left (a+b x^2\right )^{3/4} \, _2F_1\left (-\frac{3}{2},-\frac{3}{4};-\frac{1}{2};-\frac{b x^2}{a}\right )}{3 x^3 \left (\frac{b x^2}{a}+1\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/4)/x^4,x]

[Out]

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

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Maple [F]  time = 0.027, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{3}{4}}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(3/4)/x^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{2} + a\right )}^{\frac{3}{4}}}{x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/4)/x^4,x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(3/4)/x^4, x)

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Sympy [C]  time = 1.14974, size = 34, normalized size = 0.28 \begin{align*} - \frac{a^{\frac{3}{4}}{{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{2}, - \frac{3}{4} \\ - \frac{1}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{3 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(3/4)*hyper((-3/2, -3/4), (-1/2,), b*x**2*exp_polar(I*pi)/a)/(3*x**3)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/4)/x^4,x, algorithm="giac")

[Out]

Exception raised: TypeError