3.954 \(\int \frac{1}{(c x)^{11/2} \sqrt [4]{a+b x^2}} \, dx\)

Optimal. Leaf size=157 \[ -\frac{8 b^2}{15 a^2 c^5 \sqrt{c x} \sqrt [4]{a+b x^2}}+\frac{8 b^{5/2} \sqrt{c x} \sqrt [4]{\frac{a}{b x^2}+1} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{15 a^{5/2} c^6 \sqrt [4]{a+b x^2}}+\frac{4 b \left (a+b x^2\right )^{3/4}}{15 a^2 c^3 (c x)^{5/2}}-\frac{2 \left (a+b x^2\right )^{3/4}}{9 a c (c x)^{9/2}} \]

[Out]

(-8*b^2)/(15*a^2*c^5*Sqrt[c*x]*(a + b*x^2)^(1/4)) - (2*(a + b*x^2)^(3/4))/(9*a*c*(c*x)^(9/2)) + (4*b*(a + b*x^
2)^(3/4))/(15*a^2*c^3*(c*x)^(5/2)) + (8*b^(5/2)*(1 + a/(b*x^2))^(1/4)*Sqrt[c*x]*EllipticE[ArcCot[(Sqrt[b]*x)/S
qrt[a]]/2, 2])/(15*a^(5/2)*c^6*(a + b*x^2)^(1/4))

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Rubi [A]  time = 0.0678214, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {325, 316, 284, 335, 196} \[ -\frac{8 b^2}{15 a^2 c^5 \sqrt{c x} \sqrt [4]{a+b x^2}}+\frac{8 b^{5/2} \sqrt{c x} \sqrt [4]{\frac{a}{b x^2}+1} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{15 a^{5/2} c^6 \sqrt [4]{a+b x^2}}+\frac{4 b \left (a+b x^2\right )^{3/4}}{15 a^2 c^3 (c x)^{5/2}}-\frac{2 \left (a+b x^2\right )^{3/4}}{9 a c (c x)^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(11/2)*(a + b*x^2)^(1/4)),x]

[Out]

(-8*b^2)/(15*a^2*c^5*Sqrt[c*x]*(a + b*x^2)^(1/4)) - (2*(a + b*x^2)^(3/4))/(9*a*c*(c*x)^(9/2)) + (4*b*(a + b*x^
2)^(3/4))/(15*a^2*c^3*(c*x)^(5/2)) + (8*b^(5/2)*(1 + a/(b*x^2))^(1/4)*Sqrt[c*x]*EllipticE[ArcCot[(Sqrt[b]*x)/S
qrt[a]]/2, 2])/(15*a^(5/2)*c^6*(a + b*x^2)^(1/4))

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 316

Int[1/(((c_.)*(x_))^(3/2)*((a_) + (b_.)*(x_)^2)^(1/4)), x_Symbol] :> Simp[-2/(c*Sqrt[c*x]*(a + b*x^2)^(1/4)),
x] - Dist[b/c^2, Int[Sqrt[c*x]/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]

Rule 284

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

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

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{1}{(c x)^{11/2} \sqrt [4]{a+b x^2}} \, dx &=-\frac{2 \left (a+b x^2\right )^{3/4}}{9 a c (c x)^{9/2}}-\frac{(2 b) \int \frac{1}{(c x)^{7/2} \sqrt [4]{a+b x^2}} \, dx}{3 a c^2}\\ &=-\frac{2 \left (a+b x^2\right )^{3/4}}{9 a c (c x)^{9/2}}+\frac{4 b \left (a+b x^2\right )^{3/4}}{15 a^2 c^3 (c x)^{5/2}}+\frac{\left (4 b^2\right ) \int \frac{1}{(c x)^{3/2} \sqrt [4]{a+b x^2}} \, dx}{15 a^2 c^4}\\ &=-\frac{8 b^2}{15 a^2 c^5 \sqrt{c x} \sqrt [4]{a+b x^2}}-\frac{2 \left (a+b x^2\right )^{3/4}}{9 a c (c x)^{9/2}}+\frac{4 b \left (a+b x^2\right )^{3/4}}{15 a^2 c^3 (c x)^{5/2}}-\frac{\left (4 b^3\right ) \int \frac{\sqrt{c x}}{\left (a+b x^2\right )^{5/4}} \, dx}{15 a^2 c^6}\\ &=-\frac{8 b^2}{15 a^2 c^5 \sqrt{c x} \sqrt [4]{a+b x^2}}-\frac{2 \left (a+b x^2\right )^{3/4}}{9 a c (c x)^{9/2}}+\frac{4 b \left (a+b x^2\right )^{3/4}}{15 a^2 c^3 (c x)^{5/2}}-\frac{\left (4 b^2 \sqrt [4]{1+\frac{a}{b x^2}} \sqrt{c x}\right ) \int \frac{1}{\left (1+\frac{a}{b x^2}\right )^{5/4} x^2} \, dx}{15 a^2 c^6 \sqrt [4]{a+b x^2}}\\ &=-\frac{8 b^2}{15 a^2 c^5 \sqrt{c x} \sqrt [4]{a+b x^2}}-\frac{2 \left (a+b x^2\right )^{3/4}}{9 a c (c x)^{9/2}}+\frac{4 b \left (a+b x^2\right )^{3/4}}{15 a^2 c^3 (c x)^{5/2}}+\frac{\left (4 b^2 \sqrt [4]{1+\frac{a}{b x^2}} \sqrt{c x}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{a x^2}{b}\right )^{5/4}} \, dx,x,\frac{1}{x}\right )}{15 a^2 c^6 \sqrt [4]{a+b x^2}}\\ &=-\frac{8 b^2}{15 a^2 c^5 \sqrt{c x} \sqrt [4]{a+b x^2}}-\frac{2 \left (a+b x^2\right )^{3/4}}{9 a c (c x)^{9/2}}+\frac{4 b \left (a+b x^2\right )^{3/4}}{15 a^2 c^3 (c x)^{5/2}}+\frac{8 b^{5/2} \sqrt [4]{1+\frac{a}{b x^2}} \sqrt{c x} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{15 a^{5/2} c^6 \sqrt [4]{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.013287, size = 56, normalized size = 0.36 \[ -\frac{2 x \sqrt [4]{\frac{b x^2}{a}+1} \, _2F_1\left (-\frac{9}{4},\frac{1}{4};-\frac{5}{4};-\frac{b x^2}{a}\right )}{9 (c x)^{11/2} \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(11/2)*(a + b*x^2)^(1/4)),x]

[Out]

(-2*x*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[-9/4, 1/4, -5/4, -((b*x^2)/a)])/(9*(c*x)^(11/2)*(a + b*x^2)^(1/4
))

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Maple [F]  time = 0.038, size = 0, normalized size = 0. \begin{align*} \int{ \left ( cx \right ) ^{-{\frac{11}{2}}}{\frac{1}{\sqrt [4]{b{x}^{2}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(11/2)/(b*x^2+a)^(1/4),x)

[Out]

int(1/(c*x)^(11/2)/(b*x^2+a)^(1/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{1}{4}} \left (c x\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(11/2)/(b*x^2+a)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(1/4)*(c*x)^(11/2)), 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}} \sqrt{c x}}{b c^{6} x^{8} + a c^{6} x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(11/2)/(b*x^2+a)^(1/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(3/4)*sqrt(c*x)/(b*c^6*x^8 + a*c^6*x^6), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(11/2)/(b*x**2+a)**(1/4),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{1}{4}} \left (c x\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(11/2)/(b*x^2+a)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(1/4)*(c*x)^(11/2)), x)