3.950 \(\int \frac{(c x)^{5/2}}{\sqrt [4]{a+b x^2}} \, dx\)

Optimal. Leaf size=125 \[ -\frac{a^{3/2} c^2 \sqrt{c x} \sqrt [4]{\frac{a}{b x^2}+1} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{2 b^{3/2} \sqrt [4]{a+b x^2}}-\frac{a c^2 x \sqrt{c x}}{2 b \sqrt [4]{a+b x^2}}+\frac{c (c x)^{3/2} \left (a+b x^2\right )^{3/4}}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.0465621, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {321, 314, 284, 335, 196} \[ -\frac{a^{3/2} c^2 \sqrt{c x} \sqrt [4]{\frac{a}{b x^2}+1} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{2 b^{3/2} \sqrt [4]{a+b x^2}}-\frac{a c^2 x \sqrt{c x}}{2 b \sqrt [4]{a+b x^2}}+\frac{c (c x)^{3/2} \left (a+b x^2\right )^{3/4}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(5/2)/(a + b*x^2)^(1/4),x]

[Out]

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 314

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

Mathematica [C]  time = 0.0238553, size = 69, normalized size = 0.55 \[ \frac{c (c x)^{3/2} \left (-a \sqrt [4]{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{3}{4};\frac{7}{4};-\frac{b x^2}{a}\right )+a+b x^2\right )}{3 b \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(c*x)^(3/2)*(a + b*x^2 - a*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 3/4, 7/4, -((b*x^2)/a)]))/(3*b*(a +
b*x^2)^(1/4))

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Maple [F]  time = 0.031, size = 0, normalized size = 0. \begin{align*} \int{ \left ( cx \right ) ^{{\frac{5}{2}}}{\frac{1}{\sqrt [4]{b{x}^{2}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(5/2)/(b*x^2+a)^(1/4),x)

[Out]

int((c*x)^(5/2)/(b*x^2+a)^(1/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x\right )^{\frac{5}{2}}}{{\left (b x^{2} + a\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(5/2)/(b*x^2+a)^(1/4),x, algorithm="maxima")

[Out]

integrate((c*x)^(5/2)/(b*x^2 + a)^(1/4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x} c^{2} x^{2}}{{\left (b x^{2} + a\right )}^{\frac{1}{4}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(5/2)/(b*x^2+a)^(1/4),x, algorithm="fricas")

[Out]

integral(sqrt(c*x)*c^2*x^2/(b*x^2 + a)^(1/4), x)

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Sympy [C]  time = 30.3746, size = 44, normalized size = 0.35 \begin{align*} \frac{c^{\frac{5}{2}} x^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt [4]{a} \Gamma \left (\frac{11}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(5/2)/(b*x**2+a)**(1/4),x)

[Out]

c**(5/2)*x**(7/2)*gamma(7/4)*hyper((1/4, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(1/4)*gamma(11/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x\right )^{\frac{5}{2}}}{{\left (b x^{2} + a\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(5/2)/(b*x^2+a)^(1/4),x, algorithm="giac")

[Out]

integrate((c*x)^(5/2)/(b*x^2 + a)^(1/4), x)