3.949 \(\int \frac{(c x)^{9/2}}{\sqrt [4]{a+b x^2}} \, dx\)

Optimal. Leaf size=156 \[ \frac{7 a^2 c^4 x \sqrt{c x}}{20 b^2 \sqrt [4]{a+b x^2}}+\frac{7 a^{5/2} c^4 \sqrt{c x} \sqrt [4]{\frac{a}{b x^2}+1} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{20 b^{5/2} \sqrt [4]{a+b x^2}}-\frac{7 a c^3 (c x)^{3/2} \left (a+b x^2\right )^{3/4}}{30 b^2}+\frac{c (c x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 b} \]

[Out]

(7*a^2*c^4*x*Sqrt[c*x])/(20*b^2*(a + b*x^2)^(1/4)) - (7*a*c^3*(c*x)^(3/2)*(a + b*x^2)^(3/4))/(30*b^2) + (c*(c*
x)^(7/2)*(a + b*x^2)^(3/4))/(5*b) + (7*a^(5/2)*c^4*(1 + a/(b*x^2))^(1/4)*Sqrt[c*x]*EllipticE[ArcCot[(Sqrt[b]*x
)/Sqrt[a]]/2, 2])/(20*b^(5/2)*(a + b*x^2)^(1/4))

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Rubi [A]  time = 0.0671365, antiderivative size = 156, 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 = {321, 314, 284, 335, 196} \[ \frac{7 a^2 c^4 x \sqrt{c x}}{20 b^2 \sqrt [4]{a+b x^2}}+\frac{7 a^{5/2} c^4 \sqrt{c x} \sqrt [4]{\frac{a}{b x^2}+1} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{20 b^{5/2} \sqrt [4]{a+b x^2}}-\frac{7 a c^3 (c x)^{3/2} \left (a+b x^2\right )^{3/4}}{30 b^2}+\frac{c (c x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(9/2)/(a + b*x^2)^(1/4),x]

[Out]

(7*a^2*c^4*x*Sqrt[c*x])/(20*b^2*(a + b*x^2)^(1/4)) - (7*a*c^3*(c*x)^(3/2)*(a + b*x^2)^(3/4))/(30*b^2) + (c*(c*
x)^(7/2)*(a + b*x^2)^(3/4))/(5*b) + (7*a^(5/2)*c^4*(1 + a/(b*x^2))^(1/4)*Sqrt[c*x]*EllipticE[ArcCot[(Sqrt[b]*x
)/Sqrt[a]]/2, 2])/(20*b^(5/2)*(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)^{9/2}}{\sqrt [4]{a+b x^2}} \, dx &=\frac{c (c x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 b}-\frac{\left (7 a c^2\right ) \int \frac{(c x)^{5/2}}{\sqrt [4]{a+b x^2}} \, dx}{10 b}\\ &=-\frac{7 a c^3 (c x)^{3/2} \left (a+b x^2\right )^{3/4}}{30 b^2}+\frac{c (c x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 b}+\frac{\left (7 a^2 c^4\right ) \int \frac{\sqrt{c x}}{\sqrt [4]{a+b x^2}} \, dx}{20 b^2}\\ &=\frac{7 a^2 c^4 x \sqrt{c x}}{20 b^2 \sqrt [4]{a+b x^2}}-\frac{7 a c^3 (c x)^{3/2} \left (a+b x^2\right )^{3/4}}{30 b^2}+\frac{c (c x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 b}-\frac{\left (7 a^3 c^4\right ) \int \frac{\sqrt{c x}}{\left (a+b x^2\right )^{5/4}} \, dx}{40 b^2}\\ &=\frac{7 a^2 c^4 x \sqrt{c x}}{20 b^2 \sqrt [4]{a+b x^2}}-\frac{7 a c^3 (c x)^{3/2} \left (a+b x^2\right )^{3/4}}{30 b^2}+\frac{c (c x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 b}-\frac{\left (7 a^3 c^4 \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}{40 b^3 \sqrt [4]{a+b x^2}}\\ &=\frac{7 a^2 c^4 x \sqrt{c x}}{20 b^2 \sqrt [4]{a+b x^2}}-\frac{7 a c^3 (c x)^{3/2} \left (a+b x^2\right )^{3/4}}{30 b^2}+\frac{c (c x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 b}+\frac{\left (7 a^3 c^4 \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 )}{40 b^3 \sqrt [4]{a+b x^2}}\\ &=\frac{7 a^2 c^4 x \sqrt{c x}}{20 b^2 \sqrt [4]{a+b x^2}}-\frac{7 a c^3 (c x)^{3/2} \left (a+b x^2\right )^{3/4}}{30 b^2}+\frac{c (c x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 b}+\frac{7 a^{5/2} c^4 \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 )}{20 b^{5/2} \sqrt [4]{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0312011, size = 87, normalized size = 0.56 \[ \frac{c^3 (c x)^{3/2} \left (7 a^2 \sqrt [4]{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{3}{4};\frac{7}{4};-\frac{b x^2}{a}\right )-7 a^2-a b x^2+6 b^2 x^4\right )}{30 b^2 \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(9/2)/(a + b*x^2)^(1/4),x]

[Out]

(c^3*(c*x)^(3/2)*(-7*a^2 - a*b*x^2 + 6*b^2*x^4 + 7*a^2*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 3/4, 7/4,
-((b*x^2)/a)]))/(30*b^2*(a + b*x^2)^(1/4))

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Maple [F]  time = 0.036, size = 0, normalized size = 0. \begin{align*} \int{ \left ( cx \right ) ^{{\frac{9}{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)^(9/2)/(b*x^2+a)^(1/4),x)

[Out]

int((c*x)^(9/2)/(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{9}{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)^(9/2)/(b*x^2+a)^(1/4),x, algorithm="maxima")

[Out]

integrate((c*x)^(9/2)/(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^{4} x^{4}}{{\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)^(9/2)/(b*x^2+a)^(1/4),x, algorithm="fricas")

[Out]

integral(sqrt(c*x)*c^4*x^4/(b*x^2 + a)^(1/4), 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((c*x)**(9/2)/(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{\left (c x\right )^{\frac{9}{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)^(9/2)/(b*x^2+a)^(1/4),x, algorithm="giac")

[Out]

integrate((c*x)^(9/2)/(b*x^2 + a)^(1/4), x)