3.990 \(\int \frac{1}{(c x)^{13/2} (a+b x^2)^{5/4}} \, dx\)

Optimal. Leaf size=109 \[ -\frac{256 \left (a+b x^2\right )^{11/4}}{77 a^4 c (c x)^{11/2}}+\frac{64 \left (a+b x^2\right )^{7/4}}{7 a^3 c (c x)^{11/2}}-\frac{8 \left (a+b x^2\right )^{3/4}}{a^2 c (c x)^{11/2}}+\frac{2}{a c (c x)^{11/2} \sqrt [4]{a+b x^2}} \]

[Out]

2/(a*c*(c*x)^(11/2)*(a + b*x^2)^(1/4)) - (8*(a + b*x^2)^(3/4))/(a^2*c*(c*x)^(11/2)) + (64*(a + b*x^2)^(7/4))/(
7*a^3*c*(c*x)^(11/2)) - (256*(a + b*x^2)^(11/4))/(77*a^4*c*(c*x)^(11/2))

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Rubi [A]  time = 0.0374668, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {273, 264} \[ -\frac{256 \left (a+b x^2\right )^{11/4}}{77 a^4 c (c x)^{11/2}}+\frac{64 \left (a+b x^2\right )^{7/4}}{7 a^3 c (c x)^{11/2}}-\frac{8 \left (a+b x^2\right )^{3/4}}{a^2 c (c x)^{11/2}}+\frac{2}{a c (c x)^{11/2} \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(13/2)*(a + b*x^2)^(5/4)),x]

[Out]

2/(a*c*(c*x)^(11/2)*(a + b*x^2)^(1/4)) - (8*(a + b*x^2)^(3/4))/(a^2*c*(c*x)^(11/2)) + (64*(a + b*x^2)^(7/4))/(
7*a^3*c*(c*x)^(11/2)) - (256*(a + b*x^2)^(11/4))/(77*a^4*c*(c*x)^(11/2))

Rule 273

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

Rule 264

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] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{1}{(c x)^{13/2} \left (a+b x^2\right )^{5/4}} \, dx &=\frac{2}{a c (c x)^{11/2} \sqrt [4]{a+b x^2}}+\frac{12 \int \frac{1}{(c x)^{13/2} \sqrt [4]{a+b x^2}} \, dx}{a}\\ &=\frac{2}{a c (c x)^{11/2} \sqrt [4]{a+b x^2}}-\frac{8 \left (a+b x^2\right )^{3/4}}{a^2 c (c x)^{11/2}}-\frac{32 \int \frac{\left (a+b x^2\right )^{3/4}}{(c x)^{13/2}} \, dx}{a^2}\\ &=\frac{2}{a c (c x)^{11/2} \sqrt [4]{a+b x^2}}-\frac{8 \left (a+b x^2\right )^{3/4}}{a^2 c (c x)^{11/2}}+\frac{64 \left (a+b x^2\right )^{7/4}}{7 a^3 c (c x)^{11/2}}+\frac{128 \int \frac{\left (a+b x^2\right )^{7/4}}{(c x)^{13/2}} \, dx}{7 a^3}\\ &=\frac{2}{a c (c x)^{11/2} \sqrt [4]{a+b x^2}}-\frac{8 \left (a+b x^2\right )^{3/4}}{a^2 c (c x)^{11/2}}+\frac{64 \left (a+b x^2\right )^{7/4}}{7 a^3 c (c x)^{11/2}}-\frac{256 \left (a+b x^2\right )^{11/4}}{77 a^4 c (c x)^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.0122965, size = 58, normalized size = 0.53 \[ -\frac{2 x \left (-12 a^2 b x^2+7 a^3+32 a b^2 x^4+128 b^3 x^6\right )}{77 a^4 (c x)^{13/2} \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(13/2)*(a + b*x^2)^(5/4)),x]

[Out]

(-2*x*(7*a^3 - 12*a^2*b*x^2 + 32*a*b^2*x^4 + 128*b^3*x^6))/(77*a^4*(c*x)^(13/2)*(a + b*x^2)^(1/4))

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Maple [A]  time = 0.004, size = 53, normalized size = 0.5 \begin{align*} -{\frac{2\,x \left ( 128\,{b}^{3}{x}^{6}+32\,a{b}^{2}{x}^{4}-12\,{a}^{2}b{x}^{2}+7\,{a}^{3} \right ) }{77\,{a}^{4}}{\frac{1}{\sqrt [4]{b{x}^{2}+a}}} \left ( cx \right ) ^{-{\frac{13}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(13/2)/(b*x^2+a)^(5/4),x)

[Out]

-2/77*x*(128*b^3*x^6+32*a*b^2*x^4-12*a^2*b*x^2+7*a^3)/(b*x^2+a)^(1/4)/a^4/(c*x)^(13/2)

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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{5}{4}} \left (c x\right )^{\frac{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(13/2)/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(5/4)*(c*x)^(13/2)), x)

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Fricas [A]  time = 1.61597, size = 158, normalized size = 1.45 \begin{align*} -\frac{2 \,{\left (128 \, b^{3} x^{6} + 32 \, a b^{2} x^{4} - 12 \, a^{2} b x^{2} + 7 \, a^{3}\right )}{\left (b x^{2} + a\right )}^{\frac{3}{4}} \sqrt{c x}}{77 \,{\left (a^{4} b c^{7} x^{8} + a^{5} c^{7} x^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(13/2)/(b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

-2/77*(128*b^3*x^6 + 32*a*b^2*x^4 - 12*a^2*b*x^2 + 7*a^3)*(b*x^2 + a)^(3/4)*sqrt(c*x)/(a^4*b*c^7*x^8 + a^5*c^7
*x^6)

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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)**(13/2)/(b*x**2+a)**(5/4),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{5}{4}} \left (c x\right )^{\frac{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(13/2)/(b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(5/4)*(c*x)^(13/2)), x)