3.777 \(\int \frac{1}{(c x)^{17/3} (a+b x^2)^{2/3}} \, dx\)

Optimal. Leaf size=85 \[ -\frac{27 \left (a+b x^2\right )^{7/3}}{28 a^3 c (c x)^{14/3}}+\frac{9 \left (a+b x^2\right )^{4/3}}{4 a^2 c (c x)^{14/3}}-\frac{3 \sqrt [3]{a+b x^2}}{2 a c (c x)^{14/3}} \]

[Out]

(-3*(a + b*x^2)^(1/3))/(2*a*c*(c*x)^(14/3)) + (9*(a + b*x^2)^(4/3))/(4*a^2*c*(c*x)^(14/3)) - (27*(a + b*x^2)^(
7/3))/(28*a^3*c*(c*x)^(14/3))

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Rubi [A]  time = 0.0254695, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {273, 264} \[ -\frac{27 \left (a+b x^2\right )^{7/3}}{28 a^3 c (c x)^{14/3}}+\frac{9 \left (a+b x^2\right )^{4/3}}{4 a^2 c (c x)^{14/3}}-\frac{3 \sqrt [3]{a+b x^2}}{2 a c (c x)^{14/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(17/3)*(a + b*x^2)^(2/3)),x]

[Out]

(-3*(a + b*x^2)^(1/3))/(2*a*c*(c*x)^(14/3)) + (9*(a + b*x^2)^(4/3))/(4*a^2*c*(c*x)^(14/3)) - (27*(a + b*x^2)^(
7/3))/(28*a^3*c*(c*x)^(14/3))

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)^{17/3} \left (a+b x^2\right )^{2/3}} \, dx &=-\frac{3 \sqrt [3]{a+b x^2}}{2 a c (c x)^{14/3}}-\frac{6 \int \frac{\sqrt [3]{a+b x^2}}{(c x)^{17/3}} \, dx}{a}\\ &=-\frac{3 \sqrt [3]{a+b x^2}}{2 a c (c x)^{14/3}}+\frac{9 \left (a+b x^2\right )^{4/3}}{4 a^2 c (c x)^{14/3}}+\frac{9 \int \frac{\left (a+b x^2\right )^{4/3}}{(c x)^{17/3}} \, dx}{2 a^2}\\ &=-\frac{3 \sqrt [3]{a+b x^2}}{2 a c (c x)^{14/3}}+\frac{9 \left (a+b x^2\right )^{4/3}}{4 a^2 c (c x)^{14/3}}-\frac{27 \left (a+b x^2\right )^{7/3}}{28 a^3 c (c x)^{14/3}}\\ \end{align*}

Mathematica [A]  time = 0.0238583, size = 52, normalized size = 0.61 \[ -\frac{3 \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (2 a^2-3 a b x^2+9 b^2 x^4\right )}{28 a^3 c^6 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(17/3)*(a + b*x^2)^(2/3)),x]

[Out]

(-3*(c*x)^(1/3)*(a + b*x^2)^(1/3)*(2*a^2 - 3*a*b*x^2 + 9*b^2*x^4))/(28*a^3*c^6*x^5)

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Maple [A]  time = 0.004, size = 42, normalized size = 0.5 \begin{align*} -{\frac{3\,x \left ( 9\,{b}^{2}{x}^{4}-3\,ab{x}^{2}+2\,{a}^{2} \right ) }{28\,{a}^{3}}\sqrt [3]{b{x}^{2}+a} \left ( cx \right ) ^{-{\frac{17}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(17/3)/(b*x^2+a)^(2/3),x)

[Out]

-3/28*x*(b*x^2+a)^(1/3)*(9*b^2*x^4-3*a*b*x^2+2*a^2)/a^3/(c*x)^(17/3)

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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{2}{3}} \left (c x\right )^{\frac{17}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(17/3)/(b*x^2+a)^(2/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(2/3)*(c*x)^(17/3)), x)

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Fricas [A]  time = 2.03088, size = 112, normalized size = 1.32 \begin{align*} -\frac{3 \,{\left (9 \, b^{2} x^{4} - 3 \, a b x^{2} + 2 \, a^{2}\right )}{\left (b x^{2} + a\right )}^{\frac{1}{3}} \left (c x\right )^{\frac{1}{3}}}{28 \, a^{3} c^{6} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(17/3)/(b*x^2+a)^(2/3),x, algorithm="fricas")

[Out]

-3/28*(9*b^2*x^4 - 3*a*b*x^2 + 2*a^2)*(b*x^2 + a)^(1/3)*(c*x)^(1/3)/(a^3*c^6*x^5)

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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)**(17/3)/(b*x**2+a)**(2/3),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{2}{3}} \left (c x\right )^{\frac{17}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(17/3)/(b*x^2+a)^(2/3),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(2/3)*(c*x)^(17/3)), x)