3.983 \(\int \frac{1}{(c x)^{7/2} (a-b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=57 \[ \frac{8 \left (a-b x^2\right )^{5/4}}{5 a^2 c (c x)^{5/2}}-\frac{2 \sqrt [4]{a-b x^2}}{a c (c x)^{5/2}} \]

[Out]

(-2*(a - b*x^2)^(1/4))/(a*c*(c*x)^(5/2)) + (8*(a - b*x^2)^(5/4))/(5*a^2*c*(c*x)^(5/2))

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Rubi [A]  time = 0.0140473, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {273, 264} \[ \frac{8 \left (a-b x^2\right )^{5/4}}{5 a^2 c (c x)^{5/2}}-\frac{2 \sqrt [4]{a-b x^2}}{a c (c x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a - b*x^2)^(1/4))/(a*c*(c*x)^(5/2)) + (8*(a - b*x^2)^(5/4))/(5*a^2*c*(c*x)^(5/2))

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)^{7/2} \left (a-b x^2\right )^{3/4}} \, dx &=-\frac{2 \sqrt [4]{a-b x^2}}{a c (c x)^{5/2}}-\frac{4 \int \frac{\sqrt [4]{a-b x^2}}{(c x)^{7/2}} \, dx}{a}\\ &=-\frac{2 \sqrt [4]{a-b x^2}}{a c (c x)^{5/2}}+\frac{8 \left (a-b x^2\right )^{5/4}}{5 a^2 c (c x)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0177282, size = 35, normalized size = 0.61 \[ -\frac{2 x \sqrt [4]{a-b x^2} \left (a+4 b x^2\right )}{5 a^2 (c x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x*(a - b*x^2)^(1/4)*(a + 4*b*x^2))/(5*a^2*(c*x)^(7/2))

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Maple [A]  time = 0.004, size = 30, normalized size = 0.5 \begin{align*} -{\frac{2\,x \left ( 4\,b{x}^{2}+a \right ) }{5\,{a}^{2}}\sqrt [4]{-b{x}^{2}+a} \left ( cx \right ) ^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*x*(-b*x^2+a)^(1/4)*(4*b*x^2+a)/a^2/(c*x)^(7/2)

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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{3}{4}} \left (c x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(7/2)/(-b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.07214, size = 85, normalized size = 1.49 \begin{align*} -\frac{2 \,{\left (4 \, b x^{2} + a\right )}{\left (-b x^{2} + a\right )}^{\frac{1}{4}} \sqrt{c x}}{5 \, a^{2} c^{4} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(7/2)/(-b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

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

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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)**(7/2)/(-b*x**2+a)**(3/4),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{3}{4}} \left (c x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(7/2)/(-b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

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