3.929 \(\int \frac{\sqrt [4]{a+b x^2}}{(c x)^{11/2}} \, dx\)

Optimal. Leaf size=57 \[ \frac{8 \left (a+b x^2\right )^{9/4}}{45 a^2 c (c x)^{9/2}}-\frac{2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{9/2}} \]

[Out]

(-2*(a + b*x^2)^(5/4))/(5*a*c*(c*x)^(9/2)) + (8*(a + b*x^2)^(9/4))/(45*a^2*c*(c*x)^(9/2))

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Rubi [A]  time = 0.0145266, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {273, 264} \[ \frac{8 \left (a+b x^2\right )^{9/4}}{45 a^2 c (c x)^{9/2}}-\frac{2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/4)/(c*x)^(11/2),x]

[Out]

(-2*(a + b*x^2)^(5/4))/(5*a*c*(c*x)^(9/2)) + (8*(a + b*x^2)^(9/4))/(45*a^2*c*(c*x)^(9/2))

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{\sqrt [4]{a+b x^2}}{(c x)^{11/2}} \, dx &=-\frac{2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{9/2}}-\frac{4 \int \frac{\left (a+b x^2\right )^{5/4}}{(c x)^{11/2}} \, dx}{5 a}\\ &=-\frac{2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{9/2}}+\frac{8 \left (a+b x^2\right )^{9/4}}{45 a^2 c (c x)^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0148063, size = 41, normalized size = 0.72 \[ \frac{2 \sqrt{c x} \left (a+b x^2\right )^{5/4} \left (4 b x^2-5 a\right )}{45 a^2 c^6 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(1/4)/(c*x)^(11/2),x]

[Out]

(2*Sqrt[c*x]*(a + b*x^2)^(5/4)*(-5*a + 4*b*x^2))/(45*a^2*c^6*x^5)

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Maple [A]  time = 0.005, size = 31, normalized size = 0.5 \begin{align*} -{\frac{2\,x \left ( -4\,b{x}^{2}+5\,a \right ) }{45\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{4}}} \left ( cx \right ) ^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(1/4)/(c*x)^(11/2),x)

[Out]

-2/45*x*(b*x^2+a)^(5/4)*(-4*b*x^2+5*a)/a^2/(c*x)^(11/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{1}{4}}}{\left (c x\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/4)/(c*x)^(11/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(1/4)/(c*x)^(11/2), x)

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Fricas [A]  time = 1.69684, size = 105, normalized size = 1.84 \begin{align*} \frac{2 \,{\left (4 \, b^{2} x^{4} - a b x^{2} - 5 \, a^{2}\right )}{\left (b x^{2} + a\right )}^{\frac{1}{4}} \sqrt{c x}}{45 \, a^{2} c^{6} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/4)/(c*x)^(11/2),x, algorithm="fricas")

[Out]

2/45*(4*b^2*x^4 - a*b*x^2 - 5*a^2)*(b*x^2 + a)^(1/4)*sqrt(c*x)/(a^2*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((b*x**2+a)**(1/4)/(c*x)**(11/2),x)

[Out]

Timed out

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Giac [B]  time = 2.25524, size = 144, normalized size = 2.53 \begin{align*} \frac{2 \,{\left (\frac{9 \,{\left (b c^{4} x^{2} + a c^{4}\right )}^{\frac{1}{4}}{\left (b c^{2} + \frac{a c^{2}}{x^{2}}\right )} b c^{2}}{\sqrt{c x}} - \frac{5 \,{\left (b^{2} c^{8} x^{4} + 2 \, a b c^{8} x^{2} + a^{2} c^{8}\right )}{\left (b c^{4} x^{2} + a c^{4}\right )}^{\frac{1}{4}}}{\sqrt{c x} c^{4} x^{4}}\right )}}{45 \, a^{2} c^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/4)/(c*x)^(11/2),x, algorithm="giac")

[Out]

2/45*(9*(b*c^4*x^2 + a*c^4)^(1/4)*(b*c^2 + a*c^2/x^2)*b*c^2/sqrt(c*x) - 5*(b^2*c^8*x^4 + 2*a*b*c^8*x^2 + a^2*c
^8)*(b*c^4*x^2 + a*c^4)^(1/4)/(sqrt(c*x)*c^4*x^4))/(a^2*c^10)