3.930 \(\int \frac{\sqrt [4]{a+b x^2}}{(c x)^{15/2}} \, dx\)

Optimal. Leaf size=85 \[ -\frac{64 \left (a+b x^2\right )^{13/4}}{585 a^3 c (c x)^{13/2}}+\frac{16 \left (a+b x^2\right )^{9/4}}{45 a^2 c (c x)^{13/2}}-\frac{2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{13/2}} \]

[Out]

(-2*(a + b*x^2)^(5/4))/(5*a*c*(c*x)^(13/2)) + (16*(a + b*x^2)^(9/4))/(45*a^2*c*(c*x)^(13/2)) - (64*(a + b*x^2)
^(13/4))/(585*a^3*c*(c*x)^(13/2))

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Rubi [A]  time = 0.0244733, 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} \[ -\frac{64 \left (a+b x^2\right )^{13/4}}{585 a^3 c (c x)^{13/2}}+\frac{16 \left (a+b x^2\right )^{9/4}}{45 a^2 c (c x)^{13/2}}-\frac{2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{13/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/4)/(c*x)^(15/2),x]

[Out]

(-2*(a + b*x^2)^(5/4))/(5*a*c*(c*x)^(13/2)) + (16*(a + b*x^2)^(9/4))/(45*a^2*c*(c*x)^(13/2)) - (64*(a + b*x^2)
^(13/4))/(585*a^3*c*(c*x)^(13/2))

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)^{15/2}} \, dx &=-\frac{2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{13/2}}-\frac{8 \int \frac{\left (a+b x^2\right )^{5/4}}{(c x)^{15/2}} \, dx}{5 a}\\ &=-\frac{2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{13/2}}+\frac{16 \left (a+b x^2\right )^{9/4}}{45 a^2 c (c x)^{13/2}}+\frac{32 \int \frac{\left (a+b x^2\right )^{9/4}}{(c x)^{15/2}} \, dx}{45 a^2}\\ &=-\frac{2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{13/2}}+\frac{16 \left (a+b x^2\right )^{9/4}}{45 a^2 c (c x)^{13/2}}-\frac{64 \left (a+b x^2\right )^{13/4}}{585 a^3 c (c x)^{13/2}}\\ \end{align*}

Mathematica [A]  time = 0.0169091, size = 52, normalized size = 0.61 \[ -\frac{2 \sqrt{c x} \left (a+b x^2\right )^{5/4} \left (45 a^2-40 a b x^2+32 b^2 x^4\right )}{585 a^3 c^8 x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(1/4)/(c*x)^(15/2),x]

[Out]

(-2*Sqrt[c*x]*(a + b*x^2)^(5/4)*(45*a^2 - 40*a*b*x^2 + 32*b^2*x^4))/(585*a^3*c^8*x^7)

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Maple [A]  time = 0.004, size = 42, normalized size = 0.5 \begin{align*} -{\frac{2\,x \left ( 32\,{b}^{2}{x}^{4}-40\,ab{x}^{2}+45\,{a}^{2} \right ) }{585\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{4}}} \left ( cx \right ) ^{-{\frac{15}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(1/4)/(c*x)^(15/2),x)

[Out]

-2/585*x*(b*x^2+a)^(5/4)*(32*b^2*x^4-40*a*b*x^2+45*a^2)/a^3/(c*x)^(15/2)

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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{15}{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)^(15/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(1/4)/(c*x)^(15/2), x)

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Fricas [A]  time = 1.77874, size = 135, normalized size = 1.59 \begin{align*} -\frac{2 \,{\left (32 \, b^{3} x^{6} - 8 \, a b^{2} x^{4} + 5 \, a^{2} b x^{2} + 45 \, a^{3}\right )}{\left (b x^{2} + a\right )}^{\frac{1}{4}} \sqrt{c x}}{585 \, a^{3} c^{8} x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/4)/(c*x)^(15/2),x, algorithm="fricas")

[Out]

-2/585*(32*b^3*x^6 - 8*a*b^2*x^4 + 5*a^2*b*x^2 + 45*a^3)*(b*x^2 + a)^(1/4)*sqrt(c*x)/(a^3*c^8*x^7)

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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)**(15/2),x)

[Out]

Timed out

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Giac [B]  time = 3.14117, size = 244, normalized size = 2.87 \begin{align*} -\frac{2 \,{\left (\frac{117 \,{\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^{2} c^{4}}{\sqrt{c x}} - \frac{130 \,{\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}} b}{\sqrt{c x} c^{2} x^{4}} + \frac{45 \,{\left (b^{3} c^{12} x^{6} + 3 \, a b^{2} c^{12} x^{4} + 3 \, a^{2} b c^{12} x^{2} + a^{3} c^{12}\right )}{\left (b c^{4} x^{2} + a c^{4}\right )}^{\frac{1}{4}}}{\sqrt{c x} c^{6} x^{6}}\right )}}{585 \, a^{3} c^{14}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/4)/(c*x)^(15/2),x, algorithm="giac")

[Out]

-2/585*(117*(b*c^4*x^2 + a*c^4)^(1/4)*(b*c^2 + a*c^2/x^2)*b^2*c^4/sqrt(c*x) - 130*(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)*b/(sqrt(c*x)*c^2*x^4) + 45*(b^3*c^12*x^6 + 3*a*b^2*c^12*x^4 + 3*a^2*b*c^
12*x^2 + a^3*c^12)*(b*c^4*x^2 + a*c^4)^(1/4)/(sqrt(c*x)*c^6*x^6))/(a^3*c^14)