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]

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

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Rubi [A]  time = 0.02, antiderivative size = 85, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 1.04, size = 57, normalized size = 0.67 \[ -\frac {2 \, {\left (32 \, b^{3} x^{6} - 8 \, a b^{2} x^{4} + 5 \, a^{2} b x^{2} + 45 \, a^{3}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {c x}}{585 \, a^{3} c^{8} x^{7}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {15}{2}}}\,{d x} \]

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]

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

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maple [A]  time = 0.01, size = 42, normalized size = 0.49 \[ -\frac {2 \left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (32 b^{2} x^{4}-40 a b \,x^{2}+45 a^{2}\right ) x}{585 \left (c x \right )^{\frac {15}{2}} a^{3}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {15}{2}}}\,{d x} \]

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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mupad [B]  time = 5.00, size = 65, normalized size = 0.76 \[ -\frac {{\left (b\,x^2+a\right )}^{1/4}\,\left (\frac {2}{13\,c^7}+\frac {2\,b\,x^2}{117\,a\,c^7}-\frac {16\,b^2\,x^4}{585\,a^2\,c^7}+\frac {64\,b^3\,x^6}{585\,a^3\,c^7}\right )}{x^6\,\sqrt {c\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/4)/(c*x)^(15/2),x)

[Out]

-((a + b*x^2)^(1/4)*(2/(13*c^7) + (2*b*x^2)/(117*a*c^7) - (16*b^2*x^4)/(585*a^2*c^7) + (64*b^3*x^6)/(585*a^3*c
^7)))/(x^6*(c*x)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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