3.931 \(\int \frac {\sqrt [4]{a+b x^2}}{(c x)^{19/2}} \, dx\)

Optimal. Leaf size=113 \[ \frac {256 \left (a+b x^2\right )^{17/4}}{3315 a^4 c (c x)^{17/2}}-\frac {64 \left (a+b x^2\right )^{13/4}}{195 a^3 c (c x)^{17/2}}+\frac {8 \left (a+b x^2\right )^{9/4}}{15 a^2 c (c x)^{17/2}}-\frac {2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{17/2}} \]

[Out]

-2/5*(b*x^2+a)^(5/4)/a/c/(c*x)^(17/2)+8/15*(b*x^2+a)^(9/4)/a^2/c/(c*x)^(17/2)-64/195*(b*x^2+a)^(13/4)/a^3/c/(c
*x)^(17/2)+256/3315*(b*x^2+a)^(17/4)/a^4/c/(c*x)^(17/2)

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Rubi [A]  time = 0.04, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {273, 264} \[ \frac {256 \left (a+b x^2\right )^{17/4}}{3315 a^4 c (c x)^{17/2}}-\frac {64 \left (a+b x^2\right )^{13/4}}{195 a^3 c (c x)^{17/2}}+\frac {8 \left (a+b x^2\right )^{9/4}}{15 a^2 c (c x)^{17/2}}-\frac {2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{17/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/4)/(c*x)^(19/2),x]

[Out]

(-2*(a + b*x^2)^(5/4))/(5*a*c*(c*x)^(17/2)) + (8*(a + b*x^2)^(9/4))/(15*a^2*c*(c*x)^(17/2)) - (64*(a + b*x^2)^
(13/4))/(195*a^3*c*(c*x)^(17/2)) + (256*(a + b*x^2)^(17/4))/(3315*a^4*c*(c*x)^(17/2))

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)^{19/2}} \, dx &=-\frac {2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{17/2}}-\frac {12 \int \frac {\left (a+b x^2\right )^{5/4}}{(c x)^{19/2}} \, dx}{5 a}\\ &=-\frac {2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{17/2}}+\frac {8 \left (a+b x^2\right )^{9/4}}{15 a^2 c (c x)^{17/2}}+\frac {32 \int \frac {\left (a+b x^2\right )^{9/4}}{(c x)^{19/2}} \, dx}{15 a^2}\\ &=-\frac {2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{17/2}}+\frac {8 \left (a+b x^2\right )^{9/4}}{15 a^2 c (c x)^{17/2}}-\frac {64 \left (a+b x^2\right )^{13/4}}{195 a^3 c (c x)^{17/2}}-\frac {128 \int \frac {\left (a+b x^2\right )^{13/4}}{(c x)^{19/2}} \, dx}{195 a^3}\\ &=-\frac {2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{17/2}}+\frac {8 \left (a+b x^2\right )^{9/4}}{15 a^2 c (c x)^{17/2}}-\frac {64 \left (a+b x^2\right )^{13/4}}{195 a^3 c (c x)^{17/2}}+\frac {256 \left (a+b x^2\right )^{17/4}}{3315 a^4 c (c x)^{17/2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 63, normalized size = 0.56 \[ \frac {2 \left (a+b x^2\right )^{5/4} \left (-195 a^3+180 a^2 b x^2-160 a b^2 x^4+128 b^3 x^6\right )}{3315 a^4 c^9 x^8 \sqrt {c x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(1/4)/(c*x)^(19/2),x]

[Out]

(2*(a + b*x^2)^(5/4)*(-195*a^3 + 180*a^2*b*x^2 - 160*a*b^2*x^4 + 128*b^3*x^6))/(3315*a^4*c^9*x^8*Sqrt[c*x])

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fricas [A]  time = 1.16, size = 68, normalized size = 0.60 \[ \frac {2 \, {\left (128 \, b^{4} x^{8} - 32 \, a b^{3} x^{6} + 20 \, a^{2} b^{2} x^{4} - 15 \, a^{3} b x^{2} - 195 \, a^{4}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {c x}}{3315 \, a^{4} c^{10} x^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/4)/(c*x)^(19/2),x, algorithm="fricas")

[Out]

2/3315*(128*b^4*x^8 - 32*a*b^3*x^6 + 20*a^2*b^2*x^4 - 15*a^3*b*x^2 - 195*a^4)*(b*x^2 + a)^(1/4)*sqrt(c*x)/(a^4
*c^10*x^9)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {19}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/4)/(c*x)^(19/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(1/4)/(c*x)^(19/2), x)

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maple [A]  time = 0.01, size = 53, normalized size = 0.47 \[ -\frac {2 \left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (-128 b^{3} x^{6}+160 a \,b^{2} x^{4}-180 a^{2} b \,x^{2}+195 a^{3}\right ) x}{3315 \left (c x \right )^{\frac {19}{2}} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(1/4)/(c*x)^(19/2),x)

[Out]

-2/3315*x*(b*x^2+a)^(5/4)*(-128*b^3*x^6+160*a*b^2*x^4-180*a^2*b*x^2+195*a^3)/a^4/(c*x)^(19/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {19}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/4)/(c*x)^(19/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(1/4)/(c*x)^(19/2), x)

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mupad [B]  time = 5.02, size = 79, normalized size = 0.70 \[ -\frac {{\left (b\,x^2+a\right )}^{1/4}\,\left (\frac {2}{17\,c^9}+\frac {2\,b\,x^2}{221\,a\,c^9}-\frac {8\,b^2\,x^4}{663\,a^2\,c^9}+\frac {64\,b^3\,x^6}{3315\,a^3\,c^9}-\frac {256\,b^4\,x^8}{3315\,a^4\,c^9}\right )}{x^8\,\sqrt {c\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/4)/(c*x)^(19/2),x)

[Out]

-((a + b*x^2)^(1/4)*(2/(17*c^9) + (2*b*x^2)/(221*a*c^9) - (8*b^2*x^4)/(663*a^2*c^9) + (64*b^3*x^6)/(3315*a^3*c
^9) - (256*b^4*x^8)/(3315*a^4*c^9)))/(x^8*(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)**(19/2),x)

[Out]

Timed out

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