3.557 \(\int \frac{A+B x^3}{(e x)^{5/2} (a+b x^3)^{3/2}} \, dx\)

Optimal. Leaf size=67 \[ -\frac{2 (e x)^{3/2} (2 A b-a B)}{3 a^2 e^4 \sqrt{a+b x^3}}-\frac{2 A}{3 a e (e x)^{3/2} \sqrt{a+b x^3}} \]

[Out]

(-2*A)/(3*a*e*(e*x)^(3/2)*Sqrt[a + b*x^3]) - (2*(2*A*b - a*B)*(e*x)^(3/2))/(3*a^2*e^4*Sqrt[a + b*x^3])

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Rubi [A]  time = 0.029186, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {453, 264} \[ -\frac{2 (e x)^{3/2} (2 A b-a B)}{3 a^2 e^4 \sqrt{a+b x^3}}-\frac{2 A}{3 a e (e x)^{3/2} \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/((e*x)^(5/2)*(a + b*x^3)^(3/2)),x]

[Out]

(-2*A)/(3*a*e*(e*x)^(3/2)*Sqrt[a + b*x^3]) - (2*(2*A*b - a*B)*(e*x)^(3/2))/(3*a^2*e^4*Sqrt[a + b*x^3])

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[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{A+B x^3}{(e x)^{5/2} \left (a+b x^3\right )^{3/2}} \, dx &=-\frac{2 A}{3 a e (e x)^{3/2} \sqrt{a+b x^3}}-\frac{(2 A b-a B) \int \frac{\sqrt{e x}}{\left (a+b x^3\right )^{3/2}} \, dx}{a e^3}\\ &=-\frac{2 A}{3 a e (e x)^{3/2} \sqrt{a+b x^3}}-\frac{2 (2 A b-a B) (e x)^{3/2}}{3 a^2 e^4 \sqrt{a+b x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0206157, size = 45, normalized size = 0.67 \[ \frac{x \left (-2 a A+2 a B x^3-4 A b x^3\right )}{3 a^2 (e x)^{5/2} \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^3)/((e*x)^(5/2)*(a + b*x^3)^(3/2)),x]

[Out]

(x*(-2*a*A - 4*A*b*x^3 + 2*a*B*x^3))/(3*a^2*(e*x)^(5/2)*Sqrt[a + b*x^3])

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Maple [A]  time = 0.006, size = 39, normalized size = 0.6 \begin{align*} -{\frac{2\,x \left ( 2\,A{x}^{3}b-Ba{x}^{3}+Aa \right ) }{3\,{a}^{2}}{\frac{1}{\sqrt{b{x}^{3}+a}}} \left ( ex \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)/(e*x)^(5/2)/(b*x^3+a)^(3/2),x)

[Out]

-2/3*x*(2*A*b*x^3-B*a*x^3+A*a)/(b*x^3+a)^(1/2)/a^2/(e*x)^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac{3}{2}} \left (e x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/(e*x)^(5/2)/(b*x^3+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x^3 + A)/((b*x^3 + a)^(3/2)*(e*x)^(5/2)), x)

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Fricas [A]  time = 1.31067, size = 117, normalized size = 1.75 \begin{align*} \frac{2 \,{\left ({\left (B a - 2 \, A b\right )} x^{3} - A a\right )} \sqrt{b x^{3} + a} \sqrt{e x}}{3 \,{\left (a^{2} b e^{3} x^{5} + a^{3} e^{3} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/(e*x)^(5/2)/(b*x^3+a)^(3/2),x, algorithm="fricas")

[Out]

2/3*((B*a - 2*A*b)*x^3 - A*a)*sqrt(b*x^3 + a)*sqrt(e*x)/(a^2*b*e^3*x^5 + a^3*e^3*x^2)

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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**3+A)/(e*x)**(5/2)/(b*x**3+a)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac{3}{2}} \left (e x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/(e*x)^(5/2)/(b*x^3+a)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x^3 + A)/((b*x^3 + a)^(3/2)*(e*x)^(5/2)), x)