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

Optimal. Leaf size=104 \[ -\frac{4 (e x)^{3/2} (4 A b-a B)}{9 a^3 e^4 \sqrt{a+b x^3}}-\frac{2 (e x)^{3/2} (4 A b-a B)}{9 a^2 e^4 \left (a+b x^3\right )^{3/2}}-\frac{2 A}{3 a e (e x)^{3/2} \left (a+b x^3\right )^{3/2}} \]

[Out]

(-2*A)/(3*a*e*(e*x)^(3/2)*(a + b*x^3)^(3/2)) - (2*(4*A*b - a*B)*(e*x)^(3/2))/(9*a^2*e^4*(a + b*x^3)^(3/2)) - (
4*(4*A*b - a*B)*(e*x)^(3/2))/(9*a^3*e^4*Sqrt[a + b*x^3])

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Rubi [A]  time = 0.0464514, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {453, 273, 264} \[ -\frac{4 (e x)^{3/2} (4 A b-a B)}{9 a^3 e^4 \sqrt{a+b x^3}}-\frac{2 (e x)^{3/2} (4 A b-a B)}{9 a^2 e^4 \left (a+b x^3\right )^{3/2}}-\frac{2 A}{3 a e (e x)^{3/2} \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(3*a*e*(e*x)^(3/2)*(a + b*x^3)^(3/2)) - (2*(4*A*b - a*B)*(e*x)^(3/2))/(9*a^2*e^4*(a + b*x^3)^(3/2)) - (
4*(4*A*b - a*B)*(e*x)^(3/2))/(9*a^3*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 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{A+B x^3}{(e x)^{5/2} \left (a+b x^3\right )^{5/2}} \, dx &=-\frac{2 A}{3 a e (e x)^{3/2} \left (a+b x^3\right )^{3/2}}-\frac{(4 A b-a B) \int \frac{\sqrt{e x}}{\left (a+b x^3\right )^{5/2}} \, dx}{a e^3}\\ &=-\frac{2 A}{3 a e (e x)^{3/2} \left (a+b x^3\right )^{3/2}}-\frac{2 (4 A b-a B) (e x)^{3/2}}{9 a^2 e^4 \left (a+b x^3\right )^{3/2}}-\frac{(2 (4 A b-a B)) \int \frac{\sqrt{e x}}{\left (a+b x^3\right )^{3/2}} \, dx}{3 a^2 e^3}\\ &=-\frac{2 A}{3 a e (e x)^{3/2} \left (a+b x^3\right )^{3/2}}-\frac{2 (4 A b-a B) (e x)^{3/2}}{9 a^2 e^4 \left (a+b x^3\right )^{3/2}}-\frac{4 (4 A b-a B) (e x)^{3/2}}{9 a^3 e^4 \sqrt{a+b x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0367259, size = 65, normalized size = 0.62 \[ \frac{x \left (-6 a^2 \left (A-B x^3\right )+4 a b x^3 \left (B x^3-6 A\right )-16 A b^2 x^6\right )}{9 a^3 (e x)^{5/2} \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-16*A*b^2*x^6 - 6*a^2*(A - B*x^3) + 4*a*b*x^3*(-6*A + B*x^3)))/(9*a^3*(e*x)^(5/2)*(a + b*x^3)^(3/2))

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Maple [A]  time = 0.006, size = 62, normalized size = 0.6 \begin{align*} -{\frac{2\,x \left ( 8\,A{b}^{2}{x}^{6}-2\,B{x}^{6}ab+12\,aAb{x}^{3}-3\,B{x}^{3}{a}^{2}+3\,A{a}^{2} \right ) }{9\,{a}^{3}} \left ( b{x}^{3}+a \right ) ^{-{\frac{3}{2}}} \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)^(5/2),x)

[Out]

-2/9*x*(8*A*b^2*x^6-2*B*a*b*x^6+12*A*a*b*x^3-3*B*a^2*x^3+3*A*a^2)/(b*x^3+a)^(3/2)/a^3/(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{5}{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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.32226, size = 193, normalized size = 1.86 \begin{align*} \frac{2 \,{\left (2 \,{\left (B a b - 4 \, A b^{2}\right )} x^{6} + 3 \,{\left (B a^{2} - 4 \, A a b\right )} x^{3} - 3 \, A a^{2}\right )} \sqrt{b x^{3} + a} \sqrt{e x}}{9 \,{\left (a^{3} b^{2} e^{3} x^{8} + 2 \, a^{4} b e^{3} x^{5} + a^{5} 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)^(5/2),x, algorithm="fricas")

[Out]

2/9*(2*(B*a*b - 4*A*b^2)*x^6 + 3*(B*a^2 - 4*A*a*b)*x^3 - 3*A*a^2)*sqrt(b*x^3 + a)*sqrt(e*x)/(a^3*b^2*e^3*x^8 +
2*a^4*b*e^3*x^5 + a^5*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)**(5/2),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{5}{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)^(5/2),x, algorithm="giac")

[Out]

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