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

Optimal. Leaf size=79 \[ \frac{2 (e x)^{3/2} (a B+2 A b)}{9 a^2 b e \sqrt{a+b x^3}}+\frac{2 (e x)^{3/2} (A b-a B)}{9 a b e \left (a+b x^3\right )^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0311775, antiderivative size = 79, 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 = {457, 264} \[ \frac{2 (e x)^{3/2} (a B+2 A b)}{9 a^2 b e \sqrt{a+b x^3}}+\frac{2 (e x)^{3/2} (A b-a B)}{9 a b e \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(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{e x} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx &=\frac{2 (A b-a B) (e x)^{3/2}}{9 a b e \left (a+b x^3\right )^{3/2}}+\frac{\left (2 \left (3 A b+\frac{3 a B}{2}\right )\right ) \int \frac{\sqrt{e x}}{\left (a+b x^3\right )^{3/2}} \, dx}{9 a b}\\ &=\frac{2 (A b-a B) (e x)^{3/2}}{9 a b e \left (a+b x^3\right )^{3/2}}+\frac{2 (2 A b+a B) (e x)^{3/2}}{9 a^2 b e \sqrt{a+b x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0371331, size = 44, normalized size = 0.56 \[ \frac{2 x \sqrt{e x} \left (3 a A+a B x^3+2 A b x^3\right )}{9 a^2 \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 39, normalized size = 0.5 \begin{align*}{\frac{2\,x \left ( 2\,A{x}^{3}b+Ba{x}^{3}+3\,Aa \right ) }{9\,{a}^{2}}\sqrt{ex} \left ( b{x}^{3}+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*x*(2*A*b*x^3+B*a*x^3+3*A*a)*(e*x)^(1/2)/(b*x^3+a)^(3/2)/a^2

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{3} + A\right )} \sqrt{e x}}{{\left (b x^{3} + a\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)^(1/2)/(b*x^3+a)^(5/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.36708, size = 128, normalized size = 1.62 \begin{align*} \frac{2 \,{\left ({\left (B a + 2 \, A b\right )} x^{4} + 3 \, A a x\right )} \sqrt{b x^{3} + a} \sqrt{e x}}{9 \,{\left (a^{2} b^{2} x^{6} + 2 \, a^{3} b x^{3} + a^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*(e*x)^(1/2)/(b*x^3+a)^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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)**(1/2)/(b*x**3+a)**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.47911, size = 86, normalized size = 1.09 \begin{align*} \frac{2 \, x^{\frac{3}{2}}{\left (\frac{3 \, A e^{5}}{a} + \frac{{\left (B a^{5} b^{5} e^{21} + 2 \, A a^{4} b^{6} e^{21}\right )} x^{3} e^{\left (-16\right )}}{a^{6} b^{5}}\right )} e^{\frac{3}{2}}}{9 \,{\left (b x^{3} e^{4} + a e^{4}\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*(e*x)^(1/2)/(b*x^3+a)^(5/2),x, algorithm="giac")

[Out]

2/9*x^(3/2)*(3*A*e^5/a + (B*a^5*b^5*e^21 + 2*A*a^4*b^6*e^21)*x^3*e^(-16)/(a^6*b^5))*e^(3/2)/(b*x^3*e^4 + a*e^4
)^(3/2)