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

Optimal. Leaf size=134 \[ \frac{2 B \left (a+b x^3\right )^{7/2} (e x)^{m+1}}{b e (2 m+23)}-\frac{a^2 \sqrt{a+b x^3} (e x)^{m+1} (2 a B (m+1)-A b (2 m+23)) \, _2F_1\left (-\frac{5}{2},\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )}{b e (m+1) (2 m+23) \sqrt{\frac{b x^3}{a}+1}} \]

[Out]

(2*B*(e*x)^(1 + m)*(a + b*x^3)^(7/2))/(b*e*(23 + 2*m)) - (a^2*(2*a*B*(1 + m) - A*b*(23 + 2*m))*(e*x)^(1 + m)*S
qrt[a + b*x^3]*Hypergeometric2F1[-5/2, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)])/(b*e*(1 + m)*(23 + 2*m)*Sqrt[1 + (
b*x^3)/a])

________________________________________________________________________________________

Rubi [A]  time = 0.0791409, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {459, 365, 364} \[ \frac{2 B \left (a+b x^3\right )^{7/2} (e x)^{m+1}}{b e (2 m+23)}-\frac{a^2 \sqrt{a+b x^3} (e x)^{m+1} (2 a B (m+1)-A b (2 m+23)) \, _2F_1\left (-\frac{5}{2},\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )}{b e (m+1) (2 m+23) \sqrt{\frac{b x^3}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*(e*x)^(1 + m)*(a + b*x^3)^(7/2))/(b*e*(23 + 2*m)) - (a^2*(2*a*B*(1 + m) - A*b*(23 + 2*m))*(e*x)^(1 + m)*S
qrt[a + b*x^3]*Hypergeometric2F1[-5/2, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)])/(b*e*(1 + m)*(23 + 2*m)*Sqrt[1 + (
b*x^3)/a])

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
 n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
 && NeQ[m + n*(p + 1) + 1, 0]

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
 (ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int (e x)^m \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx &=\frac{2 B (e x)^{1+m} \left (a+b x^3\right )^{7/2}}{b e (23+2 m)}-\frac{\left (a B (1+m)-A b \left (\frac{23}{2}+m\right )\right ) \int (e x)^m \left (a+b x^3\right )^{5/2} \, dx}{b \left (\frac{23}{2}+m\right )}\\ &=\frac{2 B (e x)^{1+m} \left (a+b x^3\right )^{7/2}}{b e (23+2 m)}-\frac{\left (a^2 \left (a B (1+m)-A b \left (\frac{23}{2}+m\right )\right ) \sqrt{a+b x^3}\right ) \int (e x)^m \left (1+\frac{b x^3}{a}\right )^{5/2} \, dx}{b \left (\frac{23}{2}+m\right ) \sqrt{1+\frac{b x^3}{a}}}\\ &=\frac{2 B (e x)^{1+m} \left (a+b x^3\right )^{7/2}}{b e (23+2 m)}-\frac{a^2 (2 a B (1+m)-A b (23+2 m)) (e x)^{1+m} \sqrt{a+b x^3} \, _2F_1\left (-\frac{5}{2},\frac{1+m}{3};\frac{4+m}{3};-\frac{b x^3}{a}\right )}{b e (1+m) (23+2 m) \sqrt{1+\frac{b x^3}{a}}}\\ \end{align*}

Mathematica [A]  time = 0.111362, size = 113, normalized size = 0.84 \[ \frac{a^2 x \sqrt{a+b x^3} (e x)^m \left (A (m+4) \, _2F_1\left (-\frac{5}{2},\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )+B (m+1) x^3 \, _2F_1\left (-\frac{5}{2},\frac{m+4}{3};\frac{m+7}{3};-\frac{b x^3}{a}\right )\right )}{(m+1) (m+4) \sqrt{\frac{b x^3}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x*(e*x)^m*Sqrt[a + b*x^3]*(A*(4 + m)*Hypergeometric2F1[-5/2, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)] + B*(1 +
 m)*x^3*Hypergeometric2F1[-5/2, (4 + m)/3, (7 + m)/3, -((b*x^3)/a)]))/((1 + m)*(4 + m)*Sqrt[1 + (b*x^3)/a])

________________________________________________________________________________________

Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ( b{x}^{3}+a \right ) ^{{\frac{5}{2}}} \left ( B{x}^{3}+A \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B b^{2} x^{9} +{\left (2 \, B a b + A b^{2}\right )} x^{6} +{\left (B a^{2} + 2 \, A a b\right )} x^{3} + A a^{2}\right )} \sqrt{b x^{3} + a} \left (e x\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*b^2*x^9 + (2*B*a*b + A*b^2)*x^6 + (B*a^2 + 2*A*a*b)*x^3 + A*a^2)*sqrt(b*x^3 + a)*(e*x)^m, x)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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