3.6.18 \(\int (e x)^{3/2} \sqrt {a+b x^3} (A+B x^3) \, dx\) [518]

Optimal. Leaf size=581 \[ \frac {(14 A b-5 a B) (e x)^{5/2} \sqrt {a+b x^3}}{56 b e}+\frac {3 \left (1+\sqrt {3}\right ) a (14 A b-5 a B) e \sqrt {e x} \sqrt {a+b x^3}}{112 b^{5/3} \left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )}+\frac {B (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 b e}-\frac {3 \sqrt [4]{3} a^{4/3} (14 A b-5 a B) e \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} E\left (\cos ^{-1}\left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{112 b^{5/3} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}-\frac {3^{3/4} \left (1-\sqrt {3}\right ) a^{4/3} (14 A b-5 a B) e \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{224 b^{5/3} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \]

[Out]

1/7*B*(e*x)^(5/2)*(b*x^3+a)^(3/2)/b/e+1/56*(14*A*b-5*B*a)*(e*x)^(5/2)*(b*x^3+a)^(1/2)/b/e+3/112*a*(14*A*b-5*B*
a)*e*(1+3^(1/2))*(e*x)^(1/2)*(b*x^3+a)^(1/2)/b^(5/3)/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))-3/112*3^(1/4)*a^(4/3)*(14
*A*b-5*B*a)*e*(a^(1/3)+b^(1/3)*x)*((a^(1/3)+b^(1/3)*x*(1-3^(1/2)))^2/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)/
(a^(1/3)+b^(1/3)*x*(1-3^(1/2)))*(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))*EllipticE((1-(a^(1/3)+b^(1/3)*x*(1-3^(1/2)))^2
/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(e*x)^(1/2)*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(
2/3)*x^2)/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)/b^(5/3)/(b*x^3+a)^(1/2)/(b^(1/3)*x*(a^(1/3)+b^(1/3)*x)/(a^(
1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)-1/224*3^(3/4)*a^(4/3)*(14*A*b-5*B*a)*e*(a^(1/3)+b^(1/3)*x)*((a^(1/3)+b^(1
/3)*x*(1-3^(1/2)))^2/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)/(a^(1/3)+b^(1/3)*x*(1-3^(1/2)))*(a^(1/3)+b^(1/3)
*x*(1+3^(1/2)))*EllipticF((1-(a^(1/3)+b^(1/3)*x*(1-3^(1/2)))^2/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2),1/4*6^
(1/2)+1/4*2^(1/2))*(1-3^(1/2))*(e*x)^(1/2)*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/(a^(1/3)+b^(1/3)*x*(1+3^(1
/2)))^2)^(1/2)/b^(5/3)/(b*x^3+a)^(1/2)/(b^(1/3)*x*(a^(1/3)+b^(1/3)*x)/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]
time = 0.45, antiderivative size = 581, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {470, 285, 335, 314, 231, 1895} \begin {gather*} -\frac {3^{3/4} \left (1-\sqrt {3}\right ) a^{4/3} e \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} (14 A b-5 a B) F\left (\text {ArcCos}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{224 b^{5/3} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}-\frac {3 \sqrt [4]{3} a^{4/3} e \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} (14 A b-5 a B) E\left (\text {ArcCos}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{112 b^{5/3} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {3 \left (1+\sqrt {3}\right ) a e \sqrt {e x} \sqrt {a+b x^3} (14 A b-5 a B)}{112 b^{5/3} \left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )}+\frac {(e x)^{5/2} \sqrt {a+b x^3} (14 A b-5 a B)}{56 b e}+\frac {B (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((14*A*b - 5*a*B)*(e*x)^(5/2)*Sqrt[a + b*x^3])/(56*b*e) + (3*(1 + Sqrt[3])*a*(14*A*b - 5*a*B)*e*Sqrt[e*x]*Sqrt
[a + b*x^3])/(112*b^(5/3)*(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)) + (B*(e*x)^(5/2)*(a + b*x^3)^(3/2))/(7*b*e) - (
3*3^(1/4)*a^(4/3)*(14*A*b - 5*a*B)*e*Sqrt[e*x]*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/
3)*x^2)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*EllipticE[ArcCos[(a^(1/3) + (1 - Sqrt[3])*b^(1/3)*x)/(a^(1/3) +
(1 + Sqrt[3])*b^(1/3)*x)], (2 + Sqrt[3])/4])/(112*b^(5/3)*Sqrt[(b^(1/3)*x*(a^(1/3) + b^(1/3)*x))/(a^(1/3) + (
1 + Sqrt[3])*b^(1/3)*x)^2]*Sqrt[a + b*x^3]) - (3^(3/4)*(1 - Sqrt[3])*a^(4/3)*(14*A*b - 5*a*B)*e*Sqrt[e*x]*(a^(
1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*Elli
pticF[ArcCos[(a^(1/3) + (1 - Sqrt[3])*b^(1/3)*x)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)], (2 + Sqrt[3])/4])/(224*
b^(5/3)*Sqrt[(b^(1/3)*x*(a^(1/3) + b^(1/3)*x))/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*Sqrt[a + b*x^3])

Rule 231

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[x*(s +
r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*(
(s + r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^
2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x]

Rule 285

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

Rule 314

Int[(x_)^4/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Dist[(
Sqrt[3] - 1)*(s^2/(2*r^2)), Int[1/Sqrt[a + b*x^6], x], x] - Dist[1/(2*r^2), Int[((Sqrt[3] - 1)*s^2 - 2*r^2*x^4
)/Sqrt[a + b*x^6], x], x]] /; FreeQ[{a, b}, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 470

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 1895

Int[((c_) + (d_.)*(x_)^4)/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/
a, 3]]}, Simp[(1 + Sqrt[3])*d*s^3*x*(Sqrt[a + b*x^6]/(2*a*r^2*(s + (1 + Sqrt[3])*r*x^2))), x] - Simp[3^(1/4)*d
*s*x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]/(2*r^2*Sqrt[(r*x^2*(s + r*x^2))/
(s + (1 + Sqrt[3])*r*x^2)^2]*Sqrt[a + b*x^6]))*EllipticE[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r
*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b, c, d}, x] && EqQ[2*Rt[b/a, 3]^2*c - (1 - Sqrt[3])*d, 0]

Rubi steps

\begin {align*} \int (e x)^{3/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx &=\frac {B (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 b e}-\frac {\left (-7 A b+\frac {5 a B}{2}\right ) \int (e x)^{3/2} \sqrt {a+b x^3} \, dx}{7 b}\\ &=\frac {(14 A b-5 a B) (e x)^{5/2} \sqrt {a+b x^3}}{56 b e}+\frac {B (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 b e}+\frac {(3 a (14 A b-5 a B)) \int \frac {(e x)^{3/2}}{\sqrt {a+b x^3}} \, dx}{112 b}\\ &=\frac {(14 A b-5 a B) (e x)^{5/2} \sqrt {a+b x^3}}{56 b e}+\frac {B (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 b e}+\frac {(3 a (14 A b-5 a B)) \text {Subst}\left (\int \frac {x^4}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{56 b e}\\ &=\frac {(14 A b-5 a B) (e x)^{5/2} \sqrt {a+b x^3}}{56 b e}+\frac {B (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 b e}-\frac {(3 a (14 A b-5 a B)) \text {Subst}\left (\int \frac {\left (-1+\sqrt {3}\right ) a^{2/3} e^2-2 b^{2/3} x^4}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{112 b^{5/3} e}-\frac {\left (3 \left (1-\sqrt {3}\right ) a^{5/3} (14 A b-5 a B) e\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{112 b^{5/3}}\\ &=\frac {(14 A b-5 a B) (e x)^{5/2} \sqrt {a+b x^3}}{56 b e}+\frac {3 \left (1+\sqrt {3}\right ) a (14 A b-5 a B) e \sqrt {e x} \sqrt {a+b x^3}}{112 b^{5/3} \left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )}+\frac {B (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 b e}-\frac {3 \sqrt [4]{3} a^{4/3} (14 A b-5 a B) e \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} E\left (\cos ^{-1}\left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{112 b^{5/3} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}-\frac {3^{3/4} \left (1-\sqrt {3}\right ) a^{4/3} (14 A b-5 a B) e \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{224 b^{5/3} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.07, size = 94, normalized size = 0.16 \begin {gather*} \frac {x (e x)^{3/2} \sqrt {a+b x^3} \left (5 B \left (a+b x^3\right ) \sqrt {1+\frac {b x^3}{a}}+(14 A b-5 a B) \, _2F_1\left (-\frac {1}{2},\frac {5}{6};\frac {11}{6};-\frac {b x^3}{a}\right )\right )}{35 b \sqrt {1+\frac {b x^3}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^(3/2)*Sqrt[a + b*x^3]*(5*B*(a + b*x^3)*Sqrt[1 + (b*x^3)/a] + (14*A*b - 5*a*B)*Hypergeometric2F1[-1/2,
5/6, 11/6, -((b*x^3)/a)]))/(35*b*Sqrt[1 + (b*x^3)/a])

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Maple [C] Result contains complex when optimal does not.
time = 0.46, size = 5358, normalized size = 9.22

method	result	size
risch	\(\text {Expression too large to display}\)	\(1140\)
elliptic	\(\text {Expression too large to display}\)	\(1184\)
default	\(\text {Expression too large to display}\)	\(5358\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 7.58, size = 97, normalized size = 0.17 \begin {gather*} \frac {A \sqrt {a} e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{6} \\ \frac {11}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {11}{6}\right )} + \frac {B \sqrt {a} e^{\frac {3}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {11}{6} \\ \frac {17}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {17}{6}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(3/2)*(B*x**3+A)*(b*x**3+a)**(1/2),x)

[Out]

A*sqrt(a)*e**(3/2)*x**(5/2)*gamma(5/6)*hyper((-1/2, 5/6), (11/6,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(11/6)) +
B*sqrt(a)*e**(3/2)*x**(11/2)*gamma(11/6)*hyper((-1/2, 11/6), (17/6,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(17/6
))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \left (B\,x^3+A\right )\,{\left (e\,x\right )}^{3/2}\,\sqrt {b\,x^3+a} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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