3.6.29 \(\int (e x)^{3/2} (a+b x^3)^{3/2} (A+B x^3) \, dx\) [529]

Optimal. Leaf size=621 \[ \frac {9 a (4 A b-a B) (e x)^{5/2} \sqrt {a+b x^3}}{224 b e}+\frac {27 \left (1+\sqrt {3}\right ) a^2 (4 A b-a B) e \sqrt {e x} \sqrt {a+b x^3}}{448 b^{5/3} \left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )}+\frac {(4 A b-a B) (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{28 b e}+\frac {B (e x)^{5/2} \left (a+b x^3\right )^{5/2}}{10 b e}-\frac {27 \sqrt [4]{3} a^{7/3} (4 A b-a B) e \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} E\left (\cos ^{-1}\left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{448 b^{5/3} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}-\frac {9\ 3^{3/4} \left (1-\sqrt {3}\right ) a^{7/3} (4 A b-a B) e \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{896 b^{5/3} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \]

[Out]

1/28*(4*A*b-B*a)*(e*x)^(5/2)*(b*x^3+a)^(3/2)/b/e+1/10*B*(e*x)^(5/2)*(b*x^3+a)^(5/2)/b/e+9/224*a*(4*A*b-B*a)*(e
*x)^(5/2)*(b*x^3+a)^(1/2)/b/e+27/448*a^2*(4*A*b-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)))-27/448*3^(1/4)*a^(7/3)*(4*A*b-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)-9/896*3^(3/4)*a^(7/3)*(4*
A*b-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 = 0.49, antiderivative size = 621, normalized size of antiderivative = 1.00, number of steps used = 7, 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 {9\ 3^{3/4} \left (1-\sqrt {3}\right ) a^{7/3} 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}} (4 A b-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 )}{896 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 {27 \sqrt [4]{3} a^{7/3} 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}} (4 A b-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 )}{448 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 {27 \left (1+\sqrt {3}\right ) a^2 e \sqrt {e x} \sqrt {a+b x^3} (4 A b-a B)}{448 b^{5/3} \left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )}+\frac {(e x)^{5/2} \left (a+b x^3\right )^{3/2} (4 A b-a B)}{28 b e}+\frac {9 a (e x)^{5/2} \sqrt {a+b x^3} (4 A b-a B)}{224 b e}+\frac {B (e x)^{5/2} \left (a+b x^3\right )^{5/2}}{10 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(9*a*(4*A*b - a*B)*(e*x)^(5/2)*Sqrt[a + b*x^3])/(224*b*e) + (27*(1 + Sqrt[3])*a^2*(4*A*b - a*B)*e*Sqrt[e*x]*Sq
rt[a + b*x^3])/(448*b^(5/3)*(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)) + ((4*A*b - a*B)*(e*x)^(5/2)*(a + b*x^3)^(3/2
))/(28*b*e) + (B*(e*x)^(5/2)*(a + b*x^3)^(5/2))/(10*b*e) - (27*3^(1/4)*a^(7/3)*(4*A*b - 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]*Ellip
ticE[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])/(448*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]) - (9*3^(
3/4)*(1 - Sqrt[3])*a^(7/3)*(4*A*b - 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]*EllipticF[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])/(896*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} \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx &=\frac {B (e x)^{5/2} \left (a+b x^3\right )^{5/2}}{10 b e}-\frac {\left (-10 A b+\frac {5 a B}{2}\right ) \int (e x)^{3/2} \left (a+b x^3\right )^{3/2} \, dx}{10 b}\\ &=\frac {(4 A b-a B) (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{28 b e}+\frac {B (e x)^{5/2} \left (a+b x^3\right )^{5/2}}{10 b e}+\frac {(9 a (4 A b-a B)) \int (e x)^{3/2} \sqrt {a+b x^3} \, dx}{56 b}\\ &=\frac {9 a (4 A b-a B) (e x)^{5/2} \sqrt {a+b x^3}}{224 b e}+\frac {(4 A b-a B) (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{28 b e}+\frac {B (e x)^{5/2} \left (a+b x^3\right )^{5/2}}{10 b e}+\frac {\left (27 a^2 (4 A b-a B)\right ) \int \frac {(e x)^{3/2}}{\sqrt {a+b x^3}} \, dx}{448 b}\\ &=\frac {9 a (4 A b-a B) (e x)^{5/2} \sqrt {a+b x^3}}{224 b e}+\frac {(4 A b-a B) (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{28 b e}+\frac {B (e x)^{5/2} \left (a+b x^3\right )^{5/2}}{10 b e}+\frac {\left (27 a^2 (4 A b-a B)\right ) \text {Subst}\left (\int \frac {x^4}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{224 b e}\\ &=\frac {9 a (4 A b-a B) (e x)^{5/2} \sqrt {a+b x^3}}{224 b e}+\frac {(4 A b-a B) (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{28 b e}+\frac {B (e x)^{5/2} \left (a+b x^3\right )^{5/2}}{10 b e}-\frac {\left (27 a^2 (4 A b-a B)\right ) \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 )}{448 b^{5/3} e}-\frac {\left (27 \left (1-\sqrt {3}\right ) a^{8/3} (4 A b-a B) e\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{448 b^{5/3}}\\ &=\frac {9 a (4 A b-a B) (e x)^{5/2} \sqrt {a+b x^3}}{224 b e}+\frac {27 \left (1+\sqrt {3}\right ) a^2 (4 A b-a B) e \sqrt {e x} \sqrt {a+b x^3}}{448 b^{5/3} \left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )}+\frac {(4 A b-a B) (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{28 b e}+\frac {B (e x)^{5/2} \left (a+b x^3\right )^{5/2}}{10 b e}-\frac {27 \sqrt [4]{3} a^{7/3} (4 A b-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 )}{448 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 {9\ 3^{3/4} \left (1-\sqrt {3}\right ) a^{7/3} (4 A b-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 )}{896 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 = 10.08, size = 96, normalized size = 0.15 \begin {gather*} \frac {x (e x)^{3/2} \sqrt {a+b x^3} \left (B \left (a+b x^3\right )^2 \sqrt {1+\frac {b x^3}{a}}+a (4 A b-a B) \, _2F_1\left (-\frac {3}{2},\frac {5}{6};\frac {11}{6};-\frac {b x^3}{a}\right )\right )}{10 b \sqrt {1+\frac {b x^3}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^(3/2)*Sqrt[a + b*x^3]*(B*(a + b*x^3)^2*Sqrt[1 + (b*x^3)/a] + a*(4*A*b - a*B)*Hypergeometric2F1[-3/2,
5/6, 11/6, -((b*x^3)/a)]))/(10*b*Sqrt[1 + (b*x^3)/a])

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Maple [C] Result contains complex when optimal does not.
time = 0.45, size = 5790, normalized size = 9.32

method	result	size
risch	\(\text {Expression too large to display}\)	\(1164\)
elliptic	\(\text {Expression too large to display}\)	\(1279\)
default	\(\text {Expression too large to display}\)	\(5790\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integral((B*b*x^7 + (B*a + A*b)*x^4 + A*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 = 25.09, size = 199, normalized size = 0.32 \begin {gather*} \frac {A a^{\frac {3}{2}} 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 {A \sqrt {a} b 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 )} + \frac {B a^{\frac {3}{2}} 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 )} + \frac {B \sqrt {a} b e^{\frac {3}{2}} x^{\frac {17}{2}} \Gamma \left (\frac {17}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {17}{6} \\ \frac {23}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {23}{6}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**(3/2)*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))
+ A*sqrt(a)*b*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(1
7/6)) + B*a**(3/2)*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*ga
mma(17/6)) + B*sqrt(a)*b*e**(3/2)*x**(17/2)*gamma(17/6)*hyper((-1/2, 17/6), (23/6,), b*x**3*exp_polar(I*pi)/a)
/(3*gamma(23/6))

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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)^(3/2)*(B*x^3+A),x, algorithm="giac")

[Out]

integrate((B*x^3 + A)*(b*x^3 + a)^(3/2)*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}\,{\left (b\,x^3+a\right )}^{3/2} \,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)^(3/2),x)

[Out]

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

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