3.6.32 \(\int \frac {(a+b x^3)^{3/2} (A+B x^3)}{(e x)^{3/2}} \, dx\) [532]

Optimal. Leaf size=614 \[ \frac {9 (14 A b+a B) (e x)^{5/2} \sqrt {a+b x^3}}{56 e^4}+\frac {27 \left (1+\sqrt {3}\right ) a (14 A b+a B) \sqrt {e x} \sqrt {a+b x^3}}{112 b^{2/3} e^2 \left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )}+\frac {(14 A b+a B) (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 a e^4}-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}}-\frac {27 \sqrt [4]{3} a^{4/3} (14 A b+a B) \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} E\left (\cos ^{-1}\left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{112 b^{2/3} e^2 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}-\frac {9\ 3^{3/4} \left (1-\sqrt {3}\right ) a^{4/3} (14 A b+a B) \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{224 b^{2/3} e^2 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \]

[Out]

1/7*(14*A*b+B*a)*(e*x)^(5/2)*(b*x^3+a)^(3/2)/a/e^4-2*A*(b*x^3+a)^(5/2)/a/e/(e*x)^(1/2)+9/56*(14*A*b+B*a)*(e*x)
^(5/2)*(b*x^3+a)^(1/2)/e^4+27/112*a*(14*A*b+B*a)*(1+3^(1/2))*(e*x)^(1/2)*(b*x^3+a)^(1/2)/b^(2/3)/e^2/(a^(1/3)+
b^(1/3)*x*(1+3^(1/2)))-27/112*3^(1/4)*a^(4/3)*(14*A*b+B*a)*(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)))*E
llipticE((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^(2/3)/e^2/(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)-9/224*3^(3/4)*a^(4/3)*(1
4*A*b+B*a)*(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^(2/3)/e^2/(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 = 614, 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 = {464, 285, 335, 314, 231, 1895} \begin {gather*} -\frac {9\ 3^{3/4} \left (1-\sqrt {3}\right ) a^{4/3} \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}} (a B+14 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^{2/3} e^2 \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^{4/3} \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}} (a B+14 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^{2/3} e^2 \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 \sqrt {e x} \sqrt {a+b x^3} (a B+14 A b)}{112 b^{2/3} e^2 \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} (a B+14 A b)}{7 a e^4}+\frac {9 (e x)^{5/2} \sqrt {a+b x^3} (a B+14 A b)}{56 e^4}-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(9*(14*A*b + a*B)*(e*x)^(5/2)*Sqrt[a + b*x^3])/(56*e^4) + (27*(1 + Sqrt[3])*a*(14*A*b + a*B)*Sqrt[e*x]*Sqrt[a
+ b*x^3])/(112*b^(2/3)*e^2*(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)) + ((14*A*b + a*B)*(e*x)^(5/2)*(a + b*x^3)^(3/2
))/(7*a*e^4) - (2*A*(a + b*x^3)^(5/2))/(a*e*Sqrt[e*x]) - (27*3^(1/4)*a^(4/3)*(14*A*b + a*B)*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]*Elliptic
E[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^(2
/3)*e^2*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^(4/3)*(14*A*b + a*B)*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])/(224*b^(2/3)*e^2*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 464

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 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 \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{(e x)^{3/2}} \, dx &=-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}}+\frac {(14 A b+a B) \int (e x)^{3/2} \left (a+b x^3\right )^{3/2} \, dx}{a e^3}\\ &=\frac {(14 A b+a B) (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 a e^4}-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}}+\frac {(9 (14 A b+a B)) \int (e x)^{3/2} \sqrt {a+b x^3} \, dx}{14 e^3}\\ &=\frac {9 (14 A b+a B) (e x)^{5/2} \sqrt {a+b x^3}}{56 e^4}+\frac {(14 A b+a B) (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 a e^4}-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}}+\frac {(27 a (14 A b+a B)) \int \frac {(e x)^{3/2}}{\sqrt {a+b x^3}} \, dx}{112 e^3}\\ &=\frac {9 (14 A b+a B) (e x)^{5/2} \sqrt {a+b x^3}}{56 e^4}+\frac {(14 A b+a B) (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 a e^4}-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}}+\frac {(27 a (14 A b+a B)) \text {Subst}\left (\int \frac {x^4}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{56 e^4}\\ &=\frac {9 (14 A b+a B) (e x)^{5/2} \sqrt {a+b x^3}}{56 e^4}+\frac {(14 A b+a B) (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 a e^4}-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}}-\frac {(27 a (14 A b+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^{2/3} e^4}-\frac {\left (27 \left (1-\sqrt {3}\right ) a^{5/3} (14 A b+a B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{112 b^{2/3} e^2}\\ &=\frac {9 (14 A b+a B) (e x)^{5/2} \sqrt {a+b x^3}}{56 e^4}+\frac {27 \left (1+\sqrt {3}\right ) a (14 A b+a B) \sqrt {e x} \sqrt {a+b x^3}}{112 b^{2/3} e^2 \left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )}+\frac {(14 A b+a B) (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 a e^4}-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}}-\frac {27 \sqrt [4]{3} a^{4/3} (14 A b+a B) \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^{2/3} e^2 \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^{4/3} (14 A b+a B) \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^{2/3} e^2 \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 = 10.05, size = 84, normalized size = 0.14 \begin {gather*} \frac {2 x \sqrt {a+b x^3} \left (-\frac {5 A \left (a+b x^3\right )^2}{a}+\frac {(14 A b+a B) x^3 \, _2F_1\left (-\frac {3}{2},\frac {5}{6};\frac {11}{6};-\frac {b x^3}{a}\right )}{\sqrt {1+\frac {b x^3}{a}}}\right )}{5 (e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.38, size = 6142, normalized size = 10.00

method	result	size
risch	\(\text {Expression too large to display}\)	\(1140\)
elliptic	\(\text {Expression too large to display}\)	\(1233\)
default	\(\text {Expression too large to display}\)	\(6142\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 7.38, size = 202, normalized size = 0.33 \begin {gather*} \frac {A a^{\frac {3}{2}} \Gamma \left (- \frac {1}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{6} \\ \frac {5}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {5}{6}\right )} + \frac {A \sqrt {a} b 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 e^{\frac {3}{2}} \Gamma \left (\frac {11}{6}\right )} + \frac {B a^{\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 e^{\frac {3}{2}} \Gamma \left (\frac {11}{6}\right )} + \frac {B \sqrt {a} b 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 e^{\frac {3}{2}} \Gamma \left (\frac {17}{6}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**(3/2)*gamma(-1/6)*hyper((-1/2, -1/6), (5/6,), b*x**3*exp_polar(I*pi)/a)/(3*e**(3/2)*sqrt(x)*gamma(5/6)) +
A*sqrt(a)*b*x**(5/2)*gamma(5/6)*hyper((-1/2, 5/6), (11/6,), b*x**3*exp_polar(I*pi)/a)/(3*e**(3/2)*gamma(11/6)
) + B*a**(3/2)*x**(5/2)*gamma(5/6)*hyper((-1/2, 5/6), (11/6,), b*x**3*exp_polar(I*pi)/a)/(3*e**(3/2)*gamma(11/
6)) + B*sqrt(a)*b*x**(11/2)*gamma(11/6)*hyper((-1/2, 11/6), (17/6,), b*x**3*exp_polar(I*pi)/a)/(3*e**(3/2)*gam
ma(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((b*x^3+a)^(3/2)*(B*x^3+A)/(e*x)^(3/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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