3.526 \(\int \frac{\sqrt{a+b x^3} (A+B x^3)}{x^{13/2}} \, dx\)

Optimal. Leaf size=269 \[ -\frac{3^{3/4} b \sqrt{x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} (2 A b-11 a B) \text{EllipticF}\left (\cos ^{-1}\left (\frac{\sqrt [3]{a}+\left (1-\sqrt{3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{55 a^{4/3} \sqrt{\frac{\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{2 \sqrt{a+b x^3} (2 A b-11 a B)}{55 a x^{5/2}}-\frac{2 A \left (a+b x^3\right )^{3/2}}{11 a x^{11/2}} \]

[Out]

(2*(2*A*b - 11*a*B)*Sqrt[a + b*x^3])/(55*a*x^(5/2)) - (2*A*(a + b*x^3)^(3/2))/(11*a*x^(11/2)) - (3^(3/4)*b*(2*
A*b - 11*a*B)*Sqrt[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 + S
qrt[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])/(55*a^(4/3)*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])

________________________________________________________________________________________

Rubi [A]  time = 0.199148, antiderivative size = 269, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {453, 277, 329, 225} \[ -\frac{3^{3/4} b \sqrt{x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} (2 A b-11 a B) F\left (\cos ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt{3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{55 a^{4/3} \sqrt{\frac{\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{2 \sqrt{a+b x^3} (2 A b-11 a B)}{55 a x^{5/2}}-\frac{2 A \left (a+b x^3\right )^{3/2}}{11 a x^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x^3]*(A + B*x^3))/x^(13/2),x]

[Out]

(2*(2*A*b - 11*a*B)*Sqrt[a + b*x^3])/(55*a*x^(5/2)) - (2*A*(a + b*x^3)^(3/2))/(11*a*x^(11/2)) - (3^(3/4)*b*(2*
A*b - 11*a*B)*Sqrt[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 + S
qrt[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])/(55*a^(4/3)*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 453

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 277

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

Rule 329

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 225

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]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(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]), x]] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{\sqrt{a+b x^3} \left (A+B x^3\right )}{x^{13/2}} \, dx &=-\frac{2 A \left (a+b x^3\right )^{3/2}}{11 a x^{11/2}}-\frac{\left (2 \left (A b-\frac{11 a B}{2}\right )\right ) \int \frac{\sqrt{a+b x^3}}{x^{7/2}} \, dx}{11 a}\\ &=\frac{2 (2 A b-11 a B) \sqrt{a+b x^3}}{55 a x^{5/2}}-\frac{2 A \left (a+b x^3\right )^{3/2}}{11 a x^{11/2}}-\frac{(3 b (2 A b-11 a B)) \int \frac{1}{\sqrt{x} \sqrt{a+b x^3}} \, dx}{55 a}\\ &=\frac{2 (2 A b-11 a B) \sqrt{a+b x^3}}{55 a x^{5/2}}-\frac{2 A \left (a+b x^3\right )^{3/2}}{11 a x^{11/2}}-\frac{(6 b (2 A b-11 a B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^6}} \, dx,x,\sqrt{x}\right )}{55 a}\\ &=\frac{2 (2 A b-11 a B) \sqrt{a+b x^3}}{55 a x^{5/2}}-\frac{2 A \left (a+b x^3\right )^{3/2}}{11 a x^{11/2}}-\frac{3^{3/4} b (2 A b-11 a B) \sqrt{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 )}{55 a^{4/3} \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*}

Mathematica [C]  time = 0.0866127, size = 80, normalized size = 0.3 \[ \frac{2 \sqrt{a+b x^3} \left (\frac{x^3 (2 A b-11 a B) \, _2F_1\left (-\frac{5}{6},-\frac{1}{2};\frac{1}{6};-\frac{b x^3}{a}\right )}{\sqrt{\frac{b x^3}{a}+1}}-5 A \left (a+b x^3\right )\right )}{55 a x^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x^3]*(A + B*x^3))/x^(13/2),x]

[Out]

(2*Sqrt[a + b*x^3]*(-5*A*(a + b*x^3) + ((2*A*b - 11*a*B)*x^3*Hypergeometric2F1[-5/6, -1/2, 1/6, -((b*x^3)/a)])
/Sqrt[1 + (b*x^3)/a]))/(55*a*x^(11/2))

________________________________________________________________________________________

Maple [C]  time = 0.068, size = 3690, normalized size = 13.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)*(b*x^3+a)^(1/2)/x^(13/2),x)

[Out]

2/55*(b*x^3+a)^(1/2)*(-66*I*B*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1
/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)
-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2)
)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2))*x^8*a*b^2-24*
I*A*(-a*b^2)^(1/3)*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2
)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*
b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a
*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2))*x^7*b^2+132*I*B*(-a*b^2)
^(1/3)*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*
x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/
(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3))
)^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2))*x^7*a*b-5*I*A*((b*x^3+a)*x)^(1/2)*(-
a*b^2)^(1/3)*3^(1/2)*(1/b^2*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)
*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*a+12*I*A*(-a*b^2)^(2/3)*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2)
)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(
1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*Elli
pticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)
+1)/(I*3^(1/2)-3))^(1/2))*x^6*b-12*A*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/
2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-
2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))
/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2))*x^8*b^3-3*I*A*
((b*x^3+a)*x)^(1/2)*(-a*b^2)^(1/3)*3^(1/2)*(1/b^2*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*
b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*x^3*b+66*B*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2
))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^
(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*Ell
ipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2
)+1)/(I*3^(1/2)-3))^(1/2))*x^8*a*b^2+24*A*(-a*b^2)^(1/3)*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/
3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(
1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3
)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(
1/2))*x^7*b^2-132*B*(-a*b^2)^(1/3)*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)
*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*
b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(
-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2))*x^7*a*b-12*A*(-a
*b^2)^(2/3)*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(
-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1
+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(
1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2))*x^6*b+66*B*(-a*b^2)^(2/3)*(-(I*3^(1/2)-
3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)
+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2
)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I
*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2))*x^6*a-11*I*B*((b*x^3+a)*x)^(1/2)*(-a*b^2)^(1/3)*3^(1/2)*(1/b^2*x
*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2
)^(1/3)))^(1/2)*x^3*a+12*I*A*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/
2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-
2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))
/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2))*x^8*b^3+9*A*((
b*x^3+a)*x)^(1/2)*(-a*b^2)^(1/3)*(1/b^2*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3)
)*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*x^3*b+33*B*((b*x^3+a)*x)^(1/2)*(-a*b^2)^(1/3)*(1/b^2*
x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^
2)^(1/3)))^(1/2)*x^3*a-66*I*B*(-a*b^2)^(2/3)*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))
^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(I*3^(1/2)+1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)
*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*
b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(I*3^(1/2)+1)/(I*3^(1/2)-3))^(1/2)
)*x^6*a+15*A*((b*x^3+a)*x)^(1/2)*(-a*b^2)^(1/3)*(1/b^2*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x
+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*a)/x^(11/2)/(-a*b^2)^(1/3)/((b*x^3+a)*
x)^(1/2)/a/(I*3^(1/2)-3)/(1/b^2*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(
1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{3} + A\right )} \sqrt{b x^{3} + a}}{x^{\frac{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*(b*x^3+a)^(1/2)/x^(13/2),x, algorithm="maxima")

[Out]

integrate((B*x^3 + A)*sqrt(b*x^3 + a)/x^(13/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x^{3} + A\right )} \sqrt{b x^{3} + a}}{x^{\frac{13}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*(b*x^3+a)^(1/2)/x^(13/2),x, algorithm="fricas")

[Out]

integral((B*x^3 + A)*sqrt(b*x^3 + a)/x^(13/2), x)

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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)*(b*x**3+a)**(1/2)/x**(13/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{3} + A\right )} \sqrt{b x^{3} + a}}{x^{\frac{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*(b*x^3+a)^(1/2)/x^(13/2),x, algorithm="giac")

[Out]

integrate((B*x^3 + A)*sqrt(b*x^3 + a)/x^(13/2), x)