3.205 \(\int \frac {(a+b x^3)^{3/2} (A+B x^3)}{x^6} \, dx\)

Optimal. Leaf size=297 \[ \frac {9\ 3^{3/4} \sqrt {2+\sqrt {3}} b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} (2 a B+A b) F\left (\sin ^{-1}\left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{20 \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {9 b x \sqrt {a+b x^3} (2 a B+A b)}{20 a}-\frac {\left (a+b x^3\right )^{3/2} (2 a B+A b)}{4 a x^2}-\frac {A \left (a+b x^3\right )^{5/2}}{5 a x^5} \]

[Out]

-1/4*(A*b+2*B*a)*(b*x^3+a)^(3/2)/a/x^2-1/5*A*(b*x^3+a)^(5/2)/a/x^5+9/20*b*(A*b+2*B*a)*x*(b*x^3+a)^(1/2)/a+9/20
*3^(3/4)*b^(2/3)*(A*b+2*B*a)*(a^(1/3)+b^(1/3)*x)*EllipticF((b^(1/3)*x+a^(1/3)*(1-3^(1/2)))/(b^(1/3)*x+a^(1/3)*
(1+3^(1/2))),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/2))*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/(b^(1/3)*x+a^(1
/3)*(1+3^(1/2)))^2)^(1/2)/(b*x^3+a)^(1/2)/(a^(1/3)*(a^(1/3)+b^(1/3)*x)/(b^(1/3)*x+a^(1/3)*(1+3^(1/2)))^2)^(1/2
)

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Rubi [A]  time = 0.12, antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {453, 277, 195, 218} \[ \frac {9\ 3^{3/4} \sqrt {2+\sqrt {3}} b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} (2 a B+A b) F\left (\sin ^{-1}\left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{20 \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}-\frac {\left (a+b x^3\right )^{3/2} (2 a B+A b)}{4 a x^2}+\frac {9 b x \sqrt {a+b x^3} (2 a B+A b)}{20 a}-\frac {A \left (a+b x^3\right )^{5/2}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)^(3/2)*(A + B*x^3))/x^6,x]

[Out]

(9*b*(A*b + 2*a*B)*x*Sqrt[a + b*x^3])/(20*a) - ((A*b + 2*a*B)*(a + b*x^3)^(3/2))/(4*a*x^2) - (A*(a + b*x^3)^(5
/2))/(5*a*x^5) + (9*3^(3/4)*Sqrt[2 + Sqrt[3]]*b^(2/3)*(A*b + 2*a*B)*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1
/3)*b^(1/3)*x + b^(2/3)*x^2)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*a^(1/3) +
b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)], -7 - 4*Sqrt[3]])/(20*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)*x))/((1
+ Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*Sqrt[a + b*x^3])

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

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 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]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x^6} \, dx &=-\frac {A \left (a+b x^3\right )^{5/2}}{5 a x^5}+-\frac {\left (-\frac {5 A b}{2}-5 a B\right ) \int \frac {\left (a+b x^3\right )^{3/2}}{x^3} \, dx}{5 a}\\ &=-\frac {(A b+2 a B) \left (a+b x^3\right )^{3/2}}{4 a x^2}-\frac {A \left (a+b x^3\right )^{5/2}}{5 a x^5}+\frac {(9 b (A b+2 a B)) \int \sqrt {a+b x^3} \, dx}{8 a}\\ &=\frac {9 b (A b+2 a B) x \sqrt {a+b x^3}}{20 a}-\frac {(A b+2 a B) \left (a+b x^3\right )^{3/2}}{4 a x^2}-\frac {A \left (a+b x^3\right )^{5/2}}{5 a x^5}+\frac {1}{40} (27 b (A b+2 a B)) \int \frac {1}{\sqrt {a+b x^3}} \, dx\\ &=\frac {9 b (A b+2 a B) x \sqrt {a+b x^3}}{20 a}-\frac {(A b+2 a B) \left (a+b x^3\right )^{3/2}}{4 a x^2}-\frac {A \left (a+b x^3\right )^{5/2}}{5 a x^5}+\frac {9\ 3^{3/4} \sqrt {2+\sqrt {3}} b^{2/3} (A b+2 a B) \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 (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right )|-7-4 \sqrt {3}\right )}{20 \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 82, normalized size = 0.28 \[ \frac {\sqrt {a+b x^3} \left (-\frac {5 x^3 (2 a B+A b) \, _2F_1\left (-\frac {3}{2},-\frac {2}{3};\frac {1}{3};-\frac {b x^3}{a}\right )}{2 \sqrt {\frac {b x^3}{a}+1}}-\frac {2 A \left (a+b x^3\right )^2}{a}\right )}{10 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)^(3/2)*(A + B*x^3))/x^6,x]

[Out]

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

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B b x^{6} + {\left (B a + A b\right )} x^{3} + A a\right )} \sqrt {b x^{3} + a}}{x^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(3/2)*(B*x^3+A)/x^6,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {3}{2}}}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(3/2)*(B*x^3+A)/x^6,x, algorithm="giac")

[Out]

integrate((B*x^3 + A)*(b*x^3 + a)^(3/2)/x^6, x)

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maple [B]  time = 0.05, size = 626, normalized size = 2.11 \[ \left (-\frac {9 i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}}{-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, b \EllipticF \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) b}}\right )}{20 \sqrt {b \,x^{3}+a}}-\frac {13 \sqrt {b \,x^{3}+a}\, b}{20 x^{2}}-\frac {\sqrt {b \,x^{3}+a}\, a}{5 x^{5}}\right ) A +\left (-\frac {9 i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}}{-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, a \EllipticF \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) b}}\right )}{10 \sqrt {b \,x^{3}+a}}+\frac {2 \sqrt {b \,x^{3}+a}\, b x}{5}-\frac {\sqrt {b \,x^{3}+a}\, a}{2 x^{2}}\right ) B \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(3/2)*(B*x^3+A)/x^6,x)

[Out]

A*(-1/5*a*(b*x^3+a)^(1/2)/x^5-13/20*(b*x^3+a)^(1/2)*b/x^2-9/20*I*b*3^(1/2)*(-a*b^2)^(1/3)*(I*(x+1/2*(-a*b^2)^(
1/3)/b-1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*3^(1/2)/(-a*b^2)^(1/3)*b)^(1/2)*((x-(-a*b^2)^(1/3)/b)/(-3/2*(-a*b^2)^(1
/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b))^(1/2)*(-I*(x+1/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*3^(1/
2)/(-a*b^2)^(1/3)*b)^(1/2)/(b*x^3+a)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2*(-a*b^2)^(1/3)/b-1/2*I*3^(1/2)*(-a*
b^2)^(1/3)/b)*3^(1/2)/(-a*b^2)^(1/3)*b)^(1/2),(I*3^(1/2)*(-a*b^2)^(1/3)/(-3/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(
-a*b^2)^(1/3)/b)/b)^(1/2)))+B*(-1/2*(b*x^3+a)^(1/2)*a/x^2+2/5*(b*x^3+a)^(1/2)*b*x-9/10*I*a*3^(1/2)*(-a*b^2)^(1
/3)*(I*(x+1/2*(-a*b^2)^(1/3)/b-1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*3^(1/2)/(-a*b^2)^(1/3)*b)^(1/2)*((x-(-a*b^2)^(1
/3)/b)/(-3/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b))^(1/2)*(-I*(x+1/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)
*(-a*b^2)^(1/3)/b)*3^(1/2)/(-a*b^2)^(1/3)*b)^(1/2)/(b*x^3+a)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2*(-a*b^2)^(1
/3)/b-1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*3^(1/2)/(-a*b^2)^(1/3)*b)^(1/2),(I*3^(1/2)*(-a*b^2)^(1/3)/(-3/2*(-a*b^2)
^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)/b)^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {3}{2}}}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(3/2)*(B*x^3+A)/x^6,x, algorithm="maxima")

[Out]

integrate((B*x^3 + A)*(b*x^3 + a)^(3/2)/x^6, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (B\,x^3+A\right )\,{\left (b\,x^3+a\right )}^{3/2}}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^3)*(a + b*x^3)^(3/2))/x^6,x)

[Out]

int(((A + B*x^3)*(a + b*x^3)^(3/2))/x^6, x)

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sympy [A]  time = 6.18, size = 184, normalized size = 0.62 \[ \frac {A a^{\frac {3}{2}} \Gamma \left (- \frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{3}, - \frac {1}{2} \\ - \frac {2}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 x^{5} \Gamma \left (- \frac {2}{3}\right )} + \frac {A \sqrt {a} b \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {1}{2} \\ \frac {1}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 x^{2} \Gamma \left (\frac {1}{3}\right )} + \frac {B a^{\frac {3}{2}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {1}{2} \\ \frac {1}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 x^{2} \Gamma \left (\frac {1}{3}\right )} + \frac {B \sqrt {a} b x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(3/2)*(B*x**3+A)/x**6,x)

[Out]

A*a**(3/2)*gamma(-5/3)*hyper((-5/3, -1/2), (-2/3,), b*x**3*exp_polar(I*pi)/a)/(3*x**5*gamma(-2/3)) + A*sqrt(a)
*b*gamma(-2/3)*hyper((-2/3, -1/2), (1/3,), b*x**3*exp_polar(I*pi)/a)/(3*x**2*gamma(1/3)) + B*a**(3/2)*gamma(-2
/3)*hyper((-2/3, -1/2), (1/3,), b*x**3*exp_polar(I*pi)/a)/(3*x**2*gamma(1/3)) + B*sqrt(a)*b*x*gamma(1/3)*hyper
((-1/2, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(4/3))

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