Integrand size = 22, antiderivative size = 300 \[ \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=-\frac {2 (11 A b-14 a B) x^4}{33 b^2 \sqrt {a+b x^3}}+\frac {2 B x^7}{11 b \sqrt {a+b x^3}}+\frac {16 (11 A b-14 a B) x \sqrt {a+b x^3}}{165 b^3}-\frac {32 \sqrt {2+\sqrt {3}} a (11 A b-14 a B) \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt {3}\right )}{165 \sqrt [4]{3} b^{10/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \]
-2/33*(11*A*b-14*B*a)*x^4/b^2/(b*x^3+a)^(1/2)+2/11*B*x^7/b/(b*x^3+a)^(1/2) +16/165*(11*A*b-14*B*a)*x*(b*x^3+a)^(1/2)/b^3-32/495*a*(11*A*b-14*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)*3^(3/4)/ b^(10/3)/(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)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.10 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.34 \[ \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {2 x \left (-112 a^2 B+3 b^2 x^3 \left (11 A+5 B x^3\right )+a \left (88 A b-42 b B x^3\right )+8 a (-11 A b+14 a B) \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},-\frac {b x^3}{a}\right )\right )}{165 b^3 \sqrt {a+b x^3}} \]
(2*x*(-112*a^2*B + 3*b^2*x^3*(11*A + 5*B*x^3) + a*(88*A*b - 42*b*B*x^3) + 8*a*(-11*A*b + 14*a*B)*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/3, 1/2, 4/3 , -((b*x^3)/a)]))/(165*b^3*Sqrt[a + b*x^3])
Time = 0.34 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {959, 817, 843, 759}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 959 |
| \(\displaystyle \frac {(11 A b-14 a B) \int \frac {x^6}{\left (b x^3+a\right )^{3/2}}dx}{11 b}+\frac {2 B x^7}{11 b \sqrt {a+b x^3}}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {(11 A b-14 a B) \left (\frac {8 \int \frac {x^3}{\sqrt {b x^3+a}}dx}{3 b}-\frac {2 x^4}{3 b \sqrt {a+b x^3}}\right )}{11 b}+\frac {2 B x^7}{11 b \sqrt {a+b x^3}}\) |
| \(\Big \downarrow \) 843 |
| \(\displaystyle \frac {(11 A b-14 a B) \left (\frac {8 \left (\frac {2 x \sqrt {a+b x^3}}{5 b}-\frac {2 a \int \frac {1}{\sqrt {b x^3+a}}dx}{5 b}\right )}{3 b}-\frac {2 x^4}{3 b \sqrt {a+b x^3}}\right )}{11 b}+\frac {2 B x^7}{11 b \sqrt {a+b x^3}}\) |
| \(\Big \downarrow \) 759 |
| \(\displaystyle \frac {(11 A b-14 a B) \left (\frac {8 \left (\frac {2 x \sqrt {a+b x^3}}{5 b}-\frac {4 \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} b^{4/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}\right )}{3 b}-\frac {2 x^4}{3 b \sqrt {a+b x^3}}\right )}{11 b}+\frac {2 B x^7}{11 b \sqrt {a+b x^3}}\) |
(2*B*x^7)/(11*b*Sqrt[a + b*x^3]) + ((11*A*b - 14*a*B)*((-2*x^4)/(3*b*Sqrt[ a + b*x^3]) + (8*((2*x*Sqrt[a + b*x^3])/(5*b) - (4*Sqrt[2 + Sqrt[3]]*a*(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]])/(5*3^(1 /4)*b^(4/3)*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])))/(3*b)))/(11*b)
3.3.34.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* ((s + r*x)/((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]], x]] /; FreeQ[{a, b}, x] & & PosQ[a]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n *((m - n + 1)/(b*n*(p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x ] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && ! ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
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] - Simp[(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]
Time = 4.81 (sec) , antiderivative size = 406, normalized size of antiderivative = 1.35
| method | result | size |
| elliptic | \(\frac {2 x a \left (A b -B a \right )}{3 b^{3} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 B \,x^{4} \sqrt {b \,x^{3}+a}}{11 b^{2}}+\frac {2 \left (\frac {A b -B a}{b^{2}}-\frac {8 B a}{11 b^{2}}\right ) x \sqrt {b \,x^{3}+a}}{5 b}-\frac {2 i \left (-\frac {2 a \left (A b -B a \right )}{3 b^{3}}-\frac {2 \left (\frac {A b -B a}{b^{2}}-\frac {8 B a}{11 b^{2}}\right ) a}{5 b}\right ) \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}}{-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, F\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{b \left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right )}{3 b \sqrt {b \,x^{3}+a}}\) | \(406\) |
| default | \(B \left (-\frac {2 a^{2} x}{3 b^{3} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 x^{4} \sqrt {b \,x^{3}+a}}{11 b^{2}}-\frac {38 a x \sqrt {b \,x^{3}+a}}{55 b^{3}}-\frac {448 i a^{2} \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}}}}\, F\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}}}{b \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 )}}\right )}{495 b^{4} \sqrt {b \,x^{3}+a}}\right )+A \left (\frac {2 a x}{3 b^{2} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 x \sqrt {b \,x^{3}+a}}{5 b^{2}}+\frac {32 i a \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}}}}\, F\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}}}{b \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 )}}\right )}{45 b^{3} \sqrt {b \,x^{3}+a}}\right )\) | \(666\) |
| risch | \(\text {Expression too large to display}\) | \(976\) |
2/3/b^3*x*a*(A*b-B*a)/((x^3+a/b)*b)^(1/2)+2/11*B/b^2*x^4*(b*x^3+a)^(1/2)+2 /5*((A*b-B*a)/b^2-8/11*B/b^2*a)/b*x*(b*x^3+a)^(1/2)-2/3*I*(-2/3*a*(A*b-B*a )/b^3-2/5*((A*b-B*a)/b^2-8/11*B/b^2*a)/b*a)*3^(1/2)/b*(-a*b^2)^(1/3)*(I*(x +1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^( 1/3))^(1/2)*((x-1/b*(-a*b^2)^(1/3))/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b *(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^ 2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*EllipticF(1/3*3^ (1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b /(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+ 1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.41 \[ \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {2 \, {\left (16 \, {\left (14 \, B a^{3} - 11 \, A a^{2} b + {\left (14 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{3}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right ) + {\left (15 \, B b^{3} x^{7} - 3 \, {\left (14 \, B a b^{2} - 11 \, A b^{3}\right )} x^{4} - 8 \, {\left (14 \, B a^{2} b - 11 \, A a b^{2}\right )} x\right )} \sqrt {b x^{3} + a}\right )}}{165 \, {\left (b^{5} x^{3} + a b^{4}\right )}} \]
2/165*(16*(14*B*a^3 - 11*A*a^2*b + (14*B*a^2*b - 11*A*a*b^2)*x^3)*sqrt(b)* weierstrassPInverse(0, -4*a/b, x) + (15*B*b^3*x^7 - 3*(14*B*a*b^2 - 11*A*b ^3)*x^4 - 8*(14*B*a^2*b - 11*A*a*b^2)*x)*sqrt(b*x^3 + a))/(b^5*x^3 + a*b^4 )
Time = 8.75 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.27 \[ \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {A x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {3}{2}} \Gamma \left (\frac {10}{3}\right )} + \frac {B x^{10} \Gamma \left (\frac {10}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {10}{3} \\ \frac {13}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {3}{2}} \Gamma \left (\frac {13}{3}\right )} \]
A*x**7*gamma(7/3)*hyper((3/2, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3* a**(3/2)*gamma(10/3)) + B*x**10*gamma(10/3)*hyper((3/2, 10/3), (13/3,), b* x**3*exp_polar(I*pi)/a)/(3*a**(3/2)*gamma(13/3))
\[ \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x^{6}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x^{6}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\int \frac {x^6\,\left (B\,x^3+A\right )}{{\left (b\,x^3+a\right )}^{3/2}} \,d x \]