Integrand size = 24, antiderivative size = 289 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx=\frac {2 \left (c d^2-b d e+a e^2\right ) x}{3 d e^2 \sqrt {d+e x^3}}+\frac {2 c x \sqrt {d+e x^3}}{5 e^2}-\frac {2 \sqrt {2+\sqrt {3}} \left (16 c d^2-5 e (2 b d+a e)\right ) \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt {\frac {d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}\right ),-7-4 \sqrt {3}\right )}{15 \sqrt [4]{3} d e^{7/3} \sqrt {\frac {\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt {d+e x^3}} \]
2/3*(a*e^2-b*d*e+c*d^2)*x/d/e^2/(e*x^3+d)^(1/2)+2/5*c*x*(e*x^3+d)^(1/2)/e^ 2-2/45*(16*c*d^2-5*e*(a*e+2*b*d))*(d^(1/3)+e^(1/3)*x)*EllipticF((e^(1/3)*x +d^(1/3)*(1-3^(1/2)))/(e^(1/3)*x+d^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(1/2* 6^(1/2)+1/2*2^(1/2))*((d^(2/3)-d^(1/3)*e^(1/3)*x+e^(2/3)*x^2)/(e^(1/3)*x+d ^(1/3)*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/d/e^(7/3)/(e*x^3+d)^(1/2)/(d^(1/3)*(d ^(1/3)+e^(1/3)*x)/(e^(1/3)*x+d^(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 = 10.07 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.35 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx=\frac {x \left (2 \left (5 e (-b d+a e)+c d \left (8 d+3 e x^3\right )\right )+\left (-16 c d^2+5 e (2 b d+a e)\right ) \sqrt {1+\frac {e x^3}{d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},-\frac {e x^3}{d}\right )\right )}{15 d e^2 \sqrt {d+e x^3}} \]
(x*(2*(5*e*(-(b*d) + a*e) + c*d*(8*d + 3*e*x^3)) + (-16*c*d^2 + 5*e*(2*b*d + a*e))*Sqrt[1 + (e*x^3)/d]*Hypergeometric2F1[1/3, 1/2, 4/3, -((e*x^3)/d) ]))/(15*d*e^2*Sqrt[d + e*x^3])
Time = 0.37 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1739, 27, 913, 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.
\(\displaystyle \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1739 |
\(\displaystyle \frac {2 x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \sqrt {d+e x^3}}-\frac {2 \int \frac {-3 c d e x^3+2 c d^2-e (2 b d+a e)}{2 \sqrt {e x^3+d}}dx}{3 d e^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \sqrt {d+e x^3}}-\frac {\int \frac {-3 c d e x^3+2 c d^2-e (2 b d+a e)}{\sqrt {e x^3+d}}dx}{3 d e^2}\) |
\(\Big \downarrow \) 913 |
\(\displaystyle \frac {2 x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \sqrt {d+e x^3}}-\frac {\frac {1}{5} \left (16 c d^2-5 e (a e+2 b d)\right ) \int \frac {1}{\sqrt {e x^3+d}}dx-\frac {6}{5} c d x \sqrt {d+e x^3}}{3 d e^2}\) |
\(\Big \downarrow \) 759 |
\(\displaystyle \frac {2 x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \sqrt {d+e x^3}}-\frac {\frac {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt {\frac {d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (16 c d^2-5 e (a e+2 b d)\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{e} x+\left (1-\sqrt {3}\right ) \sqrt [3]{d}}{\sqrt [3]{e} x+\left (1+\sqrt {3}\right ) \sqrt [3]{d}}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt [3]{e} \sqrt {\frac {\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt {d+e x^3}}-\frac {6}{5} c d x \sqrt {d+e x^3}}{3 d e^2}\) |
(2*(c*d^2 - b*d*e + a*e^2)*x)/(3*d*e^2*Sqrt[d + e*x^3]) - ((-6*c*d*x*Sqrt[ d + e*x^3])/5 + (2*Sqrt[2 + Sqrt[3]]*(16*c*d^2 - 5*e*(2*b*d + a*e))*(d^(1/ 3) + e^(1/3)*x)*Sqrt[(d^(2/3) - d^(1/3)*e^(1/3)*x + e^(2/3)*x^2)/((1 + Sqr t[3])*d^(1/3) + e^(1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*d^(1/3) + e^ (1/3)*x)/((1 + Sqrt[3])*d^(1/3) + e^(1/3)*x)], -7 - 4*Sqrt[3]])/(5*3^(1/4) *e^(1/3)*Sqrt[(d^(1/3)*(d^(1/3) + e^(1/3)*x))/((1 + Sqrt[3])*d^(1/3) + e^( 1/3)*x)^2]*Sqrt[d + e*x^3]))/(3*d*e^2)
3.1.39.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si mp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(p + 1) + 1))), x] - Simp[(a*d - b*c*(n*( p + 1) + 1))/(b*(n*(p + 1) + 1)) Int[(a + b*x^n)^p, x], x] /; FreeQ[{a, b , c, d, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_ )), x_Symbol] :> Simp[(-(c*d^2 - b*d*e + a*e^2))*x*((d + e*x^n)^(q + 1)/(d* e^2*n*(q + 1))), x] + Simp[1/(n*(q + 1)*d*e^2) Int[(d + e*x^n)^(q + 1)*Si mp[c*d^2 - b*d*e + a*e^2*(n*(q + 1) + 1) + c*d*e*n*(q + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && N eQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[q, -1]
Time = 1.29 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.32
method | result | size |
elliptic | \(\frac {2 x \left (a \,e^{2}-b d e +c \,d^{2}\right )}{3 e^{2} d \sqrt {\left (x^{3}+\frac {d}{e}\right ) e}}+\frac {2 c x \sqrt {e \,x^{3}+d}}{5 e^{2}}-\frac {2 i \left (\frac {b e -c d}{e^{2}}+\frac {a \,e^{2}-b d e +c \,d^{2}}{3 e^{2} d}-\frac {2 c d}{5 e^{2}}\right ) \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}-\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{e}}{-\frac {3 \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}\, F\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}-\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{e \left (-\frac {3 \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right )}}\right )}{3 e \sqrt {e \,x^{3}+d}}\) | \(382\) |
default | \(\text {Expression too large to display}\) | \(934\) |
risch | \(\text {Expression too large to display}\) | \(961\) |
2/3/e^2/d*x*(a*e^2-b*d*e+c*d^2)/((x^3+1/e*d)*e)^(1/2)+2/5*c*x*(e*x^3+d)^(1 /2)/e^2-2/3*I*((b*e-c*d)/e^2+1/3*(a*e^2-b*d*e+c*d^2)/e^2/d-2/5*c/e^2*d)*3^ (1/2)/e*(-d*e^2)^(1/3)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*3^(1/2)/e*(-d*e^2) ^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)*((x-1/e*(-d*e^2)^(1/3))/(-3/2/e*(- d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)*(-I*(x+1/2/e*(-d*e^2)^ (1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)/(e*x ^3+d)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*3^(1/2) /e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2),(I*3^(1/2)/e*(-d*e^2)^( 1/3)/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.43 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx=-\frac {2 \, {\left ({\left (16 \, c d^{3} - 10 \, b d^{2} e - 5 \, a d e^{2} + {\left (16 \, c d^{2} e - 10 \, b d e^{2} - 5 \, a e^{3}\right )} x^{3}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (0, -\frac {4 \, d}{e}, x\right ) - {\left (3 \, c d e^{2} x^{4} + {\left (8 \, c d^{2} e - 5 \, b d e^{2} + 5 \, a e^{3}\right )} x\right )} \sqrt {e x^{3} + d}\right )}}{15 \, {\left (d e^{4} x^{3} + d^{2} e^{3}\right )}} \]
-2/15*((16*c*d^3 - 10*b*d^2*e - 5*a*d*e^2 + (16*c*d^2*e - 10*b*d*e^2 - 5*a *e^3)*x^3)*sqrt(e)*weierstrassPInverse(0, -4*d/e, x) - (3*c*d*e^2*x^4 + (8 *c*d^2*e - 5*b*d*e^2 + 5*a*e^3)*x)*sqrt(e*x^3 + d))/(d*e^4*x^3 + d^2*e^3)
Time = 6.25 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.41 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx=\frac {a x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {3}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {3}{2}} \Gamma \left (\frac {4}{3}\right )} + \frac {b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {3}{2} \\ \frac {7}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {3}{2}} \Gamma \left (\frac {7}{3}\right )} + \frac {c x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {3}{2}} \Gamma \left (\frac {10}{3}\right )} \]
a*x*gamma(1/3)*hyper((1/3, 3/2), (4/3,), e*x**3*exp_polar(I*pi)/d)/(3*d**( 3/2)*gamma(4/3)) + b*x**4*gamma(4/3)*hyper((4/3, 3/2), (7/3,), e*x**3*exp_ polar(I*pi)/d)/(3*d**(3/2)*gamma(7/3)) + c*x**7*gamma(7/3)*hyper((3/2, 7/3 ), (10/3,), e*x**3*exp_polar(I*pi)/d)/(3*d**(3/2)*gamma(10/3))
\[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx=\int { \frac {c x^{6} + b x^{3} + a}{{\left (e x^{3} + d\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx=\int { \frac {c x^{6} + b x^{3} + a}{{\left (e x^{3} + d\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{3/2}} \, dx=\int \frac {c\,x^6+b\,x^3+a}{{\left (e\,x^3+d\right )}^{3/2}} \,d x \]