\(\int \frac {a+b x^3+c x^6}{(d+e x^3)^{5/2}} \, dx\) [40]

Optimal result

Integrand size = 24, antiderivative size = 309 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{5/2}} \, dx=\frac {2 \left (c d^2-b d e+a e^2\right ) x}{9 d e^2 \left (d+e x^3\right )^{3/2}}-\frac {2 \left (11 c d^2-2 b d e-7 a e^2\right ) x}{27 d^2 e^2 \sqrt {d+e x^3}}+\frac {2 \sqrt {2+\sqrt {3}} \left (16 c d^2+e (2 b d+7 a e)\right ) \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt {\frac {d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}\right ),-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} d^2 e^{7/3} \sqrt {\frac {\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt {d+e x^3}} \]

[Out]

2/9*(a*e^2-b*d*e+c*d^2)*x/d/e^2/(e*x^3+d)^(3/2)-2/27*(-7*a*e^2-2*b*d*e+11*c*d^2)*x/d^2/e^2/(e*x^3+d)^(1/2)+2/8
1*(16*c*d^2+e*(7*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^2/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)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1423, 393, 224} \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{5/2}} \, dx=\frac {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt {\frac {d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (e (7 a e+2 b d)+16 c d^2\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{e} x+\left (1-\sqrt {3}\right ) \sqrt [3]{d}}{\sqrt [3]{e} x+\left (1+\sqrt {3}\right ) \sqrt [3]{d}}\right ),-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} d^2 e^{7/3} \sqrt {\frac {\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt {d+e x^3}}-\frac {2 x \left (-7 a e^2-2 b d e+11 c d^2\right )}{27 d^2 e^2 \sqrt {d+e x^3}}+\frac {2 x \left (a e^2-b d e+c d^2\right )}{9 d e^2 \left (d+e x^3\right )^{3/2}} \]

[In]

Int[(a + b*x^3 + c*x^6)/(d + e*x^3)^(5/2),x]

[Out]

(2*(c*d^2 - b*d*e + a*e^2)*x)/(9*d*e^2*(d + e*x^3)^(3/2)) - (2*(11*c*d^2 - 2*b*d*e - 7*a*e^2)*x)/(27*d^2*e^2*S
qrt[d + e*x^3]) + (2*Sqrt[2 + Sqrt[3]]*(16*c*d^2 + e*(2*b*d + 7*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 + Sqrt[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]])/(27*3^(1/4)*d^2*e^(7/3)*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])

Rule 224

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]*Sq
rt[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]

Rule 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 1423

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] + Dist[1/(n*(q + 1)*d*e^2), Int[(d + e*x^n)^(q + 1
)*Simp[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] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[q, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \left (c d^2-b d e+a e^2\right ) x}{9 d e^2 \left (d+e x^3\right )^{3/2}}-\frac {2 \int \frac {\frac {1}{2} \left (2 c d^2-e (2 b d+7 a e)\right )-\frac {9}{2} c d e x^3}{\left (d+e x^3\right )^{3/2}} \, dx}{9 d e^2} \\ & = \frac {2 \left (c d^2-b d e+a e^2\right ) x}{9 d e^2 \left (d+e x^3\right )^{3/2}}-\frac {2 \left (11 c d^2-2 b d e-7 a e^2\right ) x}{27 d^2 e^2 \sqrt {d+e x^3}}-\frac {\left (4 \left (-\frac {9}{2} c d^2 e+\frac {1}{4} e \left (2 c d^2-e (2 b d+7 a e)\right )\right )\right ) \int \frac {1}{\sqrt {d+e x^3}} \, dx}{27 d^2 e^3} \\ & = \frac {2 \left (c d^2-b d e+a e^2\right ) x}{9 d e^2 \left (d+e x^3\right )^{3/2}}-\frac {2 \left (11 c d^2-2 b d e-7 a e^2\right ) x}{27 d^2 e^2 \sqrt {d+e x^3}}+\frac {2 \sqrt {2+\sqrt {3}} \left (16 c d^2+e (2 b d+7 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}} F\left (\sin ^{-1}\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 )}{27 \sqrt [4]{3} d^2 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}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.09 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.42 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{5/2}} \, dx=\frac {-2 x \left (c d^2 \left (8 d+11 e x^3\right )+e \left (b d \left (d-2 e x^3\right )-a e \left (10 d+7 e x^3\right )\right )\right )+\left (16 c d^2+e (2 b d+7 a e)\right ) x \left (d+e x^3\right ) \sqrt {1+\frac {e x^3}{d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},-\frac {e x^3}{d}\right )}{27 d^2 e^2 \left (d+e x^3\right )^{3/2}} \]

[In]

Integrate[(a + b*x^3 + c*x^6)/(d + e*x^3)^(5/2),x]

[Out]

(-2*x*(c*d^2*(8*d + 11*e*x^3) + e*(b*d*(d - 2*e*x^3) - a*e*(10*d + 7*e*x^3))) + (16*c*d^2 + e*(2*b*d + 7*a*e))
*x*(d + e*x^3)*Sqrt[1 + (e*x^3)/d]*Hypergeometric2F1[1/3, 1/2, 4/3, -((e*x^3)/d)])/(27*d^2*e^2*(d + e*x^3)^(3/
2))

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 401, normalized size of antiderivative = 1.30

method	result	size
elliptic	\(\frac {2 x \left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {e \,x^{3}+d}}{9 d \,e^{4} \left (x^{3}+\frac {d}{e}\right )^{2}}+\frac {2 x \left (7 a \,e^{2}+2 b d e -11 c \,d^{2}\right )}{27 e^{2} d^{2} \sqrt {\left (x^{3}+\frac {d}{e}\right ) e}}-\frac {2 i \left (\frac {c}{e^{2}}+\frac {7 a \,e^{2}+2 b d e -11 c \,d^{2}}{27 e^{2} d^{2}}\right ) \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}-\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{e}}{-\frac {3 \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}\, F\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}-\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{e \left (-\frac {3 \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right )}}\right )}{3 e \sqrt {e \,x^{3}+d}}\)	\(401\)
default	\(\text {Expression too large to display}\)	\(1005\)

[In]

int((c*x^6+b*x^3+a)/(e*x^3+d)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/9/d*x/e^4*(a*e^2-b*d*e+c*d^2)*(e*x^3+d)^(1/2)/(x^3+1/e*d)^2+2/27/e^2/d^2*x*(7*a*e^2+2*b*d*e-11*c*d^2)/((x^3+
1/e*d)*e)^(1/2)-2/3*I*(c/e^2+1/27/e^2/d^2*(7*a*e^2+2*b*d*e-11*c*d^2))*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))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.61 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{5/2}} \, dx=\frac {2 \, {\left ({\left ({\left (16 \, c d^{2} e^{2} + 2 \, b d e^{3} + 7 \, a e^{4}\right )} x^{6} + 16 \, c d^{4} + 2 \, b d^{3} e + 7 \, a d^{2} e^{2} + 2 \, {\left (16 \, c d^{3} e + 2 \, b d^{2} e^{2} + 7 \, a d e^{3}\right )} x^{3}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (0, -\frac {4 \, d}{e}, x\right ) - {\left ({\left (11 \, c d^{2} e^{2} - 2 \, b d e^{3} - 7 \, a e^{4}\right )} x^{4} + {\left (8 \, c d^{3} e + b d^{2} e^{2} - 10 \, a d e^{3}\right )} x\right )} \sqrt {e x^{3} + d}\right )}}{27 \, {\left (d^{2} e^{5} x^{6} + 2 \, d^{3} e^{4} x^{3} + d^{4} e^{3}\right )}} \]

[In]

integrate((c*x^6+b*x^3+a)/(e*x^3+d)^(5/2),x, algorithm="fricas")

[Out]

2/27*(((16*c*d^2*e^2 + 2*b*d*e^3 + 7*a*e^4)*x^6 + 16*c*d^4 + 2*b*d^3*e + 7*a*d^2*e^2 + 2*(16*c*d^3*e + 2*b*d^2
*e^2 + 7*a*d*e^3)*x^3)*sqrt(e)*weierstrassPInverse(0, -4*d/e, x) - ((11*c*d^2*e^2 - 2*b*d*e^3 - 7*a*e^4)*x^4 +
(8*c*d^3*e + b*d^2*e^2 - 10*a*d*e^3)*x)*sqrt(e*x^3 + d))/(d^2*e^5*x^6 + 2*d^3*e^4*x^3 + d^4*e^3)

Sympy [A] (verification not implemented)

Time = 34.98 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.39 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{5/2}} \, dx=\frac {a x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {5}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {5}{2}} \Gamma \left (\frac {4}{3}\right )} + \frac {b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {5}{2} \\ \frac {7}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {5}{2}} \Gamma \left (\frac {7}{3}\right )} + \frac {c x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{3}, \frac {5}{2} \\ \frac {10}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {5}{2}} \Gamma \left (\frac {10}{3}\right )} \]

[In]

integrate((c*x**6+b*x**3+a)/(e*x**3+d)**(5/2),x)

[Out]

a*x*gamma(1/3)*hyper((1/3, 5/2), (4/3,), e*x**3*exp_polar(I*pi)/d)/(3*d**(5/2)*gamma(4/3)) + b*x**4*gamma(4/3)
*hyper((4/3, 5/2), (7/3,), e*x**3*exp_polar(I*pi)/d)/(3*d**(5/2)*gamma(7/3)) + c*x**7*gamma(7/3)*hyper((7/3, 5
/2), (10/3,), e*x**3*exp_polar(I*pi)/d)/(3*d**(5/2)*gamma(10/3))

Maxima [F]

\[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{5/2}} \, dx=\int { \frac {c x^{6} + b x^{3} + a}{{\left (e x^{3} + d\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((c*x^6+b*x^3+a)/(e*x^3+d)^(5/2),x, algorithm="maxima")

[Out]

integrate((c*x^6 + b*x^3 + a)/(e*x^3 + d)^(5/2), x)

Giac [F]

\[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{5/2}} \, dx=\int { \frac {c x^{6} + b x^{3} + a}{{\left (e x^{3} + d\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((c*x^6+b*x^3+a)/(e*x^3+d)^(5/2),x, algorithm="giac")

[Out]

integrate((c*x^6 + b*x^3 + a)/(e*x^3 + d)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{5/2}} \, dx=\int \frac {c\,x^6+b\,x^3+a}{{\left (e\,x^3+d\right )}^{5/2}} \,d x \]

[In]

int((a + b*x^3 + c*x^6)/(d + e*x^3)^(5/2),x)

[Out]

int((a + b*x^3 + c*x^6)/(d + e*x^3)^(5/2), x)