Integrand size = 21, antiderivative size = 285 \[ \int \frac {\left (c+d x^3\right )^2}{\sqrt {a+b x^3}} \, dx=\frac {4 d (11 b c-4 a d) x \sqrt {a+b x^3}}{55 b^2}+\frac {2 d^2 x^4 \sqrt {a+b x^3}}{11 b}+\frac {2 \sqrt {2+\sqrt {3}} \left (55 b^2 c^2-4 a d (11 b c-4 a d)\right ) \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt {3}\right )}{55 \sqrt [4]{3} b^{7/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \] Output:
4/55*d*(-4*a*d+11*b*c)*x*(b*x^3+a)^(1/2)/b^2+2/11*d^2*x^4*(b*x^3+a)^(1/2)/ b+2/165*(1/2*6^(1/2)+1/2*2^(1/2))*(55*b^2*c^2-4*a*d*(-4*a*d+11*b*c))*(a^(1 /3)+b^(1/3)*x)*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/((1+3^(1/2))*a^(1/ 3)+b^(1/3)*x)^2)^(1/2)*EllipticF(((1-3^(1/2))*a^(1/3)+b^(1/3)*x)/((1+3^(1/ 2))*a^(1/3)+b^(1/3)*x),I*3^(1/2)+2*I)*3^(3/4)/b^(7/3)/(a^(1/3)*(a^(1/3)+b^ (1/3)*x)/((1+3^(1/2))*a^(1/3)+b^(1/3)*x)^2)^(1/2)/(b*x^3+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 12.80 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.56 \[ \int \frac {\left (c+d x^3\right )^2}{\sqrt {a+b x^3}} \, dx=\frac {x \sqrt {1+\frac {b x^3}{a}} \left (40 a \left (14 c^2+7 c d x^3+2 d^2 x^6\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {10}{3},-\frac {b x^3}{a}\right )-3 b x^3 \left (11 c^2+16 c d x^3+5 d^2 x^6\right ) \operatorname {Hypergeometric2F1}\left (\frac {4}{3},\frac {3}{2},\frac {13}{3},-\frac {b x^3}{a}\right )-9 b x^3 \left (c+d x^3\right )^2 \, _3F_2\left (\frac {4}{3},\frac {3}{2},2;1,\frac {13}{3};-\frac {b x^3}{a}\right )\right )}{560 a \sqrt {a+b x^3}} \] Input:
Integrate[(c + d*x^3)^2/Sqrt[a + b*x^3],x]
Output:
(x*Sqrt[1 + (b*x^3)/a]*(40*a*(14*c^2 + 7*c*d*x^3 + 2*d^2*x^6)*Hypergeometr ic2F1[1/3, 1/2, 10/3, -((b*x^3)/a)] - 3*b*x^3*(11*c^2 + 16*c*d*x^3 + 5*d^2 *x^6)*Hypergeometric2F1[4/3, 3/2, 13/3, -((b*x^3)/a)] - 9*b*x^3*(c + d*x^3 )^2*HypergeometricPFQ[{4/3, 3/2, 2}, {1, 13/3}, -((b*x^3)/a)]))/(560*a*Sqr t[a + b*x^3])
Time = 0.57 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {933, 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.
| \(\displaystyle \int \frac {\left (c+d x^3\right )^2}{\sqrt {a+b x^3}} \, dx\) |
| \(\Big \downarrow \) 933 |
| \(\displaystyle \frac {2 \int \frac {d (17 b c-8 a d) x^3+c (11 b c-2 a d)}{2 \sqrt {b x^3+a}}dx}{11 b}+\frac {2 d x \sqrt {a+b x^3} \left (c+d x^3\right )}{11 b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {d (17 b c-8 a d) x^3+c (11 b c-2 a d)}{\sqrt {b x^3+a}}dx}{11 b}+\frac {2 d x \sqrt {a+b x^3} \left (c+d x^3\right )}{11 b}\) |
| \(\Big \downarrow \) 913 |
| \(\displaystyle \frac {\frac {\left (16 a^2 d^2-44 a b c d+55 b^2 c^2\right ) \int \frac {1}{\sqrt {b x^3+a}}dx}{5 b}+\frac {2 d x \sqrt {a+b x^3} (17 b c-8 a d)}{5 b}}{11 b}+\frac {2 d x \sqrt {a+b x^3} \left (c+d x^3\right )}{11 b}\) |
| \(\Big \downarrow \) 759 |
| \(\displaystyle \frac {\frac {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \left (16 a^2 d^2-44 a b c d+55 b^2 c^2\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} b^{4/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {2 d x \sqrt {a+b x^3} (17 b c-8 a d)}{5 b}}{11 b}+\frac {2 d x \sqrt {a+b x^3} \left (c+d x^3\right )}{11 b}\) |
Input:
Int[(c + d*x^3)^2/Sqrt[a + b*x^3],x]
Output:
(2*d*x*Sqrt[a + b*x^3]*(c + d*x^3))/(11*b) + ((2*d*(17*b*c - 8*a*d)*x*Sqrt [a + b*x^3])/(5*b) + (2*Sqrt[2 + Sqrt[3]]*(55*b^2*c^2 - 44*a*b*c*d + 16*a^ 2*d^2)*(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]))/(11*b)
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[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), x] + Simp[1/(b*(n*(p + q) + 1)) Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Sim p[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d , 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[ a, b, c, d, n, p, q, x]
Time = 1.56 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.19
| method | result | size |
| risch | \(-\frac {2 x d \left (-5 b d \,x^{3}+8 a d -22 b c \right ) \sqrt {b \,x^{3}+a}}{55 b^{2}}-\frac {2 i \left (16 a^{2} d^{2}-44 a b c d +55 b^{2} c^{2}\right ) \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}}{-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{b \left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right )}{165 b^{3} \sqrt {b \,x^{3}+a}}\) | \(339\) |
| elliptic | \(\frac {2 d^{2} x^{4} \sqrt {b \,x^{3}+a}}{11 b}+\frac {2 \left (2 c d -\frac {8 d^{2} a}{11 b}\right ) x \sqrt {b \,x^{3}+a}}{5 b}-\frac {2 i \left (c^{2}-\frac {2 a \left (2 c d -\frac {8 d^{2} a}{11 b}\right )}{5 b}\right ) \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}}{-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{b \left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right )}{3 b \sqrt {b \,x^{3}+a}}\) | \(357\) |
| default | \(\text {Expression too large to display}\) | \(913\) |
Input:
int((d*x^3+c)^2/(b*x^3+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/55*x*d*(-5*b*d*x^3+8*a*d-22*b*c)/b^2*(b*x^3+a)^(1/2)-2/165*I*(16*a^2*d^ 2-44*a*b*c*d+55*b^2*c^2)/b^3*3^(1/2)*(-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))
Time = 0.07 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.29 \[ \int \frac {\left (c+d x^3\right )^2}{\sqrt {a+b x^3}} \, dx=\frac {2 \, {\left ({\left (55 \, b^{2} c^{2} - 44 \, a b c d + 16 \, a^{2} d^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right ) + {\left (5 \, b^{2} d^{2} x^{4} + 2 \, {\left (11 \, b^{2} c d - 4 \, a b d^{2}\right )} x\right )} \sqrt {b x^{3} + a}\right )}}{55 \, b^{3}} \] Input:
integrate((d*x^3+c)^2/(b*x^3+a)^(1/2),x, algorithm="fricas")
Output:
2/55*((55*b^2*c^2 - 44*a*b*c*d + 16*a^2*d^2)*sqrt(b)*weierstrassPInverse(0 , -4*a/b, x) + (5*b^2*d^2*x^4 + 2*(11*b^2*c*d - 4*a*b*d^2)*x)*sqrt(b*x^3 + a))/b^3
Time = 1.52 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.44 \[ \int \frac {\left (c+d x^3\right )^2}{\sqrt {a+b x^3}} \, dx=\frac {c^{2} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt {a} \Gamma \left (\frac {4}{3}\right )} + \frac {2 c d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt {a} \Gamma \left (\frac {7}{3}\right )} + \frac {d^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt {a} \Gamma \left (\frac {10}{3}\right )} \] Input:
integrate((d*x**3+c)**2/(b*x**3+a)**(1/2),x)
Output:
c**2*x*gamma(1/3)*hyper((1/3, 1/2), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*s qrt(a)*gamma(4/3)) + 2*c*d*x**4*gamma(4/3)*hyper((1/2, 4/3), (7/3,), b*x** 3*exp_polar(I*pi)/a)/(3*sqrt(a)*gamma(7/3)) + d**2*x**7*gamma(7/3)*hyper(( 1/2, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*sqrt(a)*gamma(10/3))
\[ \int \frac {\left (c+d x^3\right )^2}{\sqrt {a+b x^3}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{2}}{\sqrt {b x^{3} + a}} \,d x } \] Input:
integrate((d*x^3+c)^2/(b*x^3+a)^(1/2),x, algorithm="maxima")
Output:
integrate((d*x^3 + c)^2/sqrt(b*x^3 + a), x)
\[ \int \frac {\left (c+d x^3\right )^2}{\sqrt {a+b x^3}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{2}}{\sqrt {b x^{3} + a}} \,d x } \] Input:
integrate((d*x^3+c)^2/(b*x^3+a)^(1/2),x, algorithm="giac")
Output:
integrate((d*x^3 + c)^2/sqrt(b*x^3 + a), x)
Timed out. \[ \int \frac {\left (c+d x^3\right )^2}{\sqrt {a+b x^3}} \, dx=\int \frac {{\left (d\,x^3+c\right )}^2}{\sqrt {b\,x^3+a}} \,d x \] Input:
int((c + d*x^3)^2/(a + b*x^3)^(1/2),x)
Output:
int((c + d*x^3)^2/(a + b*x^3)^(1/2), x)
\[ \int \frac {\left (c+d x^3\right )^2}{\sqrt {a+b x^3}} \, dx=\frac {-16 \sqrt {b \,x^{3}+a}\, a \,d^{2} x +44 \sqrt {b \,x^{3}+a}\, b c d x +10 \sqrt {b \,x^{3}+a}\, b \,d^{2} x^{4}+16 \left (\int \frac {\sqrt {b \,x^{3}+a}}{b \,x^{3}+a}d x \right ) a^{2} d^{2}-44 \left (\int \frac {\sqrt {b \,x^{3}+a}}{b \,x^{3}+a}d x \right ) a b c d +55 \left (\int \frac {\sqrt {b \,x^{3}+a}}{b \,x^{3}+a}d x \right ) b^{2} c^{2}}{55 b^{2}} \] Input:
int((d*x^3+c)^2/(b*x^3+a)^(1/2),x)
Output:
( - 16*sqrt(a + b*x**3)*a*d**2*x + 44*sqrt(a + b*x**3)*b*c*d*x + 10*sqrt(a + b*x**3)*b*d**2*x**4 + 16*int(sqrt(a + b*x**3)/(a + b*x**3),x)*a**2*d**2 - 44*int(sqrt(a + b*x**3)/(a + b*x**3),x)*a*b*c*d + 55*int(sqrt(a + b*x** 3)/(a + b*x**3),x)*b**2*c**2)/(55*b**2)