Integrand size = 19, antiderivative size = 307 \[ \int \sqrt [3]{a+b x^2} \left (c+d x^2\right ) \, dx=\frac {3 (11 b c-3 a d) x \sqrt [3]{a+b x^2}}{55 b}+\frac {3 d x \left (a+b x^2\right )^{4/3}}{11 b}-\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} a (11 b c-3 a d) \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),-7+4 \sqrt {3}\right )}{55 b^2 x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \] Output:
3/55*(-3*a*d+11*b*c)*x*(b*x^2+a)^(1/3)/b+3/11*d*x*(b*x^2+a)^(4/3)/b-2/55*3 ^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*a*(-3*a*d+11*b*c)*(a^(1/3)-(b*x^2+a)^(1/3 ))*((a^(2/3)+a^(1/3)*(b*x^2+a)^(1/3)+(b*x^2+a)^(2/3))/((1-3^(1/2))*a^(1/3) -(b*x^2+a)^(1/3))^2)^(1/2)*EllipticF(((1+3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3)) /((1-3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3)),2*I-I*3^(1/2))/b^2/x/(-a^(1/3)*(a^( 1/3)-(b*x^2+a)^(1/3))/((1-3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 5.06 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.24 \[ \int \sqrt [3]{a+b x^2} \left (c+d x^2\right ) \, dx=\frac {3 x \sqrt [3]{a+b x^2} \left (d \left (a+b x^2\right )+\frac {(11 b c-3 a d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{3 \sqrt [3]{1+\frac {b x^2}{a}}}\right )}{11 b} \] Input:
Integrate[(a + b*x^2)^(1/3)*(c + d*x^2),x]
Output:
(3*x*(a + b*x^2)^(1/3)*(d*(a + b*x^2) + ((11*b*c - 3*a*d)*Hypergeometric2F 1[-1/3, 1/2, 3/2, -((b*x^2)/a)])/(3*(1 + (b*x^2)/a)^(1/3))))/(11*b)
Time = 0.28 (sec) , antiderivative size = 303, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {299, 211, 234, 760}
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.
| \(\displaystyle \int \sqrt [3]{a+b x^2} \left (c+d x^2\right ) \, dx\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {(11 b c-3 a d) \int \sqrt [3]{b x^2+a}dx}{11 b}+\frac {3 d x \left (a+b x^2\right )^{4/3}}{11 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {(11 b c-3 a d) \left (\frac {2}{5} a \int \frac {1}{\left (b x^2+a\right )^{2/3}}dx+\frac {3}{5} x \sqrt [3]{a+b x^2}\right )}{11 b}+\frac {3 d x \left (a+b x^2\right )^{4/3}}{11 b}\) |
| \(\Big \downarrow \) 234 |
| \(\displaystyle \frac {(11 b c-3 a d) \left (\frac {3 a \sqrt {b x^2} \int \frac {1}{\sqrt {b x^2}}d\sqrt [3]{b x^2+a}}{5 b x}+\frac {3}{5} x \sqrt [3]{a+b x^2}\right )}{11 b}+\frac {3 d x \left (a+b x^2\right )^{4/3}}{11 b}\) |
| \(\Big \downarrow \) 760 |
| \(\displaystyle \frac {(11 b c-3 a d) \left (\frac {3}{5} x \sqrt [3]{a+b x^2}-\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} a \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right ),-7+4 \sqrt {3}\right )}{5 b x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}\right )}{11 b}+\frac {3 d x \left (a+b x^2\right )^{4/3}}{11 b}\) |
Input:
Int[(a + b*x^2)^(1/3)*(c + d*x^2),x]
Output:
(3*d*x*(a + b*x^2)^(4/3))/(11*b) + ((11*b*c - 3*a*d)*((3*x*(a + b*x^2)^(1/ 3))/5 - (2*3^(3/4)*Sqrt[2 - Sqrt[3]]*a*(a^(1/3) - (a + b*x^2)^(1/3))*Sqrt[ (a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3))/((1 - Sqrt[3])*a ^(1/3) - (a + b*x^2)^(1/3))^2]*EllipticF[ArcSin[((1 + Sqrt[3])*a^(1/3) - ( a + b*x^2)^(1/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))], -7 + 4*Sqr t[3]])/(5*b*x*Sqrt[-((a^(1/3)*(a^(1/3) - (a + b*x^2)^(1/3)))/((1 - Sqrt[3] )*a^(1/3) - (a + b*x^2)^(1/3))^2)])))/(11*b)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) Subst[Int[1/Sqrt[-a + x^3], x], x, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b }, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
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 ] && NegQ[a]
\[\int \left (b \,x^{2}+a \right )^{\frac {1}{3}} \left (x^{2} d +c \right )d x\]
Input:
int((b*x^2+a)^(1/3)*(d*x^2+c),x)
Output:
int((b*x^2+a)^(1/3)*(d*x^2+c),x)
\[ \int \sqrt [3]{a+b x^2} \left (c+d x^2\right ) \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{3}} {\left (d x^{2} + c\right )} \,d x } \] Input:
integrate((b*x^2+a)^(1/3)*(d*x^2+c),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(1/3)*(d*x^2 + c), x)
Time = 0.86 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.20 \[ \int \sqrt [3]{a+b x^2} \left (c+d x^2\right ) \, dx=\sqrt [3]{a} c x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {\sqrt [3]{a} d x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} \] Input:
integrate((b*x**2+a)**(1/3)*(d*x**2+c),x)
Output:
a**(1/3)*c*x*hyper((-1/3, 1/2), (3/2,), b*x**2*exp_polar(I*pi)/a) + a**(1/ 3)*d*x**3*hyper((-1/3, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a)/3
\[ \int \sqrt [3]{a+b x^2} \left (c+d x^2\right ) \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{3}} {\left (d x^{2} + c\right )} \,d x } \] Input:
integrate((b*x^2+a)^(1/3)*(d*x^2+c),x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(1/3)*(d*x^2 + c), x)
\[ \int \sqrt [3]{a+b x^2} \left (c+d x^2\right ) \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{3}} {\left (d x^{2} + c\right )} \,d x } \] Input:
integrate((b*x^2+a)^(1/3)*(d*x^2+c),x, algorithm="giac")
Output:
integrate((b*x^2 + a)^(1/3)*(d*x^2 + c), x)
Timed out. \[ \int \sqrt [3]{a+b x^2} \left (c+d x^2\right ) \, dx=\int {\left (b\,x^2+a\right )}^{1/3}\,\left (d\,x^2+c\right ) \,d x \] Input:
int((a + b*x^2)^(1/3)*(c + d*x^2),x)
Output:
int((a + b*x^2)^(1/3)*(c + d*x^2), x)
\[ \int \sqrt [3]{a+b x^2} \left (c+d x^2\right ) \, dx=\frac {6 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a d x +33 \left (b \,x^{2}+a \right )^{\frac {1}{3}} b c x +15 \left (b \,x^{2}+a \right )^{\frac {1}{3}} b d \,x^{3}-6 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} d +22 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x \right ) a b c}{55 b} \] Input:
int((b*x^2+a)^(1/3)*(d*x^2+c),x)
Output:
(6*(a + b*x**2)**(1/3)*a*d*x + 33*(a + b*x**2)**(1/3)*b*c*x + 15*(a + b*x* *2)**(1/3)*b*d*x**3 - 6*int((a + b*x**2)**(1/3)/(a + b*x**2),x)*a**2*d + 2 2*int((a + b*x**2)**(1/3)/(a + b*x**2),x)*a*b*c)/(55*b)