Integrand size = 19, antiderivative size = 82 \[ \int \sqrt [3]{a+b x^3} \left (c+d x^3\right ) \, dx=\frac {d x \left (a+b x^3\right )^{4/3}}{5 b}+\frac {(5 b c-a d) x \sqrt [3]{a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{5 b \sqrt [3]{1+\frac {b x^3}{a}}} \] Output:
1/5*d*x*(b*x^3+a)^(4/3)/b+1/5*(-a*d+5*b*c)*x*(b*x^3+a)^(1/3)*hypergeom([-1 /3, 1/3],[4/3],-b*x^3/a)/b/(1+b*x^3/a)^(1/3)
Time = 0.06 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.88 \[ \int \sqrt [3]{a+b x^3} \left (c+d x^3\right ) \, dx=\frac {x \sqrt [3]{a+b x^3} \left (d \left (a+b x^3\right )+\frac {(5 b c-a d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{\sqrt [3]{1+\frac {b x^3}{a}}}\right )}{5 b} \] Input:
Integrate[(a + b*x^3)^(1/3)*(c + d*x^3),x]
Output:
(x*(a + b*x^3)^(1/3)*(d*(a + b*x^3) + ((5*b*c - a*d)*Hypergeometric2F1[-1/ 3, 1/3, 4/3, -((b*x^3)/a)])/(1 + (b*x^3)/a)^(1/3)))/(5*b)
Time = 0.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {913, 779, 778}
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.
\(\displaystyle \int \sqrt [3]{a+b x^3} \left (c+d x^3\right ) \, dx\) |
\(\Big \downarrow \) 913 |
\(\displaystyle \frac {(5 b c-a d) \int \sqrt [3]{b x^3+a}dx}{5 b}+\frac {d x \left (a+b x^3\right )^{4/3}}{5 b}\) |
\(\Big \downarrow \) 779 |
\(\displaystyle \frac {\sqrt [3]{a+b x^3} (5 b c-a d) \int \sqrt [3]{\frac {b x^3}{a}+1}dx}{5 b \sqrt [3]{\frac {b x^3}{a}+1}}+\frac {d x \left (a+b x^3\right )^{4/3}}{5 b}\) |
\(\Big \downarrow \) 778 |
\(\displaystyle \frac {x \sqrt [3]{a+b x^3} (5 b c-a d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{5 b \sqrt [3]{\frac {b x^3}{a}+1}}+\frac {d x \left (a+b x^3\right )^{4/3}}{5 b}\) |
Input:
Int[(a + b*x^3)^(1/3)*(c + d*x^3),x]
Output:
(d*x*(a + b*x^3)^(4/3))/(5*b) + ((5*b*c - a*d)*x*(a + b*x^3)^(1/3)*Hyperge ometric2F1[-1/3, 1/3, 4/3, -((b*x^3)/a)])/(5*b*(1 + (b*x^3)/a)^(1/3))
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] && !IGtQ[p , 0] && !IntegerQ[1/n] && !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || GtQ[a, 0])
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x ^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) Int[(1 + b*(x^n/a))^p, x], x ] /; FreeQ[{a, b, n, p}, x] && !IGtQ[p, 0] && !IntegerQ[1/n] && !ILtQ[Si mplify[1/n + p], 0] && !(IntegerQ[p] || GtQ[a, 0])
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]
\[\int \left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (d \,x^{3}+c \right )d x\]
Input:
int((b*x^3+a)^(1/3)*(d*x^3+c),x)
Output:
int((b*x^3+a)^(1/3)*(d*x^3+c),x)
\[ \int \sqrt [3]{a+b x^3} \left (c+d x^3\right ) \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (d x^{3} + c\right )} \,d x } \] Input:
integrate((b*x^3+a)^(1/3)*(d*x^3+c),x, algorithm="fricas")
Output:
integral((b*x^3 + a)^(1/3)*(d*x^3 + c), x)
Result contains complex when optimal does not.
Time = 1.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \sqrt [3]{a+b x^3} \left (c+d x^3\right ) \, dx=\frac {\sqrt [3]{a} c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {\sqrt [3]{a} d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} \] Input:
integrate((b*x**3+a)**(1/3)*(d*x**3+c),x)
Output:
a**(1/3)*c*x*gamma(1/3)*hyper((-1/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/ a)/(3*gamma(4/3)) + a**(1/3)*d*x**4*gamma(4/3)*hyper((-1/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3))
\[ \int \sqrt [3]{a+b x^3} \left (c+d x^3\right ) \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (d x^{3} + c\right )} \,d x } \] Input:
integrate((b*x^3+a)^(1/3)*(d*x^3+c),x, algorithm="maxima")
Output:
integrate((b*x^3 + a)^(1/3)*(d*x^3 + c), x)
\[ \int \sqrt [3]{a+b x^3} \left (c+d x^3\right ) \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (d x^{3} + c\right )} \,d x } \] Input:
integrate((b*x^3+a)^(1/3)*(d*x^3+c),x, algorithm="giac")
Output:
integrate((b*x^3 + a)^(1/3)*(d*x^3 + c), x)
Timed out. \[ \int \sqrt [3]{a+b x^3} \left (c+d x^3\right ) \, dx=\int {\left (b\,x^3+a\right )}^{1/3}\,\left (d\,x^3+c\right ) \,d x \] Input:
int((a + b*x^3)^(1/3)*(c + d*x^3),x)
Output:
int((a + b*x^3)^(1/3)*(c + d*x^3), x)
\[ \int \sqrt [3]{a+b x^3} \left (c+d x^3\right ) \, dx=\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} a d x +5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b c x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b d \,x^{4}-\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} d +5 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a b c}{10 b} \] Input:
int((b*x^3+a)^(1/3)*(d*x^3+c),x)
Output:
((a + b*x**3)**(1/3)*a*d*x + 5*(a + b*x**3)**(1/3)*b*c*x + 2*(a + b*x**3)* *(1/3)*b*d*x**4 - int((a + b*x**3)**(1/3)/(a + b*x**3),x)*a**2*d + 5*int(( a + b*x**3)**(1/3)/(a + b*x**3),x)*a*b*c)/(10*b)