Integrand size = 21, antiderivative size = 133 \[ \int \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^2 \, dx=\frac {d (7 b c-2 a d) x \left (a+b x^3\right )^{7/3}}{44 b^2}+\frac {d x \left (a+b x^3\right )^{7/3} \left (c+d x^3\right )}{11 b}+\frac {a \left (44 b^2 c^2-11 a b c d+2 a^2 d^2\right ) x \sqrt [3]{a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{44 b^2 \sqrt [3]{1+\frac {b x^3}{a}}} \]
1/44*d*(-2*a*d+7*b*c)*x*(b*x^3+a)^(7/3)/b^2+1/11*d*x*(b*x^3+a)^(7/3)*(d*x^ 3+c)/b+1/44*a*(2*a^2*d^2-11*a*b*c*d+44*b^2*c^2)*x*(b*x^3+a)^(1/3)*hypergeo m([-4/3, 1/3],[4/3],-b*x^3/a)/b^2/(1+b*x^3/a)^(1/3)
Time = 12.00 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.32 \[ \int \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^2 \, dx=\frac {x \sqrt [3]{a+b x^3} \left (20 a \left (14 c^2+7 c d x^3+2 d^2 x^6\right ) \operatorname {Gamma}\left (-\frac {4}{3}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{3},\frac {10}{3},-\frac {b x^3}{a}\right )-3 b x^3 \left (11 c^2+16 c d x^3+5 d^2 x^6\right ) \operatorname {Gamma}\left (-\frac {1}{3}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {4}{3},\frac {13}{3},-\frac {b x^3}{a}\right )-9 b x^3 \left (c+d x^3\right )^2 \operatorname {Gamma}\left (-\frac {1}{3}\right ) \, _3F_2\left (-\frac {1}{3},\frac {4}{3},2;1,\frac {13}{3};-\frac {b x^3}{a}\right )\right )}{280 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Gamma}\left (-\frac {4}{3}\right )} \]
(x*(a + b*x^3)^(1/3)*(20*a*(14*c^2 + 7*c*d*x^3 + 2*d^2*x^6)*Gamma[-4/3]*Hy pergeometric2F1[-4/3, 1/3, 10/3, -((b*x^3)/a)] - 3*b*x^3*(11*c^2 + 16*c*d* x^3 + 5*d^2*x^6)*Gamma[-1/3]*Hypergeometric2F1[-1/3, 4/3, 13/3, -((b*x^3)/ a)] - 9*b*x^3*(c + d*x^3)^2*Gamma[-1/3]*HypergeometricPFQ[{-1/3, 4/3, 2}, {1, 13/3}, -((b*x^3)/a)]))/(280*(1 + (b*x^3)/a)^(1/3)*Gamma[-4/3])
Time = 0.27 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {933, 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.
| \(\displaystyle \int \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^2 \, dx\) |
| \(\Big \downarrow \) 933 |
| \(\displaystyle \frac {\int \left (b x^3+a\right )^{4/3} \left (2 d (7 b c-2 a d) x^3+c (11 b c-a d)\right )dx}{11 b}+\frac {d x \left (a+b x^3\right )^{7/3} \left (c+d x^3\right )}{11 b}\) |
| \(\Big \downarrow \) 913 |
| \(\displaystyle \frac {\frac {\left (2 a^2 d^2-11 a b c d+44 b^2 c^2\right ) \int \left (b x^3+a\right )^{4/3}dx}{4 b}+\frac {d x \left (a+b x^3\right )^{7/3} (7 b c-2 a d)}{4 b}}{11 b}+\frac {d x \left (a+b x^3\right )^{7/3} \left (c+d x^3\right )}{11 b}\) |
| \(\Big \downarrow \) 779 |
| \(\displaystyle \frac {\frac {a \sqrt [3]{a+b x^3} \left (2 a^2 d^2-11 a b c d+44 b^2 c^2\right ) \int \left (\frac {b x^3}{a}+1\right )^{4/3}dx}{4 b \sqrt [3]{\frac {b x^3}{a}+1}}+\frac {d x \left (a+b x^3\right )^{7/3} (7 b c-2 a d)}{4 b}}{11 b}+\frac {d x \left (a+b x^3\right )^{7/3} \left (c+d x^3\right )}{11 b}\) |
| \(\Big \downarrow \) 778 |
| \(\displaystyle \frac {\frac {a x \sqrt [3]{a+b x^3} \left (2 a^2 d^2-11 a b c d+44 b^2 c^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{4 b \sqrt [3]{\frac {b x^3}{a}+1}}+\frac {d x \left (a+b x^3\right )^{7/3} (7 b c-2 a d)}{4 b}}{11 b}+\frac {d x \left (a+b x^3\right )^{7/3} \left (c+d x^3\right )}{11 b}\) |
(d*x*(a + b*x^3)^(7/3)*(c + d*x^3))/(11*b) + ((d*(7*b*c - 2*a*d)*x*(a + b* x^3)^(7/3))/(4*b) + (a*(44*b^2*c^2 - 11*a*b*c*d + 2*a^2*d^2)*x*(a + b*x^3) ^(1/3)*Hypergeometric2F1[-4/3, 1/3, 4/3, -((b*x^3)/a)])/(4*b*(1 + (b*x^3)/ a)^(1/3)))/(11*b)
3.1.80.3.1 Defintions of rubi rules used
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[((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]
\[\int \left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (d \,x^{3}+c \right )^{2}d x\]
\[ \int \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^2 \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )}^{2} \,d x } \]
integral((b*d^2*x^9 + (2*b*c*d + a*d^2)*x^6 + (b*c^2 + 2*a*c*d)*x^3 + a*c^ 2)*(b*x^3 + a)^(1/3), x)
Result contains complex when optimal does not.
Time = 2.60 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.03 \[ \int \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^2 \, dx=\frac {a^{\frac {4}{3}} c^{2} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {2 a^{\frac {4}{3}} c d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {a^{\frac {4}{3}} d^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} + \frac {\sqrt [3]{a} b c^{2} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {2 \sqrt [3]{a} b c d x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} + \frac {\sqrt [3]{a} b d^{2} x^{10} \Gamma \left (\frac {10}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {10}{3} \\ \frac {13}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {13}{3}\right )} \]
a**(4/3)*c**2*x*gamma(1/3)*hyper((-1/3, 1/3), (4/3,), b*x**3*exp_polar(I*p i)/a)/(3*gamma(4/3)) + 2*a**(4/3)*c*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)) + a**(4/3)*d**2*x**7*gamma (7/3)*hyper((-1/3, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(10/3) ) + a**(1/3)*b*c**2*x**4*gamma(4/3)*hyper((-1/3, 4/3), (7/3,), b*x**3*exp_ polar(I*pi)/a)/(3*gamma(7/3)) + 2*a**(1/3)*b*c*d*x**7*gamma(7/3)*hyper((-1 /3, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(10/3)) + a**(1/3)*b* d**2*x**10*gamma(10/3)*hyper((-1/3, 10/3), (13/3,), b*x**3*exp_polar(I*pi) /a)/(3*gamma(13/3))
\[ \int \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^2 \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )}^{2} \,d x } \]
\[ \int \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^2 \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )}^{2} \,d x } \]
Timed out. \[ \int \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^2 \, dx=\int {\left (b\,x^3+a\right )}^{4/3}\,{\left (d\,x^3+c\right )}^2 \,d x \]