Integrand size = 19, antiderivative size = 167 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^q \, dx=-\frac {b (4 b c-a d (10+3 q)) x \left (c+d x^3\right )^{1+q}}{d^2 (4+3 q) (7+3 q)}+\frac {b x \left (a+b x^3\right ) \left (c+d x^3\right )^{1+q}}{d (7+3 q)}+\frac {\left (4 b^2 c^2-2 a b c d (7+3 q)+a^2 d^2 \left (28+33 q+9 q^2\right )\right ) x \left (c+d x^3\right )^{1+q} \operatorname {Hypergeometric2F1}\left (1,\frac {4}{3}+q,\frac {4}{3},-\frac {d x^3}{c}\right )}{c d^2 (4+3 q) (7+3 q)} \]
-b*(4*b*c-a*d*(10+3*q))*x*(d*x^3+c)^(1+q)/d^2/(9*q^2+33*q+28)+b*x*(b*x^3+a )*(d*x^3+c)^(1+q)/d/(7+3*q)+(4*b^2*c^2-2*a*b*c*d*(7+3*q)+a^2*d^2*(9*q^2+33 *q+28))*x*(d*x^3+c)^(1+q)*hypergeom([1, 4/3+q],[4/3],-d*x^3/c)/c/d^2/(9*q^ 2+33*q+28)
Time = 0.21 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.63 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^q \, dx=\frac {1}{14} x \left (c+d x^3\right )^q \left (1+\frac {d x^3}{c}\right )^{-q} \left (14 a^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-q,\frac {4}{3},-\frac {d x^3}{c}\right )+b x^3 \left (7 a \operatorname {Hypergeometric2F1}\left (\frac {4}{3},-q,\frac {7}{3},-\frac {d x^3}{c}\right )+2 b x^3 \operatorname {Hypergeometric2F1}\left (\frac {7}{3},-q,\frac {10}{3},-\frac {d x^3}{c}\right )\right )\right ) \]
(x*(c + d*x^3)^q*(14*a^2*Hypergeometric2F1[1/3, -q, 4/3, -((d*x^3)/c)] + b *x^3*(7*a*Hypergeometric2F1[4/3, -q, 7/3, -((d*x^3)/c)] + 2*b*x^3*Hypergeo metric2F1[7/3, -q, 10/3, -((d*x^3)/c)])))/(14*(1 + (d*x^3)/c)^q)
Time = 0.34 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {933, 25, 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.
\(\displaystyle \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^q \, dx\) |
\(\Big \downarrow \) 933 |
\(\displaystyle \frac {\int -\left (d x^3+c\right )^q \left (b (4 b c-a d (3 q+10)) x^3+a (b c-a d (3 q+7))\right )dx}{d (3 q+7)}+\frac {b x \left (a+b x^3\right ) \left (c+d x^3\right )^{q+1}}{d (3 q+7)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b x \left (a+b x^3\right ) \left (c+d x^3\right )^{q+1}}{d (3 q+7)}-\frac {\int \left (d x^3+c\right )^q \left (b (4 b c-a d (3 q+10)) x^3+a (b c-a d (3 q+7))\right )dx}{d (3 q+7)}\) |
\(\Big \downarrow \) 913 |
\(\displaystyle \frac {b x \left (a+b x^3\right ) \left (c+d x^3\right )^{q+1}}{d (3 q+7)}-\frac {\frac {b x \left (c+d x^3\right )^{q+1} (4 b c-a d (3 q+10))}{d (3 q+4)}-\frac {\left (a^2 d^2 \left (9 q^2+33 q+28\right )-2 a b c d (3 q+7)+4 b^2 c^2\right ) \int \left (d x^3+c\right )^qdx}{d (3 q+4)}}{d (3 q+7)}\) |
\(\Big \downarrow \) 779 |
\(\displaystyle \frac {b x \left (a+b x^3\right ) \left (c+d x^3\right )^{q+1}}{d (3 q+7)}-\frac {\frac {b x \left (c+d x^3\right )^{q+1} (4 b c-a d (3 q+10))}{d (3 q+4)}-\frac {\left (c+d x^3\right )^q \left (\frac {d x^3}{c}+1\right )^{-q} \left (a^2 d^2 \left (9 q^2+33 q+28\right )-2 a b c d (3 q+7)+4 b^2 c^2\right ) \int \left (\frac {d x^3}{c}+1\right )^qdx}{d (3 q+4)}}{d (3 q+7)}\) |
\(\Big \downarrow \) 778 |
\(\displaystyle \frac {b x \left (a+b x^3\right ) \left (c+d x^3\right )^{q+1}}{d (3 q+7)}-\frac {\frac {b x \left (c+d x^3\right )^{q+1} (4 b c-a d (3 q+10))}{d (3 q+4)}-\frac {x \left (c+d x^3\right )^q \left (\frac {d x^3}{c}+1\right )^{-q} \left (a^2 d^2 \left (9 q^2+33 q+28\right )-2 a b c d (3 q+7)+4 b^2 c^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-q,\frac {4}{3},-\frac {d x^3}{c}\right )}{d (3 q+4)}}{d (3 q+7)}\) |
(b*x*(a + b*x^3)*(c + d*x^3)^(1 + q))/(d*(7 + 3*q)) - ((b*(4*b*c - a*d*(10 + 3*q))*x*(c + d*x^3)^(1 + q))/(d*(4 + 3*q)) - ((4*b^2*c^2 - 2*a*b*c*d*(7 + 3*q) + a^2*d^2*(28 + 33*q + 9*q^2))*x*(c + d*x^3)^q*Hypergeometric2F1[1 /3, -q, 4/3, -((d*x^3)/c)])/(d*(4 + 3*q)*(1 + (d*x^3)/c)^q))/(d*(7 + 3*q))
3.2.34.3.1 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]
\[\int \left (b \,x^{3}+a \right )^{2} \left (d \,x^{3}+c \right )^{q}d x\]
\[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^q \, dx=\int { {\left (b x^{3} + a\right )}^{2} {\left (d x^{3} + c\right )}^{q} \,d x } \]
Result contains complex when optimal does not.
Time = 90.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.72 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^q \, dx=\frac {a^{2} c^{q} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, - q \\ \frac {4}{3} \end {matrix}\middle | {\frac {d x^{3} e^{i \pi }}{c}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {2 a b c^{q} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, - q \\ \frac {7}{3} \end {matrix}\middle | {\frac {d x^{3} e^{i \pi }}{c}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {b^{2} c^{q} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{3}, - q \\ \frac {10}{3} \end {matrix}\middle | {\frac {d x^{3} e^{i \pi }}{c}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \]
a**2*c**q*x*gamma(1/3)*hyper((1/3, -q), (4/3,), d*x**3*exp_polar(I*pi)/c)/ (3*gamma(4/3)) + 2*a*b*c**q*x**4*gamma(4/3)*hyper((4/3, -q), (7/3,), d*x** 3*exp_polar(I*pi)/c)/(3*gamma(7/3)) + b**2*c**q*x**7*gamma(7/3)*hyper((7/3 , -q), (10/3,), d*x**3*exp_polar(I*pi)/c)/(3*gamma(10/3))
\[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^q \, dx=\int { {\left (b x^{3} + a\right )}^{2} {\left (d x^{3} + c\right )}^{q} \,d x } \]
\[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^q \, dx=\int { {\left (b x^{3} + a\right )}^{2} {\left (d x^{3} + c\right )}^{q} \,d x } \]
Timed out. \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^q \, dx=\int {\left (b\,x^3+a\right )}^2\,{\left (d\,x^3+c\right )}^q \,d x \]