\(\int (a+b x^3) (c+d x^3)^q \, dx\) [135]

Optimal result

Integrand size = 17, antiderivative size = 84 \[ \int \left (a+b x^3\right ) \left (c+d x^3\right )^q \, dx=\frac {b x \left (c+d x^3\right )^{1+q}}{d (4+3 q)}-\frac {(b c-a d (4+3 q)) x \left (c+d x^3\right )^{1+q} \operatorname {Hypergeometric2F1}\left (1,\frac {4}{3}+q,\frac {4}{3},-\frac {d x^3}{c}\right )}{c d (4+3 q)} \]

[Out]

b*x*(d*x^3+c)^(1+q)/d/(4+3*q)-(b*c-a*d*(4+3*q))*x*(d*x^3+c)^(1+q)*hypergeom([1, 4/3+q],[4/3],-d*x^3/c)/c/d/(4+
3*q)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {396, 252, 251} \[ \int \left (a+b x^3\right ) \left (c+d x^3\right )^q \, dx=x \left (c+d x^3\right )^q \left (\frac {d x^3}{c}+1\right )^{-q} \left (a-\frac {b c}{3 d q+4 d}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-q,\frac {4}{3},-\frac {d x^3}{c}\right )+\frac {b x \left (c+d x^3\right )^{q+1}}{d (3 q+4)} \]

[In]

Int[(a + b*x^3)*(c + d*x^3)^q,x]

[Out]

(b*x*(c + d*x^3)^(1 + q))/(d*(4 + 3*q)) + ((a - (b*c)/(4*d + 3*d*q))*x*(c + d*x^3)^q*Hypergeometric2F1[1/3, -q
, 4/3, -((d*x^3)/c)])/(1 + (d*x^3)/c)^q

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-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])

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(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}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b x \left (c+d x^3\right )^{1+q}}{d (4+3 q)}-\left (-a+\frac {b c}{4 d+3 d q}\right ) \int \left (c+d x^3\right )^q \, dx \\ & = \frac {b x \left (c+d x^3\right )^{1+q}}{d (4+3 q)}-\left (\left (-a+\frac {b c}{4 d+3 d q}\right ) \left (c+d x^3\right )^q \left (1+\frac {d x^3}{c}\right )^{-q}\right ) \int \left (1+\frac {d x^3}{c}\right )^q \, dx \\ & = \frac {b x \left (c+d x^3\right )^{1+q}}{d (4+3 q)}+\left (a-\frac {b c}{4 d+3 d q}\right ) x \left (c+d x^3\right )^q \left (1+\frac {d x^3}{c}\right )^{-q} \, _2F_1\left (\frac {1}{3},-q;\frac {4}{3};-\frac {d x^3}{c}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.07 \[ \int \left (a+b x^3\right ) \left (c+d x^3\right )^q \, dx=\frac {x \left (c+d x^3\right )^q \left (1+\frac {d x^3}{c}\right )^{-q} \left (b \left (c+d x^3\right ) \left (1+\frac {d x^3}{c}\right )^q+(-b c+a d (4+3 q)) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-q,\frac {4}{3},-\frac {d x^3}{c}\right )\right )}{d (4+3 q)} \]

[In]

Integrate[(a + b*x^3)*(c + d*x^3)^q,x]

[Out]

(x*(c + d*x^3)^q*(b*(c + d*x^3)*(1 + (d*x^3)/c)^q + (-(b*c) + a*d*(4 + 3*q))*Hypergeometric2F1[1/3, -q, 4/3, -
((d*x^3)/c)]))/(d*(4 + 3*q)*(1 + (d*x^3)/c)^q)

Maple [F]

\[\int \left (b \,x^{3}+a \right ) \left (d \,x^{3}+c \right )^{q}d x\]

[In]

int((b*x^3+a)*(d*x^3+c)^q,x)

[Out]

int((b*x^3+a)*(d*x^3+c)^q,x)

Fricas [F]

\[ \int \left (a+b x^3\right ) \left (c+d x^3\right )^q \, dx=\int { {\left (b x^{3} + a\right )} {\left (d x^{3} + c\right )}^{q} \,d x } \]

[In]

integrate((b*x^3+a)*(d*x^3+c)^q,x, algorithm="fricas")

[Out]

integral((b*x^3 + a)*(d*x^3 + c)^q, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 34.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.89 \[ \int \left (a+b x^3\right ) \left (c+d x^3\right )^q \, dx=\frac {a c^{q} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, - q \\ \frac {4}{3} \end {matrix}\middle | {\frac {d x^{3} e^{i \pi }}{c}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {b c^{q} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, - q \\ \frac {7}{3} \end {matrix}\middle | {\frac {d x^{3} e^{i \pi }}{c}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} \]

[In]

integrate((b*x**3+a)*(d*x**3+c)**q,x)

[Out]

a*c**q*x*gamma(1/3)*hyper((1/3, -q), (4/3,), d*x**3*exp_polar(I*pi)/c)/(3*gamma(4/3)) + 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))

Maxima [F]

\[ \int \left (a+b x^3\right ) \left (c+d x^3\right )^q \, dx=\int { {\left (b x^{3} + a\right )} {\left (d x^{3} + c\right )}^{q} \,d x } \]

[In]

integrate((b*x^3+a)*(d*x^3+c)^q,x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)*(d*x^3 + c)^q, x)

Giac [F]

\[ \int \left (a+b x^3\right ) \left (c+d x^3\right )^q \, dx=\int { {\left (b x^{3} + a\right )} {\left (d x^{3} + c\right )}^{q} \,d x } \]

[In]

integrate((b*x^3+a)*(d*x^3+c)^q,x, algorithm="giac")

[Out]

integrate((b*x^3 + a)*(d*x^3 + c)^q, x)

Mupad [F(-1)]

Timed out. \[ \int \left (a+b x^3\right ) \left (c+d x^3\right )^q \, dx=\int \left (b\,x^3+a\right )\,{\left (d\,x^3+c\right )}^q \,d x \]

[In]

int((a + b*x^3)*(c + d*x^3)^q,x)

[Out]

int((a + b*x^3)*(c + d*x^3)^q, x)