3.134 \(\int (a+b x^3)^2 (c+d x^3)^q \, dx\)

Optimal. Leaf size=167 \[ \frac {x \left (c+d x^3\right )^{q+1} \left (a^2 d^2 \left (9 q^2+33 q+28\right )-2 a b c d (3 q+7)+4 b^2 c^2\right ) \, _2F_1\left (1,q+\frac {4}{3};\frac {4}{3};-\frac {d x^3}{c}\right )}{c d^2 (3 q+4) (3 q+7)}-\frac {b x \left (c+d x^3\right )^{q+1} (4 b c-a d (3 q+10))}{d^2 (3 q+4) (3 q+7)}+\frac {b x \left (a+b x^3\right ) \left (c+d x^3\right )^{q+1}}{d (3 q+7)} \]

[Out]

-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)

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 176, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {416, 388, 246, 245} \[ \frac {x \left (c+d x^3\right )^q \left (\frac {d x^3}{c}+1\right )^{-q} \left (a^2 d^2 \left (9 q^2+33 q+28\right )-2 a b c d (3 q+7)+4 b^2 c^2\right ) \, _2F_1\left (\frac {1}{3},-q;\frac {4}{3};-\frac {d x^3}{c}\right )}{d^2 (3 q+4) (3 q+7)}-\frac {b x \left (c+d x^3\right )^{q+1} (4 b c-a d (3 q+10))}{d^2 (3 q+4) (3 q+7)}+\frac {b x \left (a+b x^3\right ) \left (c+d x^3\right )^{q+1}}{d (3 q+7)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^2*(c + d*x^3)^q,x]

[Out]

-((b*(4*b*c - a*d*(10 + 3*q))*x*(c + d*x^3)^(1 + q))/(d^2*(4 + 3*q)*(7 + 3*q))) + (b*x*(a + b*x^3)*(c + d*x^3)
^(1 + q))/(d*(7 + 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*Hyp
ergeometric2F1[1/3, -q, 4/3, -((d*x^3)/c)])/(d^2*(4 + 3*q)*(7 + 3*q)*(1 + (d*x^3)/c)^q)

Rule 245

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 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[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 388

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]

Rule 416

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] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[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] /; F
reeQ[{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] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rubi steps

\begin {align*} \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^q \, dx &=\frac {b x \left (a+b x^3\right ) \left (c+d x^3\right )^{1+q}}{d (7+3 q)}+\frac {\int \left (c+d x^3\right )^q \left (-a (b c-a d (7+3 q))-b (4 b c-a d (10+3 q)) x^3\right ) \, dx}{d (7+3 q)}\\ &=-\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 ) \int \left (c+d x^3\right )^q \, dx}{d^2 (4+3 q) (7+3 q)}\\ &=-\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 (\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 ) \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}{d^2 (4+3 q) (7+3 q)}\\ &=-\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 )^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 )}{d^2 (4+3 q) (7+3 q)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 106, normalized size = 0.63 \[ \frac {1}{14} x \left (c+d x^3\right )^q \left (\frac {d x^3}{c}+1\right )^{-q} \left (14 a^2 \, _2F_1\left (\frac {1}{3},-q;\frac {4}{3};-\frac {d x^3}{c}\right )+b x^3 \left (7 a \, _2F_1\left (\frac {4}{3},-q;\frac {7}{3};-\frac {d x^3}{c}\right )+2 b x^3 \, _2F_1\left (\frac {7}{3},-q;\frac {10}{3};-\frac {d x^3}{c}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^2*(c + d*x^3)^q,x]

[Out]

(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*Hypergeometric2F1[7/3, -q, 10/3, -((d*x^3)/c)])))/(14*(1 + (d*x^3)/c)^q)

________________________________________________________________________________________

fricas [F]  time = 1.16, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )} {\left (d x^{3} + c\right )}^{q}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(d*x^3+c)^q,x, algorithm="fricas")

[Out]

integral((b^2*x^6 + 2*a*b*x^3 + a^2)*(d*x^3 + c)^q, x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{3} + a\right )}^{2} {\left (d x^{3} + c\right )}^{q}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(d*x^3+c)^q,x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^2*(d*x^3 + c)^q, x)

________________________________________________________________________________________

maple [F]  time = 0.53, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{3}+a \right )^{2} \left (d \,x^{3}+c \right )^{q}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^2*(d*x^3+c)^q,x)

[Out]

int((b*x^3+a)^2*(d*x^3+c)^q,x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{3} + a\right )}^{2} {\left (d x^{3} + c\right )}^{q}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(d*x^3+c)^q,x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^2*(d*x^3 + c)^q, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,x^3+a\right )}^2\,{\left (d\,x^3+c\right )}^q \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^2*(c + d*x^3)^q,x)

[Out]

int((a + b*x^3)^2*(c + d*x^3)^q, x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**2*(d*x**3+c)**q,x)

[Out]

Timed out

________________________________________________________________________________________