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*(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)^(1 +
q)*Hypergeometric2F1[1, 4/3 + q, 4/3, -((d*x^3)/c)])/(c*d^2*(4 + 3*q)*(7 + 3*q))

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Rubi [A]  time = 0.125828, antiderivative size = 176, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, 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 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]

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 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 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])

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 = 0.0506743, 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)

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Maple [F]  time = 0.431, size = 0, normalized size = 0. \begin{align*} \int \left ( b{x}^{3}+a \right ) ^{2} \left ( d{x}^{3}+c \right ) ^{q}\, dx \end{align*}

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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{3} + a\right )}^{2}{\left (d x^{3} + c\right )}^{q}\,{d x} \end{align*}

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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{3} + a\right )}^{2}{\left (d x^{3} + c\right )}^{q}\,{d x} \end{align*}

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)