3.185 \(\int \frac{(a+b x)^n (c+d x^3)^3}{x} \, dx\)

Optimal. Leaf size=358 \[ \frac{a^2 d \left (-3 a^3 b^3 c d+a^6 d^2+3 b^6 c^2\right ) (a+b x)^{n+1}}{b^9 (n+1)}-\frac{a d \left (-15 a^3 b^3 c d+8 a^6 d^2+6 b^6 c^2\right ) (a+b x)^{n+2}}{b^9 (n+2)}+\frac{d \left (-30 a^3 b^3 c d+28 a^6 d^2+3 b^6 c^2\right ) (a+b x)^{n+3}}{b^9 (n+3)}+\frac{2 a^2 d^2 \left (15 b^3 c-28 a^3 d\right ) (a+b x)^{n+4}}{b^9 (n+4)}-\frac{5 a d^2 \left (3 b^3 c-14 a^3 d\right ) (a+b x)^{n+5}}{b^9 (n+5)}+\frac{d^2 \left (3 b^3 c-56 a^3 d\right ) (a+b x)^{n+6}}{b^9 (n+6)}+\frac{28 a^2 d^3 (a+b x)^{n+7}}{b^9 (n+7)}-\frac{8 a d^3 (a+b x)^{n+8}}{b^9 (n+8)}+\frac{d^3 (a+b x)^{n+9}}{b^9 (n+9)}-\frac{c^3 (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{b x}{a}+1\right )}{a (n+1)} \]

[Out]

(a^2*d*(3*b^6*c^2 - 3*a^3*b^3*c*d + a^6*d^2)*(a + b*x)^(1 + n))/(b^9*(1 + n)) - (a*d*(6*b^6*c^2 - 15*a^3*b^3*c
*d + 8*a^6*d^2)*(a + b*x)^(2 + n))/(b^9*(2 + n)) + (d*(3*b^6*c^2 - 30*a^3*b^3*c*d + 28*a^6*d^2)*(a + b*x)^(3 +
n))/(b^9*(3 + n)) + (2*a^2*d^2*(15*b^3*c - 28*a^3*d)*(a + b*x)^(4 + n))/(b^9*(4 + n)) - (5*a*d^2*(3*b^3*c - 1
4*a^3*d)*(a + b*x)^(5 + n))/(b^9*(5 + n)) + (d^2*(3*b^3*c - 56*a^3*d)*(a + b*x)^(6 + n))/(b^9*(6 + n)) + (28*a
^2*d^3*(a + b*x)^(7 + n))/(b^9*(7 + n)) - (8*a*d^3*(a + b*x)^(8 + n))/(b^9*(8 + n)) + (d^3*(a + b*x)^(9 + n))/
(b^9*(9 + n)) - (c^3*(a + b*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (b*x)/a])/(a*(1 + n))

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Rubi [A]  time = 0.224214, antiderivative size = 358, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {1620, 65} \[ \frac{a^2 d \left (-3 a^3 b^3 c d+a^6 d^2+3 b^6 c^2\right ) (a+b x)^{n+1}}{b^9 (n+1)}-\frac{a d \left (-15 a^3 b^3 c d+8 a^6 d^2+6 b^6 c^2\right ) (a+b x)^{n+2}}{b^9 (n+2)}+\frac{d \left (-30 a^3 b^3 c d+28 a^6 d^2+3 b^6 c^2\right ) (a+b x)^{n+3}}{b^9 (n+3)}+\frac{2 a^2 d^2 \left (15 b^3 c-28 a^3 d\right ) (a+b x)^{n+4}}{b^9 (n+4)}-\frac{5 a d^2 \left (3 b^3 c-14 a^3 d\right ) (a+b x)^{n+5}}{b^9 (n+5)}+\frac{d^2 \left (3 b^3 c-56 a^3 d\right ) (a+b x)^{n+6}}{b^9 (n+6)}+\frac{28 a^2 d^3 (a+b x)^{n+7}}{b^9 (n+7)}-\frac{8 a d^3 (a+b x)^{n+8}}{b^9 (n+8)}+\frac{d^3 (a+b x)^{n+9}}{b^9 (n+9)}-\frac{c^3 (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{b x}{a}+1\right )}{a (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^n*(c + d*x^3)^3)/x,x]

[Out]

(a^2*d*(3*b^6*c^2 - 3*a^3*b^3*c*d + a^6*d^2)*(a + b*x)^(1 + n))/(b^9*(1 + n)) - (a*d*(6*b^6*c^2 - 15*a^3*b^3*c
*d + 8*a^6*d^2)*(a + b*x)^(2 + n))/(b^9*(2 + n)) + (d*(3*b^6*c^2 - 30*a^3*b^3*c*d + 28*a^6*d^2)*(a + b*x)^(3 +
n))/(b^9*(3 + n)) + (2*a^2*d^2*(15*b^3*c - 28*a^3*d)*(a + b*x)^(4 + n))/(b^9*(4 + n)) - (5*a*d^2*(3*b^3*c - 1
4*a^3*d)*(a + b*x)^(5 + n))/(b^9*(5 + n)) + (d^2*(3*b^3*c - 56*a^3*d)*(a + b*x)^(6 + n))/(b^9*(6 + n)) + (28*a
^2*d^3*(a + b*x)^(7 + n))/(b^9*(7 + n)) - (8*a*d^3*(a + b*x)^(8 + n))/(b^9*(8 + n)) + (d^3*(a + b*x)^(9 + n))/
(b^9*(9 + n)) - (c^3*(a + b*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (b*x)/a])/(a*(1 + n))

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

\begin{align*} \int \frac{(a+b x)^n \left (c+d x^3\right )^3}{x} \, dx &=\int \left (\frac{a^2 d \left (3 b^6 c^2-3 a^3 b^3 c d+a^6 d^2\right ) (a+b x)^n}{b^8}+\frac{c^3 (a+b x)^n}{x}-\frac{a d \left (6 b^6 c^2-15 a^3 b^3 c d+8 a^6 d^2\right ) (a+b x)^{1+n}}{b^8}+\frac{d \left (3 b^6 c^2-30 a^3 b^3 c d+28 a^6 d^2\right ) (a+b x)^{2+n}}{b^8}-\frac{2 a^2 d^2 \left (-15 b^3 c+28 a^3 d\right ) (a+b x)^{3+n}}{b^8}+\frac{5 a d^2 \left (-3 b^3 c+14 a^3 d\right ) (a+b x)^{4+n}}{b^8}+\frac{d^2 \left (3 b^3 c-56 a^3 d\right ) (a+b x)^{5+n}}{b^8}+\frac{28 a^2 d^3 (a+b x)^{6+n}}{b^8}-\frac{8 a d^3 (a+b x)^{7+n}}{b^8}+\frac{d^3 (a+b x)^{8+n}}{b^8}\right ) \, dx\\ &=\frac{a^2 d \left (3 b^6 c^2-3 a^3 b^3 c d+a^6 d^2\right ) (a+b x)^{1+n}}{b^9 (1+n)}-\frac{a d \left (6 b^6 c^2-15 a^3 b^3 c d+8 a^6 d^2\right ) (a+b x)^{2+n}}{b^9 (2+n)}+\frac{d \left (3 b^6 c^2-30 a^3 b^3 c d+28 a^6 d^2\right ) (a+b x)^{3+n}}{b^9 (3+n)}+\frac{2 a^2 d^2 \left (15 b^3 c-28 a^3 d\right ) (a+b x)^{4+n}}{b^9 (4+n)}-\frac{5 a d^2 \left (3 b^3 c-14 a^3 d\right ) (a+b x)^{5+n}}{b^9 (5+n)}+\frac{d^2 \left (3 b^3 c-56 a^3 d\right ) (a+b x)^{6+n}}{b^9 (6+n)}+\frac{28 a^2 d^3 (a+b x)^{7+n}}{b^9 (7+n)}-\frac{8 a d^3 (a+b x)^{8+n}}{b^9 (8+n)}+\frac{d^3 (a+b x)^{9+n}}{b^9 (9+n)}+c^3 \int \frac{(a+b x)^n}{x} \, dx\\ &=\frac{a^2 d \left (3 b^6 c^2-3 a^3 b^3 c d+a^6 d^2\right ) (a+b x)^{1+n}}{b^9 (1+n)}-\frac{a d \left (6 b^6 c^2-15 a^3 b^3 c d+8 a^6 d^2\right ) (a+b x)^{2+n}}{b^9 (2+n)}+\frac{d \left (3 b^6 c^2-30 a^3 b^3 c d+28 a^6 d^2\right ) (a+b x)^{3+n}}{b^9 (3+n)}+\frac{2 a^2 d^2 \left (15 b^3 c-28 a^3 d\right ) (a+b x)^{4+n}}{b^9 (4+n)}-\frac{5 a d^2 \left (3 b^3 c-14 a^3 d\right ) (a+b x)^{5+n}}{b^9 (5+n)}+\frac{d^2 \left (3 b^3 c-56 a^3 d\right ) (a+b x)^{6+n}}{b^9 (6+n)}+\frac{28 a^2 d^3 (a+b x)^{7+n}}{b^9 (7+n)}-\frac{8 a d^3 (a+b x)^{8+n}}{b^9 (8+n)}+\frac{d^3 (a+b x)^{9+n}}{b^9 (9+n)}-\frac{c^3 (a+b x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac{b x}{a}\right )}{a (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.345472, size = 332, normalized size = 0.93 \[ (a+b x)^{n+1} \left (\frac{d (a+b x)^2 \left (-30 a^3 b^3 c d+28 a^6 d^2+3 b^6 c^2\right )}{b^9 (n+3)}-\frac{a d (a+b x) \left (-15 a^3 b^3 c d+8 a^6 d^2+6 b^6 c^2\right )}{b^9 (n+2)}+\frac{a^2 d \left (-3 a^3 b^3 c d+a^6 d^2+3 b^6 c^2\right )}{b^9 (n+1)}+\frac{d^2 (a+b x)^5 \left (3 b^3 c-56 a^3 d\right )}{b^9 (n+6)}+\frac{5 a d^2 (a+b x)^4 \left (14 a^3 d-3 b^3 c\right )}{b^9 (n+5)}+\frac{2 a^2 d^2 (a+b x)^3 \left (15 b^3 c-28 a^3 d\right )}{b^9 (n+4)}+\frac{28 a^2 d^3 (a+b x)^6}{b^9 (n+7)}+\frac{d^3 (a+b x)^8}{b^9 (n+9)}-\frac{8 a d^3 (a+b x)^7}{b^9 (n+8)}-\frac{c^3 \, _2F_1\left (1,n+1;n+2;\frac{a+b x}{a}\right )}{a n+a}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*x)^(1 + n)*((a^2*d*(3*b^6*c^2 - 3*a^3*b^3*c*d + a^6*d^2))/(b^9*(1 + n)) - (a*d*(6*b^6*c^2 - 15*a^3*b^3*
c*d + 8*a^6*d^2)*(a + b*x))/(b^9*(2 + n)) + (d*(3*b^6*c^2 - 30*a^3*b^3*c*d + 28*a^6*d^2)*(a + b*x)^2)/(b^9*(3
+ n)) + (2*a^2*d^2*(15*b^3*c - 28*a^3*d)*(a + b*x)^3)/(b^9*(4 + n)) + (5*a*d^2*(-3*b^3*c + 14*a^3*d)*(a + b*x)
^4)/(b^9*(5 + n)) + (d^2*(3*b^3*c - 56*a^3*d)*(a + b*x)^5)/(b^9*(6 + n)) + (28*a^2*d^3*(a + b*x)^6)/(b^9*(7 +
n)) - (8*a*d^3*(a + b*x)^7)/(b^9*(8 + n)) + (d^3*(a + b*x)^8)/(b^9*(9 + n)) - (c^3*Hypergeometric2F1[1, 1 + n,
2 + n, (a + b*x)/a])/(a + a*n))

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Maple [F]  time = 0.027, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{n} \left ( d{x}^{3}+c \right ) ^{3}}{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (d^{3} x^{9} + 3 \, c d^{2} x^{6} + 3 \, c^{2} d x^{3} + c^{3}\right )}{\left (b x + a\right )}^{n}}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d^3*x^9 + 3*c*d^2*x^6 + 3*c^2*d*x^3 + c^3)*(b*x + a)^n/x, 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+a)**n*(d*x**3+c)**3/x,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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