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

Optimal. Leaf size=99 \[ \frac{a^2 d (a+b x)^{n+1}}{b^3 (n+1)}-\frac{2 a d (a+b x)^{n+2}}{b^3 (n+2)}+\frac{d (a+b x)^{n+3}}{b^3 (n+3)}-\frac{c (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{b x}{a}+1\right )}{a (n+1)} \]

[Out]

(a^2*d*(a + b*x)^(1 + n))/(b^3*(1 + n)) - (2*a*d*(a + b*x)^(2 + n))/(b^3*(2 + n)) + (d*(a + b*x)^(3 + n))/(b^3
*(3 + n)) - (c*(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 = 0.0578469, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {1620, 65} \[ \frac{a^2 d (a+b x)^{n+1}}{b^3 (n+1)}-\frac{2 a d (a+b x)^{n+2}}{b^3 (n+2)}+\frac{d (a+b x)^{n+3}}{b^3 (n+3)}-\frac{c (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{b x}{a}+1\right )}{a (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^n*(c + d*x^3))/x,x]

[Out]

(a^2*d*(a + b*x)^(1 + n))/(b^3*(1 + n)) - (2*a*d*(a + b*x)^(2 + n))/(b^3*(2 + n)) + (d*(a + b*x)^(3 + n))/(b^3
*(3 + n)) - (c*(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 )}{x} \, dx &=\int \left (\frac{a^2 d (a+b x)^n}{b^2}+\frac{c (a+b x)^n}{x}-\frac{2 a d (a+b x)^{1+n}}{b^2}+\frac{d (a+b x)^{2+n}}{b^2}\right ) \, dx\\ &=\frac{a^2 d (a+b x)^{1+n}}{b^3 (1+n)}-\frac{2 a d (a+b x)^{2+n}}{b^3 (2+n)}+\frac{d (a+b x)^{3+n}}{b^3 (3+n)}+c \int \frac{(a+b x)^n}{x} \, dx\\ &=\frac{a^2 d (a+b x)^{1+n}}{b^3 (1+n)}-\frac{2 a d (a+b x)^{2+n}}{b^3 (2+n)}+\frac{d (a+b x)^{3+n}}{b^3 (3+n)}-\frac{c (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 = 0.056102, size = 94, normalized size = 0.95 \[ \frac{(a+b x)^{n+1} \left (a d \left (2 a^2-2 a b (n+1) x+b^2 \left (n^2+3 n+2\right ) x^2\right )-b^3 c \left (n^2+5 n+6\right ) \, _2F_1\left (1,n+1;n+2;\frac{b x}{a}+1\right )\right )}{a b^3 (n+1) (n+2) (n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(1 + n)*(a*d*(2*a^2 - 2*a*b*(1 + n)*x + b^2*(2 + 3*n + n^2)*x^2) - b^3*c*(6 + 5*n + n^2)*Hypergeome
tric2F1[1, 1 + n, 2 + n, 1 + (b*x)/a]))/(a*b^3*(1 + n)*(2 + n)*(3 + 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 ) }{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((b*x+a)^n*(d*x^3+c)/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 )}{\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)/x,x, algorithm="maxima")

[Out]

integrate((d*x^3 + c)*(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 x^{3} + c\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)/x,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 5.25384, size = 741, normalized size = 7.48 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**n*c*n*(a/b + x)**n*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/gamma(n + 2) - b**n*c*(a/b + x)**n*lerchphi(
1 + b*x/a, 1, n + 1)*gamma(n + 1)/gamma(n + 2) + d*Piecewise((a**n*x**3/3, Eq(b, 0)), (2*a**2*log(a/b + x)/(2*
a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 3*a**2/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 4*a*b*x*log(a/b + x)
/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 4*a*b*x/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 2*b**2*x**2*log
(a/b + x)/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2), Eq(n, -3)), (-2*a**2*log(a/b + x)/(a*b**3 + b**4*x) - 2*a*
*2/(a*b**3 + b**4*x) - 2*a*b*x*log(a/b + x)/(a*b**3 + b**4*x) + b**2*x**2/(a*b**3 + b**4*x), Eq(n, -2)), (a**2
*log(a/b + x)/b**3 - a*x/b**2 + x**2/(2*b), Eq(n, -1)), (2*a**3*(a + b*x)**n/(b**3*n**3 + 6*b**3*n**2 + 11*b**
3*n + 6*b**3) - 2*a**2*b*n*x*(a + b*x)**n/(b**3*n**3 + 6*b**3*n**2 + 11*b**3*n + 6*b**3) + a*b**2*n**2*x**2*(a
+ b*x)**n/(b**3*n**3 + 6*b**3*n**2 + 11*b**3*n + 6*b**3) + a*b**2*n*x**2*(a + b*x)**n/(b**3*n**3 + 6*b**3*n**
2 + 11*b**3*n + 6*b**3) + b**3*n**2*x**3*(a + b*x)**n/(b**3*n**3 + 6*b**3*n**2 + 11*b**3*n + 6*b**3) + 3*b**3*
n*x**3*(a + b*x)**n/(b**3*n**3 + 6*b**3*n**2 + 11*b**3*n + 6*b**3) + 2*b**3*x**3*(a + b*x)**n/(b**3*n**3 + 6*b
**3*n**2 + 11*b**3*n + 6*b**3), True)) - b*b**n*c*n*x*(a/b + x)**n*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/
(a*gamma(n + 2)) - b*b**n*c*x*(a/b + x)**n*lerchphi(1 + b*x/a, 1, n + 1)*gamma(n + 1)/(a*gamma(n + 2))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{3} + c\right )}{\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)/x,x, algorithm="giac")

[Out]

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