3.354 \(\int \frac{(a+b x)^n (c+d x^2)}{x} \, dx\)

Optimal. Leaf size=77 \[ -\frac{a d (a+b x)^{n+1}}{b^2 (n+1)}+\frac{d (a+b x)^{n+2}}{b^2 (n+2)}-\frac{c (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{b x}{a}+1\right )}{a (n+1)} \]

[Out]

-((a*d*(a + b*x)^(1 + n))/(b^2*(1 + n))) + (d*(a + b*x)^(2 + n))/(b^2*(2 + n)) - (c*(a + b*x)^(1 + n)*Hypergeo
metric2F1[1, 1 + n, 2 + n, 1 + (b*x)/a])/(a*(1 + n))

________________________________________________________________________________________

Rubi [A]  time = 0.0508392, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {952, 80, 65} \[ -\frac{a d (a+b x)^{n+1}}{b^2 (n+1)}+\frac{d (a+b x)^{n+2}}{b^2 (n+2)}-\frac{c (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{b x}{a}+1\right )}{a (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^n*(c + d*x^2))/x,x]

[Out]

-((a*d*(a + b*x)^(1 + n))/(b^2*(1 + n))) + (d*(a + b*x)^(2 + n))/(b^2*(2 + n)) - (c*(a + b*x)^(1 + n)*Hypergeo
metric2F1[1, 1 + n, 2 + n, 1 + (b*x)/a])/(a*(1 + n))

Rule 952

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(c^p*(d
+ e*x)^(m + 2*p)*(f + g*x)^(n + 1))/(g*e^(2*p)*(m + n + 2*p + 1)), x] + Dist[1/(g*e^(2*p)*(m + n + 2*p + 1)),
Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + c*x^2)^p - c^p*(d + e*x)^(2*p)) - c
^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0]
&& NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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^2\right )}{x} \, dx &=\frac{d (a+b x)^{2+n}}{b^2 (2+n)}+\frac{\int \frac{(a+b x)^n \left (b^2 c (2+n)-a b d (2+n) x\right )}{x} \, dx}{b^2 (2+n)}\\ &=-\frac{a d (a+b x)^{1+n}}{b^2 (1+n)}+\frac{d (a+b x)^{2+n}}{b^2 (2+n)}+c \int \frac{(a+b x)^n}{x} \, dx\\ &=-\frac{a d (a+b x)^{1+n}}{b^2 (1+n)}+\frac{d (a+b x)^{2+n}}{b^2 (2+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 = 0.0449481, size = 64, normalized size = 0.83 \[ -\frac{(a+b x)^{n+1} \left (b^2 c (n+2) \, _2F_1\left (1,n+1;n+2;\frac{b x}{a}+1\right )+a d (a-b (n+1) x)\right )}{a b^2 (n+1) (n+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^n*(c + d*x^2))/x,x]

[Out]

-(((a + b*x)^(1 + n)*(a*d*(a - b*(1 + n)*x) + b^2*c*(2 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (b*x)/a]))/
(a*b^2*(1 + n)*(2 + n)))

________________________________________________________________________________________

Maple [F]  time = 0.365, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{n} \left ( d{x}^{2}+c \right ) }{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^n*(d*x^2+c)/x,x)

[Out]

int((b*x+a)^n*(d*x^2+c)/x,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + 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^2+c)/x,x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)*(b*x + a)^n/x, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (d x^{2} + 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^2+c)/x,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [B]  time = 6.48357, size = 347, normalized size = 4.51 \begin{align*} - \frac{b^{n} c n \left (\frac{a}{b} + x\right )^{n} \Phi \left (1 + \frac{b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{\Gamma \left (n + 2\right )} - \frac{b^{n} c \left (\frac{a}{b} + x\right )^{n} \Phi \left (1 + \frac{b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{\Gamma \left (n + 2\right )} + d \left (\begin{cases} \frac{a^{n} x^{2}}{2} & \text{for}\: b = 0 \\\frac{a \log{\left (\frac{a}{b} + x \right )}}{a b^{2} + b^{3} x} + \frac{b x \log{\left (\frac{a}{b} + x \right )}}{a b^{2} + b^{3} x} - \frac{b x}{a b^{2} + b^{3} x} & \text{for}\: n = -2 \\- \frac{a \log{\left (\frac{a}{b} + x \right )}}{b^{2}} + \frac{x}{b} & \text{for}\: n = -1 \\- \frac{a^{2} \left (a + b x\right )^{n}}{b^{2} n^{2} + 3 b^{2} n + 2 b^{2}} + \frac{a b n x \left (a + b x\right )^{n}}{b^{2} n^{2} + 3 b^{2} n + 2 b^{2}} + \frac{b^{2} n x^{2} \left (a + b x\right )^{n}}{b^{2} n^{2} + 3 b^{2} n + 2 b^{2}} + \frac{b^{2} x^{2} \left (a + b x\right )^{n}}{b^{2} n^{2} + 3 b^{2} n + 2 b^{2}} & \text{otherwise} \end{cases}\right ) - \frac{b b^{n} c n x \left (\frac{a}{b} + x\right )^{n} \Phi \left (1 + \frac{b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{a \Gamma \left (n + 2\right )} - \frac{b b^{n} c x \left (\frac{a}{b} + x\right )^{n} \Phi \left (1 + \frac{b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{a \Gamma \left (n + 2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**n*(d*x**2+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**2/2, Eq(b, 0)), (a*log(a/b + x)/(a*b**2
+ b**3*x) + b*x*log(a/b + x)/(a*b**2 + b**3*x) - b*x/(a*b**2 + b**3*x), Eq(n, -2)), (-a*log(a/b + x)/b**2 + x/
b, Eq(n, -1)), (-a**2*(a + b*x)**n/(b**2*n**2 + 3*b**2*n + 2*b**2) + a*b*n*x*(a + b*x)**n/(b**2*n**2 + 3*b**2*
n + 2*b**2) + b**2*n*x**2*(a + b*x)**n/(b**2*n**2 + 3*b**2*n + 2*b**2) + b**2*x**2*(a + b*x)**n/(b**2*n**2 + 3
*b**2*n + 2*b**2), 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))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + 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^2+c)/x,x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*(b*x + a)^n/x, x)