3.117 \(\int \frac{(e+f x)^n}{x (a+b x) (c+d x)} \, dx\)

Optimal. Leaf size=175 \[ \frac{b^2 (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{b (e+f x)}{b e-a f}\right )}{a (n+1) (b c-a d) (b e-a f)}-\frac{d^2 (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{d (e+f x)}{d e-c f}\right )}{c (n+1) (b c-a d) (d e-c f)}-\frac{(e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{f x}{e}+1\right )}{a c e (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.119487, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {180, 65, 68} \[ \frac{b^2 (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{b (e+f x)}{b e-a f}\right )}{a (n+1) (b c-a d) (b e-a f)}-\frac{d^2 (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{d (e+f x)}{d e-c f}\right )}{c (n+1) (b c-a d) (d e-c f)}-\frac{(e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{f x}{e}+1\right )}{a c e (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^n/(x*(a + b*x)*(c + d*x)),x]

[Out]

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

Rule 180

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

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])

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{(e+f x)^n}{x (a+b x) (c+d x)} \, dx &=\int \left (\frac{(e+f x)^n}{a c x}+\frac{b^2 (e+f x)^n}{a (-b c+a d) (a+b x)}+\frac{d^2 (e+f x)^n}{c (b c-a d) (c+d x)}\right ) \, dx\\ &=\frac{\int \frac{(e+f x)^n}{x} \, dx}{a c}-\frac{b^2 \int \frac{(e+f x)^n}{a+b x} \, dx}{a (b c-a d)}+\frac{d^2 \int \frac{(e+f x)^n}{c+d x} \, dx}{c (b c-a d)}\\ &=\frac{b^2 (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac{b (e+f x)}{b e-a f}\right )}{a (b c-a d) (b e-a f) (1+n)}-\frac{d^2 (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac{d (e+f x)}{d e-c f}\right )}{c (b c-a d) (d e-c f) (1+n)}-\frac{(e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac{f x}{e}\right )}{a c e (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.183661, size = 170, normalized size = 0.97 \[ -\frac{(e+f x)^{n+1} \left (b^2 c e (d e-c f) \, _2F_1\left (1,n+1;n+2;\frac{b (e+f x)}{b e-a f}\right )+(a f-b e) \left (a d^2 e \, _2F_1\left (1,n+1;n+2;\frac{d (e+f x)}{d e-c f}\right )-(b c-a d) (c f-d e) \, _2F_1\left (1,n+1;n+2;\frac{f x}{e}+1\right )\right )\right )}{a c e (n+1) (a d-b c) (a f-b e) (c f-d e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.052, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{n}}{x \left ( bx+a \right ) \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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