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

Optimal. Leaf size=222 \[ -\frac{b^3 (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{b (e+f x)}{b e-a f}\right )}{a^2 (n+1) (b c-a d) (b e-a f)}+\frac{(a d+b c) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{f x}{e}+1\right )}{a^2 c^2 e (n+1)}+\frac{d^3 (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{d (e+f x)}{d e-c f}\right )}{c^2 (n+1) (b c-a d) (d e-c f)}+\frac{f (e+f x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{f x}{e}+1\right )}{a c e^2 (n+1)} \]

[Out]

-((b^3*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (b*(e + f*x))/(b*e - a*f)])/(a^2*(b*c - a*d)*(b*e
- a*f)*(1 + n))) + (d^3*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (d*(e + f*x))/(d*e - c*f)])/(c^2*
(b*c - a*d)*(d*e - c*f)*(1 + n)) + ((b*c + a*d)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (f*x)
/e])/(a^2*c^2*e*(1 + n)) + (f*(e + f*x)^(1 + n)*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (f*x)/e])/(a*c*e^2*(1 +
n))

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Rubi [A]  time = 0.151256, antiderivative size = 222, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {180, 65, 68} \[ -\frac{b^3 (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{b (e+f x)}{b e-a f}\right )}{a^2 (n+1) (b c-a d) (b e-a f)}+\frac{(a d+b c) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{f x}{e}+1\right )}{a^2 c^2 e (n+1)}+\frac{d^3 (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{d (e+f x)}{d e-c f}\right )}{c^2 (n+1) (b c-a d) (d e-c f)}+\frac{f (e+f x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{f x}{e}+1\right )}{a c e^2 (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^3*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (b*(e + f*x))/(b*e - a*f)])/(a^2*(b*c - a*d)*(b*e
- a*f)*(1 + n))) + (d^3*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (d*(e + f*x))/(d*e - c*f)])/(c^2*
(b*c - a*d)*(d*e - c*f)*(1 + n)) + ((b*c + a*d)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (f*x)
/e])/(a^2*c^2*e*(1 + n)) + (f*(e + f*x)^(1 + n)*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (f*x)/e])/(a*c*e^2*(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^2 (a+b x) (c+d x)} \, dx &=\int \left (\frac{(e+f x)^n}{a c x^2}+\frac{(-b c-a d) (e+f x)^n}{a^2 c^2 x}-\frac{b^3 (e+f x)^n}{a^2 (-b c+a d) (a+b x)}-\frac{d^3 (e+f x)^n}{c^2 (b c-a d) (c+d x)}\right ) \, dx\\ &=\frac{\int \frac{(e+f x)^n}{x^2} \, dx}{a c}+\frac{b^3 \int \frac{(e+f x)^n}{a+b x} \, dx}{a^2 (b c-a d)}-\frac{d^3 \int \frac{(e+f x)^n}{c+d x} \, dx}{c^2 (b c-a d)}-\frac{(b c+a d) \int \frac{(e+f x)^n}{x} \, dx}{a^2 c^2}\\ &=-\frac{b^3 (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac{b (e+f x)}{b e-a f}\right )}{a^2 (b c-a d) (b e-a f) (1+n)}+\frac{d^3 (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac{d (e+f x)}{d e-c f}\right )}{c^2 (b c-a d) (d e-c f) (1+n)}+\frac{(b c+a d) (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac{f x}{e}\right )}{a^2 c^2 e (1+n)}+\frac{f (e+f x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac{f x}{e}\right )}{a c e^2 (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.267072, size = 177, normalized size = 0.8 \[ \frac{(e+f x)^{n+1} \left (\frac{\frac{e (a d+b c) \, _2F_1\left (1,n+1;n+2;\frac{f x}{e}+1\right )+a c f \, _2F_1\left (2,n+1;n+2;\frac{f x}{e}+1\right )}{a^2 e^2}-\frac{d^3 \, _2F_1\left (1,n+1;n+2;\frac{d (e+f x)}{d e-c f}\right )}{(b c-a d) (c f-d e)}}{c^2}-\frac{b^3 \, _2F_1\left (1,n+1;n+2;\frac{b (e+f x)}{b e-a f}\right )}{a^2 (b c-a d) (b e-a f)}\right )}{n+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e + f*x)^(1 + n)*(-((b^3*Hypergeometric2F1[1, 1 + n, 2 + n, (b*(e + f*x))/(b*e - a*f)])/(a^2*(b*c - a*d)*(b*
e - a*f))) + (-((d^3*Hypergeometric2F1[1, 1 + n, 2 + n, (d*(e + f*x))/(d*e - c*f)])/((b*c - a*d)*(-(d*e) + c*f
))) + ((b*c + a*d)*e*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (f*x)/e] + a*c*f*Hypergeometric2F1[2, 1 + n, 2 + n
, 1 + (f*x)/e])/(a^2*e^2))/c^2))/(1 + n)

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Maple [F]  time = 0.082, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{n}}{{x}^{2} \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^2/(b*x+a)/(d*x+c),x)

[Out]

int((f*x+e)^n/x^2/(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^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^n/x^2/(b*x+a)/(d*x+c),x, algorithm="maxima")

[Out]

integrate((f*x + e)^n/((b*x + a)*(d*x + c)*x^2), 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^{4} + a c x^{2} +{\left (b c + a d\right )} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^n/x^2/(b*x+a)/(d*x+c),x, algorithm="fricas")

[Out]

integral((f*x + e)^n/(b*d*x^4 + a*c*x^2 + (b*c + a*d)*x^3), 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**2/(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^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^n/x^2/(b*x+a)/(d*x+c),x, algorithm="giac")

[Out]

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