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

Optimal. Leaf size=140 \[ \frac{b x^{m+1} (e+f x)^n \left (\frac{f x}{e}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac{f x}{e},-\frac{b x}{a}\right )}{a (m+1) (b c-a d)}-\frac{d x^{m+1} (e+f x)^n \left (\frac{f x}{e}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac{f x}{e},-\frac{d x}{c}\right )}{c (m+1) (b c-a d)} \]

[Out]

(b*x^(1 + m)*(e + f*x)^n*AppellF1[1 + m, -n, 1, 2 + m, -((f*x)/e), -((b*x)/a)])/(a*(b*c - a*d)*(1 + m)*(1 + (f
*x)/e)^n) - (d*x^(1 + m)*(e + f*x)^n*AppellF1[1 + m, -n, 1, 2 + m, -((f*x)/e), -((d*x)/c)])/(c*(b*c - a*d)*(1
+ m)*(1 + (f*x)/e)^n)

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Rubi [A]  time = 0.122773, antiderivative size = 140, 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, 135, 133} \[ \frac{b x^{m+1} (e+f x)^n \left (\frac{f x}{e}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac{f x}{e},-\frac{b x}{a}\right )}{a (m+1) (b c-a d)}-\frac{d x^{m+1} (e+f x)^n \left (\frac{f x}{e}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac{f x}{e},-\frac{d x}{c}\right )}{c (m+1) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x^(1 + m)*(e + f*x)^n*AppellF1[1 + m, -n, 1, 2 + m, -((f*x)/e), -((b*x)/a)])/(a*(b*c - a*d)*(1 + m)*(1 + (f
*x)/e)^n) - (d*x^(1 + m)*(e + f*x)^n*AppellF1[1 + m, -n, 1, 2 + m, -((f*x)/e), -((d*x)/c)])/(c*(b*c - a*d)*(1
+ m)*(1 + (f*x)/e)^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 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int \frac{x^m (e+f x)^n}{(a+b x) (c+d x)} \, dx &=\int \left (\frac{b x^m (e+f x)^n}{(b c-a d) (a+b x)}-\frac{d x^m (e+f x)^n}{(b c-a d) (c+d x)}\right ) \, dx\\ &=\frac{b \int \frac{x^m (e+f x)^n}{a+b x} \, dx}{b c-a d}-\frac{d \int \frac{x^m (e+f x)^n}{c+d x} \, dx}{b c-a d}\\ &=\frac{\left (b (e+f x)^n \left (1+\frac{f x}{e}\right )^{-n}\right ) \int \frac{x^m \left (1+\frac{f x}{e}\right )^n}{a+b x} \, dx}{b c-a d}-\frac{\left (d (e+f x)^n \left (1+\frac{f x}{e}\right )^{-n}\right ) \int \frac{x^m \left (1+\frac{f x}{e}\right )^n}{c+d x} \, dx}{b c-a d}\\ &=\frac{b x^{1+m} (e+f x)^n \left (1+\frac{f x}{e}\right )^{-n} F_1\left (1+m;-n,1;2+m;-\frac{f x}{e},-\frac{b x}{a}\right )}{a (b c-a d) (1+m)}-\frac{d x^{1+m} (e+f x)^n \left (1+\frac{f x}{e}\right )^{-n} F_1\left (1+m;-n,1;2+m;-\frac{f x}{e},-\frac{d x}{c}\right )}{c (b c-a d) (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.216623, size = 104, normalized size = 0.74 \[ \frac{x^{m+1} (e+f x)^n \left (\frac{f x}{e}+1\right )^{-n} \left (a d F_1\left (m+1;-n,1;m+2;-\frac{f x}{e},-\frac{d x}{c}\right )-b c F_1\left (m+1;-n,1;m+2;-\frac{f x}{e},-\frac{b x}{a}\right )\right )}{a c (m+1) (a d-b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*(e + f*x)^n*(-(b*c*AppellF1[1 + m, -n, 1, 2 + m, -((f*x)/e), -((b*x)/a)]) + a*d*AppellF1[1 + m, -n,
1, 2 + m, -((f*x)/e), -((d*x)/c)]))/(a*c*(-(b*c) + a*d)*(1 + m)*(1 + (f*x)/e)^n)

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Maple [F]  time = 0.066, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{n}{x}^{m}}{ \left ( bx+a \right ) \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^m*(f*x+e)^n/(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} x^{m}}{{\left (b x + a\right )}{\left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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