3.3059 \(\int (a+b x)^m (c+d x)^{-m} (e+f x)^p \, dx\)

Optimal. Leaf size=121 \[ \frac {(a+b x)^{m+1} (c+d x)^{-m} (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+1;m,-p;m+2;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {140, 139, 138} \[ \frac {(a+b x)^{m+1} (c+d x)^{-m} (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+1;m,-p;m+2;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 138

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

Rule 139

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

Rule 140

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

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 119, normalized size = 0.98 \[ \frac {(a+b x)^{m+1} (c+d x)^{-m} (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+1;m,-p;m+2;\frac {d (a+b x)}{a d-b c},\frac {f (a+b x)}{a f-b e}\right )}{b (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{m} {\left (f x + e\right )}^{p}}{{\left (d x + c\right )}^{m}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x + a)^m*(f*x + e)^p/(d*x + c)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{m} {\left (f x + e\right )}^{p}}{{\left (d x + c\right )}^{m}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int \left (b x +a \right )^{m} \left (d x +c \right )^{-m} \left (f x +e \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{m} {\left (f x + e\right )}^{p}}{{\left (d x + c\right )}^{m}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e+f\,x\right )}^p\,{\left (a+b\,x\right )}^m}{{\left (c+d\,x\right )}^m} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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