3.3079 \(\int (a+b x)^m (c+d x)^{-2-m} (e+f x)^p \, dx\)

Optimal. Leaf size=131 \[ \frac {b (a+b x)^{m+1} (c+d x)^{-m} (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+1;m+2,-p;m+2;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{(m+1) (b c-a d)^2} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {140, 139, 138} \[ \frac {b (a+b x)^{m+1} (c+d x)^{-m} (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+1;m+2,-p;m+2;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{(m+1) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*(e + f*x)^p*AppellF1[1 + m, 2 + m, -p, 2 + m, -((d*(a + b*x
))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/((b*c - a*d)^2*(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)^{-2-m} (e+f x)^p \, dx &=\frac {\left (b^2 (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 )^{-2-m} (e+f x)^p \, dx}{(b c-a d)^2}\\ &=\frac {\left (b^2 (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 )^{-2-m} \left (\frac {b e}{b e-a f}+\frac {b f x}{b e-a f}\right )^p \, dx}{(b c-a d)^2}\\ &=\frac {b (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;2+m,-p;2+m;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{(b c-a d)^2 (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 129, normalized size = 0.98 \[ \frac {b (a+b x)^{m+1} (c+d x)^{-m} (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+1;m+2,-p;m+2;\frac {d (a+b x)}{a d-b c},\frac {f (a+b x)}{a f-b e}\right )}{(m+1) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.22, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 2} {\left (f x + e\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-2-m)*(f*x+e)^p,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 2} {\left (f x + e\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-2-m)*(f*x+e)^p,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.27, size = 0, normalized size = 0.00 \[ \int \left (b x +a \right )^{m} \left (d x +c \right )^{-m -2} \left (f x +e \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 2} {\left (f x + e\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-2-m)*(f*x+e)^p,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e+f\,x\right )}^p\,{\left (a+b\,x\right )}^m}{{\left (c+d\,x\right )}^{m+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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