3.32.45 \(\int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{n+p} \, dx\) [3145]

Optimal. Leaf size=139 \[ \frac {(a+b x)^{1+m} (c+d x)^{-m-n} \left (\frac {b (c+d x)}{b c-a d}\right )^{m+n} (e+f x)^{n+p} \left (\frac {b (e+f x)}{b e-a f}\right )^{-n-p} F_1\left (1+m;m+n,-n-p;2+m;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {145, 144, 143} \begin {gather*} \frac {(a+b x)^{m+1} (c+d x)^{-m-n} (e+f x)^{n+p} \left (\frac {b (c+d x)}{b c-a d}\right )^{m+n} \left (\frac {b (e+f x)}{b e-a f}\right )^{-n-p} F_1\left (m+1;m+n,-n-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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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 144

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 145

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-n} (e+f x)^{n+p} \, dx &=\left ((c+d x)^{-m-n} \left (\frac {b (c+d x)}{b c-a d}\right )^{m+n}\right ) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{-m-n} (e+f x)^{n+p} \, dx\\ &=\left ((c+d x)^{-m-n} \left (\frac {b (c+d x)}{b c-a d}\right )^{m+n} (e+f x)^{n+p} \left (\frac {b (e+f x)}{b e-a f}\right )^{-n-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-n} \left (\frac {b e}{b e-a f}+\frac {b f x}{b e-a f}\right )^{n+p} \, dx\\ &=\frac {(a+b x)^{1+m} (c+d x)^{-m-n} \left (\frac {b (c+d x)}{b c-a d}\right )^{m+n} (e+f x)^{n+p} \left (\frac {b (e+f x)}{b e-a f}\right )^{-n-p} F_1\left (1+m;m+n,-n-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 = 0.24, size = 137, normalized size = 0.99 \begin {gather*} \frac {(a+b x)^{1+m} (c+d x)^{-m-n} \left (\frac {b (c+d x)}{b c-a d}\right )^{m+n} (e+f x)^{n+p} \left (\frac {b (e+f x)}{b e-a f}\right )^{-n-p} F_1\left (1+m;m+n,-n-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 {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \left (b x +a \right )^{m} \left (d x +c \right )^{-m -n} \left (f x +e \right )^{n +p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6438 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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