3.31.63 \(\int (a+b x)^m (c+d x)^{-m} (e+f x) \, dx\) [3063]

Optimal. Leaf size=135 \[ \frac {f (a+b x)^{1+m} (c+d x)^{1-m}}{2 b d}-\frac {(a d f (1-m)-b (2 d e-c f (1+m))) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{2 b^2 d (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 134, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {81, 72, 71} \begin {gather*} \frac {(a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m (-a d f (1-m)-b c f (m+1)+2 b d e) \, _2F_1\left (m,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right )}{2 b^2 d (m+1)}+\frac {f (a+b x)^{m+1} (c+d x)^{1-m}}{2 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), 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*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rubi steps

\begin {align*} \int (a+b x)^m (c+d x)^{-m} (e+f x) \, dx &=\frac {f (a+b x)^{1+m} (c+d x)^{1-m}}{2 b d}+\frac {(2 b d e-f (a d (1-m)+b c (1+m))) \int (a+b x)^m (c+d x)^{-m} \, dx}{2 b d}\\ &=\frac {f (a+b x)^{1+m} (c+d x)^{1-m}}{2 b d}+\frac {\left ((2 b d e-f (a d (1-m)+b c (1+m))) (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} \, dx}{2 b d}\\ &=\frac {f (a+b x)^{1+m} (c+d x)^{1-m}}{2 b d}+\frac {(2 b d e-a d f (1-m)-b c f (1+m)) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{2 b^2 d (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 109, normalized size = 0.81 \begin {gather*} \frac {(a+b x)^{1+m} (c+d x)^{-m} \left (b f (c+d x)-\frac {(-2 b d e-a d f (-1+m)+b c f (1+m)) \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,1+m;2+m;\frac {d (a+b x)}{-b c+a d}\right )}{1+m}\right )}{2 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (b x +a \right )^{m} \left (f x +e \right ) \left (d x +c \right )^{-m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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