3.3082 \(\int (a+b x)^m (c+d x)^{-2-m} (e+f x) \, dx\)

Optimal. Leaf size=124 \[ \frac{(a+b x)^{m+1} (d e-c f) (c+d x)^{-m-1}}{d (m+1) (b c-a d)}-\frac{f (a+b x)^m (c+d x)^{-m} \left (-\frac{d (a+b x)}{b c-a d}\right )^{-m} \, _2F_1\left (-m,-m;1-m;\frac{b (c+d x)}{b c-a d}\right )}{d^2 m} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0571682, antiderivative size = 124, normalized size of antiderivative = 1., 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 = {79, 70, 69} \[ \frac{(a+b x)^{m+1} (d e-c f) (c+d x)^{-m-1}}{d (m+1) (b c-a d)}-\frac{f (a+b x)^m (c+d x)^{-m} \left (-\frac{d (a+b x)}{b c-a d}\right )^{-m} \, _2F_1\left (-m,-m;1-m;\frac{b (c+d x)}{b c-a d}\right )}{d^2 m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 79

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

Rule 70

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 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), 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]))

Rubi steps

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

Mathematica [A]  time = 0.155028, size = 114, normalized size = 0.92 \[ \frac{(a+b x)^m (c+d x)^{-m} \left (\frac{d (a+b x) (d e-c f)}{(m+1) (c+d x) (b c-a d)}-\frac{f \left (\frac{d (a+b x)}{a d-b c}\right )^{-m} \, _2F_1\left (-m,-m;1-m;\frac{b (c+d x)}{b c-a d}\right )}{m}\right )}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.047, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{-2-m} \left ( fx+e \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (f x + e\right )}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (f x + e\right )}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (f x + e\right )}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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