3.3135 \(\int \frac{(a+b x)^m (c+d x)^{3-m}}{(e+f x)^2} \, dx\)

Optimal. Leaf size=397 \[ \frac{d^2 (a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (a^2 d^2 f^2 (1-m) m+2 a b d f m (2 d e-c f (3-m))+b^2 \left (-\left (c^2 f^2 \left (m^2-5 m+6\right )-4 c d e f (3-m)+6 d^2 e^2\right )\right )\right ) \, _2F_1\left (m,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{2 b^2 f^4 m (m+1) (b c-a d)}+\frac{d^2 (a+b x)^{m+1} (c+d x)^{1-m}}{2 b f^2}+\frac{(a+b x)^m (d e-c f)^2 (c+d x)^{-m} (a d f (3-m)-b (3 d e-c f m)) \, _2F_1\left (1,m;m+1;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{f^4 m (b e-a f)}+\frac{3 b d (a+b x)^m (d e-c f)^2 (c+d x)^{1-m}}{f^4 m (b c-a d)}-\frac{(a+b x)^m (d e-c f)^2 (c+d x)^{1-m}}{f^3 (e+f x)} \]

[Out]

(3*b*d*(d*e - c*f)^2*(a + b*x)^m*(c + d*x)^(1 - m))/((b*c - a*d)*f^4*m) + (d^2*(a + b*x)^(1 + m)*(c + d*x)^(1
- m))/(2*b*f^2) - ((d*e - c*f)^2*(a + b*x)^m*(c + d*x)^(1 - m))/(f^3*(e + f*x)) + ((d*e - c*f)^2*(a*d*f*(3 - m
) - b*(3*d*e - c*f*m))*(a + b*x)^m*Hypergeometric2F1[1, m, 1 + m, ((d*e - c*f)*(a + b*x))/((b*e - a*f)*(c + d*
x))])/(f^4*(b*e - a*f)*m*(c + d*x)^m) + (d^2*(2*a*b*d*f*(2*d*e - c*f*(3 - m))*m + a^2*d^2*f^2*(1 - m)*m - b^2*
(6*d^2*e^2 - 4*c*d*e*f*(3 - m) + c^2*f^2*(6 - 5*m + m^2)))*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*Hyp
ergeometric2F1[m, 1 + m, 2 + m, -((d*(a + b*x))/(b*c - a*d))])/(2*b^2*(b*c - a*d)*f^4*m*(1 + m)*(c + d*x)^m)

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Rubi [C]  time = 0.0535479, antiderivative size = 113, normalized size of antiderivative = 0.28, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {137, 136} \[ \frac{(b c-a d)^3 (a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m F_1\left (m+1;m-3,2;m+2;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b^2 (m+1) (b e-a f)^2} \]

Warning: Unable to verify antiderivative.

[In]

Int[((a + b*x)^m*(c + d*x)^(3 - m))/(e + f*x)^2,x]

[Out]

((b*c - a*d)^3*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*AppellF1[1 + m, -3 + m, 2, 2 + m, -((d*(a + b*x
))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b^2*(b*e - a*f)^2*(1 + m)*(c + d*x)^m)

Rule 137

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}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

Rule 136

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

Rubi steps

\begin{align*} \int \frac{(a+b x)^m (c+d x)^{3-m}}{(e+f x)^2} \, dx &=\frac{\left ((b c-a d)^3 (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m\right ) \int \frac{(a+b x)^m \left (\frac{b c}{b c-a d}+\frac{b d x}{b c-a d}\right )^{3-m}}{(e+f x)^2} \, dx}{b^3}\\ &=\frac{(b c-a d)^3 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m F_1\left (1+m;-3+m,2;2+m;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b^2 (b e-a f)^2 (1+m)}\\ \end{align*}

Mathematica [C]  time = 0.208886, size = 111, normalized size = 0.28 \[ \frac{(b c-a d)^3 (a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m F_1\left (m+1;m-3,2;m+2;\frac{d (a+b x)}{a d-b c},\frac{f (a+b x)}{a f-b e}\right )}{b^2 (m+1) (b e-a f)^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + b*x)^m*(c + d*x)^(3 - m))/(e + f*x)^2,x]

[Out]

((b*c - a*d)^3*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*AppellF1[1 + m, -3 + m, 2, 2 + m, (d*(a + b*x))
/(-(b*c) + a*d), (f*(a + b*x))/(-(b*e) + a*f)])/(b^2*(b*e - a*f)^2*(1 + m)*(c + d*x)^m)

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Maple [F]  time = 0.07, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx+c \right ) ^{3-m} \left ( bx+a \right ) ^{m}}{ \left ( fx+e \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^(3-m)/(f*x+e)^2,x)

[Out]

int((b*x+a)^m*(d*x+c)^(3-m)/(f*x+e)^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m + 3}}{{\left (f x + e\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(3-m)/(f*x+e)^2,x, algorithm="maxima")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m + 3)/(f*x + e)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m + 3}}{f^{2} x^{2} + 2 \, e f x + e^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(3-m)/(f*x+e)^2,x, algorithm="fricas")

[Out]

integral((b*x + a)^m*(d*x + c)^(-m + 3)/(f^2*x^2 + 2*e*f*x + e^2), x)

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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)**(3-m)/(f*x+e)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m + 3}}{{\left (f x + e\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(3-m)/(f*x+e)^2,x, algorithm="giac")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m + 3)/(f*x + e)^2, x)