3.3127 \(\int \frac{(a+b x)^m (c+d x)^{2-m}}{e+f x} \, dx\)

Optimal. Leaf size=370 \[ \frac{d (a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (a^2 d^2 f^2 (1-m) m+2 a b d f m (d e-c f (2-m))+b^2 \left (-\left (c^2 f^2 \left (m^2-3 m+2\right )-2 c d e f (2-m)+2 d^2 e^2\right )\right )\right ) \, _2F_1\left (m,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{2 b^2 f^3 m (m+1) (b c-a d)}-\frac{d (a+b x)^{m+1} (c+d x)^{-m} \left (-a^2 d^2 f^2 m+2 a b c d f^2 m+b^2 \left (-\left (c^2 f^2 (m+2)-4 c d e f+2 d^2 e^2\right )\right )\right )}{2 b^2 f^3 m (b c-a d)}+\frac{d^2 (a+b x)^{m+2} (c+d x)^{-m}}{2 b^2 f}+\frac{(a+b x)^m (d e-c f)^2 (c+d x)^{-m} \, _2F_1\left (1,-m;1-m;\frac{(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{f^3 m} \]

[Out]

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

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Rubi [A]  time = 0.193405, antiderivative size = 319, normalized size of antiderivative = 0.86, number of steps used = 10, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {105, 70, 69, 131} \[ \frac{d (b c-a d) (a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m-1,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{b^2 f (m+1)}-\frac{(a+b x)^m (d e-c f)^2 (c+d x)^{-m} \, _2F_1\left (1,m;m+1;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{f^3 m}+\frac{(a+b x)^m (d e-c f)^2 (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m;m+1;-\frac{d (a+b x)}{b c-a d}\right )}{f^3 m}-\frac{d (a+b x)^{m+1} (d e-c f) (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{b f^2 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -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]))

Rule 131

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*c -
a*d)^n*(a + b*x)^(m + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))
)])/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)), x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0]

Rubi steps

\begin{align*} \int \frac{(a+b x)^m (c+d x)^{2-m}}{e+f x} \, dx &=\frac{d \int (a+b x)^m (c+d x)^{1-m} \, dx}{f}-\frac{(d e-c f) \int \frac{(a+b x)^m (c+d x)^{1-m}}{e+f x} \, dx}{f}\\ &=-\frac{(d (d e-c f)) \int (a+b x)^m (c+d x)^{-m} \, dx}{f^2}+\frac{(d e-c f)^2 \int \frac{(a+b x)^m (c+d x)^{-m}}{e+f x} \, dx}{f^2}+\frac{\left (d (b c-a d) (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 )^{1-m} \, dx}{b f}\\ &=\frac{d (b c-a d) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-1+m,1+m;2+m;-\frac{d (a+b x)}{b c-a d}\right )}{b^2 f (1+m)}+\frac{\left (b (d e-c f)^2\right ) \int (a+b x)^{-1+m} (c+d x)^{-m} \, dx}{f^3}-\frac{\left ((b e-a f) (d e-c f)^2\right ) \int \frac{(a+b x)^{-1+m} (c+d x)^{-m}}{e+f x} \, dx}{f^3}-\frac{\left (d (d e-c f) (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}{f^2}\\ &=-\frac{(d e-c f)^2 (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,m;1+m;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{f^3 m}+\frac{d (b c-a d) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-1+m,1+m;2+m;-\frac{d (a+b x)}{b c-a d}\right )}{b^2 f (1+m)}-\frac{d (d e-c f) (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 )}{b f^2 (1+m)}+\frac{\left (b (d e-c f)^2 (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m\right ) \int (a+b x)^{-1+m} \left (\frac{b c}{b c-a d}+\frac{b d x}{b c-a d}\right )^{-m} \, dx}{f^3}\\ &=-\frac{(d e-c f)^2 (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,m;1+m;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{f^3 m}+\frac{d (b c-a d) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-1+m,1+m;2+m;-\frac{d (a+b x)}{b c-a d}\right )}{b^2 f (1+m)}+\frac{(d e-c f)^2 (a+b x)^m (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m;1+m;-\frac{d (a+b x)}{b c-a d}\right )}{f^3 m}-\frac{d (d e-c f) (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 )}{b f^2 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.248993, size = 258, normalized size = 0.7 \[ \frac{(a+b x)^m (c+d x)^{-m} \left (-b (d e-c f) \left (b (m+1) (d e-c f) \left (\, _2F_1\left (1,m;m+1;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )-\left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m;m+1;\frac{d (a+b x)}{a d-b c}\right )\right )+d f m (a+b x) \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m+1;m+2;\frac{d (a+b x)}{a d-b c}\right )\right )-d f^2 m (a+b x) (a d-b c) \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m-1,m+1;m+2;\frac{d (a+b x)}{a d-b c}\right )\right )}{b^2 f^3 m (m+1)} \]

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

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Maple [F]  time = 0.065, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{2-m}}{fx+e}}\, 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)

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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 + 2}}{f x + e}\,{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((b*x + a)^m*(d*x + c)^(-m + 2)/(f*x + e), 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 + 2}}{f x + e}, 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((b*x + a)^m*(d*x + c)^(-m + 2)/(f*x + e), 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)**(2-m)/(f*x+e),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 + 2}}{f x + e}\,{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((b*x + a)^m*(d*x + c)^(-m + 2)/(f*x + e), x)