3.3134 \(\int \frac{(a+b x)^m (c+d x)^{3-m}}{e+f x} \, dx\)

Optimal. Leaf size=488 \[ -\frac{(b c-a d)^2 (a+b x)^{m-2} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (3 a^2 b d^2 f^2 (1-m) m (d e-c f (3-m))+a^3 d^3 f^3 m \left (m^2-3 m+2\right )+3 a b^2 d f m \left (c^2 f^2 \left (m^2-5 m+6\right )-2 c d e f (3-m)+2 d^2 e^2\right )+b^3 \left (-\left (3 c^2 d e f^2 \left (m^2-5 m+6\right )-c^3 f^3 \left (-m^3+6 m^2-11 m+6\right )-6 c d^2 e^2 f (3-m)+6 d^3 e^3\right )\right )\right ) \, _2F_1\left (m-3,m-2;m-1;-\frac{d (a+b x)}{b c-a d}\right )}{6 b^3 d^2 f^4 (2-m) (3-m)}-\frac{b (a+b x)^{m-2} (c+d x)^{4-m} (b (3 d e-c f (1-m))-a d f (m+2))}{6 d^2 f^2}-\frac{(b e-a f)^3 (a+b x)^{m-3} (c+d x)^{3-m} \, _2F_1\left (1,m-3;m-2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{f^4 (3-m)}+\frac{b (b e-a f)^3 (a+b x)^{m-3} (c+d x)^{4-m}}{f^4 (3-m) (b c-a d)}+\frac{b (a+b x)^{m-1} (c+d x)^{4-m}}{3 d f} \]

[Out]

(b*(b*e - a*f)^3*(a + b*x)^(-3 + m)*(c + d*x)^(4 - m))/((b*c - a*d)*f^4*(3 - m)) - (b*(b*(3*d*e - c*f*(1 - m))
- a*d*f*(2 + m))*(a + b*x)^(-2 + m)*(c + d*x)^(4 - m))/(6*d^2*f^2) + (b*(a + b*x)^(-1 + m)*(c + d*x)^(4 - m))
/(3*d*f) - ((b*e - a*f)^3*(a + b*x)^(-3 + m)*(c + d*x)^(3 - m)*Hypergeometric2F1[1, -3 + m, -2 + m, ((d*e - c*
f)*(a + b*x))/((b*e - a*f)*(c + d*x))])/(f^4*(3 - m)) - ((b*c - a*d)^2*(3*a^2*b*d^2*f^2*(d*e - c*f*(3 - m))*(1
- m)*m + a^3*d^3*f^3*m*(2 - 3*m + m^2) + 3*a*b^2*d*f*m*(2*d^2*e^2 - 2*c*d*e*f*(3 - m) + c^2*f^2*(6 - 5*m + m^
2)) - b^3*(6*d^3*e^3 - 6*c*d^2*e^2*f*(3 - m) + 3*c^2*d*e*f^2*(6 - 5*m + m^2) - c^3*f^3*(6 - 11*m + 6*m^2 - m^3
)))*(a + b*x)^(-2 + m)*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[-3 + m, -2 + m, -1 + m, -((d*(a + b*x))
/(b*c - a*d))])/(6*b^3*d^2*f^4*(2 - m)*(3 - m)*(c + d*x)^m)

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Rubi [A]  time = 0.338886, antiderivative size = 417, normalized size of antiderivative = 0.85, number of steps used = 13, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {105, 70, 69, 131} \[ -\frac{d (b c-a d) (a+b x)^{m+1} (d e-c f) (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m-1,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{b^2 f^2 (m+1)}+\frac{d (b c-a d)^2 (a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m-2,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{b^3 f (m+1)}+\frac{(a+b x)^m (d e-c f)^3 (c+d x)^{-m} \, _2F_1\left (1,m;m+1;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{f^4 m}-\frac{(a+b x)^m (d e-c f)^3 (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m;m+1;-\frac{d (a+b x)}{b c-a d}\right )}{f^4 m}+\frac{d (a+b x)^{m+1} (d e-c f)^2 (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{b f^3 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d*e - c*f)^3*(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*m*(c + d*x)^m) + (d*(b*c - a*d)^2*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[-2 + m,
1 + m, 2 + m, -((d*(a + b*x))/(b*c - a*d))])/(b^3*f*(1 + m)*(c + d*x)^m) - (d*(b*c - a*d)*(d*e - c*f)*(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^2*(1 + m)*(c + d*x)^m) - ((d*e - c*f)^3*(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^4*m*(c + d*x)^m) + (d*(d*e - c*f)^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))])/(b*f^3*(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)^{3-m}}{e+f x} \, dx &=\frac{d \int (a+b x)^m (c+d x)^{2-m} \, dx}{f}-\frac{(d e-c f) \int \frac{(a+b x)^m (c+d x)^{2-m}}{e+f x} \, dx}{f}\\ &=-\frac{(d (d e-c f)) \int (a+b x)^m (c+d x)^{1-m} \, dx}{f^2}+\frac{(d e-c f)^2 \int \frac{(a+b x)^m (c+d x)^{1-m}}{e+f x} \, dx}{f^2}+\frac{\left (d (b c-a d)^2 (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 )^{2-m} \, dx}{b^2 f}\\ &=\frac{d (b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-2+m,1+m;2+m;-\frac{d (a+b x)}{b c-a d}\right )}{b^3 f (1+m)}+\frac{\left (d (d e-c f)^2\right ) \int (a+b x)^m (c+d x)^{-m} \, dx}{f^3}-\frac{(d e-c f)^3 \int \frac{(a+b x)^m (c+d x)^{-m}}{e+f x} \, dx}{f^3}-\frac{\left (d (b c-a 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 )^{1-m} \, dx}{b f^2}\\ &=\frac{d (b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-2+m,1+m;2+m;-\frac{d (a+b x)}{b c-a d}\right )}{b^3 f (1+m)}-\frac{d (b c-a 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 (-1+m,1+m;2+m;-\frac{d (a+b x)}{b c-a d}\right )}{b^2 f^2 (1+m)}-\frac{\left (b (d e-c f)^3\right ) \int (a+b x)^{-1+m} (c+d x)^{-m} \, dx}{f^4}+\frac{\left ((b e-a f) (d e-c f)^3\right ) \int \frac{(a+b x)^{-1+m} (c+d x)^{-m}}{e+f x} \, dx}{f^4}+\frac{\left (d (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)^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)^3 (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^4 m}+\frac{d (b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-2+m,1+m;2+m;-\frac{d (a+b x)}{b c-a d}\right )}{b^3 f (1+m)}-\frac{d (b c-a 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 (-1+m,1+m;2+m;-\frac{d (a+b x)}{b c-a d}\right )}{b^2 f^2 (1+m)}+\frac{d (d e-c f)^2 (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^3 (1+m)}-\frac{\left (b (d e-c f)^3 (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^4}\\ &=\frac{(d e-c f)^3 (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^4 m}+\frac{d (b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-2+m,1+m;2+m;-\frac{d (a+b x)}{b c-a d}\right )}{b^3 f (1+m)}-\frac{d (b c-a 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 (-1+m,1+m;2+m;-\frac{d (a+b x)}{b c-a d}\right )}{b^2 f^2 (1+m)}-\frac{(d e-c f)^3 (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^4 m}+\frac{d (d e-c f)^2 (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^3 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.466151, size = 337, normalized size = 0.69 \[ \frac{(a+b x)^m (c+d x)^{-m} \left (d f^3 m (a+b x) (b c-a d)^2 \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m-2,m+1;m+2;\frac{d (a+b x)}{a d-b c}\right )-b (d e-c f) \left (-b (d e-c f) \left (b (m+1) (d e-c f) \left (\, _2F_1\left (1,m;m+1;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )-\left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m;m+1;\frac{d (a+b x)}{a d-b c}\right )\right )+d f m (a+b x) \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m+1;m+2;\frac{d (a+b x)}{a d-b c}\right )\right )-d f^2 m (a+b x) (a d-b c) \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m-1,m+1;m+2;\frac{d (a+b x)}{a d-b c}\right )\right )\right )}{b^3 f^4 m (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.069, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx+c \right ) ^{3-m} \left ( bx+a \right ) ^{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)^(3-m)/(f*x+e),x)

[Out]

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

[Out]

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

[Out]

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

[Out]

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