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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.34, 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 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 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 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 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 = 0.46, 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
)

________________________________________________________________________________________

fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 3}}{f x + e}, x\right ) \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 3}}{f x + e}\,{d x} \]

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)

________________________________________________________________________________________

maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-m +3}}{f x +e}\, dx \]

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)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 3}}{f x + e}\,{d x} \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{3-m}}{e+f\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]

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]

Exception raised: HeuristicGCDFailed

________________________________________________________________________________________