3.3129 \(\int \frac {(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx\)

Optimal. Leaf size=362 \[ -\frac {(a+b x)^{m-1} (c+d x)^{1-m} \left (-a^2 d^2 f^2 \left (m^2-3 m+2\right )+2 a b d f (2-m) (d e-c f m)-\left (b^2 \left (-c^2 f^2 (1-m) m-2 c d e f m+2 d^2 e^2\right )\right )\right ) \, _2F_1\left (1,m-1;m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{2 f^3 (1-m) (b e-a f) (d e-c f)}+\frac {(a+b x)^{m-1} (c+d x)^{2-m} (a d f (2-m)-b (3 d e-c f (m+1)))}{2 f^2 (e+f x) (d e-c f)}+\frac {(b e-a f) (a+b x)^{m-1} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {d (b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m-1,m-1;m;-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)} \]

[Out]

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

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Rubi [C]  time = 0.05, antiderivative size = 110, normalized size of antiderivative = 0.30, 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} \[ \frac {(b c-a d)^2 (a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m F_1\left (m+1;m-2,3;m+2;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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])

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]

Rubi steps

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

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Mathematica [C]  time = 0.31, size = 108, normalized size = 0.30 \[ \frac {(b c-a d)^2 (a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m F_1\left (m+1;m-2,3;m+2;\frac {d (a+b x)}{a d-b c},\frac {f (a+b x)}{a f-b e}\right )}{(m+1) (b e-a f)^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{f^{3} x^{3} + 3 \, e f^{2} x^{2} + 3 \, e^{2} f x + e^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (f x + e\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.25, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-m +2}}{\left (f x +e \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (f x + e\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{2-m}}{{\left (e+f\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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