3.32.35 \(\int \frac {(a+b x)^m (c+d x)^{3-m}}{(e+f x)^2} \, dx\) [3135]

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

[Out]

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

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Rubi [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 0.04, 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 = {142, 141} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[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 141

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)/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(
b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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 142

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)^{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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 0.22, size = 111, normalized size = 0.28 \begin {gather*} \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 {gather*}

Antiderivative was successfully verified.

[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 = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{3-m}}{\left (f x +e \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^(3-m)/(f*x+e)^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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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*f*x*e + e^2), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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