3.3112 \(\int \frac{(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx\)

Optimal. Leaf size=110 \[ \frac{(b c-a d)^2 (a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m F_1\left (m+1;m-2,3;m+2;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^3} \]

[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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Rubi [A]  time = 0.21213, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077 \[ \frac{(b c-a d)^2 (a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m F_1\left (m+1;m-2,3;m+2;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^3} \]

Antiderivative was successfully verified.

[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)

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Rubi in Sympy [A]  time = 29.5629, size = 85, normalized size = 0.77 \[ - \frac{\left (\frac{b \left (- c - d x\right )}{a d - b c}\right )^{m} \left (a + b x\right )^{m + 1} \left (c + d x\right )^{- m} \left (a d - b c\right )^{2} \operatorname{appellf_{1}}{\left (m + 1,3,m - 2,m + 2,\frac{f \left (a + b x\right )}{a f - b e},\frac{d \left (a + b x\right )}{a d - b c} \right )}}{\left (m + 1\right ) \left (a f - b e\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-(b*(-c - d*x)/(a*d - b*c))**m*(a + b*x)**(m + 1)*(c + d*x)**(-m)*(a*d - b*c)**2
*appellf1(m + 1, 3, m - 2, m + 2, f*(a + b*x)/(a*f - b*e), d*(a + b*x)/(a*d - b*
c))/((m + 1)*(a*f - b*e)**3)

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Mathematica [B]  time = 2.5513, size = 304, normalized size = 2.76 \[ -\frac{(m+2) (b c-a d) (b e-a f)^4 (a+b x)^{m+1} (c+d x)^{2-m} F_1\left (m+1;m-2,3;m+2;\frac{d (a+b x)}{a d-b c},\frac{f (a+b x)}{a f-b e}\right )}{b (m+1) (e+f x)^3 (a f-b e)^3 \left ((m+2) (b c-a d) (b e-a f) F_1\left (m+1;m-2,3;m+2;\frac{d (a+b x)}{a d-b c},\frac{f (a+b x)}{a f-b e}\right )+(a+b x) \left ((3 a d f-3 b c f) F_1\left (m+2;m-2,4;m+3;\frac{d (a+b x)}{a d-b c},\frac{f (a+b x)}{a f-b e}\right )-d (m-2) (b e-a f) F_1\left (m+2;m-1,3;m+3;\frac{d (a+b x)}{a d-b c},\frac{f (a+b x)}{a f-b e}\right )\right )\right )} \]

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)*(b*e - a*f)^4*(2 + m)*(a + b*x)^(1 + m)*(c + d*x)^(2 - m)*AppellF
1[1 + m, -2 + m, 3, 2 + m, (d*(a + b*x))/(-(b*c) + a*d), (f*(a + b*x))/(-(b*e) +
a*f)])/(b*(-(b*e) + a*f)^3*(1 + m)*(e + f*x)^3*((b*c - a*d)*(b*e - a*f)*(2 + 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)] + (a + b*x)*((-3*b*c*f + 3*a*d*f)*AppellF1[2 + m, -2 + m, 4, 3 +
m, (d*(a + b*x))/(-(b*c) + a*d), (f*(a + b*x))/(-(b*e) + a*f)] - d*(b*e - a*f)*(
-2 + m)*AppellF1[2 + m, -1 + m, 3, 3 + m, (d*(a + b*x))/(-(b*c) + a*d), (f*(a +
b*x))/(-(b*e) + a*f)]))))

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Maple [F]  time = 0.127, size = 0, normalized size = 0. \[ \int{\frac{ \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{2-m}}{ \left ( fx+e \right ) ^{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x + a)^m*(d*x + c)^(-m + 2)/(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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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m + 2}}{f^{3} x^{3} + 3 \, e f^{2} x^{2} + 3 \, e^{2} f x + e^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x + a)^m*(d*x + c)^(-m + 2)/(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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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