3.3045 \(\int (a+b x)^m (c+d x)^{-m} (e+f x)^2 \, dx\)

Optimal. Leaf size=250 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (a^2 d^2 f^2 \left (m^2-3 m+2\right )-2 a b d f (1-m) (3 d e-c f (m+1))+b^2 \left (c^2 f^2 \left (m^2+3 m+2\right )-6 c d e f (m+1)+6 d^2 e^2\right )\right ) \, _2F_1\left (m,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{6 b^3 d^2 (m+1)}-\frac{f (a+b x)^{m+1} (c+d x)^{1-m} (a d f (2-m)-b (4 d e-c f (m+2)))}{6 b^2 d^2}+\frac{f (e+f x) (a+b x)^{m+1} (c+d x)^{1-m}}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.553645, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (a^2 d^2 f^2 \left (m^2-3 m+2\right )-2 a b d f (1-m) (3 d e-c f (m+1))+b^2 \left (c^2 f^2 \left (m^2+3 m+2\right )-6 c d e f (m+1)+6 d^2 e^2\right )\right ) \, _2F_1\left (m,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{6 b^3 d^2 (m+1)}+\frac{f (a+b x)^{m+1} (c+d x)^{1-m} (-a d f (2-m)-b c f (m+2)+4 b d e)}{6 b^2 d^2}+\frac{f (e+f x) (a+b x)^{m+1} (c+d x)^{1-m}}{3 b d} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 49.9279, size = 212, normalized size = 0.85 \[ \frac{f \left (a + b x\right )^{m + 1} \left (c + d x\right )^{- m + 1} \left (e + f x\right )}{3 b d} - \frac{f \left (a + b x\right )^{m + 1} \left (c + d x\right )^{- m + 1} \left (- 4 b d e + f \left (a d \left (- m + 2\right ) + b c \left (m + 2\right )\right )\right )}{6 b^{2} d^{2}} - \frac{\left (\frac{b \left (- c - d x\right )}{a d - b c}\right )^{m} \left (a + b x\right )^{m + 1} \left (c + d x\right )^{- m} \left (2 b d \left (- 3 b d e^{2} + f \left (a c f + e \left (a d \left (- m + 1\right ) + b c \left (m + 1\right )\right )\right )\right ) - f \left (a d \left (- m + 1\right ) + b c \left (m + 1\right )\right ) \left (- 4 b d e + f \left (a d \left (- m + 2\right ) + b c \left (m + 2\right )\right )\right )\right ){{}_{2}F_{1}\left (\begin{matrix} m, m + 1 \\ m + 2 \end{matrix}\middle |{\frac{d \left (a + b x\right )}{a d - b c}} \right )}}{6 b^{3} d^{2} \left (m + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

f*(a + b*x)**(m + 1)*(c + d*x)**(-m + 1)*(e + f*x)/(3*b*d) - f*(a + b*x)**(m + 1
)*(c + d*x)**(-m + 1)*(-4*b*d*e + f*(a*d*(-m + 2) + b*c*(m + 2)))/(6*b**2*d**2)
- (b*(-c - d*x)/(a*d - b*c))**m*(a + b*x)**(m + 1)*(c + d*x)**(-m)*(2*b*d*(-3*b*
d*e**2 + f*(a*c*f + e*(a*d*(-m + 1) + b*c*(m + 1)))) - f*(a*d*(-m + 1) + b*c*(m
+ 1))*(-4*b*d*e + f*(a*d*(-m + 2) + b*c*(m + 2))))*hyper((m, m + 1), (m + 2,), d
*(a + b*x)/(a*d - b*c))/(6*b**3*d**2*(m + 1))

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Mathematica [C]  time = 0.858867, size = 320, normalized size = 1.28 \[ (a+b x)^m (c+d x)^{-m} \left (\frac{3 a c e f x^2 F_1\left (2;-m,m;3;-\frac{b x}{a},-\frac{d x}{c}\right )}{3 a c F_1\left (2;-m,m;3;-\frac{b x}{a},-\frac{d x}{c}\right )+m x \left (b c F_1\left (3;1-m,m;4;-\frac{b x}{a},-\frac{d x}{c}\right )-a d F_1\left (3;-m,m+1;4;-\frac{b x}{a},-\frac{d x}{c}\right )\right )}+\frac{4 a c f^2 x^3 F_1\left (3;-m,m;4;-\frac{b x}{a},-\frac{d x}{c}\right )}{12 a c F_1\left (3;-m,m;4;-\frac{b x}{a},-\frac{d x}{c}\right )+3 b c m x F_1\left (4;1-m,m;5;-\frac{b x}{a},-\frac{d x}{c}\right )-3 a d m x F_1\left (4;-m,m+1;5;-\frac{b x}{a},-\frac{d x}{c}\right )}-\frac{e^2 (c+d x) \left (\frac{d (a+b x)}{a d-b c}\right )^{-m} \, _2F_1\left (1-m,-m;2-m;\frac{b (c+d x)}{b c-a d}\right )}{d (m-1)}\right ) \]

Warning: Unable to verify antiderivative.

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

[Out]

((a + b*x)^m*((3*a*c*e*f*x^2*AppellF1[2, -m, m, 3, -((b*x)/a), -((d*x)/c)])/(3*a
*c*AppellF1[2, -m, m, 3, -((b*x)/a), -((d*x)/c)] + m*x*(b*c*AppellF1[3, 1 - m, m
, 4, -((b*x)/a), -((d*x)/c)] - a*d*AppellF1[3, -m, 1 + m, 4, -((b*x)/a), -((d*x)
/c)])) + (4*a*c*f^2*x^3*AppellF1[3, -m, m, 4, -((b*x)/a), -((d*x)/c)])/(12*a*c*A
ppellF1[3, -m, m, 4, -((b*x)/a), -((d*x)/c)] + 3*b*c*m*x*AppellF1[4, 1 - m, m, 5
, -((b*x)/a), -((d*x)/c)] - 3*a*d*m*x*AppellF1[4, -m, 1 + m, 5, -((b*x)/a), -((d
*x)/c)]) - (e^2*(c + d*x)*Hypergeometric2F1[1 - m, -m, 2 - m, (b*(c + d*x))/(b*c
- a*d)])/(d*(-1 + m)*((d*(a + b*x))/(-(b*c) + a*d))^m)))/(c + d*x)^m

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Maple [F]  time = 0.093, size = 0, normalized size = 0. \[ \int{\frac{ \left ( bx+a \right ) ^{m} \left ( fx+e \right ) ^{2}}{ \left ( dx+c \right ) ^{m}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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