3.2642 \(\int \frac{(A+B x) (d+e x)^m}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=211 \[ -\frac{(d+e x)^{m+1} \left (B-\frac{b B-2 A c}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-b e+\sqrt{b^2-4 a c} e}\right )}{(m+1) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}-\frac{(d+e x)^{m+1} \left (\frac{b B-2 A c}{\sqrt{b^2-4 a c}}+B\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )} \]

[Out]

-(((B - (b*B - 2*A*c)/Sqrt[b^2 - 4*a*c])*(d + e*x)^(1 + m)*Hypergeometric2F1[1,
1 + m, 2 + m, (2*c*(d + e*x))/(2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e)])/((2*c*d - (b
- Sqrt[b^2 - 4*a*c])*e)*(1 + m))) - ((B + (b*B - 2*A*c)/Sqrt[b^2 - 4*a*c])*(d +
e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (2*c*(d + e*x))/(2*c*d - (b + S
qrt[b^2 - 4*a*c])*e)])/((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(1 + m))

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Rubi [A]  time = 0.676599, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08 \[ -\frac{(d+e x)^{m+1} \left (B-\frac{b B-2 A c}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-b e+\sqrt{b^2-4 a c} e}\right )}{(m+1) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}-\frac{(d+e x)^{m+1} \left (\frac{b B-2 A c}{\sqrt{b^2-4 a c}}+B\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )} \]

Antiderivative was successfully verified.

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

[Out]

-(((B - (b*B - 2*A*c)/Sqrt[b^2 - 4*a*c])*(d + e*x)^(1 + m)*Hypergeometric2F1[1,
1 + m, 2 + m, (2*c*(d + e*x))/(2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e)])/((2*c*d - (b
- Sqrt[b^2 - 4*a*c])*e)*(1 + m))) - ((B + (b*B - 2*A*c)/Sqrt[b^2 - 4*a*c])*(d +
e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (2*c*(d + e*x))/(2*c*d - (b + S
qrt[b^2 - 4*a*c])*e)])/((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(1 + m))

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Rubi in Sympy [A]  time = 65.6956, size = 211, normalized size = 1. \[ \frac{\left (d + e x\right )^{m + 1} \left (2 A c - B b + B \sqrt{- 4 a c + b^{2}}\right ){{}_{2}F_{1}\left (\begin{matrix} 1, m + 1 \\ m + 2 \end{matrix}\middle |{\frac{c \left (- 2 d - 2 e x\right )}{b e - 2 c d - e \sqrt{- 4 a c + b^{2}}}} \right )}}{\left (m + 1\right ) \sqrt{- 4 a c + b^{2}} \left (b e - 2 c d - e \sqrt{- 4 a c + b^{2}}\right )} + \frac{\left (d + e x\right )^{m + 1} \left (2 A c - B b - B \sqrt{- 4 a c + b^{2}}\right ){{}_{2}F_{1}\left (\begin{matrix} 1, m + 1 \\ m + 2 \end{matrix}\middle |{\frac{c \left (- 2 d - 2 e x\right )}{b e - 2 c d + e \sqrt{- 4 a c + b^{2}}}} \right )}}{\left (m + 1\right ) \sqrt{- 4 a c + b^{2}} \left (2 c d - e \left (b + \sqrt{- 4 a c + b^{2}}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

(d + e*x)**(m + 1)*(2*A*c - B*b + B*sqrt(-4*a*c + b**2))*hyper((1, m + 1), (m +
2,), c*(-2*d - 2*e*x)/(b*e - 2*c*d - e*sqrt(-4*a*c + b**2)))/((m + 1)*sqrt(-4*a*
c + b**2)*(b*e - 2*c*d - e*sqrt(-4*a*c + b**2))) + (d + e*x)**(m + 1)*(2*A*c - B
*b - B*sqrt(-4*a*c + b**2))*hyper((1, m + 1), (m + 2,), c*(-2*d - 2*e*x)/(b*e -
2*c*d + e*sqrt(-4*a*c + b**2)))/((m + 1)*sqrt(-4*a*c + b**2)*(2*c*d - e*(b + sqr
t(-4*a*c + b**2))))

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Mathematica [A]  time = 0.848389, size = 305, normalized size = 1.45 \[ \frac{2^{-m-1} (d+e x)^m \left (\left (B \sqrt{e^2 \left (b^2-4 a c\right )}+2 A c e-b B e\right ) \left (\frac{c (d+e x)}{-\sqrt{e^2 \left (b^2-4 a c\right )}+b e+2 c e x}\right )^{-m} \, _2F_1\left (-m,-m;1-m;\frac{2 c d-b e+\sqrt{\left (b^2-4 a c\right ) e^2}}{-b e-2 c x e+\sqrt{\left (b^2-4 a c\right ) e^2}}\right )+\left (B \sqrt{e^2 \left (b^2-4 a c\right )}-2 A c e+b B e\right ) \left (\frac{c (d+e x)}{\sqrt{e^2 \left (b^2-4 a c\right )}+b e+2 c e x}\right )^{-m} \, _2F_1\left (-m,-m;1-m;\frac{-2 c d+b e+\sqrt{\left (b^2-4 a c\right ) e^2}}{b e+2 c x e+\sqrt{\left (b^2-4 a c\right ) e^2}}\right )\right )}{c m \sqrt{e^2 \left (b^2-4 a c\right )}} \]

Antiderivative was successfully verified.

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

[Out]

(2^(-1 - m)*(d + e*x)^m*(((-(b*B*e) + 2*A*c*e + B*Sqrt[(b^2 - 4*a*c)*e^2])*Hyper
geometric2F1[-m, -m, 1 - m, (2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2])/(-(b*e) + Sq
rt[(b^2 - 4*a*c)*e^2] - 2*c*e*x)])/((c*(d + e*x))/(b*e - Sqrt[(b^2 - 4*a*c)*e^2]
+ 2*c*e*x))^m + ((b*B*e - 2*A*c*e + B*Sqrt[(b^2 - 4*a*c)*e^2])*Hypergeometric2F
1[-m, -m, 1 - m, (-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])/(b*e + Sqrt[(b^2 - 4*a
*c)*e^2] + 2*c*e*x)])/((c*(d + e*x))/(b*e + Sqrt[(b^2 - 4*a*c)*e^2] + 2*c*e*x))^
m))/(c*Sqrt[(b^2 - 4*a*c)*e^2]*m)

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Maple [F]  time = 0.437, size = 0, normalized size = 0. \[ \int{\frac{ \left ( Bx+A \right ) \left ( ex+d \right ) ^{m}}{c{x}^{2}+bx+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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