3.1086 \(\int \frac{(e x)^m (A+B x)}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=172 \[ \frac{(e x)^{m+1} \left (B-\frac{b B-2 A c}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,m+1;m+2;-\frac{2 c x}{b-\sqrt{b^2-4 a c}}\right )}{e (m+1) \left (b-\sqrt{b^2-4 a c}\right )}+\frac{(e x)^{m+1} \left (\frac{b B-2 A c}{\sqrt{b^2-4 a c}}+B\right ) \, _2F_1\left (1,m+1;m+2;-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e (m+1) \left (\sqrt{b^2-4 a c}+b\right )} \]

[Out]

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

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Rubi [A]  time = 0.660674, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087 \[ \frac{(e x)^{m+1} \left (B-\frac{b B-2 A c}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,m+1;m+2;-\frac{2 c x}{b-\sqrt{b^2-4 a c}}\right )}{e (m+1) \left (b-\sqrt{b^2-4 a c}\right )}+\frac{(e x)^{m+1} \left (\frac{b B-2 A c}{\sqrt{b^2-4 a c}}+B\right ) \, _2F_1\left (1,m+1;m+2;-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e (m+1) \left (\sqrt{b^2-4 a c}+b\right )} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 43.6951, size = 168, normalized size = 0.98 \[ - \frac{\left (e x\right )^{m + 1} \left (2 A c - B b - B \sqrt{- 4 a c + b^{2}}\right ){{}_{2}F_{1}\left (\begin{matrix} 1, m + 1 \\ m + 2 \end{matrix}\middle |{- \frac{2 c x}{b + \sqrt{- 4 a c + b^{2}}}} \right )}}{e \left (b + \sqrt{- 4 a c + b^{2}}\right ) \left (m + 1\right ) \sqrt{- 4 a c + b^{2}}} + \frac{\left (e x\right )^{m + 1} \left (2 A c - B b + B \sqrt{- 4 a c + b^{2}}\right ){{}_{2}F_{1}\left (\begin{matrix} 1, m + 1 \\ m + 2 \end{matrix}\middle |{- \frac{2 c x}{b - \sqrt{- 4 a c + b^{2}}}} \right )}}{e \left (b - \sqrt{- 4 a c + b^{2}}\right ) \left (m + 1\right ) \sqrt{- 4 a c + b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.73985, size = 233, normalized size = 1.35 \[ \frac{2^{-m-1} (e x)^m \left (\left (B \sqrt{b^2-4 a c}+2 A c-b B\right ) \left (\frac{c x}{-\sqrt{b^2-4 a c}+b+2 c x}\right )^{-m} \, _2F_1\left (-m,-m;1-m;\frac{b-\sqrt{b^2-4 a c}}{b+2 c x-\sqrt{b^2-4 a c}}\right )+\left (B \sqrt{b^2-4 a c}-2 A c+b B\right ) \left (\frac{c x}{\sqrt{b^2-4 a c}+b+2 c x}\right )^{-m} \, _2F_1\left (-m,-m;1-m;\frac{b+\sqrt{b^2-4 a c}}{b+2 c x+\sqrt{b^2-4 a c}}\right )\right )}{c m \sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

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

[Out]

(2^(-1 - m)*(e*x)^m*(((-(b*B) + 2*A*c + B*Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[-
m, -m, 1 - m, (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c] + 2*c*x)])/((c*x)/(
b - Sqrt[b^2 - 4*a*c] + 2*c*x))^m + ((b*B - 2*A*c + B*Sqrt[b^2 - 4*a*c])*Hyperge
ometric2F1[-m, -m, 1 - m, (b + Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x
)])/((c*x)/(b + Sqrt[b^2 - 4*a*c] + 2*c*x))^m))/(c*Sqrt[b^2 - 4*a*c]*m)

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Maple [F]  time = 0.151, size = 0, normalized size = 0. \[ \int{\frac{ \left ( ex \right ) ^{m} \left ( Bx+A \right ) }{c{x}^{2}+bx+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Integral((e*x)**m*(A + B*x)/(a + b*x + c*x**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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