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

Optimal. Leaf size=73 \[ \frac{x^{m+1} (A b-a B)}{a b (a+b x)}-\frac{x^{m+1} (A b m-a B (m+1)) \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )}{a^2 b (m+1)} \]

[Out]

((A*b - a*B)*x^(1 + m))/(a*b*(a + b*x)) - ((A*b*m - a*B*(1 + m))*x^(1 + m)*Hyper
geometric2F1[1, 1 + m, 2 + m, -((b*x)/a)])/(a^2*b*(1 + m))

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Rubi [A]  time = 0.0854355, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111 \[ \frac{x^{m+1} (A b-a B)}{a b (a+b x)}-\frac{x^{m+1} (A b m-a B (m+1)) \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )}{a^2 b (m+1)} \]

Antiderivative was successfully verified.

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

[Out]

((A*b - a*B)*x^(1 + m))/(a*b*(a + b*x)) - ((A*b*m - a*B*(1 + m))*x^(1 + m)*Hyper
geometric2F1[1, 1 + m, 2 + m, -((b*x)/a)])/(a^2*b*(1 + m))

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Rubi in Sympy [A]  time = 21.7948, size = 56, normalized size = 0.77 \[ \frac{x^{m + 1} \left (A b - B a\right )}{a b \left (a + b x\right )} - \frac{x^{m + 1} \left (A b m - B a \left (m + 1\right )\right ){{}_{2}F_{1}\left (\begin{matrix} 1, m + 1 \\ m + 2 \end{matrix}\middle |{- \frac{b x}{a}} \right )}}{a^{2} b \left (m + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.0753611, size = 60, normalized size = 0.82 \[ \frac{x^{m+1} \left ((A b-a B) \, _2F_1\left (2,m+1;m+2;-\frac{b x}{a}\right )+a B \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )\right )}{a^2 b (m+1)} \]

Antiderivative was successfully verified.

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

[Out]

(x^(1 + m)*(a*B*Hypergeometric2F1[1, 1 + m, 2 + m, -((b*x)/a)] + (A*b - a*B)*Hyp
ergeometric2F1[2, 1 + m, 2 + m, -((b*x)/a)]))/(a^2*b*(1 + m))

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Maple [F]  time = 0.116, size = 0, normalized size = 0. \[ \int{\frac{{x}^{m} \left ( Bx+A \right ) }{{b}^{2}{x}^{2}+2\,abx+{a}^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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