3.1658 \(\int \frac{(b+2 c x) (d+e x)^m}{\left (a+b x+c x^2\right )^2} \, dx\)

Optimal. Leaf size=357 \[ -\frac{c e m (d+e x)^{m+1} \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-b e+\sqrt{b^2-4 a c} e}\right )}{(m+1) \sqrt{b^2-4 a c} \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )}+\frac{c e m (d+e x)^{m+1} \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \sqrt{b^2-4 a c} \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right ) \left (a e^2-b d e+c d^2\right )}-\frac{(d+e x)^{m+1} \left (\left (b^2-4 a c\right ) (c d-b e)-c e x \left (b^2-4 a c\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )} \]

[Out]

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

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Rubi [A]  time = 1.33452, antiderivative size = 357, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115 \[ -\frac{c e m (d+e x)^{m+1} \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-b e+\sqrt{b^2-4 a c} e}\right )}{(m+1) \sqrt{b^2-4 a c} \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )}+\frac{c e m (d+e x)^{m+1} \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \sqrt{b^2-4 a c} \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right ) \left (a e^2-b d e+c d^2\right )}-\frac{(d+e x)^{m+1} (-b e+c d-c e x)}{\left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

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

[Out]

-(((d + e*x)^(1 + m)*(c*d - b*e - c*e*x))/((c*d^2 - b*d*e + a*e^2)*(a + b*x + c*
x^2))) - (c*e*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*m*(d + e*x)^(1 + m)*Hypergeome
tric2F1[1, 1 + m, 2 + m, (2*c*(d + e*x))/(2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e)])/(
Sqrt[b^2 - 4*a*c]*(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*(c*d^2 - b*d*e + a*e^2)*(1
+ m)) + (c*e*(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*m*(d + e*x)^(1 + m)*Hypergeome
tric2F1[1, 1 + m, 2 + m, (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(
Sqrt[b^2 - 4*a*c]*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(c*d^2 - b*d*e + a*e^2)*(1
+ m))

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Rubi in Sympy [A]  time = 154.265, size = 306, normalized size = 0.86 \[ \frac{c e m \left (d + e x\right )^{m + 1} \left (e \left (- 4 a c + b^{2}\right ) - \sqrt{- 4 a c + b^{2}} \left (b e - 2 c d\right )\right ){{}_{2}F_{1}\left (\begin{matrix} 1, m + 1 \\ m + 2 \end{matrix}\middle |{\frac{c \left (- 2 d - 2 e x\right )}{b e - 2 c d + e \sqrt{- 4 a c + b^{2}}}} \right )}}{\left (m + 1\right ) \left (- 4 a c + b^{2}\right ) \left (2 c d - e \left (b + \sqrt{- 4 a c + b^{2}}\right )\right ) \left (a e^{2} - b d e + c d^{2}\right )} + \frac{c e m \left (d + e x\right )^{m + 1} \left (e \left (- 4 a c + b^{2}\right ) + \sqrt{- 4 a c + b^{2}} \left (b e - 2 c d\right )\right ){{}_{2}F_{1}\left (\begin{matrix} 1, m + 1 \\ m + 2 \end{matrix}\middle |{\frac{c \left (- 2 d - 2 e x\right )}{b e - 2 c d - e \sqrt{- 4 a c + b^{2}}}} \right )}}{\left (m + 1\right ) \left (- 4 a c + b^{2}\right ) \left (2 c d - e \left (b - \sqrt{- 4 a c + b^{2}}\right )\right ) \left (a e^{2} - b d e + c d^{2}\right )} + \frac{\left (d + e x\right )^{m + 1} \left (b e - c d + c e x\right )}{\left (a + b x + c x^{2}\right ) \left (a e^{2} - b d e + c d^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

c*e*m*(d + e*x)**(m + 1)*(e*(-4*a*c + b**2) - sqrt(-4*a*c + b**2)*(b*e - 2*c*d))
*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)*(-4*a*c + b**2)*(2*c*d - e*(b + sqrt(-4*a*c + b**2)))*(a*e**2 - b*
d*e + c*d**2)) + c*e*m*(d + e*x)**(m + 1)*(e*(-4*a*c + b**2) + sqrt(-4*a*c + b**
2)*(b*e - 2*c*d))*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)*(-4*a*c + b**2)*(2*c*d - e*(b - sqrt(-4*a*c + b**
2)))*(a*e**2 - b*d*e + c*d**2)) + (d + e*x)**(m + 1)*(b*e - c*d + c*e*x)/((a + b
*x + c*x**2)*(a*e**2 - b*d*e + c*d**2))

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Mathematica [A]  time = 0.178095, size = 0, normalized size = 0. \[ \int \frac{(b+2 c x) (d+e x)^m}{\left (a+b x+c x^2\right )^2} \, dx \]

Verification is Not applicable to the result.

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

[Out]

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

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Maple [F]  time = 0.191, size = 0, normalized size = 0. \[ \int{\frac{ \left ( 2\,cx+b \right ) \left ( ex+d \right ) ^{m}}{ \left ( c{x}^{2}+bx+a \right ) ^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (2 \, c x + b\right )}{\left (e x + d\right )}^{m}}{c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((2*c*x + b)*(e*x + d)^m/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*
x^2 + a^2), 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((2*c*x+b)*(e*x+d)**m/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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