### 3.2554 $$\int \frac{(d+e x)^m}{(a+b x+c x^2)^2} \, dx$$

Optimal. Leaf size=425 $\frac{c (d+e x)^{m+1} \left (-2 c e \left (d m \sqrt{b^2-4 a c}-2 a e (1-m)+2 b d\right )+b e^2 m \left (\sqrt{b^2-4 a c}+b\right )+4 c^2 d^2\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \left (b^2-4 a c\right )^{3/2} \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )}-\frac{c (d+e x)^{m+1} \left (\frac{-4 c e (b d-a e (1-m))+b^2 e^2 m+4 c^2 d^2}{\sqrt{b^2-4 a c}}+e m (2 c d-b e)\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \left (b^2-4 a c\right ) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right ) \left (a e^2-b d e+c d^2\right )}-\frac{(d+e x)^{m+1} \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )}$

[Out]

-(((d + e*x)^(1 + m)*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x))/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*(a
+ b*x + c*x^2))) + (c*(4*c^2*d^2 + b*(b + Sqrt[b^2 - 4*a*c])*e^2*m - 2*c*e*(2*b*d - 2*a*e*(1 - m) + Sqrt[b^2 -
4*a*c]*d*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)])/((b^2 - 4*a*c)^(3/2)*(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*e)*m + (4*c^2*d^2 - 4*c*e*(b*d - a*e*(1 - m)) + b^2*e^2*m)/Sqrt[b^2 - 4*a*c])*(d + e*x)^(1 + m)*Hype
rgeometric2F1[1, 1 + m, 2 + m, (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/((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 = 1.06873, antiderivative size = 425, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.15, Rules used = {740, 830, 68} $\frac{c (d+e x)^{m+1} \left (-2 c e \left (d m \sqrt{b^2-4 a c}-2 a e (1-m)+2 b d\right )+b e^2 m \left (\sqrt{b^2-4 a c}+b\right )+4 c^2 d^2\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \left (b^2-4 a c\right )^{3/2} \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )}-\frac{c (d+e x)^{m+1} \left (\frac{-4 c e (b d-a e (1-m))+b^2 e^2 m+4 c^2 d^2}{\sqrt{b^2-4 a c}}+e m (2 c d-b e)\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \left (b^2-4 a c\right ) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right ) \left (a e^2-b d e+c d^2\right )}-\frac{(d+e x)^{m+1} \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(((d + e*x)^(1 + m)*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x))/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*(a
+ b*x + c*x^2))) + (c*(4*c^2*d^2 + b*(b + Sqrt[b^2 - 4*a*c])*e^2*m - 2*c*e*(2*b*d - 2*a*e*(1 - m) + Sqrt[b^2 -
4*a*c]*d*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)])/((b^2 - 4*a*c)^(3/2)*(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*e)*m + (4*c^2*d^2 - 4*c*e*(b*d - a*e*(1 - m)) + b^2*e^2*m)/Sqrt[b^2 - 4*a*c])*(d + e*x)^(1 + m)*Hype
rgeometric2F1[1, 1 + m, 2 + m, (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/((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))

Rule 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b
*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 830

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m, (f + g*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !RationalQ[m]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{(d+e x)^m}{\left (a+b x+c x^2\right )^2} \, dx &=-\frac{(d+e x)^{1+m} \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}-\frac{\int \frac{(d+e x)^m \left (2 c^2 d^2+b^2 e^2 m+c e (2 a e (1-m)-b d (2+m))-c e (2 c d-b e) m x\right )}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{(d+e x)^{1+m} \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}-\frac{\int \left (\frac{\left (-c e (2 c d-b e) m+\frac{c \left (4 c^2 d^2-4 b c d e+4 a c e^2+b^2 e^2 m-4 a c e^2 m\right )}{\sqrt{b^2-4 a c}}\right ) (d+e x)^m}{b-\sqrt{b^2-4 a c}+2 c x}+\frac{\left (-c e (2 c d-b e) m-\frac{c \left (4 c^2 d^2-4 b c d e+4 a c e^2+b^2 e^2 m-4 a c e^2 m\right )}{\sqrt{b^2-4 a c}}\right ) (d+e x)^m}{b+\sqrt{b^2-4 a c}+2 c x}\right ) \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{(d+e x)^{1+m} \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}+\frac{\left (c \left (e (2 c d-b e) m-\frac{4 c^2 d^2-4 c e (b d-a e (1-m))+b^2 e^2 m}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{(d+e x)^m}{b-\sqrt{b^2-4 a c}+2 c x} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}+\frac{\left (c \left (e (2 c d-b e) m+\frac{4 c^2 d^2-4 c e (b d-a e (1-m))+b^2 e^2 m}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{(d+e x)^m}{b+\sqrt{b^2-4 a c}+2 c x} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{(d+e x)^{1+m} \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}+\frac{c \left (4 c^2 d^2+b \left (b+\sqrt{b^2-4 a c}\right ) e^2 m-2 c e \left (2 b d-2 a e (1-m)+\sqrt{b^2-4 a c} d m\right )\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{\left (b^2-4 a c\right )^{3/2} \left (2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right ) (1+m)}-\frac{c \left (e (2 c d-b e) m+\frac{4 c^2 d^2-4 c e (b d-a e (1-m))+b^2 e^2 m}{\sqrt{b^2-4 a c}}\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{\left (b^2-4 a c\right ) \left (2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right ) (1+m)}\\ \end{align*}

Mathematica [A]  time = 1.28903, size = 339, normalized size = 0.8 $\frac{(d+e x)^{m+1} \left (-\frac{c \left (\frac{4 c e (a e (m-1)+b d)-b^2 e^2 m-4 c^2 d^2}{\sqrt{b^2-4 a c}}+e m (2 c d-b e)\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d+\left (\sqrt{b^2-4 a c}-b\right ) e}\right )}{(m+1) \left (e \left (\sqrt{b^2-4 a c}-b\right )+2 c d\right )}-\frac{c \left (\frac{-4 c e (a e (m-1)+b d)+b^2 e^2 m+4 c^2 d^2}{\sqrt{b^2-4 a c}}+e m (2 c d-b e)\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}+\frac{-2 c (a e+c d x)+b^2 e+b c (e x-d)}{a+x (b+c x)}\right )}{\left (b^2-4 a c\right ) \left (e (a e-b d)+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [F]  time = 1.241, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{m}}{ \left ( c{x}^{2}+bx+a \right ) ^{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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. \begin{align*} \int \frac{{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((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. \begin{align*}{\rm integral}\left (\frac{{\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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((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. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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