3.1087 \(\int \frac{(e x)^m (A+B x)}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=318 \[ -\frac{c (e x)^{m+1} \left (A b m \left (\sqrt{b^2-4 a c}+b\right )-2 a \left (B m \sqrt{b^2-4 a c}-2 A c (1-m)+b B\right )\right ) \, _2F_1\left (1,m+1;m+2;-\frac{2 c x}{b-\sqrt{b^2-4 a c}}\right )}{a e (m+1) \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt{b^2-4 a c}\right )}-\frac{c (e x)^{m+1} \left (\frac{2 a (b B-2 A c (1-m))-A b^2 m}{\sqrt{b^2-4 a c}}+m (A b-2 a B)\right ) \, _2F_1\left (1,m+1;m+2;-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{a e (m+1) \left (b^2-4 a c\right ) \left (\sqrt{b^2-4 a c}+b\right )}+\frac{(e x)^{m+1} \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{a e \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \]

[Out]

((e*x)^(1 + m)*(A*b^2 - a*b*B - 2*a*A*c + (A*b - 2*a*B)*c*x))/(a*(b^2 - 4*a*c)*e*(a + b*x + c*x^2)) - (c*(A*b*
(b + Sqrt[b^2 - 4*a*c])*m - 2*a*(b*B - 2*A*c*(1 - m) + B*Sqrt[b^2 - 4*a*c]*m))*(e*x)^(1 + m)*Hypergeometric2F1
[1, 1 + m, 2 + m, (-2*c*x)/(b - Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)^(3/2)*(b - Sqrt[b^2 - 4*a*c])*e*(1 + m))
- (c*((A*b - 2*a*B)*m + (2*a*(b*B - 2*A*c*(1 - m)) - A*b^2*m)/Sqrt[b^2 - 4*a*c])*(e*x)^(1 + m)*Hypergeometric
2F1[1, 1 + m, 2 + m, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b + Sqrt[b^2 - 4*a*c])*e*(1 + m))

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Rubi [A]  time = 0.882518, antiderivative size = 317, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {822, 830, 64} \[ -\frac{c (e x)^{m+1} \left (A b m \left (\sqrt{b^2-4 a c}+b\right )-2 a \left (B m \sqrt{b^2-4 a c}-2 A c (1-m)+b B\right )\right ) \, _2F_1\left (1,m+1;m+2;-\frac{2 c x}{b-\sqrt{b^2-4 a c}}\right )}{a e (m+1) \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt{b^2-4 a c}\right )}-\frac{c (e x)^{m+1} \left (\frac{-4 a A c (1-m)+2 a b B-A b^2 m}{\sqrt{b^2-4 a c}}+m (A b-2 a B)\right ) \, _2F_1\left (1,m+1;m+2;-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{a e (m+1) \left (b^2-4 a c\right ) \left (\sqrt{b^2-4 a c}+b\right )}+\frac{(e x)^{m+1} \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{a e \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x)^(1 + m)*(A*b^2 - a*b*B - 2*a*A*c + (A*b - 2*a*B)*c*x))/(a*(b^2 - 4*a*c)*e*(a + b*x + c*x^2)) - (c*(A*b*
(b + Sqrt[b^2 - 4*a*c])*m - 2*a*(b*B - 2*A*c*(1 - m) + B*Sqrt[b^2 - 4*a*c]*m))*(e*x)^(1 + m)*Hypergeometric2F1
[1, 1 + m, 2 + m, (-2*c*x)/(b - Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)^(3/2)*(b - Sqrt[b^2 - 4*a*c])*e*(1 + m))
- (c*((A*b - 2*a*B)*m + (2*a*b*B - 4*a*A*c*(1 - m) - A*b^2*m)/Sqrt[b^2 - 4*a*c])*(e*x)^(1 + m)*Hypergeometric
2F1[1, 1 + m, 2 + m, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b + Sqrt[b^2 - 4*a*c])*e*(1 + m))

Rule 822

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

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 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

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

Mathematica [A]  time = 0.807335, size = 253, normalized size = 0.8 \[ \frac{(e x)^m \left (\frac{c x \left (-\frac{\left (\frac{A b^2 m-2 a (2 A c (m-1)+b B)}{\sqrt{b^2-4 a c}}+m (A b-2 a B)\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )}{b-\sqrt{b^2-4 a c}}-\frac{\left (\frac{4 a A c (m-1)+2 a b B-A b^2 m}{\sqrt{b^2-4 a c}}+m (A b-2 a B)\right ) \, _2F_1\left (1,m+1;m+2;-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c}+b}\right )}{m+1}+\frac{x \left (A \left (-2 a c+b^2+b c x\right )-a B (b+2 c x)\right )}{a+x (b+c x)}\right )}{a \left (b^2-4 a c\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.089, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m} \left ( Bx+A \right ) }{ \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)^m*(B*x+A)/(c*x^2+b*x+a)^2,x)

[Out]

int((e*x)^m*(B*x+A)/(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 (B x + A\right )} \left (e x\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)^m*(B*x+A)/(c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

integrate((B*x + A)*(e*x)^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 (B x + A\right )} \left (e x\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)^m*(B*x+A)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

integral((B*x + A)*(e*x)^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)**m*(B*x+A)/(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 (B x + A\right )} \left (e x\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)^m*(B*x+A)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

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