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)*Hyperg
eometric2F1[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 = 0.305122, 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, Rules used = {830, 64} \[ \frac{(e x)^{m+1} \left (B-\frac{b B-2 A c}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,m+1;m+2;-\frac{2 c x}{b-\sqrt{b^2-4 a c}}\right )}{e (m+1) \left (b-\sqrt{b^2-4 a c}\right )}+\frac{(e x)^{m+1} \left (\frac{b B-2 A c}{\sqrt{b^2-4 a c}}+B\right ) \, _2F_1\left (1,m+1;m+2;-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e (m+1) \left (\sqrt{b^2-4 a c}+b\right )} \]

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)*Hyperg
eometric2F1[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))

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

Mathematica [A]  time = 0.277708, size = 135, normalized size = 0.78 \[ \frac{x (e x)^m \left (\left (A \left (\sqrt{b^2-4 a c}+b\right )-2 a B\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )+\left (A \left (\sqrt{b^2-4 a c}-b\right )+2 a B\right ) \, _2F_1\left (1,m+1;m+2;-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )\right )}{2 a (m+1) \sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.123, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m} \left ( Bx+A \right ) }{c{x}^{2}+bx+a}}\, 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),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. \begin{align*} \int \frac{{\left (B x + A\right )} \left (e x\right )^{m}}{c x^{2} + b x + a}\,{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),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. \begin{align*}{\rm integral}\left (\frac{{\left (B x + A\right )} \left (e x\right )^{m}}{c x^{2} + b x + a}, 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),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. \begin{align*} \int \frac{\left (e x\right )^{m} \left (A + B x\right )}{a + b x + c x^{2}}\, dx \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),x)

[Out]

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

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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}}{c x^{2} + b x + a}\,{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),x, algorithm="giac")

[Out]

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