\(\int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx\) [1086]

Optimal result

Integrand size = 23, antiderivative size = 172 \[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\frac {\left (B-\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) e (1+m)}+\frac {\left (B+\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) e (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {844, 66} \[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\frac {(e x)^{m+1} \left (B-\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{e (m+1) \left (b-\sqrt {b^2-4 a c}\right )}+\frac {(e x)^{m+1} \left (\frac {b B-2 A c}{\sqrt {b^2-4 a c}}+B\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{e (m+1) \left (\sqrt {b^2-4 a c}+b\right )} \]

[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 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], 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])))

Rule 844

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.78 \[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\frac {x (e x)^m \left (\left (-2 a B+A \left (b+\sqrt {b^2-4 a c}\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+\left (2 a B+A \left (-b+\sqrt {b^2-4 a c}\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )\right )}{2 a \sqrt {b^2-4 a c} (1+m)} \]

[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))

Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (B x +A \right )}{c \,x^{2}+b x +a}d x\]

[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)

Fricas [F]

\[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\int { \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{c x^{2} + b x + a} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\int \frac {\left (e x\right )^{m} \left (A + B x\right )}{a + b x + c x^{2}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\int { \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{c x^{2} + b x + a} \,d x } \]

[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)

Giac [F]

\[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\int { \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{c x^{2} + b x + a} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x\right )}{c\,x^2+b\,x+a} \,d x \]

[In]

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

[Out]

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