\(\int \frac {(e x)^m (A+B x)}{(a+b x+c x^2)^2} \, dx\) [1087]

Optimal result

Integrand size = 23, antiderivative size = 318 \[ \int \frac {(e x)^m (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=\frac {(e x)^{1+m} \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{a \left (b^2-4 a c\right ) e \left (a+b x+c x^2\right )}-\frac {c \left (A b \left (b+\sqrt {b^2-4 a c}\right ) m-2 a \left (b B-2 A c (1-m)+B \sqrt {b^2-4 a c} m\right )\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) e (1+m)}-\frac {c \left ((A b-2 a B) m+\frac {2 a (b B-2 A c (1-m))-A b^2 m}{\sqrt {b^2-4 a c}}\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b+\sqrt {b^2-4 a c}\right ) e (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {836, 844, 66} \[ \int \frac {(e x)^m (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {c (e x)^{m+1} \left (A b m \left (\sqrt {b^2-4 a c}+b\right )-2 a \left (B m \sqrt {b^2-4 a c}-2 A c (1-m)+b B\right )\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{a e (m+1) \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {c (e x)^{m+1} \left (\frac {-4 a A c (1-m)+2 a b B-A b^2 m}{\sqrt {b^2-4 a c}}+m (A b-2 a B)\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{a e (m+1) \left (b^2-4 a c\right ) \left (\sqrt {b^2-4 a c}+b\right )}+\frac {(e x)^{m+1} \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{a e \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \]

[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 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 836

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 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}& = \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] (verified)

Time = 0.58 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.80 \[ \int \frac {(e x)^m (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=\frac {(e x)^m \left (\frac {x \left (-a B (b+2 c x)+A \left (b^2-2 a c+b c x\right )\right )}{a+x (b+c x)}+\frac {c x \left (-\frac {\left ((A b-2 a B) m+\frac {-2 a (b B+2 A c (-1+m))+A b^2 m}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )}{b-\sqrt {b^2-4 a c}}-\frac {\left ((A b-2 a B) m+\frac {2 a b B+4 a A c (-1+m)-A b^2 m}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{b+\sqrt {b^2-4 a c}}\right )}{1+m}\right )}{a \left (b^2-4 a c\right )} \]

[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))

Maple [F]

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

[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)

Fricas [F]

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

[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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

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

[In]

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

[Out]

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