3.2369 \(\int \frac{A+B x}{(d+e x)^2 (a+b x+c x^2)} \, dx\)

Optimal. Leaf size=255 \[ -\frac{\tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \left (-b \left (a B e^2+2 A c d e+B c d^2\right )+2 c \left (-a A e^2+2 a B d e+A c d^2\right )+A b^2 e^2\right )}{\sqrt{b^2-4 a c} \left (a e^2-b d e+c d^2\right )^2}-\frac{\log \left (a+b x+c x^2\right ) \left (A e (2 c d-b e)-B \left (c d^2-a e^2\right )\right )}{2 \left (a e^2-b d e+c d^2\right )^2}+\frac{B d-A e}{(d+e x) \left (a e^2-b d e+c d^2\right )}+\frac{\log (d+e x) \left (A e (2 c d-b e)-B \left (c d^2-a e^2\right )\right )}{\left (a e^2-b d e+c d^2\right )^2} \]

[Out]

(B*d - A*e)/((c*d^2 - b*d*e + a*e^2)*(d + e*x)) - ((A*b^2*e^2 + 2*c*(A*c*d^2 + 2*a*B*d*e - a*A*e^2) - b*(B*c*d
^2 + 2*A*c*d*e + a*B*e^2))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(Sqrt[b^2 - 4*a*c]*(c*d^2 - b*d*e + a*e^2)^
2) + ((A*e*(2*c*d - b*e) - B*(c*d^2 - a*e^2))*Log[d + e*x])/(c*d^2 - b*d*e + a*e^2)^2 - ((A*e*(2*c*d - b*e) -
B*(c*d^2 - a*e^2))*Log[a + b*x + c*x^2])/(2*(c*d^2 - b*d*e + a*e^2)^2)

________________________________________________________________________________________

Rubi [A]  time = 0.444042, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {800, 634, 618, 206, 628} \[ -\frac{\tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \left (-b \left (a B e^2+2 A c d e+B c d^2\right )+2 c \left (-a A e^2+2 a B d e+A c d^2\right )+A b^2 e^2\right )}{\sqrt{b^2-4 a c} \left (a e^2-b d e+c d^2\right )^2}-\frac{\log \left (a+b x+c x^2\right ) \left (A e (2 c d-b e)-B \left (c d^2-a e^2\right )\right )}{2 \left (a e^2-b d e+c d^2\right )^2}+\frac{B d-A e}{(d+e x) \left (a e^2-b d e+c d^2\right )}+\frac{\log (d+e x) \left (A e (2 c d-b e)-B \left (c d^2-a e^2\right )\right )}{\left (a e^2-b d e+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*d - A*e)/((c*d^2 - b*d*e + a*e^2)*(d + e*x)) - ((A*b^2*e^2 + 2*c*(A*c*d^2 + 2*a*B*d*e - a*A*e^2) - b*(B*c*d
^2 + 2*A*c*d*e + a*B*e^2))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(Sqrt[b^2 - 4*a*c]*(c*d^2 - b*d*e + a*e^2)^
2) + ((A*e*(2*c*d - b*e) - B*(c*d^2 - a*e^2))*Log[d + e*x])/(c*d^2 - b*d*e + a*e^2)^2 - ((A*e*(2*c*d - b*e) -
B*(c*d^2 - a*e^2))*Log[a + b*x + c*x^2])/(2*(c*d^2 - b*d*e + a*e^2)^2)

Rule 800

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] && IntegerQ[m]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [A]  time = 0.426515, size = 219, normalized size = 0.86 \[ \frac{\frac{2 \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right ) \left (-b \left (a B e^2+2 A c d e+B c d^2\right )+2 c \left (-a A e^2+2 a B d e+A c d^2\right )+A b^2 e^2\right )}{\sqrt{4 a c-b^2}}-2 \log (d+e x) \left (B \left (c d^2-a e^2\right )+A e (b e-2 c d)\right )+\log (a+x (b+c x)) \left (B \left (c d^2-a e^2\right )+A e (b e-2 c d)\right )+\frac{2 (B d-A e) \left (e (a e-b d)+c d^2\right )}{d+e x}}{2 \left (e (a e-b d)+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(B*d - A*e)*(c*d^2 + e*(-(b*d) + a*e)))/(d + e*x) + (2*(A*b^2*e^2 + 2*c*(A*c*d^2 + 2*a*B*d*e - a*A*e^2) -
b*(B*c*d^2 + 2*A*c*d*e + a*B*e^2))*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] - 2*(A*e*(-2*c*d
+ b*e) + B*(c*d^2 - a*e^2))*Log[d + e*x] + (A*e*(-2*c*d + b*e) + B*(c*d^2 - a*e^2))*Log[a + x*(b + c*x)])/(2*
(c*d^2 + e*(-(b*d) + a*e))^2)

________________________________________________________________________________________

Maple [B]  time = 0.01, size = 729, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a*e^2-b*d*e+c*d^2)/(e*x+d)*A*e+1/(a*e^2-b*d*e+c*d^2)/(e*x+d)*B*d-1/(a*e^2-b*d*e+c*d^2)^2*ln(e*x+d)*A*b*e^2
+2/(a*e^2-b*d*e+c*d^2)^2*ln(e*x+d)*A*c*d*e+1/(a*e^2-b*d*e+c*d^2)^2*ln(e*x+d)*a*B*e^2-1/(a*e^2-b*d*e+c*d^2)^2*l
n(e*x+d)*B*c*d^2+1/2/(a*e^2-b*d*e+c*d^2)^2*ln(c*x^2+b*x+a)*A*b*e^2-1/(a*e^2-b*d*e+c*d^2)^2*c*ln(c*x^2+b*x+a)*A
*d*e-1/2/(a*e^2-b*d*e+c*d^2)^2*ln(c*x^2+b*x+a)*a*B*e^2+1/2/(a*e^2-b*d*e+c*d^2)^2*c*ln(c*x^2+b*x+a)*B*d^2-2/(a*
e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*A*e^2*c+1/(a*e^2-b*d*e+c*d^2)^2/(4*
a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*A*b^2*e^2-2/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)^(1/2)*arctan(
(2*c*x+b)/(4*a*c-b^2)^(1/2))*A*b*c*d*e+2/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^
(1/2))*A*c^2*d^2-1/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*B*e^2*a*b+4/(a*
e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*B*a*c*d*e-1/(a*e^2-b*d*e+c*d^2)^2/(4*
a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*B*b*c*d^2

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [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((B*x+A)/(e*x+d)^2/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.10627, size = 517, normalized size = 2.03 \begin{align*} -\frac{{\left (B b c d^{2} e^{2} - 2 \, A c^{2} d^{2} e^{2} - 4 \, B a c d e^{3} + 2 \, A b c d e^{3} + B a b e^{4} - A b^{2} e^{4} + 2 \, A a c e^{4}\right )} \arctan \left (\frac{{\left (2 \, c d - \frac{2 \, c d^{2}}{x e + d} - b e + \frac{2 \, b d e}{x e + d} - \frac{2 \, a e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt{-b^{2} + 4 \, a c}}\right ) e^{\left (-2\right )}}{{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{{\left (B c d^{2} - 2 \, A c d e - B a e^{2} + A b e^{2}\right )} \log \left (c - \frac{2 \, c d}{x e + d} + \frac{c d^{2}}{{\left (x e + d\right )}^{2}} + \frac{b e}{x e + d} - \frac{b d e}{{\left (x e + d\right )}^{2}} + \frac{a e^{2}}{{\left (x e + d\right )}^{2}}\right )}{2 \,{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )}} + \frac{\frac{B d e^{2}}{x e + d} - \frac{A e^{3}}{x e + d}}{c d^{2} e^{2} - b d e^{3} + a e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(B*b*c*d^2*e^2 - 2*A*c^2*d^2*e^2 - 4*B*a*c*d*e^3 + 2*A*b*c*d*e^3 + B*a*b*e^4 - A*b^2*e^4 + 2*A*a*c*e^4)*arcta
n((2*c*d - 2*c*d^2/(x*e + d) - b*e + 2*b*d*e/(x*e + d) - 2*a*e^2/(x*e + d))*e^(-1)/sqrt(-b^2 + 4*a*c))*e^(-2)/
((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + 2*a*c*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*sqrt(-b^2 + 4*a*c)) + 1/2*(B*c*
d^2 - 2*A*c*d*e - B*a*e^2 + A*b*e^2)*log(c - 2*c*d/(x*e + d) + c*d^2/(x*e + d)^2 + b*e/(x*e + d) - b*d*e/(x*e
+ d)^2 + a*e^2/(x*e + d)^2)/(c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + 2*a*c*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4) + (B
*d*e^2/(x*e + d) - A*e^3/(x*e + d))/(c*d^2*e^2 - b*d*e^3 + a*e^4)