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

Optimal. Leaf size=146 \[ \frac {\tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) (-2 a B e+A b e-2 A c d+b B d)}{\sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )}+\frac {(B d-A e) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )}-\frac {(B d-A e) \log (d+e x)}{a e^2-b d e+c d^2} \]

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Rubi [A]  time = 0.22, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {800, 634, 618, 206, 628} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) (-2 a B e+A b e-2 A c d+b B d)}{\sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )}+\frac {(B d-A e) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )}-\frac {(B d-A e) \log (d+e x)}{a e^2-b d e+c d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 125, normalized size = 0.86 \begin {gather*} \frac {\sqrt {4 a c-b^2} (B d-A e) (2 \log (d+e x)-\log (a+x (b+c x)))+2 \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right ) (-2 a B e+A b e-2 A c d+b B d)}{2 \sqrt {4 a c-b^2} \left (e (b d-a e)-c d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x}{(d+e x) \left (a+b x+c x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 4.81, size = 409, normalized size = 2.80 \begin {gather*} \left [\frac {\sqrt {b^{2} - 4 \, a c} {\left ({\left (B b - 2 \, A c\right )} d - {\left (2 \, B a - A b\right )} e\right )} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + {\left ({\left (B b^{2} - 4 \, B a c\right )} d - {\left (A b^{2} - 4 \, A a c\right )} e\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left ({\left (B b^{2} - 4 \, B a c\right )} d - {\left (A b^{2} - 4 \, A a c\right )} e\right )} \log \left (e x + d\right )}{2 \, {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} - {\left (b^{3} - 4 \, a b c\right )} d e + {\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\right )}}, \frac {2 \, \sqrt {-b^{2} + 4 \, a c} {\left ({\left (B b - 2 \, A c\right )} d - {\left (2 \, B a - A b\right )} e\right )} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + {\left ({\left (B b^{2} - 4 \, B a c\right )} d - {\left (A b^{2} - 4 \, A a c\right )} e\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left ({\left (B b^{2} - 4 \, B a c\right )} d - {\left (A b^{2} - 4 \, A a c\right )} e\right )} \log \left (e x + d\right )}{2 \, {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} - {\left (b^{3} - 4 \, a b c\right )} d e + {\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(b^2 - 4*a*c)*((B*b - 2*A*c)*d - (2*B*a - A*b)*e)*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c + sqrt(b^2
- 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) + ((B*b^2 - 4*B*a*c)*d - (A*b^2 - 4*A*a*c)*e)*log(c*x^2 + b*x + a) -
2*((B*b^2 - 4*B*a*c)*d - (A*b^2 - 4*A*a*c)*e)*log(e*x + d))/((b^2*c - 4*a*c^2)*d^2 - (b^3 - 4*a*b*c)*d*e + (a*
b^2 - 4*a^2*c)*e^2), 1/2*(2*sqrt(-b^2 + 4*a*c)*((B*b - 2*A*c)*d - (2*B*a - A*b)*e)*arctan(-sqrt(-b^2 + 4*a*c)*
(2*c*x + b)/(b^2 - 4*a*c)) + ((B*b^2 - 4*B*a*c)*d - (A*b^2 - 4*A*a*c)*e)*log(c*x^2 + b*x + a) - 2*((B*b^2 - 4*
B*a*c)*d - (A*b^2 - 4*A*a*c)*e)*log(e*x + d))/((b^2*c - 4*a*c^2)*d^2 - (b^3 - 4*a*b*c)*d*e + (a*b^2 - 4*a^2*c)
*e^2)]

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giac [A]  time = 0.16, size = 155, normalized size = 1.06 \begin {gather*} \frac {{\left (B d - A e\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (c d^{2} - b d e + a e^{2}\right )}} - \frac {{\left (B d e - A e^{2}\right )} \log \left ({\left | x e + d \right |}\right )}{c d^{2} e - b d e^{2} + a e^{3}} - \frac {{\left (B b d - 2 \, A c d - 2 \, B a e + A b e\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 343, normalized size = 2.35 \begin {gather*} -\frac {A b e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}}+\frac {2 A c d \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}}+\frac {2 B a e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}}-\frac {B b d \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}}+\frac {A e \ln \left (e x +d \right )}{a \,e^{2}-b d e +c \,d^{2}}-\frac {A e \ln \left (c \,x^{2}+b x +a \right )}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right )}-\frac {B d \ln \left (e x +d \right )}{a \,e^{2}-b d e +c \,d^{2}}+\frac {B d \ln \left (c \,x^{2}+b x +a \right )}{2 a \,e^{2}-2 b d e +2 c \,d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 8.70, size = 1027, normalized size = 7.03 \begin {gather*} \frac {\ln \left (d+e\,x\right )\,\left (A\,e-B\,d\right )}{c\,d^2-b\,d\,e+a\,e^2}-\frac {\ln \left (B^2\,c\,e\,x-\frac {\left (B\,a\,c\,e^2-A\,b\,c\,e^2-A\,c^2\,d\,e+c\,e\,x\,\left (B\,b\,e-3\,A\,c\,e+B\,c\,d\right )+B\,b\,c\,d\,e+\frac {c\,e\,\left (b^2\,d\,e+2\,x\,b^2\,e^2+b\,c\,d^2-2\,x\,b\,c\,d\,e+a\,b\,e^2+2\,x\,c^2\,d^2-8\,a\,c\,d\,e-6\,a\,x\,c\,e^2\right )\,\left (\frac {A\,b^2\,e}{2}-\frac {B\,b^2\,d}{2}+\frac {A\,b\,e\,\sqrt {b^2-4\,a\,c}}{2}-A\,c\,d\,\sqrt {b^2-4\,a\,c}-B\,a\,e\,\sqrt {b^2-4\,a\,c}+\frac {B\,b\,d\,\sqrt {b^2-4\,a\,c}}{2}-2\,A\,a\,c\,e+2\,B\,a\,c\,d\right )}{\left (4\,a\,c-b^2\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}\right )\,\left (\frac {A\,b^2\,e}{2}-\frac {B\,b^2\,d}{2}+\frac {A\,b\,e\,\sqrt {b^2-4\,a\,c}}{2}-A\,c\,d\,\sqrt {b^2-4\,a\,c}-B\,a\,e\,\sqrt {b^2-4\,a\,c}+\frac {B\,b\,d\,\sqrt {b^2-4\,a\,c}}{2}-2\,A\,a\,c\,e+2\,B\,a\,c\,d\right )}{\left (4\,a\,c-b^2\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}+A\,B\,c\,e\right )\,\left (b\,\left (\frac {A\,e\,\sqrt {b^2-4\,a\,c}}{2}+\frac {B\,d\,\sqrt {b^2-4\,a\,c}}{2}\right )+b^2\,\left (\frac {A\,e}{2}-\frac {B\,d}{2}\right )-A\,c\,d\,\sqrt {b^2-4\,a\,c}-B\,a\,e\,\sqrt {b^2-4\,a\,c}-2\,A\,a\,c\,e+2\,B\,a\,c\,d\right )}{-4\,a^2\,c\,e^2+a\,b^2\,e^2+4\,a\,b\,c\,d\,e-4\,a\,c^2\,d^2-b^3\,d\,e+b^2\,c\,d^2}-\frac {\ln \left (B^2\,c\,e\,x-\frac {\left (B\,a\,c\,e^2-A\,b\,c\,e^2-A\,c^2\,d\,e+c\,e\,x\,\left (B\,b\,e-3\,A\,c\,e+B\,c\,d\right )+B\,b\,c\,d\,e+\frac {c\,e\,\left (b^2\,d\,e+2\,x\,b^2\,e^2+b\,c\,d^2-2\,x\,b\,c\,d\,e+a\,b\,e^2+2\,x\,c^2\,d^2-8\,a\,c\,d\,e-6\,a\,x\,c\,e^2\right )\,\left (\frac {A\,b^2\,e}{2}-\frac {B\,b^2\,d}{2}-\frac {A\,b\,e\,\sqrt {b^2-4\,a\,c}}{2}+A\,c\,d\,\sqrt {b^2-4\,a\,c}+B\,a\,e\,\sqrt {b^2-4\,a\,c}-\frac {B\,b\,d\,\sqrt {b^2-4\,a\,c}}{2}-2\,A\,a\,c\,e+2\,B\,a\,c\,d\right )}{\left (4\,a\,c-b^2\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}\right )\,\left (\frac {A\,b^2\,e}{2}-\frac {B\,b^2\,d}{2}-\frac {A\,b\,e\,\sqrt {b^2-4\,a\,c}}{2}+A\,c\,d\,\sqrt {b^2-4\,a\,c}+B\,a\,e\,\sqrt {b^2-4\,a\,c}-\frac {B\,b\,d\,\sqrt {b^2-4\,a\,c}}{2}-2\,A\,a\,c\,e+2\,B\,a\,c\,d\right )}{\left (4\,a\,c-b^2\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}+A\,B\,c\,e\right )\,\left (b^2\,\left (\frac {A\,e}{2}-\frac {B\,d}{2}\right )-b\,\left (\frac {A\,e\,\sqrt {b^2-4\,a\,c}}{2}+\frac {B\,d\,\sqrt {b^2-4\,a\,c}}{2}\right )+A\,c\,d\,\sqrt {b^2-4\,a\,c}+B\,a\,e\,\sqrt {b^2-4\,a\,c}-2\,A\,a\,c\,e+2\,B\,a\,c\,d\right )}{-4\,a^2\,c\,e^2+a\,b^2\,e^2+4\,a\,b\,c\,d\,e-4\,a\,c^2\,d^2-b^3\,d\,e+b^2\,c\,d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(d + e*x)*(A*e - B*d))/(a*e^2 + c*d^2 - b*d*e) - (log(B^2*c*e*x - ((B*a*c*e^2 - A*b*c*e^2 - A*c^2*d*e + c*
e*x*(B*b*e - 3*A*c*e + B*c*d) + B*b*c*d*e + (c*e*(2*b^2*e^2*x + 2*c^2*d^2*x + a*b*e^2 + b*c*d^2 + b^2*d*e - 6*
a*c*e^2*x - 8*a*c*d*e - 2*b*c*d*e*x)*((A*b^2*e)/2 - (B*b^2*d)/2 + (A*b*e*(b^2 - 4*a*c)^(1/2))/2 - A*c*d*(b^2 -
4*a*c)^(1/2) - B*a*e*(b^2 - 4*a*c)^(1/2) + (B*b*d*(b^2 - 4*a*c)^(1/2))/2 - 2*A*a*c*e + 2*B*a*c*d))/((4*a*c -
b^2)*(a*e^2 + c*d^2 - b*d*e)))*((A*b^2*e)/2 - (B*b^2*d)/2 + (A*b*e*(b^2 - 4*a*c)^(1/2))/2 - A*c*d*(b^2 - 4*a*c
)^(1/2) - B*a*e*(b^2 - 4*a*c)^(1/2) + (B*b*d*(b^2 - 4*a*c)^(1/2))/2 - 2*A*a*c*e + 2*B*a*c*d))/((4*a*c - b^2)*(
a*e^2 + c*d^2 - b*d*e)) + A*B*c*e)*(b*((A*e*(b^2 - 4*a*c)^(1/2))/2 + (B*d*(b^2 - 4*a*c)^(1/2))/2) + b^2*((A*e)
/2 - (B*d)/2) - A*c*d*(b^2 - 4*a*c)^(1/2) - B*a*e*(b^2 - 4*a*c)^(1/2) - 2*A*a*c*e + 2*B*a*c*d))/(a*b^2*e^2 - 4
*a*c^2*d^2 - 4*a^2*c*e^2 + b^2*c*d^2 - b^3*d*e + 4*a*b*c*d*e) - (log(B^2*c*e*x - ((B*a*c*e^2 - A*b*c*e^2 - A*c
^2*d*e + c*e*x*(B*b*e - 3*A*c*e + B*c*d) + B*b*c*d*e + (c*e*(2*b^2*e^2*x + 2*c^2*d^2*x + a*b*e^2 + b*c*d^2 + b
^2*d*e - 6*a*c*e^2*x - 8*a*c*d*e - 2*b*c*d*e*x)*((A*b^2*e)/2 - (B*b^2*d)/2 - (A*b*e*(b^2 - 4*a*c)^(1/2))/2 + A
*c*d*(b^2 - 4*a*c)^(1/2) + B*a*e*(b^2 - 4*a*c)^(1/2) - (B*b*d*(b^2 - 4*a*c)^(1/2))/2 - 2*A*a*c*e + 2*B*a*c*d))
/((4*a*c - b^2)*(a*e^2 + c*d^2 - b*d*e)))*((A*b^2*e)/2 - (B*b^2*d)/2 - (A*b*e*(b^2 - 4*a*c)^(1/2))/2 + A*c*d*(
b^2 - 4*a*c)^(1/2) + B*a*e*(b^2 - 4*a*c)^(1/2) - (B*b*d*(b^2 - 4*a*c)^(1/2))/2 - 2*A*a*c*e + 2*B*a*c*d))/((4*a
*c - b^2)*(a*e^2 + c*d^2 - b*d*e)) + A*B*c*e)*(b^2*((A*e)/2 - (B*d)/2) - b*((A*e*(b^2 - 4*a*c)^(1/2))/2 + (B*d
*(b^2 - 4*a*c)^(1/2))/2) + A*c*d*(b^2 - 4*a*c)^(1/2) + B*a*e*(b^2 - 4*a*c)^(1/2) - 2*A*a*c*e + 2*B*a*c*d))/(a*
b^2*e^2 - 4*a*c^2*d^2 - 4*a^2*c*e^2 + b^2*c*d^2 - b^3*d*e + 4*a*b*c*d*e)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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