3.1336 \(\int \frac {A+B x}{(d+e x)^2 (a+c x^2)} \, dx\)

Optimal. Leaf size=173 \[ \frac {\log \left (a+c x^2\right ) \left (-a B e^2-2 A c d e+B c d^2\right )}{2 \left (a e^2+c d^2\right )^2}+\frac {B d-A e}{(d+e x) \left (a e^2+c d^2\right )}-\frac {\log (d+e x) \left (-a B e^2-2 A c d e+B c d^2\right )}{\left (a e^2+c d^2\right )^2}+\frac {\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (-a A e^2+2 a B d e+A c d^2\right )}{\sqrt {a} \left (a e^2+c d^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {801, 635, 205, 260} \[ \frac {\log \left (a+c x^2\right ) \left (-a B e^2-2 A c d e+B c d^2\right )}{2 \left (a e^2+c d^2\right )^2}+\frac {B d-A e}{(d+e x) \left (a e^2+c d^2\right )}-\frac {\log (d+e x) \left (-a B e^2-2 A c d e+B c d^2\right )}{\left (a e^2+c d^2\right )^2}+\frac {\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (-a A e^2+2 a B d e+A c d^2\right )}{\sqrt {a} \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

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

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 148, normalized size = 0.86 \[ \frac {\log \left (a+c x^2\right ) \left (-a B e^2-2 A c d e+B c d^2\right )+\frac {2 \left (a e^2+c d^2\right ) (B d-A e)}{d+e x}+\log (d+e x) \left (2 a B e^2+4 A c d e-2 B c d^2\right )+\frac {2 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (-a A e^2+2 a B d e+A c d^2\right )}{\sqrt {a}}}{2 \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 9.62, size = 562, normalized size = 3.25 \[ \left [\frac {2 \, B c d^{3} - 2 \, A c d^{2} e + 2 \, B a d e^{2} - 2 \, A a e^{3} - {\left (A c d^{3} + 2 \, B a d^{2} e - A a d e^{2} + {\left (A c d^{2} e + 2 \, B a d e^{2} - A a e^{3}\right )} x\right )} \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{2} - 2 \, a x \sqrt {-\frac {c}{a}} - a}{c x^{2} + a}\right ) + {\left (B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} + {\left (B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x\right )} \log \left (c x^{2} + a\right ) - 2 \, {\left (B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} + {\left (B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4} + {\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x\right )}}, \frac {2 \, B c d^{3} - 2 \, A c d^{2} e + 2 \, B a d e^{2} - 2 \, A a e^{3} + 2 \, {\left (A c d^{3} + 2 \, B a d^{2} e - A a d e^{2} + {\left (A c d^{2} e + 2 \, B a d e^{2} - A a e^{3}\right )} x\right )} \sqrt {\frac {c}{a}} \arctan \left (x \sqrt {\frac {c}{a}}\right ) + {\left (B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} + {\left (B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x\right )} \log \left (c x^{2} + a\right ) - 2 \, {\left (B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} + {\left (B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4} + {\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*B*c*d^3 - 2*A*c*d^2*e + 2*B*a*d*e^2 - 2*A*a*e^3 - (A*c*d^3 + 2*B*a*d^2*e - A*a*d*e^2 + (A*c*d^2*e + 2*
B*a*d*e^2 - A*a*e^3)*x)*sqrt(-c/a)*log((c*x^2 - 2*a*x*sqrt(-c/a) - a)/(c*x^2 + a)) + (B*c*d^3 - 2*A*c*d^2*e -
B*a*d*e^2 + (B*c*d^2*e - 2*A*c*d*e^2 - B*a*e^3)*x)*log(c*x^2 + a) - 2*(B*c*d^3 - 2*A*c*d^2*e - B*a*d*e^2 + (B*
c*d^2*e - 2*A*c*d*e^2 - B*a*e^3)*x)*log(e*x + d))/(c^2*d^5 + 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e + 2*a*c*d^
2*e^3 + a^2*e^5)*x), 1/2*(2*B*c*d^3 - 2*A*c*d^2*e + 2*B*a*d*e^2 - 2*A*a*e^3 + 2*(A*c*d^3 + 2*B*a*d^2*e - A*a*d
*e^2 + (A*c*d^2*e + 2*B*a*d*e^2 - A*a*e^3)*x)*sqrt(c/a)*arctan(x*sqrt(c/a)) + (B*c*d^3 - 2*A*c*d^2*e - B*a*d*e
^2 + (B*c*d^2*e - 2*A*c*d*e^2 - B*a*e^3)*x)*log(c*x^2 + a) - 2*(B*c*d^3 - 2*A*c*d^2*e - B*a*d*e^2 + (B*c*d^2*e
- 2*A*c*d*e^2 - B*a*e^3)*x)*log(e*x + d))/(c^2*d^5 + 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e + 2*a*c*d^2*e^3 +
a^2*e^5)*x)]

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giac [A]  time = 0.16, size = 229, normalized size = 1.32 \[ \frac {{\left (A c^{2} d^{2} e^{2} + 2 \, B a c d e^{3} - A a c e^{4}\right )} \arctan \left (\frac {{\left (c d - \frac {c d^{2}}{x e + d} - \frac {a e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt {a c}}\right ) e^{\left (-2\right )}}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {a c}} + \frac {{\left (B c d^{2} - 2 \, A c d e - B a e^{2}\right )} \log \left (c - \frac {2 \, c d}{x e + d} + \frac {c d^{2}}{{\left (x e + d\right )}^{2}} + \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right )}{2 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}} + \frac {\frac {B d e^{2}}{x e + d} - \frac {A e^{3}}{x e + d}}{c d^{2} e^{2} + a e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 312, normalized size = 1.80 \[ -\frac {A a c \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {a c}}+\frac {A \,c^{2} d^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {a c}}+\frac {2 B a c d e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {a c}}-\frac {A c d e \ln \left (c \,x^{2}+a \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2}}+\frac {2 A c d e \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2}}-\frac {B a \,e^{2} \ln \left (c \,x^{2}+a \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2}}+\frac {B a \,e^{2} \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2}}+\frac {B c \,d^{2} \ln \left (c \,x^{2}+a \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2}}-\frac {B c \,d^{2} \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2}}-\frac {A e}{\left (a \,e^{2}+c \,d^{2}\right ) \left (e x +d \right )}+\frac {B d}{\left (a \,e^{2}+c \,d^{2}\right ) \left (e x +d \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.31, size = 216, normalized size = 1.25 \[ \frac {{\left (B c d^{2} - 2 \, A c d e - B a e^{2}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}} - \frac {{\left (B c d^{2} - 2 \, A c d e - B a e^{2}\right )} \log \left (e x + d\right )}{c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}} + \frac {{\left (A c^{2} d^{2} + 2 \, B a c d e - A a c e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {a c}} + \frac {B d - A e}{c d^{3} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.32, size = 810, normalized size = 4.68 \[ \frac {\ln \left (3\,B\,a^3\,e^4+3\,B\,a\,c^2\,d^4+A\,c^3\,d^4\,x+A\,a^2\,e^4\,\sqrt {-a\,c}+A\,c^2\,d^4\,\sqrt {-a\,c}+14\,A\,d^2\,e^2\,{\left (-a\,c\right )}^{3/2}+A\,a^2\,c\,e^4\,x-8\,B\,a^2\,d\,e^3\,\sqrt {-a\,c}-3\,B\,a^2\,e^4\,x\,\sqrt {-a\,c}-3\,B\,c^2\,d^4\,x\,\sqrt {-a\,c}-10\,B\,a^2\,c\,d^2\,e^2+8\,A\,d\,e^3\,x\,{\left (-a\,c\right )}^{3/2}-8\,A\,a\,c^2\,d^3\,e+8\,A\,a^2\,c\,d\,e^3+8\,B\,a\,c\,d^3\,e\,\sqrt {-a\,c}+8\,B\,a\,c^2\,d^3\,e\,x-8\,B\,a^2\,c\,d\,e^3\,x+8\,A\,c^2\,d^3\,e\,x\,\sqrt {-a\,c}-14\,A\,a\,c^2\,d^2\,e^2\,x+10\,B\,a\,c\,d^2\,e^2\,x\,\sqrt {-a\,c}\right )\,\left (c\,\left (\frac {B\,a\,d^2}{2}+\frac {A\,d^2\,\sqrt {-a\,c}}{2}-A\,a\,d\,e\right )-e^2\,\left (\frac {B\,a^2}{2}+\frac {A\,a\,\sqrt {-a\,c}}{2}\right )+B\,a\,d\,e\,\sqrt {-a\,c}\right )}{a^3\,e^4+2\,a^2\,c\,d^2\,e^2+a\,c^2\,d^4}-\frac {\ln \left (d+e\,x\right )\,\left (c\,\left (B\,d^2-2\,A\,d\,e\right )-B\,a\,e^2\right )}{a^2\,e^4+2\,a\,c\,d^2\,e^2+c^2\,d^4}-\frac {\ln \left (3\,B\,a^3\,e^4+8\,B\,d^3\,e\,{\left (-a\,c\right )}^{3/2}+3\,B\,a\,c^2\,d^4+A\,c^3\,d^4\,x-A\,a^2\,e^4\,\sqrt {-a\,c}-A\,c^2\,d^4\,\sqrt {-a\,c}+A\,a^2\,c\,e^4\,x+8\,B\,a^2\,d\,e^3\,\sqrt {-a\,c}+3\,B\,a^2\,e^4\,x\,\sqrt {-a\,c}+3\,B\,c^2\,d^4\,x\,\sqrt {-a\,c}+10\,B\,d^2\,e^2\,x\,{\left (-a\,c\right )}^{3/2}-10\,B\,a^2\,c\,d^2\,e^2-8\,A\,a\,c^2\,d^3\,e+8\,A\,a^2\,c\,d\,e^3+8\,B\,a\,c^2\,d^3\,e\,x-8\,B\,a^2\,c\,d\,e^3\,x+14\,A\,a\,c\,d^2\,e^2\,\sqrt {-a\,c}-8\,A\,c^2\,d^3\,e\,x\,\sqrt {-a\,c}-14\,A\,a\,c^2\,d^2\,e^2\,x+8\,A\,a\,c\,d\,e^3\,x\,\sqrt {-a\,c}\right )\,\left (e^2\,\left (\frac {B\,a^2}{2}-\frac {A\,a\,\sqrt {-a\,c}}{2}\right )+c\,\left (\frac {A\,d^2\,\sqrt {-a\,c}}{2}-\frac {B\,a\,d^2}{2}+A\,a\,d\,e\right )+B\,a\,d\,e\,\sqrt {-a\,c}\right )}{a^3\,e^4+2\,a^2\,c\,d^2\,e^2+a\,c^2\,d^4}-\frac {A\,e-B\,d}{\left (c\,d^2+a\,e^2\right )\,\left (d+e\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(3*B*a^3*e^4 + 3*B*a*c^2*d^4 + A*c^3*d^4*x + A*a^2*e^4*(-a*c)^(1/2) + A*c^2*d^4*(-a*c)^(1/2) + 14*A*d^2*e^
2*(-a*c)^(3/2) + A*a^2*c*e^4*x - 8*B*a^2*d*e^3*(-a*c)^(1/2) - 3*B*a^2*e^4*x*(-a*c)^(1/2) - 3*B*c^2*d^4*x*(-a*c
)^(1/2) - 10*B*a^2*c*d^2*e^2 + 8*A*d*e^3*x*(-a*c)^(3/2) - 8*A*a*c^2*d^3*e + 8*A*a^2*c*d*e^3 + 8*B*a*c*d^3*e*(-
a*c)^(1/2) + 8*B*a*c^2*d^3*e*x - 8*B*a^2*c*d*e^3*x + 8*A*c^2*d^3*e*x*(-a*c)^(1/2) - 14*A*a*c^2*d^2*e^2*x + 10*
B*a*c*d^2*e^2*x*(-a*c)^(1/2))*(c*((B*a*d^2)/2 + (A*d^2*(-a*c)^(1/2))/2 - A*a*d*e) - e^2*((B*a^2)/2 + (A*a*(-a*
c)^(1/2))/2) + B*a*d*e*(-a*c)^(1/2)))/(a^3*e^4 + a*c^2*d^4 + 2*a^2*c*d^2*e^2) - (log(d + e*x)*(c*(B*d^2 - 2*A*
d*e) - B*a*e^2))/(a^2*e^4 + c^2*d^4 + 2*a*c*d^2*e^2) - (log(3*B*a^3*e^4 + 8*B*d^3*e*(-a*c)^(3/2) + 3*B*a*c^2*d
^4 + A*c^3*d^4*x - A*a^2*e^4*(-a*c)^(1/2) - A*c^2*d^4*(-a*c)^(1/2) + A*a^2*c*e^4*x + 8*B*a^2*d*e^3*(-a*c)^(1/2
) + 3*B*a^2*e^4*x*(-a*c)^(1/2) + 3*B*c^2*d^4*x*(-a*c)^(1/2) + 10*B*d^2*e^2*x*(-a*c)^(3/2) - 10*B*a^2*c*d^2*e^2
- 8*A*a*c^2*d^3*e + 8*A*a^2*c*d*e^3 + 8*B*a*c^2*d^3*e*x - 8*B*a^2*c*d*e^3*x + 14*A*a*c*d^2*e^2*(-a*c)^(1/2) -
8*A*c^2*d^3*e*x*(-a*c)^(1/2) - 14*A*a*c^2*d^2*e^2*x + 8*A*a*c*d*e^3*x*(-a*c)^(1/2))*(e^2*((B*a^2)/2 - (A*a*(-
a*c)^(1/2))/2) + c*((A*d^2*(-a*c)^(1/2))/2 - (B*a*d^2)/2 + A*a*d*e) + B*a*d*e*(-a*c)^(1/2)))/(a^3*e^4 + a*c^2*
d^4 + 2*a^2*c*d^2*e^2) - (A*e - B*d)/((a*e^2 + c*d^2)*(d + e*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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