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

Optimal. Leaf size=195 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (a B e \left (c d^2-a e^2\right )-A c d \left (3 a e^2+c d^2\right )\right )}{2 a^{3/2} \sqrt {c} \left (a e^2+c d^2\right )^2}-\frac {a (B d-A e)-x (a B e+A c d)}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}+\frac {e^2 \log \left (a+c x^2\right ) (B d-A e)}{2 \left (a e^2+c d^2\right )^2}-\frac {e^2 (B d-A e) \log (d+e x)}{\left (a e^2+c d^2\right )^2} \]

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Rubi [A]  time = 0.28, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {823, 801, 635, 205, 260} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (a B e \left (c d^2-a e^2\right )-A c d \left (3 a e^2+c d^2\right )\right )}{2 a^{3/2} \sqrt {c} \left (a e^2+c d^2\right )^2}-\frac {a (B d-A e)-x (a B e+A c d)}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}+\frac {e^2 \log \left (a+c x^2\right ) (B d-A e)}{2 \left (a e^2+c d^2\right )^2}-\frac {e^2 (B d-A e) \log (d+e x)}{\left (a e^2+c d^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(B*d - A*e) - (A*c*d + a*B*e)*x)/(2*a*(c*d^2 + a*e^2)*(a + c*x^2)) - ((a*B*e*(c*d^2 - a*e^2) - A*c*d*(c*d^
2 + 3*a*e^2))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(2*a^(3/2)*Sqrt[c]*(c*d^2 + a*e^2)^2) - (e^2*(B*d - A*e)*Log[d + e*
x])/(c*d^2 + a*e^2)^2 + (e^2*(B*d - A*e)*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]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rubi steps

\begin {align*} \int \frac {A+B x}{(d+e x) \left (a+c x^2\right )^2} \, dx &=-\frac {a (B d-A e)-(A c d+a B e) x}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\int \frac {c \left (a B d e-A \left (c d^2+2 a e^2\right )\right )-c e (A c d+a B e) x}{(d+e x) \left (a+c x^2\right )} \, dx}{2 a c \left (c d^2+a e^2\right )}\\ &=-\frac {a (B d-A e)-(A c d+a B e) x}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\int \left (-\frac {2 a c e^3 (-B d+A e)}{\left (c d^2+a e^2\right ) (d+e x)}+\frac {c \left (a B e \left (c d^2-a e^2\right )-A c d \left (c d^2+3 a e^2\right )-2 a c e^2 (B d-A e) x\right )}{\left (c d^2+a e^2\right ) \left (a+c x^2\right )}\right ) \, dx}{2 a c \left (c d^2+a e^2\right )}\\ &=-\frac {a (B d-A e)-(A c d+a B e) x}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {e^2 (B d-A e) \log (d+e x)}{\left (c d^2+a e^2\right )^2}-\frac {\int \frac {a B e \left (c d^2-a e^2\right )-A c d \left (c d^2+3 a e^2\right )-2 a c e^2 (B d-A e) x}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )^2}\\ &=-\frac {a (B d-A e)-(A c d+a B e) x}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {e^2 (B d-A e) \log (d+e x)}{\left (c d^2+a e^2\right )^2}+\frac {\left (c e^2 (B d-A e)\right ) \int \frac {x}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^2}-\frac {\left (a B e \left (c d^2-a e^2\right )-A c d \left (c d^2+3 a e^2\right )\right ) \int \frac {1}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )^2}\\ &=-\frac {a (B d-A e)-(A c d+a B e) x}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\left (a B e \left (c d^2-a e^2\right )-A c d \left (c d^2+3 a e^2\right )\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {c} \left (c d^2+a e^2\right )^2}-\frac {e^2 (B d-A e) \log (d+e x)}{\left (c d^2+a e^2\right )^2}+\frac {e^2 (B d-A e) \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 = 0.16, size = 158, normalized size = 0.81 \begin {gather*} \frac {\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (3 a e^2+c d^2\right )+a B e \left (a e^2-c d^2\right )\right )}{a^{3/2} \sqrt {c}}+\frac {\left (a e^2+c d^2\right ) (a (A e-B d+B e x)+A c d x)}{a \left (a+c x^2\right )}+e^2 \log \left (a+c x^2\right ) (B d-A e)+2 e^2 (A e-B d) \log (d+e x)}{2 \left (a e^2+c d^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((c*d^2 + a*e^2)*(A*c*d*x + a*(-(B*d) + A*e + B*e*x)))/(a*(a + c*x^2)) + ((a*B*e*(-(c*d^2) + a*e^2) + A*c*d*(
c*d^2 + 3*a*e^2))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(a^(3/2)*Sqrt[c]) + 2*e^2*(-(B*d) + A*e)*Log[d + e*x] + e^2*(B*
d - A*e)*Log[a + c*x^2])/(2*(c*d^2 + a*e^2)^2)

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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+c x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 16.50, size = 795, normalized size = 4.08 \begin {gather*} \left [-\frac {2 \, B a^{2} c^{2} d^{3} - 2 \, A a^{2} c^{2} d^{2} e + 2 \, B a^{3} c d e^{2} - 2 \, A a^{3} c e^{3} + {\left (A a c^{2} d^{3} - B a^{2} c d^{2} e + 3 \, A a^{2} c d e^{2} + B a^{3} e^{3} + {\left (A c^{3} d^{3} - B a c^{2} d^{2} e + 3 \, A a c^{2} d e^{2} + B a^{2} c e^{3}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) - 2 \, {\left (A a c^{3} d^{3} + B a^{2} c^{2} d^{2} e + A a^{2} c^{2} d e^{2} + B a^{3} c e^{3}\right )} x - 2 \, {\left (B a^{3} c d e^{2} - A a^{3} c e^{3} + {\left (B a^{2} c^{2} d e^{2} - A a^{2} c^{2} e^{3}\right )} x^{2}\right )} \log \left (c x^{2} + a\right ) + 4 \, {\left (B a^{3} c d e^{2} - A a^{3} c e^{3} + {\left (B a^{2} c^{2} d e^{2} - A a^{2} c^{2} e^{3}\right )} x^{2}\right )} \log \left (e x + d\right )}{4 \, {\left (a^{3} c^{3} d^{4} + 2 \, a^{4} c^{2} d^{2} e^{2} + a^{5} c e^{4} + {\left (a^{2} c^{4} d^{4} + 2 \, a^{3} c^{3} d^{2} e^{2} + a^{4} c^{2} e^{4}\right )} x^{2}\right )}}, -\frac {B a^{2} c^{2} d^{3} - A a^{2} c^{2} d^{2} e + B a^{3} c d e^{2} - A a^{3} c e^{3} - {\left (A a c^{2} d^{3} - B a^{2} c d^{2} e + 3 \, A a^{2} c d e^{2} + B a^{3} e^{3} + {\left (A c^{3} d^{3} - B a c^{2} d^{2} e + 3 \, A a c^{2} d e^{2} + B a^{2} c e^{3}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - {\left (A a c^{3} d^{3} + B a^{2} c^{2} d^{2} e + A a^{2} c^{2} d e^{2} + B a^{3} c e^{3}\right )} x - {\left (B a^{3} c d e^{2} - A a^{3} c e^{3} + {\left (B a^{2} c^{2} d e^{2} - A a^{2} c^{2} e^{3}\right )} x^{2}\right )} \log \left (c x^{2} + a\right ) + 2 \, {\left (B a^{3} c d e^{2} - A a^{3} c e^{3} + {\left (B a^{2} c^{2} d e^{2} - A a^{2} c^{2} e^{3}\right )} x^{2}\right )} \log \left (e x + d\right )}{2 \, {\left (a^{3} c^{3} d^{4} + 2 \, a^{4} c^{2} d^{2} e^{2} + a^{5} c e^{4} + {\left (a^{2} c^{4} d^{4} + 2 \, a^{3} c^{3} d^{2} e^{2} + a^{4} c^{2} e^{4}\right )} x^{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+a)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 268, normalized size = 1.37 \begin {gather*} \frac {{\left (B d e^{2} - A e^{3}\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 d e^{3} - A e^{4}\right )} \log \left ({\left | x e + d \right |}\right )}{c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}} + \frac {{\left (A c^{2} d^{3} - B a c d^{2} e + 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, {\left (a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}\right )} \sqrt {a c}} - \frac {B a c d^{3} - A a c d^{2} e + B a^{2} d e^{2} - A a^{2} e^{3} - {\left (A c^{2} d^{3} + B a c d^{2} e + A a c d e^{2} + B a^{2} e^{3}\right )} x}{2 \, {\left (c d^{2} + a e^{2}\right )}^{2} {\left (c x^{2} + a\right )} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 495, normalized size = 2.54 \begin {gather*} \frac {A \,c^{2} d^{3} x}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (c \,x^{2}+a \right ) a}+\frac {A \,c^{2} d^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {a c}\, a}+\frac {A c d \,e^{2} x}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (c \,x^{2}+a \right )}+\frac {3 A c d \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {a c}}+\frac {B a \,e^{3} x}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (c \,x^{2}+a \right )}+\frac {B a \,e^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {a c}}+\frac {B c \,d^{2} e x}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (c \,x^{2}+a \right )}-\frac {B c \,d^{2} e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {a c}}+\frac {A a \,e^{3}}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (c \,x^{2}+a \right )}+\frac {A c \,d^{2} e}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (c \,x^{2}+a \right )}-\frac {A \,e^{3} \ln \left (c \,x^{2}+a \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2}}+\frac {A \,e^{3} \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2}}-\frac {B a d \,e^{2}}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (c \,x^{2}+a \right )}-\frac {B c \,d^{3}}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (c \,x^{2}+a \right )}+\frac {B d \,e^{2} \ln \left (c \,x^{2}+a \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2}}-\frac {B d \,e^{2} \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.34, size = 243, normalized size = 1.25 \begin {gather*} \frac {{\left (B d e^{2} - A e^{3}\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 d e^{2} - A e^{3}\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^{3} - B a c d^{2} e + 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, {\left (a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}\right )} \sqrt {a c}} - \frac {B a d - A a e - {\left (A c d + B a e\right )} x}{2 \, {\left (a^{2} c d^{2} + a^{3} e^{2} + {\left (a c^{2} d^{2} + a^{2} c e^{2}\right )} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.76, size = 1086, normalized size = 5.57 \begin {gather*} \frac {\frac {A\,e-B\,d}{2\,\left (c\,d^2+a\,e^2\right )}+\frac {x\,\left (A\,c\,d+B\,a\,e\right )}{2\,a\,\left (c\,d^2+a\,e^2\right )}}{c\,x^2+a}-\frac {\ln \left (A\,c^3\,d^5\,\sqrt {-a^3\,c}-B\,a^3\,e^5\,\sqrt {-a^3\,c}-6\,A\,a^4\,c\,e^5+B\,a^4\,c\,e^5\,x+2\,A\,a^2\,c^3\,d^4\,e+12\,A\,a^3\,c^2\,d^2\,e^3-8\,B\,a^3\,c^2\,d^3\,e^2+8\,B\,a^4\,c\,d\,e^4-A\,a\,c^4\,d^5\,x-2\,A\,a^2\,c^3\,d^3\,e^2\,x-14\,B\,a^3\,c^2\,d^2\,e^3\,x+2\,A\,a\,c^2\,d^3\,e^2\,\sqrt {-a^3\,c}+14\,B\,a^2\,c\,d^2\,e^3\,\sqrt {-a^3\,c}+15\,A\,a^3\,c^2\,d\,e^4\,x+B\,a^2\,c^3\,d^4\,e\,x-15\,A\,a^2\,c\,d\,e^4\,\sqrt {-a^3\,c}-B\,a\,c^2\,d^4\,e\,\sqrt {-a^3\,c}-6\,A\,a^2\,c\,e^5\,x\,\sqrt {-a^3\,c}+2\,A\,c^3\,d^4\,e\,x\,\sqrt {-a^3\,c}+8\,B\,a^2\,c\,d\,e^4\,x\,\sqrt {-a^3\,c}+12\,A\,a\,c^2\,d^2\,e^3\,x\,\sqrt {-a^3\,c}-8\,B\,a\,c^2\,d^3\,e^2\,x\,\sqrt {-a^3\,c}\right )\,\left (c\,\left (a\,\left (\frac {3\,A\,d\,e^2\,\sqrt {-a^3\,c}}{4}-\frac {B\,d^2\,e\,\sqrt {-a^3\,c}}{4}\right )+a^3\,\left (\frac {A\,e^3}{2}-\frac {B\,d\,e^2}{2}\right )\right )+\frac {A\,c^2\,d^3\,\sqrt {-a^3\,c}}{4}+\frac {B\,a^2\,e^3\,\sqrt {-a^3\,c}}{4}\right )}{a^5\,c\,e^4+2\,a^4\,c^2\,d^2\,e^2+a^3\,c^3\,d^4}+\frac {\ln \left (A\,c^3\,d^5\,\sqrt {-a^3\,c}-B\,a^3\,e^5\,\sqrt {-a^3\,c}+6\,A\,a^4\,c\,e^5-B\,a^4\,c\,e^5\,x-2\,A\,a^2\,c^3\,d^4\,e-12\,A\,a^3\,c^2\,d^2\,e^3+8\,B\,a^3\,c^2\,d^3\,e^2-8\,B\,a^4\,c\,d\,e^4+A\,a\,c^4\,d^5\,x+2\,A\,a^2\,c^3\,d^3\,e^2\,x+14\,B\,a^3\,c^2\,d^2\,e^3\,x+2\,A\,a\,c^2\,d^3\,e^2\,\sqrt {-a^3\,c}+14\,B\,a^2\,c\,d^2\,e^3\,\sqrt {-a^3\,c}-15\,A\,a^3\,c^2\,d\,e^4\,x-B\,a^2\,c^3\,d^4\,e\,x-15\,A\,a^2\,c\,d\,e^4\,\sqrt {-a^3\,c}-B\,a\,c^2\,d^4\,e\,\sqrt {-a^3\,c}-6\,A\,a^2\,c\,e^5\,x\,\sqrt {-a^3\,c}+2\,A\,c^3\,d^4\,e\,x\,\sqrt {-a^3\,c}+8\,B\,a^2\,c\,d\,e^4\,x\,\sqrt {-a^3\,c}+12\,A\,a\,c^2\,d^2\,e^3\,x\,\sqrt {-a^3\,c}-8\,B\,a\,c^2\,d^3\,e^2\,x\,\sqrt {-a^3\,c}\right )\,\left (c\,\left (a\,\left (\frac {3\,A\,d\,e^2\,\sqrt {-a^3\,c}}{4}-\frac {B\,d^2\,e\,\sqrt {-a^3\,c}}{4}\right )-a^3\,\left (\frac {A\,e^3}{2}-\frac {B\,d\,e^2}{2}\right )\right )+\frac {A\,c^2\,d^3\,\sqrt {-a^3\,c}}{4}+\frac {B\,a^2\,e^3\,\sqrt {-a^3\,c}}{4}\right )}{a^5\,c\,e^4+2\,a^4\,c^2\,d^2\,e^2+a^3\,c^3\,d^4}+\frac {\ln \left (d+e\,x\right )\,\left (A\,e^3-B\,d\,e^2\right )}{a^2\,e^4+2\,a\,c\,d^2\,e^2+c^2\,d^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((A*e - B*d)/(2*(a*e^2 + c*d^2)) + (x*(A*c*d + B*a*e))/(2*a*(a*e^2 + c*d^2)))/(a + c*x^2) - (log(A*c^3*d^5*(-a
^3*c)^(1/2) - B*a^3*e^5*(-a^3*c)^(1/2) - 6*A*a^4*c*e^5 + B*a^4*c*e^5*x + 2*A*a^2*c^3*d^4*e + 12*A*a^3*c^2*d^2*
e^3 - 8*B*a^3*c^2*d^3*e^2 + 8*B*a^4*c*d*e^4 - A*a*c^4*d^5*x - 2*A*a^2*c^3*d^3*e^2*x - 14*B*a^3*c^2*d^2*e^3*x +
2*A*a*c^2*d^3*e^2*(-a^3*c)^(1/2) + 14*B*a^2*c*d^2*e^3*(-a^3*c)^(1/2) + 15*A*a^3*c^2*d*e^4*x + B*a^2*c^3*d^4*e
*x - 15*A*a^2*c*d*e^4*(-a^3*c)^(1/2) - B*a*c^2*d^4*e*(-a^3*c)^(1/2) - 6*A*a^2*c*e^5*x*(-a^3*c)^(1/2) + 2*A*c^3
*d^4*e*x*(-a^3*c)^(1/2) + 8*B*a^2*c*d*e^4*x*(-a^3*c)^(1/2) + 12*A*a*c^2*d^2*e^3*x*(-a^3*c)^(1/2) - 8*B*a*c^2*d
^3*e^2*x*(-a^3*c)^(1/2))*(c*(a*((3*A*d*e^2*(-a^3*c)^(1/2))/4 - (B*d^2*e*(-a^3*c)^(1/2))/4) + a^3*((A*e^3)/2 -
(B*d*e^2)/2)) + (A*c^2*d^3*(-a^3*c)^(1/2))/4 + (B*a^2*e^3*(-a^3*c)^(1/2))/4))/(a^5*c*e^4 + a^3*c^3*d^4 + 2*a^4
*c^2*d^2*e^2) + (log(A*c^3*d^5*(-a^3*c)^(1/2) - B*a^3*e^5*(-a^3*c)^(1/2) + 6*A*a^4*c*e^5 - B*a^4*c*e^5*x - 2*A
*a^2*c^3*d^4*e - 12*A*a^3*c^2*d^2*e^3 + 8*B*a^3*c^2*d^3*e^2 - 8*B*a^4*c*d*e^4 + A*a*c^4*d^5*x + 2*A*a^2*c^3*d^
3*e^2*x + 14*B*a^3*c^2*d^2*e^3*x + 2*A*a*c^2*d^3*e^2*(-a^3*c)^(1/2) + 14*B*a^2*c*d^2*e^3*(-a^3*c)^(1/2) - 15*A
*a^3*c^2*d*e^4*x - B*a^2*c^3*d^4*e*x - 15*A*a^2*c*d*e^4*(-a^3*c)^(1/2) - B*a*c^2*d^4*e*(-a^3*c)^(1/2) - 6*A*a^
2*c*e^5*x*(-a^3*c)^(1/2) + 2*A*c^3*d^4*e*x*(-a^3*c)^(1/2) + 8*B*a^2*c*d*e^4*x*(-a^3*c)^(1/2) + 12*A*a*c^2*d^2*
e^3*x*(-a^3*c)^(1/2) - 8*B*a*c^2*d^3*e^2*x*(-a^3*c)^(1/2))*(c*(a*((3*A*d*e^2*(-a^3*c)^(1/2))/4 - (B*d^2*e*(-a^
3*c)^(1/2))/4) - a^3*((A*e^3)/2 - (B*d*e^2)/2)) + (A*c^2*d^3*(-a^3*c)^(1/2))/4 + (B*a^2*e^3*(-a^3*c)^(1/2))/4)
)/(a^5*c*e^4 + a^3*c^3*d^4 + 2*a^4*c^2*d^2*e^2) + (log(d + e*x)*(A*e^3 - B*d*e^2))/(a^2*e^4 + c^2*d^4 + 2*a*c*
d^2*e^2)

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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+a)**2,x)

[Out]

Timed out

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