3.5.31 \(\int \frac {1}{(d+e x)^2 (a+c x^2)^2} \, dx\)

Optimal. Leaf size=205 \[ \frac {\sqrt {c} \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} \left (a e^2+c d^2\right )^3}+\frac {a e+c d x}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}+\frac {e \left (c d^2-3 a e^2\right )}{2 a (d+e x) \left (a e^2+c d^2\right )^2}-\frac {2 c d e^3 \log \left (a+c x^2\right )}{\left (a e^2+c d^2\right )^3}+\frac {4 c d e^3 \log (d+e x)}{\left (a e^2+c d^2\right )^3} \]

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Rubi [A]  time = 0.19, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {741, 801, 635, 205, 260} \begin {gather*} \frac {\sqrt {c} \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} \left (a e^2+c d^2\right )^3}+\frac {a e+c d x}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}-\frac {2 c d e^3 \log \left (a+c x^2\right )}{\left (a e^2+c d^2\right )^3}+\frac {e \left (c d^2-3 a e^2\right )}{2 a (d+e x) \left (a e^2+c d^2\right )^2}+\frac {4 c d e^3 \log (d+e x)}{\left (a e^2+c d^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^2*(a + c*x^2)^2),x]

[Out]

(e*(c*d^2 - 3*a*e^2))/(2*a*(c*d^2 + a*e^2)^2*(d + e*x)) + (a*e + c*d*x)/(2*a*(c*d^2 + a*e^2)*(d + e*x)*(a + c*
x^2)) + (Sqrt[c]*(c^2*d^4 + 6*a*c*d^2*e^2 - 3*a^2*e^4)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(2*a^(3/2)*(c*d^2 + a*e^2)
^3) + (4*c*d*e^3*Log[d + e*x])/(c*d^2 + a*e^2)^3 - (2*c*d*e^3*Log[a + c*x^2])/(c*d^2 + a*e^2)^3

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 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

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 {1}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx &=\frac {a e+c d x}{2 a \left (c d^2+a e^2\right ) (d+e x) \left (a+c x^2\right )}-\frac {\int \frac {-c d^2-3 a e^2-2 c d e x}{(d+e x)^2 \left (a+c x^2\right )} \, dx}{2 a \left (c d^2+a e^2\right )}\\ &=\frac {a e+c d x}{2 a \left (c d^2+a e^2\right ) (d+e x) \left (a+c x^2\right )}-\frac {\int \left (\frac {c d^2 e^2-3 a e^4}{\left (c d^2+a e^2\right ) (d+e x)^2}-\frac {8 a c d e^4}{\left (c d^2+a e^2\right )^2 (d+e x)}-\frac {c \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4-8 a c d e^3 x\right )}{\left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}\right ) \, dx}{2 a \left (c d^2+a e^2\right )}\\ &=\frac {e \left (c d^2-3 a e^2\right )}{2 a \left (c d^2+a e^2\right )^2 (d+e x)}+\frac {a e+c d x}{2 a \left (c d^2+a e^2\right ) (d+e x) \left (a+c x^2\right )}+\frac {4 c d e^3 \log (d+e x)}{\left (c d^2+a e^2\right )^3}+\frac {c \int \frac {c^2 d^4+6 a c d^2 e^2-3 a^2 e^4-8 a c d e^3 x}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )^3}\\ &=\frac {e \left (c d^2-3 a e^2\right )}{2 a \left (c d^2+a e^2\right )^2 (d+e x)}+\frac {a e+c d x}{2 a \left (c d^2+a e^2\right ) (d+e x) \left (a+c x^2\right )}+\frac {4 c d e^3 \log (d+e x)}{\left (c d^2+a e^2\right )^3}-\frac {\left (4 c^2 d e^3\right ) \int \frac {x}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^3}+\frac {\left (c \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )\right ) \int \frac {1}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )^3}\\ &=\frac {e \left (c d^2-3 a e^2\right )}{2 a \left (c d^2+a e^2\right )^2 (d+e x)}+\frac {a e+c d x}{2 a \left (c d^2+a e^2\right ) (d+e x) \left (a+c x^2\right )}+\frac {\sqrt {c} \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} \left (c d^2+a e^2\right )^3}+\frac {4 c d e^3 \log (d+e x)}{\left (c d^2+a e^2\right )^3}-\frac {2 c d e^3 \log \left (a+c x^2\right )}{\left (c d^2+a e^2\right )^3}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 162, normalized size = 0.79 \begin {gather*} \frac {\frac {\sqrt {c} \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{a^{3/2}}+\frac {c \left (a e^2+c d^2\right ) \left (a e (2 d-e x)+c d^2 x\right )}{a \left (a+c x^2\right )}-\frac {2 e^3 \left (a e^2+c d^2\right )}{d+e x}-4 c d e^3 \log \left (a+c x^2\right )+8 c d e^3 \log (d+e x)}{2 \left (a e^2+c d^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^2*(a + c*x^2)^2),x]

[Out]

((-2*e^3*(c*d^2 + a*e^2))/(d + e*x) + (c*(c*d^2 + a*e^2)*(c*d^2*x + a*e*(2*d - e*x)))/(a*(a + c*x^2)) + (Sqrt[
c]*(c^2*d^4 + 6*a*c*d^2*e^2 - 3*a^2*e^4)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/a^(3/2) + 8*c*d*e^3*Log[d + e*x] - 4*c*d
*e^3*Log[a + c*x^2])/(2*(c*d^2 + a*e^2)^3)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/((d + e*x)^2*(a + c*x^2)^2),x]

[Out]

IntegrateAlgebraic[1/((d + e*x)^2*(a + c*x^2)^2), x]

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fricas [B]  time = 2.38, size = 1111, normalized size = 5.42 \begin {gather*} \left [\frac {4 \, a c^{2} d^{4} e - 4 \, a^{3} e^{5} + 2 \, {\left (c^{3} d^{4} e - 2 \, a c^{2} d^{2} e^{3} - 3 \, a^{2} c e^{5}\right )} x^{2} - {\left (a c^{2} d^{5} + 6 \, a^{2} c d^{3} e^{2} - 3 \, a^{3} d e^{4} + {\left (c^{3} d^{4} e + 6 \, a c^{2} d^{2} e^{3} - 3 \, a^{2} c e^{5}\right )} x^{3} + {\left (c^{3} d^{5} + 6 \, a c^{2} d^{3} e^{2} - 3 \, a^{2} c d e^{4}\right )} x^{2} + {\left (a c^{2} d^{4} e + 6 \, a^{2} c d^{2} e^{3} - 3 \, a^{3} e^{5}\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 ) + 2 \, {\left (c^{3} d^{5} + 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} x - 8 \, {\left (a c^{2} d e^{4} x^{3} + a c^{2} d^{2} e^{3} x^{2} + a^{2} c d e^{4} x + a^{2} c d^{2} e^{3}\right )} \log \left (c x^{2} + a\right ) + 16 \, {\left (a c^{2} d e^{4} x^{3} + a c^{2} d^{2} e^{3} x^{2} + a^{2} c d e^{4} x + a^{2} c d^{2} e^{3}\right )} \log \left (e x + d\right )}{4 \, {\left (a^{2} c^{3} d^{7} + 3 \, a^{3} c^{2} d^{5} e^{2} + 3 \, a^{4} c d^{3} e^{4} + a^{5} d e^{6} + {\left (a c^{4} d^{6} e + 3 \, a^{2} c^{3} d^{4} e^{3} + 3 \, a^{3} c^{2} d^{2} e^{5} + a^{4} c e^{7}\right )} x^{3} + {\left (a c^{4} d^{7} + 3 \, a^{2} c^{3} d^{5} e^{2} + 3 \, a^{3} c^{2} d^{3} e^{4} + a^{4} c d e^{6}\right )} x^{2} + {\left (a^{2} c^{3} d^{6} e + 3 \, a^{3} c^{2} d^{4} e^{3} + 3 \, a^{4} c d^{2} e^{5} + a^{5} e^{7}\right )} x\right )}}, \frac {2 \, a c^{2} d^{4} e - 2 \, a^{3} e^{5} + {\left (c^{3} d^{4} e - 2 \, a c^{2} d^{2} e^{3} - 3 \, a^{2} c e^{5}\right )} x^{2} + {\left (a c^{2} d^{5} + 6 \, a^{2} c d^{3} e^{2} - 3 \, a^{3} d e^{4} + {\left (c^{3} d^{4} e + 6 \, a c^{2} d^{2} e^{3} - 3 \, a^{2} c e^{5}\right )} x^{3} + {\left (c^{3} d^{5} + 6 \, a c^{2} d^{3} e^{2} - 3 \, a^{2} c d e^{4}\right )} x^{2} + {\left (a c^{2} d^{4} e + 6 \, a^{2} c d^{2} e^{3} - 3 \, a^{3} e^{5}\right )} x\right )} \sqrt {\frac {c}{a}} \arctan \left (x \sqrt {\frac {c}{a}}\right ) + {\left (c^{3} d^{5} + 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} x - 4 \, {\left (a c^{2} d e^{4} x^{3} + a c^{2} d^{2} e^{3} x^{2} + a^{2} c d e^{4} x + a^{2} c d^{2} e^{3}\right )} \log \left (c x^{2} + a\right ) + 8 \, {\left (a c^{2} d e^{4} x^{3} + a c^{2} d^{2} e^{3} x^{2} + a^{2} c d e^{4} x + a^{2} c d^{2} e^{3}\right )} \log \left (e x + d\right )}{2 \, {\left (a^{2} c^{3} d^{7} + 3 \, a^{3} c^{2} d^{5} e^{2} + 3 \, a^{4} c d^{3} e^{4} + a^{5} d e^{6} + {\left (a c^{4} d^{6} e + 3 \, a^{2} c^{3} d^{4} e^{3} + 3 \, a^{3} c^{2} d^{2} e^{5} + a^{4} c e^{7}\right )} x^{3} + {\left (a c^{4} d^{7} + 3 \, a^{2} c^{3} d^{5} e^{2} + 3 \, a^{3} c^{2} d^{3} e^{4} + a^{4} c d e^{6}\right )} x^{2} + {\left (a^{2} c^{3} d^{6} e + 3 \, a^{3} c^{2} d^{4} e^{3} + 3 \, a^{4} c d^{2} e^{5} + a^{5} e^{7}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(c*x^2+a)^2,x, algorithm="fricas")

[Out]

[1/4*(4*a*c^2*d^4*e - 4*a^3*e^5 + 2*(c^3*d^4*e - 2*a*c^2*d^2*e^3 - 3*a^2*c*e^5)*x^2 - (a*c^2*d^5 + 6*a^2*c*d^3
*e^2 - 3*a^3*d*e^4 + (c^3*d^4*e + 6*a*c^2*d^2*e^3 - 3*a^2*c*e^5)*x^3 + (c^3*d^5 + 6*a*c^2*d^3*e^2 - 3*a^2*c*d*
e^4)*x^2 + (a*c^2*d^4*e + 6*a^2*c*d^2*e^3 - 3*a^3*e^5)*x)*sqrt(-c/a)*log((c*x^2 - 2*a*x*sqrt(-c/a) - a)/(c*x^2
+ a)) + 2*(c^3*d^5 + 2*a*c^2*d^3*e^2 + a^2*c*d*e^4)*x - 8*(a*c^2*d*e^4*x^3 + a*c^2*d^2*e^3*x^2 + a^2*c*d*e^4*
x + a^2*c*d^2*e^3)*log(c*x^2 + a) + 16*(a*c^2*d*e^4*x^3 + a*c^2*d^2*e^3*x^2 + a^2*c*d*e^4*x + a^2*c*d^2*e^3)*l
og(e*x + d))/(a^2*c^3*d^7 + 3*a^3*c^2*d^5*e^2 + 3*a^4*c*d^3*e^4 + a^5*d*e^6 + (a*c^4*d^6*e + 3*a^2*c^3*d^4*e^3
+ 3*a^3*c^2*d^2*e^5 + a^4*c*e^7)*x^3 + (a*c^4*d^7 + 3*a^2*c^3*d^5*e^2 + 3*a^3*c^2*d^3*e^4 + a^4*c*d*e^6)*x^2
+ (a^2*c^3*d^6*e + 3*a^3*c^2*d^4*e^3 + 3*a^4*c*d^2*e^5 + a^5*e^7)*x), 1/2*(2*a*c^2*d^4*e - 2*a^3*e^5 + (c^3*d^
4*e - 2*a*c^2*d^2*e^3 - 3*a^2*c*e^5)*x^2 + (a*c^2*d^5 + 6*a^2*c*d^3*e^2 - 3*a^3*d*e^4 + (c^3*d^4*e + 6*a*c^2*d
^2*e^3 - 3*a^2*c*e^5)*x^3 + (c^3*d^5 + 6*a*c^2*d^3*e^2 - 3*a^2*c*d*e^4)*x^2 + (a*c^2*d^4*e + 6*a^2*c*d^2*e^3 -
3*a^3*e^5)*x)*sqrt(c/a)*arctan(x*sqrt(c/a)) + (c^3*d^5 + 2*a*c^2*d^3*e^2 + a^2*c*d*e^4)*x - 4*(a*c^2*d*e^4*x^
3 + a*c^2*d^2*e^3*x^2 + a^2*c*d*e^4*x + a^2*c*d^2*e^3)*log(c*x^2 + a) + 8*(a*c^2*d*e^4*x^3 + a*c^2*d^2*e^3*x^2
+ a^2*c*d*e^4*x + a^2*c*d^2*e^3)*log(e*x + d))/(a^2*c^3*d^7 + 3*a^3*c^2*d^5*e^2 + 3*a^4*c*d^3*e^4 + a^5*d*e^6
+ (a*c^4*d^6*e + 3*a^2*c^3*d^4*e^3 + 3*a^3*c^2*d^2*e^5 + a^4*c*e^7)*x^3 + (a*c^4*d^7 + 3*a^2*c^3*d^5*e^2 + 3*
a^3*c^2*d^3*e^4 + a^4*c*d*e^6)*x^2 + (a^2*c^3*d^6*e + 3*a^3*c^2*d^4*e^3 + 3*a^4*c*d^2*e^5 + a^5*e^7)*x)]

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giac [B]  time = 0.17, size = 386, normalized size = 1.88 \begin {gather*} -\frac {2 \, c d e^{3} \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 )}{c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}} + \frac {{\left (c^{3} d^{4} e^{2} + 6 \, a c^{2} d^{2} e^{4} - 3 \, a^{2} c e^{6}\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 )}}{2 \, {\left (a c^{3} d^{6} + 3 \, a^{2} c^{2} d^{4} e^{2} + 3 \, a^{3} c d^{2} e^{4} + a^{4} e^{6}\right )} \sqrt {a c}} - \frac {e^{7}}{{\left (c^{2} d^{4} e^{4} + 2 \, a c d^{2} e^{6} + a^{2} e^{8}\right )} {\left (x e + d\right )}} + \frac {\frac {c^{3} d^{3} e - 3 \, a c^{2} d e^{3}}{c d^{2} + a e^{2}} - \frac {{\left (c^{3} d^{4} e^{2} - 6 \, a c^{2} d^{2} e^{4} + a^{2} c e^{6}\right )} e^{\left (-1\right )}}{{\left (c d^{2} + a e^{2}\right )} {\left (x e + d\right )}}}{2 \, {\left (c d^{2} + a e^{2}\right )}^{2} a {\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(c*x^2+a)^2,x, algorithm="giac")

[Out]

-2*c*d*e^3*log(c - 2*c*d/(x*e + d) + c*d^2/(x*e + d)^2 + a*e^2/(x*e + d)^2)/(c^3*d^6 + 3*a*c^2*d^4*e^2 + 3*a^2
*c*d^2*e^4 + a^3*e^6) + 1/2*(c^3*d^4*e^2 + 6*a*c^2*d^2*e^4 - 3*a^2*c*e^6)*arctan((c*d - c*d^2/(x*e + d) - a*e^
2/(x*e + d))*e^(-1)/sqrt(a*c))*e^(-2)/((a*c^3*d^6 + 3*a^2*c^2*d^4*e^2 + 3*a^3*c*d^2*e^4 + a^4*e^6)*sqrt(a*c))
- e^7/((c^2*d^4*e^4 + 2*a*c*d^2*e^6 + a^2*e^8)*(x*e + d)) + 1/2*((c^3*d^3*e - 3*a*c^2*d*e^3)/(c*d^2 + a*e^2) -
(c^3*d^4*e^2 - 6*a*c^2*d^2*e^4 + a^2*c*e^6)*e^(-1)/((c*d^2 + a*e^2)*(x*e + d)))/((c*d^2 + a*e^2)^2*a*(c - 2*c
*d/(x*e + d) + c*d^2/(x*e + d)^2 + a*e^2/(x*e + d)^2))

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maple [A]  time = 0.07, size = 314, normalized size = 1.53 \begin {gather*} -\frac {a c \,e^{4} x}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3} \left (c \,x^{2}+a \right )}-\frac {3 a c \,e^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3} \sqrt {a c}}+\frac {c^{3} d^{4} x}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3} \left (c \,x^{2}+a \right ) a}+\frac {c^{3} d^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3} \sqrt {a c}\, a}+\frac {3 c^{2} d^{2} e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3} \sqrt {a c}}+\frac {a c d \,e^{3}}{\left (a \,e^{2}+c \,d^{2}\right )^{3} \left (c \,x^{2}+a \right )}+\frac {c^{2} d^{3} e}{\left (a \,e^{2}+c \,d^{2}\right )^{3} \left (c \,x^{2}+a \right )}-\frac {2 c d \,e^{3} \ln \left (c \,x^{2}+a \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}+\frac {4 c d \,e^{3} \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}-\frac {e^{3}}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \left (e x +d \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^2/(c*x^2+a)^2,x)

[Out]

-1/2*c/(a*e^2+c*d^2)^3/(c*x^2+a)*a*x*e^4+1/2*c^3/(a*e^2+c*d^2)^3/(c*x^2+a)/a*x*d^4+c/(a*e^2+c*d^2)^3/(c*x^2+a)
*d*e^3*a+c^2/(a*e^2+c*d^2)^3/(c*x^2+a)*d^3*e-2*c*d*e^3*ln(c*x^2+a)/(a*e^2+c*d^2)^3-3/2*c/(a*e^2+c*d^2)^3*a/(a*
c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*e^4+3*c^2/(a*e^2+c*d^2)^3/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*d^2*e^2+1/2
*c^3/(a*e^2+c*d^2)^3/a/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*d^4-e^3/(a*e^2+c*d^2)^2/(e*x+d)+4*c*d*e^3*ln(e*x+
d)/(a*e^2+c*d^2)^3

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maxima [B]  time = 3.06, size = 394, normalized size = 1.92 \begin {gather*} -\frac {2 \, c d e^{3} \log \left (c x^{2} + a\right )}{c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}} + \frac {4 \, c d e^{3} \log \left (e x + d\right )}{c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}} + \frac {{\left (c^{3} d^{4} + 6 \, a c^{2} d^{2} e^{2} - 3 \, a^{2} c e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, {\left (a c^{3} d^{6} + 3 \, a^{2} c^{2} d^{4} e^{2} + 3 \, a^{3} c d^{2} e^{4} + a^{4} e^{6}\right )} \sqrt {a c}} + \frac {2 \, a c d^{2} e - 2 \, a^{2} e^{3} + {\left (c^{2} d^{2} e - 3 \, a c e^{3}\right )} x^{2} + {\left (c^{2} d^{3} + a c d e^{2}\right )} x}{2 \, {\left (a^{2} c^{2} d^{5} + 2 \, a^{3} c d^{3} e^{2} + a^{4} d e^{4} + {\left (a c^{3} d^{4} e + 2 \, a^{2} c^{2} d^{2} e^{3} + a^{3} c e^{5}\right )} x^{3} + {\left (a c^{3} d^{5} + 2 \, a^{2} c^{2} d^{3} e^{2} + a^{3} c d e^{4}\right )} x^{2} + {\left (a^{2} c^{2} d^{4} e + 2 \, a^{3} c d^{2} e^{3} + a^{4} e^{5}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(c*x^2+a)^2,x, algorithm="maxima")

[Out]

-2*c*d*e^3*log(c*x^2 + a)/(c^3*d^6 + 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 + a^3*e^6) + 4*c*d*e^3*log(e*x + d)/(c^
3*d^6 + 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 + a^3*e^6) + 1/2*(c^3*d^4 + 6*a*c^2*d^2*e^2 - 3*a^2*c*e^4)*arctan(c*
x/sqrt(a*c))/((a*c^3*d^6 + 3*a^2*c^2*d^4*e^2 + 3*a^3*c*d^2*e^4 + a^4*e^6)*sqrt(a*c)) + 1/2*(2*a*c*d^2*e - 2*a^
2*e^3 + (c^2*d^2*e - 3*a*c*e^3)*x^2 + (c^2*d^3 + a*c*d*e^2)*x)/(a^2*c^2*d^5 + 2*a^3*c*d^3*e^2 + a^4*d*e^4 + (a
*c^3*d^4*e + 2*a^2*c^2*d^2*e^3 + a^3*c*e^5)*x^3 + (a*c^3*d^5 + 2*a^2*c^2*d^3*e^2 + a^3*c*d*e^4)*x^2 + (a^2*c^2
*d^4*e + 2*a^3*c*d^2*e^3 + a^4*e^5)*x)

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mupad [B]  time = 1.31, size = 819, normalized size = 4.00 \begin {gather*} \frac {4\,c\,d\,e^3\,\ln \left (d+e\,x\right )}{{\left (c\,d^2+a\,e^2\right )}^3}-\frac {\ln \left (c^5\,d^{12}\,{\left (-a^3\,c\right )}^{3/2}-9\,a^9\,e^{12}\,\sqrt {-a^3\,c}+a^4\,c^7\,d^{12}\,x-1119\,a\,d^4\,e^8\,{\left (-a^3\,c\right )}^{5/2}-612\,c\,d^6\,e^6\,{\left (-a^3\,c\right )}^{5/2}+558\,a^5\,d^2\,e^{10}\,{\left (-a^3\,c\right )}^{3/2}+9\,a^{10}\,c\,e^{12}\,x+55\,a^2\,c^3\,d^8\,e^4\,{\left (-a^3\,c\right )}^{3/2}+14\,a^5\,c^6\,d^{10}\,e^2\,x+55\,a^6\,c^5\,d^8\,e^4\,x+612\,a^7\,c^4\,d^6\,e^6\,x+1119\,a^8\,c^3\,d^4\,e^8\,x+558\,a^9\,c^2\,d^2\,e^{10}\,x+14\,a\,c^4\,d^{10}\,e^2\,{\left (-a^3\,c\right )}^{3/2}\right )\,\left (c\,\left (2\,a^3\,d\,e^3+\frac {3\,a\,d^2\,e^2\,\sqrt {-a^3\,c}}{2}\right )-\frac {3\,a^2\,e^4\,\sqrt {-a^3\,c}}{4}+\frac {c^2\,d^4\,\sqrt {-a^3\,c}}{4}\right )}{a^6\,e^6+3\,a^5\,c\,d^2\,e^4+3\,a^4\,c^2\,d^4\,e^2+a^3\,c^3\,d^6}-\frac {\ln \left (9\,a^9\,e^{12}\,\sqrt {-a^3\,c}-c^5\,d^{12}\,{\left (-a^3\,c\right )}^{3/2}+a^4\,c^7\,d^{12}\,x+1119\,a\,d^4\,e^8\,{\left (-a^3\,c\right )}^{5/2}+612\,c\,d^6\,e^6\,{\left (-a^3\,c\right )}^{5/2}-558\,a^5\,d^2\,e^{10}\,{\left (-a^3\,c\right )}^{3/2}+9\,a^{10}\,c\,e^{12}\,x-55\,a^2\,c^3\,d^8\,e^4\,{\left (-a^3\,c\right )}^{3/2}+14\,a^5\,c^6\,d^{10}\,e^2\,x+55\,a^6\,c^5\,d^8\,e^4\,x+612\,a^7\,c^4\,d^6\,e^6\,x+1119\,a^8\,c^3\,d^4\,e^8\,x+558\,a^9\,c^2\,d^2\,e^{10}\,x-14\,a\,c^4\,d^{10}\,e^2\,{\left (-a^3\,c\right )}^{3/2}\right )\,\left (c\,\left (2\,a^3\,d\,e^3-\frac {3\,a\,d^2\,e^2\,\sqrt {-a^3\,c}}{2}\right )+\frac {3\,a^2\,e^4\,\sqrt {-a^3\,c}}{4}-\frac {c^2\,d^4\,\sqrt {-a^3\,c}}{4}\right )}{a^6\,e^6+3\,a^5\,c\,d^2\,e^4+3\,a^4\,c^2\,d^4\,e^2+a^3\,c^3\,d^6}-\frac {\frac {a\,e^3-c\,d^2\,e}{{\left (c\,d^2+a\,e^2\right )}^2}-\frac {c\,d\,x}{2\,a\,\left (c\,d^2+a\,e^2\right )}+\frac {c\,x^2\,\left (3\,a\,e^3-c\,d^2\,e\right )}{2\,a\,\left (a^2\,e^4+2\,a\,c\,d^2\,e^2+c^2\,d^4\right )}}{c\,e\,x^3+c\,d\,x^2+a\,e\,x+a\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + c*x^2)^2*(d + e*x)^2),x)

[Out]

(4*c*d*e^3*log(d + e*x))/(a*e^2 + c*d^2)^3 - (log(c^5*d^12*(-a^3*c)^(3/2) - 9*a^9*e^12*(-a^3*c)^(1/2) + a^4*c^
7*d^12*x - 1119*a*d^4*e^8*(-a^3*c)^(5/2) - 612*c*d^6*e^6*(-a^3*c)^(5/2) + 558*a^5*d^2*e^10*(-a^3*c)^(3/2) + 9*
a^10*c*e^12*x + 55*a^2*c^3*d^8*e^4*(-a^3*c)^(3/2) + 14*a^5*c^6*d^10*e^2*x + 55*a^6*c^5*d^8*e^4*x + 612*a^7*c^4
*d^6*e^6*x + 1119*a^8*c^3*d^4*e^8*x + 558*a^9*c^2*d^2*e^10*x + 14*a*c^4*d^10*e^2*(-a^3*c)^(3/2))*(c*(2*a^3*d*e
^3 + (3*a*d^2*e^2*(-a^3*c)^(1/2))/2) - (3*a^2*e^4*(-a^3*c)^(1/2))/4 + (c^2*d^4*(-a^3*c)^(1/2))/4))/(a^6*e^6 +
a^3*c^3*d^6 + 3*a^5*c*d^2*e^4 + 3*a^4*c^2*d^4*e^2) - (log(9*a^9*e^12*(-a^3*c)^(1/2) - c^5*d^12*(-a^3*c)^(3/2)
+ a^4*c^7*d^12*x + 1119*a*d^4*e^8*(-a^3*c)^(5/2) + 612*c*d^6*e^6*(-a^3*c)^(5/2) - 558*a^5*d^2*e^10*(-a^3*c)^(3
/2) + 9*a^10*c*e^12*x - 55*a^2*c^3*d^8*e^4*(-a^3*c)^(3/2) + 14*a^5*c^6*d^10*e^2*x + 55*a^6*c^5*d^8*e^4*x + 612
*a^7*c^4*d^6*e^6*x + 1119*a^8*c^3*d^4*e^8*x + 558*a^9*c^2*d^2*e^10*x - 14*a*c^4*d^10*e^2*(-a^3*c)^(3/2))*(c*(2
*a^3*d*e^3 - (3*a*d^2*e^2*(-a^3*c)^(1/2))/2) + (3*a^2*e^4*(-a^3*c)^(1/2))/4 - (c^2*d^4*(-a^3*c)^(1/2))/4))/(a^
6*e^6 + a^3*c^3*d^6 + 3*a^5*c*d^2*e^4 + 3*a^4*c^2*d^4*e^2) - ((a*e^3 - c*d^2*e)/(a*e^2 + c*d^2)^2 - (c*d*x)/(2
*a*(a*e^2 + c*d^2)) + (c*x^2*(3*a*e^3 - c*d^2*e))/(2*a*(a^2*e^4 + c^2*d^4 + 2*a*c*d^2*e^2)))/(a*d + a*e*x + c*
d*x^2 + c*e*x^3)

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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(1/(e*x+d)**2/(c*x**2+a)**2,x)

[Out]

Timed out

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