3.517 \(\int \frac{1}{(d+e x) (a+c x^2)^3} \, dx\)

Optimal. Leaf size=212 \[ \frac{\sqrt{c} d \left (15 a^2 e^4+10 a c d^2 e^2+3 c^2 d^4\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{8 a^{5/2} \left (a e^2+c d^2\right )^3}+\frac{4 a^2 e^3+c d x \left (7 a e^2+3 c d^2\right )}{8 a^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )^2}+\frac{a e+c d x}{4 a \left (a+c x^2\right )^2 \left (a e^2+c d^2\right )}-\frac{e^5 \log \left (a+c x^2\right )}{2 \left (a e^2+c d^2\right )^3}+\frac{e^5 \log (d+e x)}{\left (a e^2+c d^2\right )^3} \]

[Out]

(a*e + c*d*x)/(4*a*(c*d^2 + a*e^2)*(a + c*x^2)^2) + (4*a^2*e^3 + c*d*(3*c*d^2 + 7*a*e^2)*x)/(8*a^2*(c*d^2 + a*
e^2)^2*(a + c*x^2)) + (Sqrt[c]*d*(3*c^2*d^4 + 10*a*c*d^2*e^2 + 15*a^2*e^4)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(
5/2)*(c*d^2 + a*e^2)^3) + (e^5*Log[d + e*x])/(c*d^2 + a*e^2)^3 - (e^5*Log[a + c*x^2])/(2*(c*d^2 + a*e^2)^3)

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Rubi [A]  time = 0.220795, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {741, 823, 801, 635, 205, 260} \[ \frac{\sqrt{c} d \left (15 a^2 e^4+10 a c d^2 e^2+3 c^2 d^4\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{8 a^{5/2} \left (a e^2+c d^2\right )^3}+\frac{4 a^2 e^3+c d x \left (7 a e^2+3 c d^2\right )}{8 a^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )^2}+\frac{a e+c d x}{4 a \left (a+c x^2\right )^2 \left (a e^2+c d^2\right )}-\frac{e^5 \log \left (a+c x^2\right )}{2 \left (a e^2+c d^2\right )^3}+\frac{e^5 \log (d+e x)}{\left (a e^2+c d^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*(a + c*x^2)^3),x]

[Out]

(a*e + c*d*x)/(4*a*(c*d^2 + a*e^2)*(a + c*x^2)^2) + (4*a^2*e^3 + c*d*(3*c*d^2 + 7*a*e^2)*x)/(8*a^2*(c*d^2 + a*
e^2)^2*(a + c*x^2)) + (Sqrt[c]*d*(3*c^2*d^4 + 10*a*c*d^2*e^2 + 15*a^2*e^4)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(
5/2)*(c*d^2 + a*e^2)^3) + (e^5*Log[d + e*x])/(c*d^2 + a*e^2)^3 - (e^5*Log[a + c*x^2])/(2*(c*d^2 + a*e^2)^3)

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 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])

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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{(d+e x) \left (a+c x^2\right )^3} \, dx &=\frac{a e+c d x}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}-\frac{\int \frac{-3 c d^2-4 a e^2-3 c d e x}{(d+e x) \left (a+c x^2\right )^2} \, dx}{4 a \left (c d^2+a e^2\right )}\\ &=\frac{a e+c d x}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}+\frac{4 a^2 e^3+c d \left (3 c d^2+7 a e^2\right ) x}{8 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac{\int \frac{c \left (3 c^2 d^4+7 a c d^2 e^2+8 a^2 e^4\right )+c^2 d e \left (3 c d^2+7 a e^2\right ) x}{(d+e x) \left (a+c x^2\right )} \, dx}{8 a^2 c \left (c d^2+a e^2\right )^2}\\ &=\frac{a e+c d x}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}+\frac{4 a^2 e^3+c d \left (3 c d^2+7 a e^2\right ) x}{8 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac{\int \left (\frac{8 a^2 c e^6}{\left (c d^2+a e^2\right ) (d+e x)}+\frac{c^2 \left (3 c^2 d^5+10 a c d^3 e^2+15 a^2 d e^4-8 a^2 e^5 x\right )}{\left (c d^2+a e^2\right ) \left (a+c x^2\right )}\right ) \, dx}{8 a^2 c \left (c d^2+a e^2\right )^2}\\ &=\frac{a e+c d x}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}+\frac{4 a^2 e^3+c d \left (3 c d^2+7 a e^2\right ) x}{8 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac{e^5 \log (d+e x)}{\left (c d^2+a e^2\right )^3}+\frac{c \int \frac{3 c^2 d^5+10 a c d^3 e^2+15 a^2 d e^4-8 a^2 e^5 x}{a+c x^2} \, dx}{8 a^2 \left (c d^2+a e^2\right )^3}\\ &=\frac{a e+c d x}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}+\frac{4 a^2 e^3+c d \left (3 c d^2+7 a e^2\right ) x}{8 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac{e^5 \log (d+e x)}{\left (c d^2+a e^2\right )^3}-\frac{\left (c e^5\right ) \int \frac{x}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^3}+\frac{\left (c d \left (3 c^2 d^4+10 a c d^2 e^2+15 a^2 e^4\right )\right ) \int \frac{1}{a+c x^2} \, dx}{8 a^2 \left (c d^2+a e^2\right )^3}\\ &=\frac{a e+c d x}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}+\frac{4 a^2 e^3+c d \left (3 c d^2+7 a e^2\right ) x}{8 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac{\sqrt{c} d \left (3 c^2 d^4+10 a c d^2 e^2+15 a^2 e^4\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{8 a^{5/2} \left (c d^2+a e^2\right )^3}+\frac{e^5 \log (d+e x)}{\left (c d^2+a e^2\right )^3}-\frac{e^5 \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^3}\\ \end{align*}

Mathematica [A]  time = 0.146377, size = 180, normalized size = 0.85 \[ \frac{\frac{\left (a e^2+c d^2\right ) \left (4 a^2 e^3+7 a c d e^2 x+3 c^2 d^3 x\right )}{a^2 \left (a+c x^2\right )}+\frac{\sqrt{c} d \left (15 a^2 e^4+10 a c d^2 e^2+3 c^2 d^4\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{a^{5/2}}+\frac{2 \left (a e^2+c d^2\right )^2 (a e+c d x)}{a \left (a+c x^2\right )^2}-4 e^5 \log \left (a+c x^2\right )+8 e^5 \log (d+e x)}{8 \left (a e^2+c d^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)*(a + c*x^2)^3),x]

[Out]

((2*(c*d^2 + a*e^2)^2*(a*e + c*d*x))/(a*(a + c*x^2)^2) + ((c*d^2 + a*e^2)*(4*a^2*e^3 + 3*c^2*d^3*x + 7*a*c*d*e
^2*x))/(a^2*(a + c*x^2)) + (Sqrt[c]*d*(3*c^2*d^4 + 10*a*c*d^2*e^2 + 15*a^2*e^4)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/a
^(5/2) + 8*e^5*Log[d + e*x] - 4*e^5*Log[a + c*x^2])/(8*(c*d^2 + a*e^2)^3)

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Maple [B]  time = 0.059, size = 532, normalized size = 2.5 \begin{align*}{\frac{7\,{c}^{2}d{x}^{3}{e}^{4}}{8\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3} \left ( c{x}^{2}+a \right ) ^{2}}}+{\frac{5\,{c}^{3}{d}^{3}{x}^{3}{e}^{2}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3} \left ( c{x}^{2}+a \right ) ^{2}a}}+{\frac{3\,{c}^{4}{d}^{5}{x}^{3}}{8\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3} \left ( c{x}^{2}+a \right ) ^{2}{a}^{2}}}+{\frac{c{x}^{2}a{e}^{5}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3} \left ( c{x}^{2}+a \right ) ^{2}}}+{\frac{{c}^{2}{x}^{2}{d}^{2}{e}^{3}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3} \left ( c{x}^{2}+a \right ) ^{2}}}+{\frac{9\,acdx{e}^{4}}{8\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3} \left ( c{x}^{2}+a \right ) ^{2}}}+{\frac{7\,{c}^{2}{d}^{3}x{e}^{2}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3} \left ( c{x}^{2}+a \right ) ^{2}}}+{\frac{5\,{c}^{3}{d}^{5}x}{8\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3} \left ( c{x}^{2}+a \right ) ^{2}a}}+{\frac{3\,{a}^{2}{e}^{5}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3} \left ( c{x}^{2}+a \right ) ^{2}}}+{\frac{ac{d}^{2}{e}^{3}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3} \left ( c{x}^{2}+a \right ) ^{2}}}+{\frac{{c}^{2}{d}^{4}e}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3} \left ( c{x}^{2}+a \right ) ^{2}}}-{\frac{{e}^{5}\ln \left ( c{x}^{2}+a \right ) }{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}+{\frac{15\,d{e}^{4}c}{8\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{5\,{d}^{3}{e}^{2}{c}^{2}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}a}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{3\,{c}^{3}{d}^{5}}{8\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}{a}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{{e}^{5}\ln \left ( ex+d \right ) }{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)/(c*x^2+a)^3,x)

[Out]

7/8*c^2/(a*e^2+c*d^2)^3/(c*x^2+a)^2*d*x^3*e^4+5/4*c^3/(a*e^2+c*d^2)^3/(c*x^2+a)^2*d^3/a*x^3*e^2+3/8*c^4/(a*e^2
+c*d^2)^3/(c*x^2+a)^2*d^5/a^2*x^3+1/2*c/(a*e^2+c*d^2)^3/(c*x^2+a)^2*x^2*a*e^5+1/2*c^2/(a*e^2+c*d^2)^3/(c*x^2+a
)^2*x^2*d^2*e^3+9/8*c/(a*e^2+c*d^2)^3/(c*x^2+a)^2*d*a*x*e^4+7/4*c^2/(a*e^2+c*d^2)^3/(c*x^2+a)^2*d^3*x*e^2+5/8*
c^3/(a*e^2+c*d^2)^3/(c*x^2+a)^2*d^5/a*x+3/4/(a*e^2+c*d^2)^3/(c*x^2+a)^2*a^2*e^5+c/(a*e^2+c*d^2)^3/(c*x^2+a)^2*
e^3*a*d^2+1/4*c^2/(a*e^2+c*d^2)^3/(c*x^2+a)^2*e*d^4-1/2*e^5*ln(c*x^2+a)/(a*e^2+c*d^2)^3+15/8*c/(a*e^2+c*d^2)^3
/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*d*e^4+5/4*c^2/(a*e^2+c*d^2)^3/a/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*d^3*e
^2+3/8*c^3/(a*e^2+c*d^2)^3/a^2/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*d^5+e^5*ln(e*x+d)/(a*e^2+c*d^2)^3

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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(1/(e*x+d)/(c*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 24.1499, size = 2045, normalized size = 9.65 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/16*(4*a^2*c^2*d^4*e + 16*a^3*c*d^2*e^3 + 12*a^4*e^5 + 2*(3*c^4*d^5 + 10*a*c^3*d^3*e^2 + 7*a^2*c^2*d*e^4)*x^
3 + 8*(a^2*c^2*d^2*e^3 + a^3*c*e^5)*x^2 + (3*a^2*c^2*d^5 + 10*a^3*c*d^3*e^2 + 15*a^4*d*e^4 + (3*c^4*d^5 + 10*a
*c^3*d^3*e^2 + 15*a^2*c^2*d*e^4)*x^4 + 2*(3*a*c^3*d^5 + 10*a^2*c^2*d^3*e^2 + 15*a^3*c*d*e^4)*x^2)*sqrt(-c/a)*l
og((c*x^2 + 2*a*x*sqrt(-c/a) - a)/(c*x^2 + a)) + 2*(5*a*c^3*d^5 + 14*a^2*c^2*d^3*e^2 + 9*a^3*c*d*e^4)*x - 8*(a
^2*c^2*e^5*x^4 + 2*a^3*c*e^5*x^2 + a^4*e^5)*log(c*x^2 + a) + 16*(a^2*c^2*e^5*x^4 + 2*a^3*c*e^5*x^2 + a^4*e^5)*
log(e*x + d))/(a^4*c^3*d^6 + 3*a^5*c^2*d^4*e^2 + 3*a^6*c*d^2*e^4 + a^7*e^6 + (a^2*c^5*d^6 + 3*a^3*c^4*d^4*e^2
+ 3*a^4*c^3*d^2*e^4 + a^5*c^2*e^6)*x^4 + 2*(a^3*c^4*d^6 + 3*a^4*c^3*d^4*e^2 + 3*a^5*c^2*d^2*e^4 + a^6*c*e^6)*x
^2), 1/8*(2*a^2*c^2*d^4*e + 8*a^3*c*d^2*e^3 + 6*a^4*e^5 + (3*c^4*d^5 + 10*a*c^3*d^3*e^2 + 7*a^2*c^2*d*e^4)*x^3
+ 4*(a^2*c^2*d^2*e^3 + a^3*c*e^5)*x^2 + (3*a^2*c^2*d^5 + 10*a^3*c*d^3*e^2 + 15*a^4*d*e^4 + (3*c^4*d^5 + 10*a*
c^3*d^3*e^2 + 15*a^2*c^2*d*e^4)*x^4 + 2*(3*a*c^3*d^5 + 10*a^2*c^2*d^3*e^2 + 15*a^3*c*d*e^4)*x^2)*sqrt(c/a)*arc
tan(x*sqrt(c/a)) + (5*a*c^3*d^5 + 14*a^2*c^2*d^3*e^2 + 9*a^3*c*d*e^4)*x - 4*(a^2*c^2*e^5*x^4 + 2*a^3*c*e^5*x^2
+ a^4*e^5)*log(c*x^2 + a) + 8*(a^2*c^2*e^5*x^4 + 2*a^3*c*e^5*x^2 + a^4*e^5)*log(e*x + d))/(a^4*c^3*d^6 + 3*a^
5*c^2*d^4*e^2 + 3*a^6*c*d^2*e^4 + a^7*e^6 + (a^2*c^5*d^6 + 3*a^3*c^4*d^4*e^2 + 3*a^4*c^3*d^2*e^4 + a^5*c^2*e^6
)*x^4 + 2*(a^3*c^4*d^6 + 3*a^4*c^3*d^4*e^2 + 3*a^5*c^2*d^2*e^4 + a^6*c*e^6)*x^2)]

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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(1/(e*x+d)/(c*x**2+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.37224, size = 462, normalized size = 2.18 \begin{align*} -\frac{e^{5} \log \left (c x^{2} + a\right )}{2 \,{\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )}} + \frac{e^{6} \log \left ({\left | x e + d \right |}\right )}{c^{3} d^{6} e + 3 \, a c^{2} d^{4} e^{3} + 3 \, a^{2} c d^{2} e^{5} + a^{3} e^{7}} + \frac{{\left (3 \, c^{3} d^{5} + 10 \, a c^{2} d^{3} e^{2} + 15 \, a^{2} c d e^{4}\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{8 \,{\left (a^{2} c^{3} d^{6} + 3 \, a^{3} c^{2} d^{4} e^{2} + 3 \, a^{4} c d^{2} e^{4} + a^{5} e^{6}\right )} \sqrt{a c}} + \frac{2 \, a^{2} c^{2} d^{4} e + 8 \, a^{3} c d^{2} e^{3} + 6 \, a^{4} e^{5} +{\left (3 \, c^{4} d^{5} + 10 \, a c^{3} d^{3} e^{2} + 7 \, a^{2} c^{2} d e^{4}\right )} x^{3} + 4 \,{\left (a^{2} c^{2} d^{2} e^{3} + a^{3} c e^{5}\right )} x^{2} +{\left (5 \, a c^{3} d^{5} + 14 \, a^{2} c^{2} d^{3} e^{2} + 9 \, a^{3} c d e^{4}\right )} x}{8 \,{\left (c d^{2} + a e^{2}\right )}^{3}{\left (c x^{2} + a\right )}^{2} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*x^2+a)^3,x, algorithm="giac")

[Out]

-1/2*e^5*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) + e^6*log(abs(x*e + d))/(c^3*d
^6*e + 3*a*c^2*d^4*e^3 + 3*a^2*c*d^2*e^5 + a^3*e^7) + 1/8*(3*c^3*d^5 + 10*a*c^2*d^3*e^2 + 15*a^2*c*d*e^4)*arct
an(c*x/sqrt(a*c))/((a^2*c^3*d^6 + 3*a^3*c^2*d^4*e^2 + 3*a^4*c*d^2*e^4 + a^5*e^6)*sqrt(a*c)) + 1/8*(2*a^2*c^2*d
^4*e + 8*a^3*c*d^2*e^3 + 6*a^4*e^5 + (3*c^4*d^5 + 10*a*c^3*d^3*e^2 + 7*a^2*c^2*d*e^4)*x^3 + 4*(a^2*c^2*d^2*e^3
+ a^3*c*e^5)*x^2 + (5*a*c^3*d^5 + 14*a^2*c^2*d^3*e^2 + 9*a^3*c*d*e^4)*x)/((c*d^2 + a*e^2)^3*(c*x^2 + a)^2*a^2
)