Integrand size = 18, antiderivative size = 328 \[ \int \frac {a+b \arctan \left (c x^2\right )}{(d+e x)^2} \, dx=\frac {b c^2 d^3 \arctan \left (c x^2\right )}{e \left (c^2 d^4+e^4\right )}-\frac {a+b \arctan \left (c x^2\right )}{e (d+e x)}+\frac {b \sqrt {c} \left (c d^2-e^2\right ) \arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \left (c^2 d^4+e^4\right )}-\frac {b \sqrt {c} \left (c d^2-e^2\right ) \arctan \left (1+\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \left (c^2 d^4+e^4\right )}-\frac {2 b c d e \log (d+e x)}{c^2 d^4+e^4}-\frac {b \sqrt {c} \left (c d^2+e^2\right ) \log \left (1-\sqrt {2} \sqrt {c} x+c x^2\right )}{2 \sqrt {2} \left (c^2 d^4+e^4\right )}+\frac {b \sqrt {c} \left (c d^2+e^2\right ) \log \left (1+\sqrt {2} \sqrt {c} x+c x^2\right )}{2 \sqrt {2} \left (c^2 d^4+e^4\right )}+\frac {b c d e \log \left (1+c^2 x^4\right )}{2 \left (c^2 d^4+e^4\right )} \]
b*c^2*d^3*arctan(c*x^2)/e/(c^2*d^4+e^4)+(-a-b*arctan(c*x^2))/e/(e*x+d)-2*b *c*d*e*ln(e*x+d)/(c^2*d^4+e^4)+1/2*b*c*d*e*ln(c^2*x^4+1)/(c^2*d^4+e^4)-1/2 *b*(c*d^2-e^2)*arctan(-1+x*2^(1/2)*c^(1/2))*c^(1/2)/(c^2*d^4+e^4)*2^(1/2)- 1/2*b*(c*d^2-e^2)*arctan(1+x*2^(1/2)*c^(1/2))*c^(1/2)/(c^2*d^4+e^4)*2^(1/2 )-1/4*b*(c*d^2+e^2)*ln(1+c*x^2-x*2^(1/2)*c^(1/2))*c^(1/2)/(c^2*d^4+e^4)*2^ (1/2)+1/4*b*(c*d^2+e^2)*ln(1+c*x^2+x*2^(1/2)*c^(1/2))*c^(1/2)/(c^2*d^4+e^4 )*2^(1/2)
Time = 0.76 (sec) , antiderivative size = 321, normalized size of antiderivative = 0.98 \[ \int \frac {a+b \arctan \left (c x^2\right )}{(d+e x)^2} \, dx=-\frac {4 a \left (c^2 d^4+e^4\right )+4 b \left (c^2 d^4+e^4\right ) \arctan \left (c x^2\right )+2 b \sqrt {c} \left (2 c^{3/2} d^3-\sqrt {2} c d^2 e+\sqrt {2} e^3\right ) (d+e x) \arctan \left (1-\sqrt {2} \sqrt {c} x\right )+2 b \sqrt {c} \left (2 c^{3/2} d^3+\sqrt {2} c d^2 e-\sqrt {2} e^3\right ) (d+e x) \arctan \left (1+\sqrt {2} \sqrt {c} x\right )+8 b c d e^2 (d+e x) \log (d+e x)+\sqrt {2} b \sqrt {c} e \left (c d^2+e^2\right ) (d+e x) \log \left (1-\sqrt {2} \sqrt {c} x+c x^2\right )-\sqrt {2} b \sqrt {c} e \left (c d^2+e^2\right ) (d+e x) \log \left (1+\sqrt {2} \sqrt {c} x+c x^2\right )-2 b c d e^2 (d+e x) \log \left (1+c^2 x^4\right )}{4 e \left (c^2 d^4+e^4\right ) (d+e x)} \]
-1/4*(4*a*(c^2*d^4 + e^4) + 4*b*(c^2*d^4 + e^4)*ArcTan[c*x^2] + 2*b*Sqrt[c ]*(2*c^(3/2)*d^3 - Sqrt[2]*c*d^2*e + Sqrt[2]*e^3)*(d + e*x)*ArcTan[1 - Sqr t[2]*Sqrt[c]*x] + 2*b*Sqrt[c]*(2*c^(3/2)*d^3 + Sqrt[2]*c*d^2*e - Sqrt[2]*e ^3)*(d + e*x)*ArcTan[1 + Sqrt[2]*Sqrt[c]*x] + 8*b*c*d*e^2*(d + e*x)*Log[d + e*x] + Sqrt[2]*b*Sqrt[c]*e*(c*d^2 + e^2)*(d + e*x)*Log[1 - Sqrt[2]*Sqrt[ c]*x + c*x^2] - Sqrt[2]*b*Sqrt[c]*e*(c*d^2 + e^2)*(d + e*x)*Log[1 + Sqrt[2 ]*Sqrt[c]*x + c*x^2] - 2*b*c*d*e^2*(d + e*x)*Log[1 + c^2*x^4])/(e*(c^2*d^4 + e^4)*(d + e*x))
Time = 0.71 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5395, 7276, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {a+b \arctan \left (c x^2\right )}{(d+e x)^2} \, dx\) |
| \(\Big \downarrow \) 5395 |
| \(\displaystyle \frac {2 b c \int \frac {x}{(d+e x) \left (c^2 x^4+1\right )}dx}{e}-\frac {a+b \arctan \left (c x^2\right )}{e (d+e x)}\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle \frac {2 b c \int \left (\frac {c^2 x d^3-c^2 e x^2 d^2+c^2 e^2 x^3 d+e^3}{\left (c^2 d^4+e^4\right ) \left (c^2 x^4+1\right )}-\frac {d e^3}{\left (c^2 d^4+e^4\right ) (d+e x)}\right )dx}{e}-\frac {a+b \arctan \left (c x^2\right )}{e (d+e x)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 b c \left (\frac {c d^3 \arctan \left (c x^2\right )}{2 \left (c^2 d^4+e^4\right )}+\frac {e \arctan \left (1-\sqrt {2} \sqrt {c} x\right ) \left (c d^2-e^2\right )}{2 \sqrt {2} \sqrt {c} \left (c^2 d^4+e^4\right )}-\frac {e \arctan \left (\sqrt {2} \sqrt {c} x+1\right ) \left (c d^2-e^2\right )}{2 \sqrt {2} \sqrt {c} \left (c^2 d^4+e^4\right )}+\frac {d e^2 \log \left (c^2 x^4+1\right )}{4 \left (c^2 d^4+e^4\right )}-\frac {d e^2 \log (d+e x)}{c^2 d^4+e^4}-\frac {e \left (c d^2+e^2\right ) \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{4 \sqrt {2} \sqrt {c} \left (c^2 d^4+e^4\right )}+\frac {e \left (c d^2+e^2\right ) \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{4 \sqrt {2} \sqrt {c} \left (c^2 d^4+e^4\right )}\right )}{e}-\frac {a+b \arctan \left (c x^2\right )}{e (d+e x)}\) |
-((a + b*ArcTan[c*x^2])/(e*(d + e*x))) + (2*b*c*((c*d^3*ArcTan[c*x^2])/(2* (c^2*d^4 + e^4)) + (e*(c*d^2 - e^2)*ArcTan[1 - Sqrt[2]*Sqrt[c]*x])/(2*Sqrt [2]*Sqrt[c]*(c^2*d^4 + e^4)) - (e*(c*d^2 - e^2)*ArcTan[1 + Sqrt[2]*Sqrt[c] *x])/(2*Sqrt[2]*Sqrt[c]*(c^2*d^4 + e^4)) - (d*e^2*Log[d + e*x])/(c^2*d^4 + e^4) - (e*(c*d^2 + e^2)*Log[1 - Sqrt[2]*Sqrt[c]*x + c*x^2])/(4*Sqrt[2]*Sq rt[c]*(c^2*d^4 + e^4)) + (e*(c*d^2 + e^2)*Log[1 + Sqrt[2]*Sqrt[c]*x + c*x^ 2])/(4*Sqrt[2]*Sqrt[c]*(c^2*d^4 + e^4)) + (d*e^2*Log[1 + c^2*x^4])/(4*(c^2 *d^4 + e^4))))/e
3.1.24.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))*((d_) + (e_.)*(x_))^(m_.), x_Sy mbol] :> Simp[(d + e*x)^(m + 1)*((a + b*ArcTan[c*x^n])/(e*(m + 1))), x] - S imp[b*c*(n/(e*(m + 1))) Int[x^(n - 1)*((d + e*x)^(m + 1)/(1 + c^2*x^(2*n) )), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[m, -1]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 1.40 (sec) , antiderivative size = 297, normalized size of antiderivative = 0.91
| method | result | size |
| default | \(-\frac {a}{\left (e x +d \right ) e}+b \left (-\frac {\arctan \left (c \,x^{2}\right )}{\left (e x +d \right ) e}+\frac {2 c \left (-\frac {d \,e^{2} \ln \left (e x +d \right )}{c^{2} d^{4}+e^{4}}+\frac {\frac {e^{3} \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{8}+\frac {c^{2} d^{3} \arctan \left (x^{2} \sqrt {c^{2}}\right )}{2 \sqrt {c^{2}}}-\frac {d^{2} e \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{8 \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+\frac {d \,e^{2} \ln \left (c^{2} x^{4}+1\right )}{4}}{c^{2} d^{4}+e^{4}}\right )}{e}\right )\) | \(297\) |
| parts | \(-\frac {a}{\left (e x +d \right ) e}+b \left (-\frac {\arctan \left (c \,x^{2}\right )}{\left (e x +d \right ) e}+\frac {2 c \left (-\frac {d \,e^{2} \ln \left (e x +d \right )}{c^{2} d^{4}+e^{4}}+\frac {\frac {e^{3} \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{8}+\frac {c^{2} d^{3} \arctan \left (x^{2} \sqrt {c^{2}}\right )}{2 \sqrt {c^{2}}}-\frac {d^{2} e \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{8 \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+\frac {d \,e^{2} \ln \left (c^{2} x^{4}+1\right )}{4}}{c^{2} d^{4}+e^{4}}\right )}{e}\right )\) | \(297\) |
-a/(e*x+d)/e+b*(-1/(e*x+d)/e*arctan(c*x^2)+2*c/e*(-d*e^2/(c^2*d^4+e^4)*ln( e*x+d)+1/(c^2*d^4+e^4)*(1/8*e^3*(1/c^2)^(1/4)*2^(1/2)*(ln((x^2+(1/c^2)^(1/ 4)*x*2^(1/2)+(1/c^2)^(1/2))/(x^2-(1/c^2)^(1/4)*x*2^(1/2)+(1/c^2)^(1/2)))+2 *arctan(2^(1/2)/(1/c^2)^(1/4)*x+1)+2*arctan(2^(1/2)/(1/c^2)^(1/4)*x-1))+1/ 2*c^2*d^3/(c^2)^(1/2)*arctan(x^2*(c^2)^(1/2))-1/8*d^2*e/(1/c^2)^(1/4)*2^(1 /2)*(ln((x^2-(1/c^2)^(1/4)*x*2^(1/2)+(1/c^2)^(1/2))/(x^2+(1/c^2)^(1/4)*x*2 ^(1/2)+(1/c^2)^(1/2)))+2*arctan(2^(1/2)/(1/c^2)^(1/4)*x+1)+2*arctan(2^(1/2 )/(1/c^2)^(1/4)*x-1))+1/4*d*e^2*ln(c^2*x^4+1))))
Result contains complex when optimal does not.
Time = 74.55 (sec) , antiderivative size = 2478078, normalized size of antiderivative = 7555.12 \[ \int \frac {a+b \arctan \left (c x^2\right )}{(d+e x)^2} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {a+b \arctan \left (c x^2\right )}{(d+e x)^2} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.88 \[ \int \frac {a+b \arctan \left (c x^2\right )}{(d+e x)^2} \, dx=-\frac {1}{4} \, {\left ({\left (\frac {8 \, d e \log \left (e x + d\right )}{c^{2} d^{4} + e^{4}} - \frac {\frac {\sqrt {2} {\left (c d^{2} e + \sqrt {2} \sqrt {c} d e^{2} + e^{3}\right )} \log \left (c x^{2} + \sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}} - \frac {\sqrt {2} {\left (c d^{2} e - \sqrt {2} \sqrt {c} d e^{2} + e^{3}\right )} \log \left (c x^{2} - \sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}} - \frac {2 \, {\left (2 \, c^{2} d^{3} + \sqrt {2} c^{\frac {3}{2}} d^{2} e - \sqrt {2} \sqrt {c} e^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x + \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{c} + \frac {2 \, {\left (2 \, c^{2} d^{3} - \sqrt {2} c^{\frac {3}{2}} d^{2} e + \sqrt {2} \sqrt {c} e^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x - \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{c}}{c^{2} d^{4} e + e^{5}}\right )} c + \frac {4 \, \arctan \left (c x^{2}\right )}{e^{2} x + d e}\right )} b - \frac {a}{e^{2} x + d e} \]
-1/4*((8*d*e*log(e*x + d)/(c^2*d^4 + e^4) - (sqrt(2)*(c*d^2*e + sqrt(2)*sq rt(c)*d*e^2 + e^3)*log(c*x^2 + sqrt(2)*sqrt(c)*x + 1)/sqrt(c) - sqrt(2)*(c *d^2*e - sqrt(2)*sqrt(c)*d*e^2 + e^3)*log(c*x^2 - sqrt(2)*sqrt(c)*x + 1)/s qrt(c) - 2*(2*c^2*d^3 + sqrt(2)*c^(3/2)*d^2*e - sqrt(2)*sqrt(c)*e^3)*arcta n(1/2*sqrt(2)*(2*c*x + sqrt(2)*sqrt(c))/sqrt(c))/c + 2*(2*c^2*d^3 - sqrt(2 )*c^(3/2)*d^2*e + sqrt(2)*sqrt(c)*e^3)*arctan(1/2*sqrt(2)*(2*c*x - sqrt(2) *sqrt(c))/sqrt(c))/c)/(c^2*d^4*e + e^5))*c + 4*arctan(c*x^2)/(e^2*x + d*e) )*b - a/(e^2*x + d*e)
\[ \int \frac {a+b \arctan \left (c x^2\right )}{(d+e x)^2} \, dx=\int { \frac {b \arctan \left (c x^{2}\right ) + a}{{\left (e x + d\right )}^{2}} \,d x } \]
Time = 0.70 (sec) , antiderivative size = 883, normalized size of antiderivative = 2.69 \[ \int \frac {a+b \arctan \left (c x^2\right )}{(d+e x)^2} \, dx=\left (\sum _{k=1}^4\ln \left (\frac {{\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )}^4\,c^8\,e^9\,x\,320-{\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )}^4\,c^{10}\,d^5\,e^4\,128+16\,b^4\,c^{10}\,e\,x-\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )\,b^3\,c^9\,e^3\,8+{\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )}^4\,c^8\,d\,e^8\,384+\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )\,b^3\,c^{11}\,d^3\,x\,8-{\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )}^3\,b\,c^9\,d^2\,e^5\,320-{\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )}^4\,c^{10}\,d^4\,e^5\,x\,192+{\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )}^3\,b\,c^{11}\,d^5\,e^2\,x\,32+{\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )}^2\,b^2\,c^{10}\,d^2\,e^3\,x\,64-{\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )}^3\,b\,c^9\,d\,e^6\,x\,416}{e^2}\right )\,\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )\right )-\frac {a}{x\,e^2+d\,e}-\frac {b\,\mathrm {atan}\left (c\,x^2\right )}{x\,e^2+d\,e}-\frac {2\,b\,c\,d\,e\,\ln \left (d+e\,x\right )}{c^2\,d^4+e^4} \]
symsum(log((320*root(16*c^2*d^4*e^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z, k)^4*c^8*e^9*x - 128*root(16*c^2*d^4*e ^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z, k)^4*c^10*d^5*e^4 + 16*b^4*c^10*e*x - 8*root(16*c^2*d^4*e^4*z^4 + 16*e^ 8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z, k)*b^3*c^9* e^3 + 384*root(16*c^2*d^4*e^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2* c^2*d^2*e^2*z^2 + b^4*c^2, z, k)^4*c^8*d*e^8 + 8*root(16*c^2*d^4*e^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z, k)*b^ 3*c^11*d^3*x - 320*root(16*c^2*d^4*e^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z, k)^3*b*c^9*d^2*e^5 - 192*root(16*c^ 2*d^4*e^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^ 4*c^2, z, k)^4*c^10*d^4*e^5*x + 32*root(16*c^2*d^4*e^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z, k)^3*b*c^11*d^5*e^2 *x + 64*root(16*c^2*d^4*e^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2*c^ 2*d^2*e^2*z^2 + b^4*c^2, z, k)^2*b^2*c^10*d^2*e^3*x - 416*root(16*c^2*d^4* e^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z, k)^3*b*c^9*d*e^6*x)/e^2)*root(16*c^2*d^4*e^4*z^4 + 16*e^8*z^4 - 32*b*c *d*e^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z, k), k, 1, 4) - a/(d*e + e ^2*x) - (b*atan(c*x^2))/(d*e + e^2*x) - (2*b*c*d*e*log(d + e*x))/(e^4 + c^ 2*d^4)