Integrand size = 26, antiderivative size = 278 \[ \int x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=-\frac {2 a e x}{5 c^4}-\frac {77 b e x^2}{300 c^3}+\frac {2 a e x^3}{15 c^2}+\frac {9 b e x^4}{200 c}-\frac {2}{25} a e x^5+\frac {2 a e \arctan (c x)}{5 c^5}-\frac {2 b e x \arctan (c x)}{5 c^4}+\frac {2 b e x^3 \arctan (c x)}{15 c^2}-\frac {2}{25} b e x^5 \arctan (c x)+\frac {b e \arctan (c x)^2}{5 c^5}+\frac {137 b e \log \left (1+c^2 x^2\right )}{300 c^5}+\frac {b e \log ^2\left (1+c^2 x^2\right )}{20 c^5}+\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^5} \]
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Time = 0.47 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.577, Rules used = {4946, 272, 45, 5141, 6857, 1816, 649, 209, 266, 5036, 4930, 5004, 2525, 2437, 2338} \[ \int x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {1}{5} x^5 (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )+\frac {2 a e \arctan (c x)}{5 c^5}-\frac {2 a e x}{5 c^4}+\frac {2 a e x^3}{15 c^2}-\frac {2}{25} a e x^5+\frac {b e \arctan (c x)^2}{5 c^5}-\frac {2 b e x \arctan (c x)}{5 c^4}+\frac {2 b e x^3 \arctan (c x)}{15 c^2}-\frac {2}{25} b e x^5 \arctan (c x)-\frac {77 b e x^2}{300 c^3}-\frac {b x^4 \left (e \log \left (c^2 x^2+1\right )+d\right )}{20 c}-\frac {b \log \left (c^2 x^2+1\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{10 c^5}+\frac {b e \log ^2\left (c^2 x^2+1\right )}{20 c^5}+\frac {137 b e \log \left (c^2 x^2+1\right )}{300 c^5}+\frac {b x^2 \left (e \log \left (c^2 x^2+1\right )+d\right )}{10 c^3}+\frac {9 b e x^4}{200 c} \]
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Rule 45
Rule 209
Rule 266
Rule 272
Rule 649
Rule 1816
Rule 2338
Rule 2437
Rule 2525
Rule 4930
Rule 4946
Rule 5004
Rule 5036
Rule 5141
Rule 6857
Rubi steps \begin{align*} \text {integral}& = \frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^5}-\left (2 c^2 e\right ) \int \left (\frac {2 b x^3-b c^2 x^5+4 a c^3 x^6+4 b c^3 x^6 \arctan (c x)}{20 c^3 \left (1+c^2 x^2\right )}-\frac {b x \log \left (1+c^2 x^2\right )}{10 c^5 \left (1+c^2 x^2\right )}\right ) \, dx \\ & = \frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^5}+\frac {(b e) \int \frac {x \log \left (1+c^2 x^2\right )}{1+c^2 x^2} \, dx}{5 c^3}-\frac {e \int \frac {2 b x^3-b c^2 x^5+4 a c^3 x^6+4 b c^3 x^6 \arctan (c x)}{1+c^2 x^2} \, dx}{10 c} \\ & = \frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^5}+\frac {(b e) \text {Subst}\left (\int \frac {\log \left (1+c^2 x\right )}{1+c^2 x} \, dx,x,x^2\right )}{10 c^3}-\frac {e \int \left (\frac {x^3 \left (2 b-b c^2 x^2+4 a c^3 x^3\right )}{1+c^2 x^2}+\frac {4 b c^3 x^6 \arctan (c x)}{1+c^2 x^2}\right ) \, dx}{10 c} \\ & = \frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^5}+\frac {(b e) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+c^2 x^2\right )}{10 c^5}-\frac {e \int \frac {x^3 \left (2 b-b c^2 x^2+4 a c^3 x^3\right )}{1+c^2 x^2} \, dx}{10 c}-\frac {1}{5} \left (2 b c^2 e\right ) \int \frac {x^6 \arctan (c x)}{1+c^2 x^2} \, dx \\ & = \frac {b e \log ^2\left (1+c^2 x^2\right )}{20 c^5}+\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^5}-\frac {1}{5} (2 b e) \int x^4 \arctan (c x) \, dx+\frac {1}{5} (2 b e) \int \frac {x^4 \arctan (c x)}{1+c^2 x^2} \, dx-\frac {e \int \left (\frac {4 a}{c^3}+\frac {3 b x}{c^2}-\frac {4 a x^2}{c}-b x^3+4 a c x^4-\frac {4 a+3 b c x}{c^3 \left (1+c^2 x^2\right )}\right ) \, dx}{10 c} \\ & = -\frac {2 a e x}{5 c^4}-\frac {3 b e x^2}{20 c^3}+\frac {2 a e x^3}{15 c^2}+\frac {b e x^4}{40 c}-\frac {2}{25} a e x^5-\frac {2}{25} b e x^5 \arctan (c x)+\frac {b e \log ^2\left (1+c^2 x^2\right )}{20 c^5}+\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^5}+\frac {e \int \frac {4 a+3 b c x}{1+c^2 x^2} \, dx}{10 c^4}+\frac {(2 b e) \int x^2 \arctan (c x) \, dx}{5 c^2}-\frac {(2 b e) \int \frac {x^2 \arctan (c x)}{1+c^2 x^2} \, dx}{5 c^2}+\frac {1}{25} (2 b c e) \int \frac {x^5}{1+c^2 x^2} \, dx \\ & = -\frac {2 a e x}{5 c^4}-\frac {3 b e x^2}{20 c^3}+\frac {2 a e x^3}{15 c^2}+\frac {b e x^4}{40 c}-\frac {2}{25} a e x^5+\frac {2 b e x^3 \arctan (c x)}{15 c^2}-\frac {2}{25} b e x^5 \arctan (c x)+\frac {b e \log ^2\left (1+c^2 x^2\right )}{20 c^5}+\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^5}+\frac {(2 a e) \int \frac {1}{1+c^2 x^2} \, dx}{5 c^4}-\frac {(2 b e) \int \arctan (c x) \, dx}{5 c^4}+\frac {(2 b e) \int \frac {\arctan (c x)}{1+c^2 x^2} \, dx}{5 c^4}+\frac {(3 b e) \int \frac {x}{1+c^2 x^2} \, dx}{10 c^3}-\frac {(2 b e) \int \frac {x^3}{1+c^2 x^2} \, dx}{15 c}+\frac {1}{25} (b c e) \text {Subst}\left (\int \frac {x^2}{1+c^2 x} \, dx,x,x^2\right ) \\ & = -\frac {2 a e x}{5 c^4}-\frac {3 b e x^2}{20 c^3}+\frac {2 a e x^3}{15 c^2}+\frac {b e x^4}{40 c}-\frac {2}{25} a e x^5+\frac {2 a e \arctan (c x)}{5 c^5}-\frac {2 b e x \arctan (c x)}{5 c^4}+\frac {2 b e x^3 \arctan (c x)}{15 c^2}-\frac {2}{25} b e x^5 \arctan (c x)+\frac {b e \arctan (c x)^2}{5 c^5}+\frac {3 b e \log \left (1+c^2 x^2\right )}{20 c^5}+\frac {b e \log ^2\left (1+c^2 x^2\right )}{20 c^5}+\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^5}+\frac {(2 b e) \int \frac {x}{1+c^2 x^2} \, dx}{5 c^3}-\frac {(b e) \text {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^2\right )}{15 c}+\frac {1}{25} (b c e) \text {Subst}\left (\int \left (-\frac {1}{c^4}+\frac {x}{c^2}+\frac {1}{c^4 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right ) \\ & = -\frac {2 a e x}{5 c^4}-\frac {19 b e x^2}{100 c^3}+\frac {2 a e x^3}{15 c^2}+\frac {9 b e x^4}{200 c}-\frac {2}{25} a e x^5+\frac {2 a e \arctan (c x)}{5 c^5}-\frac {2 b e x \arctan (c x)}{5 c^4}+\frac {2 b e x^3 \arctan (c x)}{15 c^2}-\frac {2}{25} b e x^5 \arctan (c x)+\frac {b e \arctan (c x)^2}{5 c^5}+\frac {39 b e \log \left (1+c^2 x^2\right )}{100 c^5}+\frac {b e \log ^2\left (1+c^2 x^2\right )}{20 c^5}+\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^5}-\frac {(b e) \text {Subst}\left (\int \left (\frac {1}{c^2}-\frac {1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{15 c} \\ & = -\frac {2 a e x}{5 c^4}-\frac {77 b e x^2}{300 c^3}+\frac {2 a e x^3}{15 c^2}+\frac {9 b e x^4}{200 c}-\frac {2}{25} a e x^5+\frac {2 a e \arctan (c x)}{5 c^5}-\frac {2 b e x \arctan (c x)}{5 c^4}+\frac {2 b e x^3 \arctan (c x)}{15 c^2}-\frac {2}{25} b e x^5 \arctan (c x)+\frac {b e \arctan (c x)^2}{5 c^5}+\frac {137 b e \log \left (1+c^2 x^2\right )}{300 c^5}+\frac {b e \log ^2\left (1+c^2 x^2\right )}{20 c^5}+\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^5} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.77 \[ \int x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {c x \left (b c x \left (-30 d \left (-2+c^2 x^2\right )+e \left (-154+27 c^2 x^2\right )\right )+8 a \left (15 c^4 d x^4-2 e \left (15-5 c^2 x^2+3 c^4 x^4\right )\right )\right )+120 b e \arctan (c x)^2+\left (-60 b d+120 a c^5 e x^5+2 b e \left (137+30 c^2 x^2-15 c^4 x^4\right )\right ) \log \left (1+c^2 x^2\right )-30 b e \log ^2\left (1+c^2 x^2\right )+8 \arctan (c x) \left (30 a e+15 b c^5 d x^5-2 b c e x \left (15-5 c^2 x^2+3 c^4 x^4\right )+15 b c^5 e x^5 \log \left (1+c^2 x^2\right )\right )}{600 c^5} \]
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Time = 2.69 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(\frac {120 a \,c^{5} d \,x^{5}+80 x^{3} \arctan \left (c x \right ) b \,c^{3} e +154 e b -154 b \,c^{2} e \,x^{2}-48 a \,c^{5} e \,x^{5}+27 b \,c^{4} e \,x^{4}+80 a \,c^{3} e \,x^{3}+274 \ln \left (c^{2} x^{2}+1\right ) b e -48 x^{5} \arctan \left (c x \right ) b \,c^{5} e -240 x a c e -30 e b \ln \left (c^{2} x^{2}+1\right )^{2}-30 b \,c^{4} d \,x^{4}+240 e a \arctan \left (c x \right )+120 e b \arctan \left (c x \right )^{2}-60 \ln \left (c^{2} x^{2}+1\right ) b d +120 e a \ln \left (c^{2} x^{2}+1\right ) x^{5} c^{5}+60 x^{2} \ln \left (c^{2} x^{2}+1\right ) b \,c^{2} e -240 e b \arctan \left (c x \right ) x c -30 e b \ln \left (c^{2} x^{2}+1\right ) x^{4} c^{4}+120 b \arctan \left (c x \right ) x^{5} c^{5} d +120 e b \ln \left (c^{2} x^{2}+1\right ) \arctan \left (c x \right ) x^{5} c^{5}+60 c^{2} x^{2} b d -60 b d}{600 c^{5}}\) | \(289\) |
default | \(\text {Expression too large to display}\) | \(4787\) |
parts | \(\text {Expression too large to display}\) | \(4787\) |
risch | \(\text {Expression too large to display}\) | \(23634\) |
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Time = 0.26 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.79 \[ \int x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {80 \, a c^{3} e x^{3} + 24 \, {\left (5 \, a c^{5} d - 2 \, a c^{5} e\right )} x^{5} - 3 \, {\left (10 \, b c^{4} d - 9 \, b c^{4} e\right )} x^{4} - 240 \, a c e x + 120 \, b e \arctan \left (c x\right )^{2} - 30 \, b e \log \left (c^{2} x^{2} + 1\right )^{2} + 2 \, {\left (30 \, b c^{2} d - 77 \, b c^{2} e\right )} x^{2} + 8 \, {\left (10 \, b c^{3} e x^{3} + 3 \, {\left (5 \, b c^{5} d - 2 \, b c^{5} e\right )} x^{5} - 30 \, b c e x + 30 \, a e\right )} \arctan \left (c x\right ) + 2 \, {\left (60 \, b c^{5} e x^{5} \arctan \left (c x\right ) + 60 \, a c^{5} e x^{5} - 15 \, b c^{4} e x^{4} + 30 \, b c^{2} e x^{2} - 30 \, b d + 137 \, b e\right )} \log \left (c^{2} x^{2} + 1\right )}{600 \, c^{5}} \]
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Time = 1.63 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.22 \[ \int x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\begin {cases} \frac {a d x^{5}}{5} + \frac {a e x^{5} \log {\left (c^{2} x^{2} + 1 \right )}}{5} - \frac {2 a e x^{5}}{25} + \frac {2 a e x^{3}}{15 c^{2}} - \frac {2 a e x}{5 c^{4}} + \frac {2 a e \operatorname {atan}{\left (c x \right )}}{5 c^{5}} + \frac {b d x^{5} \operatorname {atan}{\left (c x \right )}}{5} + \frac {b e x^{5} \log {\left (c^{2} x^{2} + 1 \right )} \operatorname {atan}{\left (c x \right )}}{5} - \frac {2 b e x^{5} \operatorname {atan}{\left (c x \right )}}{25} - \frac {b d x^{4}}{20 c} - \frac {b e x^{4} \log {\left (c^{2} x^{2} + 1 \right )}}{20 c} + \frac {9 b e x^{4}}{200 c} + \frac {2 b e x^{3} \operatorname {atan}{\left (c x \right )}}{15 c^{2}} + \frac {b d x^{2}}{10 c^{3}} + \frac {b e x^{2} \log {\left (c^{2} x^{2} + 1 \right )}}{10 c^{3}} - \frac {77 b e x^{2}}{300 c^{3}} - \frac {2 b e x \operatorname {atan}{\left (c x \right )}}{5 c^{4}} - \frac {b d \log {\left (c^{2} x^{2} + 1 \right )}}{10 c^{5}} - \frac {b e \log {\left (c^{2} x^{2} + 1 \right )}^{2}}{20 c^{5}} + \frac {137 b e \log {\left (c^{2} x^{2} + 1 \right )}}{300 c^{5}} + \frac {b e \operatorname {atan}^{2}{\left (c x \right )}}{5 c^{5}} & \text {for}\: c \neq 0 \\\frac {a d x^{5}}{5} & \text {otherwise} \end {cases} \]
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Time = 0.27 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.92 \[ \int x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {1}{5} \, a d x^{5} + \frac {1}{75} \, {\left (15 \, x^{5} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{2} {\left (\frac {3 \, c^{4} x^{5} - 5 \, c^{2} x^{3} + 15 \, x}{c^{6}} - \frac {15 \, \arctan \left (c x\right )}{c^{7}}\right )}\right )} b e \arctan \left (c x\right ) + \frac {1}{20} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b d + \frac {1}{75} \, {\left (15 \, x^{5} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{2} {\left (\frac {3 \, c^{4} x^{5} - 5 \, c^{2} x^{3} + 15 \, x}{c^{6}} - \frac {15 \, \arctan \left (c x\right )}{c^{7}}\right )}\right )} a e + \frac {{\left (27 \, c^{4} x^{4} - 154 \, c^{2} x^{2} - 120 \, \arctan \left (c x\right )^{2} - 2 \, {\left (15 \, c^{4} x^{4} - 30 \, c^{2} x^{2} - 137\right )} \log \left (c^{2} x^{2} + 1\right ) - 30 \, \log \left (c^{2} x^{2} + 1\right )^{2}\right )} b e}{600 \, c^{5}} \]
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\[ \int x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )} x^{4} \,d x } \]
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Time = 3.72 (sec) , antiderivative size = 276, normalized size of antiderivative = 0.99 \[ \int x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {a\,d\,x^5}{5}-\frac {2\,a\,e\,x^5}{25}-\frac {b\,e\,{\ln \left (c^2\,x^2+1\right )}^2}{20\,c^5}-\frac {2\,a\,e\,x}{5\,c^4}+\frac {2\,a\,e\,\mathrm {atan}\left (c\,x\right )}{5\,c^5}+\frac {b\,d\,x^5\,\mathrm {atan}\left (c\,x\right )}{5}-\frac {2\,b\,e\,x^5\,\mathrm {atan}\left (c\,x\right )}{25}-\frac {b\,d\,\ln \left (c^2\,x^2+1\right )}{10\,c^5}+\frac {137\,b\,e\,\ln \left (c^2\,x^2+1\right )}{300\,c^5}+\frac {2\,a\,e\,x^3}{15\,c^2}-\frac {b\,d\,x^4}{20\,c}+\frac {b\,d\,x^2}{10\,c^3}+\frac {9\,b\,e\,x^4}{200\,c}-\frac {77\,b\,e\,x^2}{300\,c^3}+\frac {a\,e\,x^5\,\ln \left (c^2\,x^2+1\right )}{5}+\frac {b\,e\,{\mathrm {atan}\left (c\,x\right )}^2}{5\,c^5}+\frac {2\,b\,e\,x^3\,\mathrm {atan}\left (c\,x\right )}{15\,c^2}+\frac {b\,e\,x^5\,\mathrm {atan}\left (c\,x\right )\,\ln \left (c^2\,x^2+1\right )}{5}-\frac {b\,e\,x^4\,\ln \left (c^2\,x^2+1\right )}{20\,c}+\frac {b\,e\,x^2\,\ln \left (c^2\,x^2+1\right )}{10\,c^3}-\frac {2\,b\,e\,x\,\mathrm {atan}\left (c\,x\right )}{5\,c^4} \]
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