Integrand size = 26, antiderivative size = 278 \[ \int x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=-\frac {2 a e x}{5 c^4}-\frac {77 b e x^2}{300 c^3}+\frac {2 a e x^3}{15 c^2}+\frac {9 b e x^4}{200 c}-\frac {2}{25} a e x^5+\frac {2 a e \arctan (c x)}{5 c^5}-\frac {2 b e x \arctan (c x)}{5 c^4}+\frac {2 b e x^3 \arctan (c x)}{15 c^2}-\frac {2}{25} b e x^5 \arctan (c x)+\frac {b e \arctan (c x)^2}{5 c^5}+\frac {137 b e \log \left (1+c^2 x^2\right )}{300 c^5}+\frac {b e \log ^2\left (1+c^2 x^2\right )}{20 c^5}+\frac {b x^2 \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^3}-\frac {b x^4 \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 c}+\frac {1}{5} x^5 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {b \log \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 c^5} \]
-2/5*a*e*x/c^4-77/300*b*e*x^2/c^3+2/15*a*e*x^3/c^2+9/200*b*e*x^4/c-2/25*a* e*x^5+2/5*a*e*arctan(c*x)/c^5-2/5*b*e*x*arctan(c*x)/c^4+2/15*b*e*x^3*arcta n(c*x)/c^2-2/25*b*e*x^5*arctan(c*x)+1/5*b*e*arctan(c*x)^2/c^5+137/300*b*e* ln(c^2*x^2+1)/c^5+1/20*b*e*ln(c^2*x^2+1)^2/c^5+1/10*b*x^2*(d+e*ln(c^2*x^2+ 1))/c^3-1/20*b*x^4*(d+e*ln(c^2*x^2+1))/c+1/5*x^5*(a+b*arctan(c*x))*(d+e*ln (c^2*x^2+1))-1/10*b*ln(c^2*x^2+1)*(d+e*ln(c^2*x^2+1))/c^5
Time = 0.13 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.77 \[ \int x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {c x \left (b c x \left (-30 d \left (-2+c^2 x^2\right )+e \left (-154+27 c^2 x^2\right )\right )+8 a \left (15 c^4 d x^4-2 e \left (15-5 c^2 x^2+3 c^4 x^4\right )\right )\right )+120 b e \arctan (c x)^2+\left (-60 b d+120 a c^5 e x^5+2 b e \left (137+30 c^2 x^2-15 c^4 x^4\right )\right ) \log \left (1+c^2 x^2\right )-30 b e \log ^2\left (1+c^2 x^2\right )+8 \arctan (c x) \left (30 a e+15 b c^5 d x^5-2 b c e x \left (15-5 c^2 x^2+3 c^4 x^4\right )+15 b c^5 e x^5 \log \left (1+c^2 x^2\right )\right )}{600 c^5} \]
(c*x*(b*c*x*(-30*d*(-2 + c^2*x^2) + e*(-154 + 27*c^2*x^2)) + 8*a*(15*c^4*d *x^4 - 2*e*(15 - 5*c^2*x^2 + 3*c^4*x^4))) + 120*b*e*ArcTan[c*x]^2 + (-60*b *d + 120*a*c^5*e*x^5 + 2*b*e*(137 + 30*c^2*x^2 - 15*c^4*x^4))*Log[1 + c^2* x^2] - 30*b*e*Log[1 + c^2*x^2]^2 + 8*ArcTan[c*x]*(30*a*e + 15*b*c^5*d*x^5 - 2*b*c*e*x*(15 - 5*c^2*x^2 + 3*c^4*x^4) + 15*b*c^5*e*x^5*Log[1 + c^2*x^2] ))/(600*c^5)
Time = 0.86 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {5556, 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.
| \(\displaystyle \int x^4 (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right ) \, dx\) |
| \(\Big \downarrow \) 5556 |
| \(\displaystyle -2 c^2 e \int \left (\frac {4 a c^3 x^6+4 b c^3 \arctan (c x) x^6-b c^2 x^5+2 b x^3}{20 c^3 \left (c^2 x^2+1\right )}-\frac {b x \log \left (c^2 x^2+1\right )}{10 c^5 \left (c^2 x^2+1\right )}\right )dx+\frac {1}{5} x^5 (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )-\frac {b x^4 \left (e \log \left (c^2 x^2+1\right )+d\right )}{20 c}-\frac {b \log \left (c^2 x^2+1\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{10 c^5}+\frac {b x^2 \left (e \log \left (c^2 x^2+1\right )+d\right )}{10 c^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} x^5 (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )-2 c^2 e \left (-\frac {a \arctan (c x)}{5 c^7}+\frac {a x}{5 c^6}-\frac {a x^3}{15 c^4}+\frac {a x^5}{25 c^2}-\frac {b \arctan (c x)^2}{10 c^7}+\frac {b x \arctan (c x)}{5 c^6}-\frac {b x^3 \arctan (c x)}{15 c^4}+\frac {b x^5 \arctan (c x)}{25 c^2}+\frac {77 b x^2}{600 c^5}-\frac {9 b x^4}{400 c^3}-\frac {b \log ^2\left (c^2 x^2+1\right )}{40 c^7}-\frac {137 b \log \left (c^2 x^2+1\right )}{600 c^7}\right )-\frac {b x^4 \left (e \log \left (c^2 x^2+1\right )+d\right )}{20 c}-\frac {b \log \left (c^2 x^2+1\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{10 c^5}+\frac {b x^2 \left (e \log \left (c^2 x^2+1\right )+d\right )}{10 c^3}\) |
(b*x^2*(d + e*Log[1 + c^2*x^2]))/(10*c^3) - (b*x^4*(d + e*Log[1 + c^2*x^2] ))/(20*c) + (x^5*(a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/5 - (b*Log[ 1 + c^2*x^2]*(d + e*Log[1 + c^2*x^2]))/(10*c^5) - 2*c^2*e*((a*x)/(5*c^6) + (77*b*x^2)/(600*c^5) - (a*x^3)/(15*c^4) - (9*b*x^4)/(400*c^3) + (a*x^5)/( 25*c^2) - (a*ArcTan[c*x])/(5*c^7) + (b*x*ArcTan[c*x])/(5*c^6) - (b*x^3*Arc Tan[c*x])/(15*c^4) + (b*x^5*ArcTan[c*x])/(25*c^2) - (b*ArcTan[c*x]^2)/(10* c^7) - (137*b*Log[1 + c^2*x^2])/(600*c^7) - (b*Log[1 + c^2*x^2]^2)/(40*c^7 ))
3.13.86.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*( e_.))*(x_)^(m_.), x_Symbol] :> With[{u = IntHide[x^m*(a + b*ArcTan[c*x]), x ]}, Simp[(d + e*Log[f + g*x^2]) u, x] - Simp[2*e*g Int[ExpandIntegrand[ x*(u/(f + g*x^2)), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && Intege rQ[m] && NeQ[m, -1]
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 | \(\text {Expression too large to display}\) | \(23634\) |
1/600*(120*a*c^5*d*x^5+80*x^3*arctan(c*x)*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(c^2*x^2+1)*b*e-48*x^5* arctan(c*x)*b*c^5*e-240*x*a*c*e-30*e*b*ln(c^2*x^2+1)^2-30*b*c^4*d*x^4+240* e*a*arctan(c*x)+120*e*b*arctan(c*x)^2-60*ln(c^2*x^2+1)*b*d+120*e*a*ln(c^2* x^2+1)*x^5*c^5+60*x^2*ln(c^2*x^2+1)*b*c^2*e-240*e*b*arctan(c*x)*x*c-30*e*b *ln(c^2*x^2+1)*x^4*c^4+120*b*arctan(c*x)*x^5*c^5*d+120*e*b*ln(c^2*x^2+1)*a rctan(c*x)*x^5*c^5+60*c^2*x^2*b*d-60*b*d)/c^5
Time = 0.26 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.79 \[ \int x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {80 \, a c^{3} e x^{3} + 24 \, {\left (5 \, a c^{5} d - 2 \, a c^{5} e\right )} x^{5} - 3 \, {\left (10 \, b c^{4} d - 9 \, b c^{4} e\right )} x^{4} - 240 \, a c e x + 120 \, b e \arctan \left (c x\right )^{2} - 30 \, b e \log \left (c^{2} x^{2} + 1\right )^{2} + 2 \, {\left (30 \, b c^{2} d - 77 \, b c^{2} e\right )} x^{2} + 8 \, {\left (10 \, b c^{3} e x^{3} + 3 \, {\left (5 \, b c^{5} d - 2 \, b c^{5} e\right )} x^{5} - 30 \, b c e x + 30 \, a e\right )} \arctan \left (c x\right ) + 2 \, {\left (60 \, b c^{5} e x^{5} \arctan \left (c x\right ) + 60 \, a c^{5} e x^{5} - 15 \, b c^{4} e x^{4} + 30 \, b c^{2} e x^{2} - 30 \, b d + 137 \, b e\right )} \log \left (c^{2} x^{2} + 1\right )}{600 \, c^{5}} \]
1/600*(80*a*c^3*e*x^3 + 24*(5*a*c^5*d - 2*a*c^5*e)*x^5 - 3*(10*b*c^4*d - 9 *b*c^4*e)*x^4 - 240*a*c*e*x + 120*b*e*arctan(c*x)^2 - 30*b*e*log(c^2*x^2 + 1)^2 + 2*(30*b*c^2*d - 77*b*c^2*e)*x^2 + 8*(10*b*c^3*e*x^3 + 3*(5*b*c^5*d - 2*b*c^5*e)*x^5 - 30*b*c*e*x + 30*a*e)*arctan(c*x) + 2*(60*b*c^5*e*x^5*a rctan(c*x) + 60*a*c^5*e*x^5 - 15*b*c^4*e*x^4 + 30*b*c^2*e*x^2 - 30*b*d + 1 37*b*e)*log(c^2*x^2 + 1))/c^5
Time = 1.63 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.22 \[ \int x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\begin {cases} \frac {a d x^{5}}{5} + \frac {a e x^{5} \log {\left (c^{2} x^{2} + 1 \right )}}{5} - \frac {2 a e x^{5}}{25} + \frac {2 a e x^{3}}{15 c^{2}} - \frac {2 a e x}{5 c^{4}} + \frac {2 a e \operatorname {atan}{\left (c x \right )}}{5 c^{5}} + \frac {b d x^{5} \operatorname {atan}{\left (c x \right )}}{5} + \frac {b e x^{5} \log {\left (c^{2} x^{2} + 1 \right )} \operatorname {atan}{\left (c x \right )}}{5} - \frac {2 b e x^{5} \operatorname {atan}{\left (c x \right )}}{25} - \frac {b d x^{4}}{20 c} - \frac {b e x^{4} \log {\left (c^{2} x^{2} + 1 \right )}}{20 c} + \frac {9 b e x^{4}}{200 c} + \frac {2 b e x^{3} \operatorname {atan}{\left (c x \right )}}{15 c^{2}} + \frac {b d x^{2}}{10 c^{3}} + \frac {b e x^{2} \log {\left (c^{2} x^{2} + 1 \right )}}{10 c^{3}} - \frac {77 b e x^{2}}{300 c^{3}} - \frac {2 b e x \operatorname {atan}{\left (c x \right )}}{5 c^{4}} - \frac {b d \log {\left (c^{2} x^{2} + 1 \right )}}{10 c^{5}} - \frac {b e \log {\left (c^{2} x^{2} + 1 \right )}^{2}}{20 c^{5}} + \frac {137 b e \log {\left (c^{2} x^{2} + 1 \right )}}{300 c^{5}} + \frac {b e \operatorname {atan}^{2}{\left (c x \right )}}{5 c^{5}} & \text {for}\: c \neq 0 \\\frac {a d x^{5}}{5} & \text {otherwise} \end {cases} \]
Piecewise((a*d*x**5/5 + a*e*x**5*log(c**2*x**2 + 1)/5 - 2*a*e*x**5/25 + 2* a*e*x**3/(15*c**2) - 2*a*e*x/(5*c**4) + 2*a*e*atan(c*x)/(5*c**5) + b*d*x** 5*atan(c*x)/5 + b*e*x**5*log(c**2*x**2 + 1)*atan(c*x)/5 - 2*b*e*x**5*atan( c*x)/25 - b*d*x**4/(20*c) - b*e*x**4*log(c**2*x**2 + 1)/(20*c) + 9*b*e*x** 4/(200*c) + 2*b*e*x**3*atan(c*x)/(15*c**2) + b*d*x**2/(10*c**3) + b*e*x**2 *log(c**2*x**2 + 1)/(10*c**3) - 77*b*e*x**2/(300*c**3) - 2*b*e*x*atan(c*x) /(5*c**4) - b*d*log(c**2*x**2 + 1)/(10*c**5) - b*e*log(c**2*x**2 + 1)**2/( 20*c**5) + 137*b*e*log(c**2*x**2 + 1)/(300*c**5) + b*e*atan(c*x)**2/(5*c** 5), Ne(c, 0)), (a*d*x**5/5, True))
Time = 0.27 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.92 \[ \int x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {1}{5} \, a d x^{5} + \frac {1}{75} \, {\left (15 \, x^{5} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{2} {\left (\frac {3 \, c^{4} x^{5} - 5 \, c^{2} x^{3} + 15 \, x}{c^{6}} - \frac {15 \, \arctan \left (c x\right )}{c^{7}}\right )}\right )} b e \arctan \left (c x\right ) + \frac {1}{20} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b d + \frac {1}{75} \, {\left (15 \, x^{5} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{2} {\left (\frac {3 \, c^{4} x^{5} - 5 \, c^{2} x^{3} + 15 \, x}{c^{6}} - \frac {15 \, \arctan \left (c x\right )}{c^{7}}\right )}\right )} a e + \frac {{\left (27 \, c^{4} x^{4} - 154 \, c^{2} x^{2} - 120 \, \arctan \left (c x\right )^{2} - 2 \, {\left (15 \, c^{4} x^{4} - 30 \, c^{2} x^{2} - 137\right )} \log \left (c^{2} x^{2} + 1\right ) - 30 \, \log \left (c^{2} x^{2} + 1\right )^{2}\right )} b e}{600 \, c^{5}} \]
1/5*a*d*x^5 + 1/75*(15*x^5*log(c^2*x^2 + 1) - 2*c^2*((3*c^4*x^5 - 5*c^2*x^ 3 + 15*x)/c^6 - 15*arctan(c*x)/c^7))*b*e*arctan(c*x) + 1/20*(4*x^5*arctan( c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*b*d + 1/75*(15* x^5*log(c^2*x^2 + 1) - 2*c^2*((3*c^4*x^5 - 5*c^2*x^3 + 15*x)/c^6 - 15*arct an(c*x)/c^7))*a*e + 1/600*(27*c^4*x^4 - 154*c^2*x^2 - 120*arctan(c*x)^2 - 2*(15*c^4*x^4 - 30*c^2*x^2 - 137)*log(c^2*x^2 + 1) - 30*log(c^2*x^2 + 1)^2 )*b*e/c^5
\[ \int x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )} x^{4} \,d x } \]
Time = 3.72 (sec) , antiderivative size = 276, normalized size of antiderivative = 0.99 \[ \int x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {a\,d\,x^5}{5}-\frac {2\,a\,e\,x^5}{25}-\frac {b\,e\,{\ln \left (c^2\,x^2+1\right )}^2}{20\,c^5}-\frac {2\,a\,e\,x}{5\,c^4}+\frac {2\,a\,e\,\mathrm {atan}\left (c\,x\right )}{5\,c^5}+\frac {b\,d\,x^5\,\mathrm {atan}\left (c\,x\right )}{5}-\frac {2\,b\,e\,x^5\,\mathrm {atan}\left (c\,x\right )}{25}-\frac {b\,d\,\ln \left (c^2\,x^2+1\right )}{10\,c^5}+\frac {137\,b\,e\,\ln \left (c^2\,x^2+1\right )}{300\,c^5}+\frac {2\,a\,e\,x^3}{15\,c^2}-\frac {b\,d\,x^4}{20\,c}+\frac {b\,d\,x^2}{10\,c^3}+\frac {9\,b\,e\,x^4}{200\,c}-\frac {77\,b\,e\,x^2}{300\,c^3}+\frac {a\,e\,x^5\,\ln \left (c^2\,x^2+1\right )}{5}+\frac {b\,e\,{\mathrm {atan}\left (c\,x\right )}^2}{5\,c^5}+\frac {2\,b\,e\,x^3\,\mathrm {atan}\left (c\,x\right )}{15\,c^2}+\frac {b\,e\,x^5\,\mathrm {atan}\left (c\,x\right )\,\ln \left (c^2\,x^2+1\right )}{5}-\frac {b\,e\,x^4\,\ln \left (c^2\,x^2+1\right )}{20\,c}+\frac {b\,e\,x^2\,\ln \left (c^2\,x^2+1\right )}{10\,c^3}-\frac {2\,b\,e\,x\,\mathrm {atan}\left (c\,x\right )}{5\,c^4} \]
(a*d*x^5)/5 - (2*a*e*x^5)/25 - (b*e*log(c^2*x^2 + 1)^2)/(20*c^5) - (2*a*e* x)/(5*c^4) + (2*a*e*atan(c*x))/(5*c^5) + (b*d*x^5*atan(c*x))/5 - (2*b*e*x^ 5*atan(c*x))/25 - (b*d*log(c^2*x^2 + 1))/(10*c^5) + (137*b*e*log(c^2*x^2 + 1))/(300*c^5) + (2*a*e*x^3)/(15*c^2) - (b*d*x^4)/(20*c) + (b*d*x^2)/(10*c ^3) + (9*b*e*x^4)/(200*c) - (77*b*e*x^2)/(300*c^3) + (a*e*x^5*log(c^2*x^2 + 1))/5 + (b*e*atan(c*x)^2)/(5*c^5) + (2*b*e*x^3*atan(c*x))/(15*c^2) + (b* e*x^5*atan(c*x)*log(c^2*x^2 + 1))/5 - (b*e*x^4*log(c^2*x^2 + 1))/(20*c) + (b*e*x^2*log(c^2*x^2 + 1))/(10*c^3) - (2*b*e*x*atan(c*x))/(5*c^4)