Integrand size = 26, antiderivative size = 221 \[ \int x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {b (2 d-3 e) x}{8 c^3}-\frac {2 b e x}{3 c^3}-\frac {b (2 d-e) x^3}{24 c}+\frac {b e x^3}{18 c}-\frac {b (2 d-3 e) \arctan (c x)}{8 c^4}+\frac {2 b e \arctan (c x)}{3 c^4}+\frac {e x^2 (a+b \arctan (c x))}{4 c^2}-\frac {1}{8} e x^4 (a+b \arctan (c x))+\frac {b e x \log \left (1+c^2 x^2\right )}{4 c^3}-\frac {b e x^3 \log \left (1+c^2 x^2\right )}{12 c}-\frac {e (a+b \arctan (c x)) \log \left (1+c^2 x^2\right )}{4 c^4}+\frac {1}{4} x^4 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \]
1/8*b*(2*d-3*e)*x/c^3-2/3*b*e*x/c^3-1/24*b*(2*d-e)*x^3/c+1/18*b*e*x^3/c-1/ 8*b*(2*d-3*e)*arctan(c*x)/c^4+2/3*b*e*arctan(c*x)/c^4+1/4*e*x^2*(a+b*arcta n(c*x))/c^2-1/8*e*x^4*(a+b*arctan(c*x))+1/4*b*e*x*ln(c^2*x^2+1)/c^3-1/12*b *e*x^3*ln(c^2*x^2+1)/c-1/4*e*(a+b*arctan(c*x))*ln(c^2*x^2+1)/c^4+1/4*x^4*( a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))
Time = 0.10 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.74 \[ \int x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {c x \left (18 a c^3 d x^3-6 b d \left (-3+c^2 x^2\right )-9 a c e x \left (-2+c^2 x^2\right )+b e \left (-75+7 c^2 x^2\right )\right )-6 e \left (b c x \left (-3+c^2 x^2\right )+a \left (3-3 c^4 x^4\right )\right ) \log \left (1+c^2 x^2\right )+3 b \arctan (c x) \left (e \left (25+6 c^2 x^2-3 c^4 x^4\right )+6 d \left (-1+c^4 x^4\right )+6 e \left (-1+c^4 x^4\right ) \log \left (1+c^2 x^2\right )\right )}{72 c^4} \]
(c*x*(18*a*c^3*d*x^3 - 6*b*d*(-3 + c^2*x^2) - 9*a*c*e*x*(-2 + c^2*x^2) + b *e*(-75 + 7*c^2*x^2)) - 6*e*(b*c*x*(-3 + c^2*x^2) + a*(3 - 3*c^4*x^4))*Log [1 + c^2*x^2] + 3*b*ArcTan[c*x]*(e*(25 + 6*c^2*x^2 - 3*c^4*x^4) + 6*d*(-1 + c^4*x^4) + 6*e*(-1 + c^4*x^4)*Log[1 + c^2*x^2]))/(72*c^4)
Time = 0.46 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {5554, 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.
| \(\displaystyle \int x^3 (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right ) \, dx\) |
| \(\Big \downarrow \) 5554 |
| \(\displaystyle -b c \int \left (\frac {x^2 \left (c^2 (2 d-e) x^2+2 e\right )}{8 c^2 \left (c^2 x^2+1\right )}-\frac {e \left (1-c^2 x^2\right ) \log \left (c^2 x^2+1\right )}{4 c^4}\right )dx+\frac {1}{4} x^4 (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )+\frac {e x^2 (a+b \arctan (c x))}{4 c^2}-\frac {e \log \left (c^2 x^2+1\right ) (a+b \arctan (c x))}{4 c^4}-\frac {1}{8} e x^4 (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} x^4 (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )+\frac {e x^2 (a+b \arctan (c x))}{4 c^2}-\frac {e \log \left (c^2 x^2+1\right ) (a+b \arctan (c x))}{4 c^4}-\frac {1}{8} e x^4 (a+b \arctan (c x))-b c \left (\frac {(2 d-3 e) \arctan (c x)}{8 c^5}-\frac {2 e \arctan (c x)}{3 c^5}-\frac {x (2 d-3 e)}{8 c^4}+\frac {2 e x}{3 c^4}+\frac {x^3 (2 d-e)}{24 c^2}-\frac {e x^3}{18 c^2}+\frac {e x^3 \log \left (c^2 x^2+1\right )}{12 c^2}-\frac {e x \log \left (c^2 x^2+1\right )}{4 c^4}\right )\) |
(e*x^2*(a + b*ArcTan[c*x]))/(4*c^2) - (e*x^4*(a + b*ArcTan[c*x]))/8 - (e*( a + b*ArcTan[c*x])*Log[1 + c^2*x^2])/(4*c^4) + (x^4*(a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/4 - b*c*(-1/8*((2*d - 3*e)*x)/c^4 + (2*e*x)/(3*c^4 ) + ((2*d - e)*x^3)/(24*c^2) - (e*x^3)/(18*c^2) + ((2*d - 3*e)*ArcTan[c*x] )/(8*c^5) - (2*e*ArcTan[c*x])/(3*c^5) - (e*x*Log[1 + c^2*x^2])/(4*c^4) + ( e*x^3*Log[1 + c^2*x^2])/(12*c^2))
3.13.87.3.1 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*(d + e*Log[f + g*x^2]) , x]}, Simp[(a + b*ArcTan[c*x]) u, x] - Simp[b*c Int[ExpandIntegrand[u/ (1 + c^2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[(m + 1)/2, 0]
Time = 1.67 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.10
| method | result | size |
| parallelrisch | \(\frac {18 e b \ln \left (c^{2} x^{2}+1\right ) \arctan \left (c x \right ) x^{4} c^{4}+18 x^{4} \arctan \left (c x \right ) b \,c^{4} d -9 x^{4} \arctan \left (c x \right ) b \,c^{4} e +18 e a \ln \left (c^{2} x^{2}+1\right ) x^{4} c^{4}+18 c^{4} a d \,x^{4}-9 x^{4} a \,c^{4} e -6 e b \ln \left (c^{2} x^{2}+1\right ) x^{3} c^{3}-6 b \,c^{3} d \,x^{3}+7 b \,c^{3} e \,x^{3}+18 \arctan \left (c x \right ) b \,c^{2} e \,x^{2}+18 a \,c^{2} e \,x^{2}+18 \ln \left (c^{2} x^{2}+1\right ) b c e x +18 b c d x -75 b c e x -18 \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right ) b e -18 \arctan \left (c x \right ) b d +75 e b \arctan \left (c x \right )-18 \ln \left (c^{2} x^{2}+1\right ) a e -18 e a}{72 c^{4}}\) | \(243\) |
| default | \(\text {Expression too large to display}\) | \(3752\) |
| parts | \(\text {Expression too large to display}\) | \(3752\) |
| risch | \(\text {Expression too large to display}\) | \(22188\) |
1/72*(18*e*b*ln(c^2*x^2+1)*arctan(c*x)*x^4*c^4+18*x^4*arctan(c*x)*b*c^4*d- 9*x^4*arctan(c*x)*b*c^4*e+18*e*a*ln(c^2*x^2+1)*x^4*c^4+18*c^4*a*d*x^4-9*x^ 4*a*c^4*e-6*e*b*ln(c^2*x^2+1)*x^3*c^3-6*b*c^3*d*x^3+7*b*c^3*e*x^3+18*arcta n(c*x)*b*c^2*e*x^2+18*a*c^2*e*x^2+18*ln(c^2*x^2+1)*b*c*e*x+18*b*c*d*x-75*b *c*e*x-18*arctan(c*x)*ln(c^2*x^2+1)*b*e-18*arctan(c*x)*b*d+75*e*b*arctan(c *x)-18*ln(c^2*x^2+1)*a*e-18*e*a)/c^4
Time = 0.27 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.81 \[ \int x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {18 \, a c^{2} e x^{2} + 9 \, {\left (2 \, a c^{4} d - a c^{4} e\right )} x^{4} - {\left (6 \, b c^{3} d - 7 \, b c^{3} e\right )} x^{3} + 3 \, {\left (6 \, b c d - 25 \, b c e\right )} x + 3 \, {\left (6 \, b c^{2} e x^{2} + 3 \, {\left (2 \, b c^{4} d - b c^{4} e\right )} x^{4} - 6 \, b d + 25 \, b e\right )} \arctan \left (c x\right ) + 6 \, {\left (3 \, a c^{4} e x^{4} - b c^{3} e x^{3} + 3 \, b c e x - 3 \, a e + 3 \, {\left (b c^{4} e x^{4} - b e\right )} \arctan \left (c x\right )\right )} \log \left (c^{2} x^{2} + 1\right )}{72 \, c^{4}} \]
1/72*(18*a*c^2*e*x^2 + 9*(2*a*c^4*d - a*c^4*e)*x^4 - (6*b*c^3*d - 7*b*c^3* e)*x^3 + 3*(6*b*c*d - 25*b*c*e)*x + 3*(6*b*c^2*e*x^2 + 3*(2*b*c^4*d - b*c^ 4*e)*x^4 - 6*b*d + 25*b*e)*arctan(c*x) + 6*(3*a*c^4*e*x^4 - b*c^3*e*x^3 + 3*b*c*e*x - 3*a*e + 3*(b*c^4*e*x^4 - b*e)*arctan(c*x))*log(c^2*x^2 + 1))/c ^4
Time = 1.08 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.26 \[ \int x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\begin {cases} \frac {a d x^{4}}{4} + \frac {a e x^{4} \log {\left (c^{2} x^{2} + 1 \right )}}{4} - \frac {a e x^{4}}{8} + \frac {a e x^{2}}{4 c^{2}} - \frac {a e \log {\left (c^{2} x^{2} + 1 \right )}}{4 c^{4}} + \frac {b d x^{4} \operatorname {atan}{\left (c x \right )}}{4} + \frac {b e x^{4} \log {\left (c^{2} x^{2} + 1 \right )} \operatorname {atan}{\left (c x \right )}}{4} - \frac {b e x^{4} \operatorname {atan}{\left (c x \right )}}{8} - \frac {b d x^{3}}{12 c} - \frac {b e x^{3} \log {\left (c^{2} x^{2} + 1 \right )}}{12 c} + \frac {7 b e x^{3}}{72 c} + \frac {b e x^{2} \operatorname {atan}{\left (c x \right )}}{4 c^{2}} + \frac {b d x}{4 c^{3}} + \frac {b e x \log {\left (c^{2} x^{2} + 1 \right )}}{4 c^{3}} - \frac {25 b e x}{24 c^{3}} - \frac {b d \operatorname {atan}{\left (c x \right )}}{4 c^{4}} - \frac {b e \log {\left (c^{2} x^{2} + 1 \right )} \operatorname {atan}{\left (c x \right )}}{4 c^{4}} + \frac {25 b e \operatorname {atan}{\left (c x \right )}}{24 c^{4}} & \text {for}\: c \neq 0 \\\frac {a d x^{4}}{4} & \text {otherwise} \end {cases} \]
Piecewise((a*d*x**4/4 + a*e*x**4*log(c**2*x**2 + 1)/4 - a*e*x**4/8 + a*e*x **2/(4*c**2) - a*e*log(c**2*x**2 + 1)/(4*c**4) + b*d*x**4*atan(c*x)/4 + b* e*x**4*log(c**2*x**2 + 1)*atan(c*x)/4 - b*e*x**4*atan(c*x)/8 - b*d*x**3/(1 2*c) - b*e*x**3*log(c**2*x**2 + 1)/(12*c) + 7*b*e*x**3/(72*c) + b*e*x**2*a tan(c*x)/(4*c**2) + b*d*x/(4*c**3) + b*e*x*log(c**2*x**2 + 1)/(4*c**3) - 2 5*b*e*x/(24*c**3) - b*d*atan(c*x)/(4*c**4) - b*e*log(c**2*x**2 + 1)*atan(c *x)/(4*c**4) + 25*b*e*atan(c*x)/(24*c**4), Ne(c, 0)), (a*d*x**4/4, True))
Time = 0.29 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.01 \[ \int x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {1}{4} \, a d x^{4} + \frac {1}{72} \, b c e {\left (\frac {7 \, c^{2} x^{3} - 6 \, {\left (c^{2} x^{3} - 3 \, x\right )} \log \left (c^{2} x^{2} + 1\right ) - 75 \, x}{c^{4}} + \frac {75 \, \arctan \left (c x\right )}{c^{5}}\right )} + \frac {1}{8} \, {\left (2 \, x^{4} \log \left (c^{2} x^{2} + 1\right ) - c^{2} {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b e \arctan \left (c x\right ) + \frac {1}{12} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b d + \frac {1}{8} \, {\left (2 \, x^{4} \log \left (c^{2} x^{2} + 1\right ) - c^{2} {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} a e \]
1/4*a*d*x^4 + 1/72*b*c*e*((7*c^2*x^3 - 6*(c^2*x^3 - 3*x)*log(c^2*x^2 + 1) - 75*x)/c^4 + 75*arctan(c*x)/c^5) + 1/8*(2*x^4*log(c^2*x^2 + 1) - c^2*((c^ 2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*b*e*arctan(c*x) + 1/12*(3*x^ 4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*b*d + 1/8*(2* x^4*log(c^2*x^2 + 1) - c^2*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6 ))*a*e
\[ \int x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )} x^{3} \,d x } \]
Time = 1.94 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.34 \[ \int x^3 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {a\,d\,x^4}{4}-\frac {a\,e\,x^4}{8}+\frac {b\,d\,x}{4\,c^3}-\frac {25\,b\,e\,x}{24\,c^3}+\frac {b\,d\,x^4\,\mathrm {atan}\left (c\,x\right )}{4}-\frac {b\,e\,x^4\,\mathrm {atan}\left (c\,x\right )}{8}-\frac {a\,e\,\ln \left (c^2\,x^2+1\right )}{4\,c^4}+\frac {a\,e\,x^2}{4\,c^2}-\frac {b\,d\,x^3}{12\,c}-\frac {b\,d\,\mathrm {atan}\left (\frac {6\,b\,c\,d\,x}{6\,b\,d-25\,b\,e}-\frac {25\,b\,c\,e\,x}{6\,b\,d-25\,b\,e}\right )}{4\,c^4}+\frac {7\,b\,e\,x^3}{72\,c}+\frac {25\,b\,e\,\mathrm {atan}\left (\frac {6\,b\,c\,d\,x}{6\,b\,d-25\,b\,e}-\frac {25\,b\,c\,e\,x}{6\,b\,d-25\,b\,e}\right )}{24\,c^4}+\frac {a\,e\,x^4\,\ln \left (c^2\,x^2+1\right )}{4}+\frac {b\,e\,x\,\ln \left (c^2\,x^2+1\right )}{4\,c^3}-\frac {b\,e\,\mathrm {atan}\left (c\,x\right )\,\ln \left (c^2\,x^2+1\right )}{4\,c^4}+\frac {b\,e\,x^2\,\mathrm {atan}\left (c\,x\right )}{4\,c^2}+\frac {b\,e\,x^4\,\mathrm {atan}\left (c\,x\right )\,\ln \left (c^2\,x^2+1\right )}{4}-\frac {b\,e\,x^3\,\ln \left (c^2\,x^2+1\right )}{12\,c} \]
(a*d*x^4)/4 - (a*e*x^4)/8 + (b*d*x)/(4*c^3) - (25*b*e*x)/(24*c^3) + (b*d*x ^4*atan(c*x))/4 - (b*e*x^4*atan(c*x))/8 - (a*e*log(c^2*x^2 + 1))/(4*c^4) + (a*e*x^2)/(4*c^2) - (b*d*x^3)/(12*c) - (b*d*atan((6*b*c*d*x)/(6*b*d - 25* b*e) - (25*b*c*e*x)/(6*b*d - 25*b*e)))/(4*c^4) + (7*b*e*x^3)/(72*c) + (25* b*e*atan((6*b*c*d*x)/(6*b*d - 25*b*e) - (25*b*c*e*x)/(6*b*d - 25*b*e)))/(2 4*c^4) + (a*e*x^4*log(c^2*x^2 + 1))/4 + (b*e*x*log(c^2*x^2 + 1))/(4*c^3) - (b*e*atan(c*x)*log(c^2*x^2 + 1))/(4*c^4) + (b*e*x^2*atan(c*x))/(4*c^2) + (b*e*x^4*atan(c*x)*log(c^2*x^2 + 1))/4 - (b*e*x^3*log(c^2*x^2 + 1))/(12*c)