Integrand size = 20, antiderivative size = 125 \[ \int (d+i c d x)^4 (a+b \arctan (c x)) \, dx=-\frac {8}{5} i b d^4 x-\frac {2 b d^4 (1+i c x)^2}{5 c}-\frac {2 b d^4 (1+i c x)^3}{15 c}-\frac {b d^4 (1+i c x)^4}{20 c}-\frac {i d^4 (1+i c x)^5 (a+b \arctan (c x))}{5 c}-\frac {16 b d^4 \log (1-i c x)}{5 c} \]
-8/5*I*b*d^4*x-2/5*b*d^4*(1+I*c*x)^2/c-2/15*b*d^4*(1+I*c*x)^3/c-1/20*b*d^4 *(1+I*c*x)^4/c-1/5*I*d^4*(1+I*c*x)^5*(a+b*arctan(c*x))/c-16/5*b*d^4*ln(1-I *c*x)/c
Time = 0.02 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.62 \[ \int (d+i c d x)^4 (a+b \arctan (c x)) \, dx=\frac {d^4 \left (12 (-i+c x)^5 (a+b \arctan (c x))-b \left (35+180 i c x-66 c^2 x^2-20 i c^3 x^3+3 c^4 x^4+192 \log (i+c x)\right )\right )}{60 c} \]
(d^4*(12*(-I + c*x)^5*(a + b*ArcTan[c*x]) - b*(35 + (180*I)*c*x - 66*c^2*x ^2 - (20*I)*c^3*x^3 + 3*c^4*x^4 + 192*Log[I + c*x])))/(60*c)
Time = 0.26 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5387, 27, 456, 49, 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.
\(\displaystyle \int (d+i c d x)^4 (a+b \arctan (c x)) \, dx\) |
\(\Big \downarrow \) 5387 |
\(\displaystyle \frac {i b \int \frac {d^5 (i c x+1)^5}{c^2 x^2+1}dx}{5 d}-\frac {i d^4 (1+i c x)^5 (a+b \arctan (c x))}{5 c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} i b d^4 \int \frac {(i c x+1)^5}{c^2 x^2+1}dx-\frac {i d^4 (1+i c x)^5 (a+b \arctan (c x))}{5 c}\) |
\(\Big \downarrow \) 456 |
\(\displaystyle \frac {1}{5} i b d^4 \int \frac {(i c x+1)^4}{1-i c x}dx-\frac {i d^4 (1+i c x)^5 (a+b \arctan (c x))}{5 c}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{5} i b d^4 \int \left (-(i c x+1)^3-2 (i c x+1)^2-4 (i c x+1)+\frac {16}{1-i c x}-8\right )dx-\frac {i d^4 (1+i c x)^5 (a+b \arctan (c x))}{5 c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} i b d^4 \left (\frac {i (1+i c x)^4}{4 c}+\frac {2 i (1+i c x)^3}{3 c}+\frac {2 i (1+i c x)^2}{c}+\frac {16 i \log (c x+i)}{c}-8 x\right )-\frac {i d^4 (1+i c x)^5 (a+b \arctan (c x))}{5 c}\) |
((-1/5*I)*d^4*(1 + I*c*x)^5*(a + b*ArcTan[c*x]))/c + (I/5)*b*d^4*(-8*x + ( (2*I)*(1 + I*c*x)^2)/c + (((2*I)/3)*(1 + I*c*x)^3)/c + ((I/4)*(1 + I*c*x)^ 4)/c + ((16*I)*Log[I + c*x])/c)
3.1.34.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ (c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !Integ erQ[n]))
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)*((a + b*ArcTan[c*x])/(e*(q + 1))), x] - Simp[b*( c/(e*(q + 1))) Int[(d + e*x)^(q + 1)/(1 + c^2*x^2), x], x] /; FreeQ[{a, b , c, d, e, q}, x] && NeQ[q, -1]
Time = 1.17 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.16
method | result | size |
derivativedivides | \(\frac {-\frac {i d^{4} a \left (i c x +1\right )^{5}}{5}+d^{4} b \left (\frac {c^{5} x^{5} \arctan \left (c x \right )}{5}-i \arctan \left (c x \right ) c^{4} x^{4}-2 c^{3} x^{3} \arctan \left (c x \right )+2 i \arctan \left (c x \right ) c^{2} x^{2}+c x \arctan \left (c x \right )-\frac {i \arctan \left (c x \right )}{5}+\frac {i \left (-15 c x +\frac {i c^{4} x^{4}}{4}+\frac {5 c^{3} x^{3}}{3}-\frac {11 i c^{2} x^{2}}{2}+8 i \ln \left (c^{2} x^{2}+1\right )+16 \arctan \left (c x \right )\right )}{5}\right )}{c}\) | \(145\) |
default | \(\frac {-\frac {i d^{4} a \left (i c x +1\right )^{5}}{5}+d^{4} b \left (\frac {c^{5} x^{5} \arctan \left (c x \right )}{5}-i \arctan \left (c x \right ) c^{4} x^{4}-2 c^{3} x^{3} \arctan \left (c x \right )+2 i \arctan \left (c x \right ) c^{2} x^{2}+c x \arctan \left (c x \right )-\frac {i \arctan \left (c x \right )}{5}+\frac {i \left (-15 c x +\frac {i c^{4} x^{4}}{4}+\frac {5 c^{3} x^{3}}{3}-\frac {11 i c^{2} x^{2}}{2}+8 i \ln \left (c^{2} x^{2}+1\right )+16 \arctan \left (c x \right )\right )}{5}\right )}{c}\) | \(145\) |
parts | \(-\frac {i d^{4} a \left (i c x +1\right )^{5}}{5 c}+\frac {d^{4} b \left (\frac {c^{5} x^{5} \arctan \left (c x \right )}{5}-i \arctan \left (c x \right ) c^{4} x^{4}-2 c^{3} x^{3} \arctan \left (c x \right )+2 i \arctan \left (c x \right ) c^{2} x^{2}+c x \arctan \left (c x \right )-\frac {i \arctan \left (c x \right )}{5}+\frac {i \left (-15 c x +\frac {i c^{4} x^{4}}{4}+\frac {5 c^{3} x^{3}}{3}-\frac {11 i c^{2} x^{2}}{2}+8 i \ln \left (c^{2} x^{2}+1\right )+16 \arctan \left (c x \right )\right )}{5}\right )}{c}\) | \(147\) |
parallelrisch | \(-\frac {-12 c^{5} b \,d^{4} \arctan \left (c x \right ) x^{5}+180 i b \,d^{4} x c -12 a \,c^{5} d^{4} x^{5}+60 i x^{4} \arctan \left (c x \right ) b \,c^{4} d^{4}+3 b \,c^{4} d^{4} x^{4}-180 i b \,d^{4} \arctan \left (c x \right )+120 x^{3} \arctan \left (c x \right ) b \,d^{4} c^{3}-20 i x^{3} b \,c^{3} d^{4}+120 a \,c^{3} d^{4} x^{3}-120 i x^{2} a \,c^{2} d^{4}-66 b \,c^{2} d^{4} x^{2}+60 i x^{4} a \,c^{4} d^{4}-60 b c \,d^{4} x \arctan \left (c x \right )-120 i b \,d^{4} \arctan \left (c x \right ) x^{2} c^{2}-60 a c \,d^{4} x +96 b \,d^{4} \ln \left (c^{2} x^{2}+1\right )}{60 c}\) | \(216\) |
risch | \(-i d^{4} c^{2} b \,x^{3} \ln \left (-i c x +1\right )-i d^{4} a \,c^{3} x^{4}+\frac {d^{4} a \,c^{4} x^{5}}{5}+2 i a c \,d^{4} x^{2}+\frac {d^{4} c^{3} x^{4} b \ln \left (-i c x +1\right )}{2}+\frac {31 i d^{4} b \arctan \left (c x \right )}{10 c}-\frac {d^{4} b \,c^{3} x^{4}}{20}-\frac {i d^{4} \left (c x -i\right )^{5} b \ln \left (i c x +1\right )}{10 c}-2 d^{4} a \,c^{2} x^{3}-3 i b \,d^{4} x -d^{4} c \,x^{2} b \ln \left (-i c x +1\right )+\frac {i d^{4} b \,c^{2} x^{3}}{3}+\frac {11 b c \,d^{4} x^{2}}{10}+\frac {i b \,d^{4} x \ln \left (-i c x +1\right )}{2}+\frac {i d^{4} c^{4} b \,x^{5} \ln \left (-i c x +1\right )}{10}+x a \,d^{4}-\frac {31 b \,d^{4} \ln \left (c^{2} x^{2}+1\right )}{20 c}\) | \(254\) |
1/c*(-1/5*I*d^4*a*(1+I*c*x)^5+d^4*b*(1/5*c^5*x^5*arctan(c*x)-I*arctan(c*x) *c^4*x^4-2*c^3*x^3*arctan(c*x)+2*I*arctan(c*x)*c^2*x^2+c*x*arctan(c*x)-1/5 *I*arctan(c*x)+1/5*I*(-15*c*x+1/4*I*c^4*x^4+5/3*c^3*x^3-11/2*I*c^2*x^2+8*I *ln(c^2*x^2+1)+16*arctan(c*x))))
Time = 0.27 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.50 \[ \int (d+i c d x)^4 (a+b \arctan (c x)) \, dx=\frac {12 \, a c^{5} d^{4} x^{5} - 3 \, {\left (20 i \, a + b\right )} c^{4} d^{4} x^{4} - 20 \, {\left (6 \, a - i \, b\right )} c^{3} d^{4} x^{3} - 6 \, {\left (-20 i \, a - 11 \, b\right )} c^{2} d^{4} x^{2} + 60 \, {\left (a - 3 i \, b\right )} c d^{4} x - 186 \, b d^{4} \log \left (\frac {c x + i}{c}\right ) - 6 \, b d^{4} \log \left (\frac {c x - i}{c}\right ) - 6 \, {\left (-i \, b c^{5} d^{4} x^{5} - 5 \, b c^{4} d^{4} x^{4} + 10 i \, b c^{3} d^{4} x^{3} + 10 \, b c^{2} d^{4} x^{2} - 5 i \, b c d^{4} x\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{60 \, c} \]
1/60*(12*a*c^5*d^4*x^5 - 3*(20*I*a + b)*c^4*d^4*x^4 - 20*(6*a - I*b)*c^3*d ^4*x^3 - 6*(-20*I*a - 11*b)*c^2*d^4*x^2 + 60*(a - 3*I*b)*c*d^4*x - 186*b*d ^4*log((c*x + I)/c) - 6*b*d^4*log((c*x - I)/c) - 6*(-I*b*c^5*d^4*x^5 - 5*b *c^4*d^4*x^4 + 10*I*b*c^3*d^4*x^3 + 10*b*c^2*d^4*x^2 - 5*I*b*c*d^4*x)*log( -(c*x + I)/(c*x - I)))/c
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (110) = 220\).
Time = 2.21 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.53 \[ \int (d+i c d x)^4 (a+b \arctan (c x)) \, dx=\frac {a c^{4} d^{4} x^{5}}{5} + \frac {b d^{4} \left (- \frac {\log {\left (41 b c d^{4} x - 41 i b d^{4} \right )}}{10} - \frac {43 \log {\left (41 b c d^{4} x + 41 i b d^{4} \right )}}{20}\right )}{c} + x^{4} \left (- i a c^{3} d^{4} - \frac {b c^{3} d^{4}}{20}\right ) + x^{3} \left (- 2 a c^{2} d^{4} + \frac {i b c^{2} d^{4}}{3}\right ) + x^{2} \cdot \left (2 i a c d^{4} + \frac {11 b c d^{4}}{10}\right ) + x \left (a d^{4} - 3 i b d^{4}\right ) + \left (- \frac {i b c^{4} d^{4} x^{5}}{10} - \frac {b c^{3} d^{4} x^{4}}{2} + i b c^{2} d^{4} x^{3} + b c d^{4} x^{2} - \frac {i b d^{4} x}{2}\right ) \log {\left (i c x + 1 \right )} + \frac {\left (2 i b c^{5} d^{4} x^{5} + 10 b c^{4} d^{4} x^{4} - 20 i b c^{3} d^{4} x^{3} - 20 b c^{2} d^{4} x^{2} + 10 i b c d^{4} x - 19 b d^{4}\right ) \log {\left (- i c x + 1 \right )}}{20 c} \]
a*c**4*d**4*x**5/5 + b*d**4*(-log(41*b*c*d**4*x - 41*I*b*d**4)/10 - 43*log (41*b*c*d**4*x + 41*I*b*d**4)/20)/c + x**4*(-I*a*c**3*d**4 - b*c**3*d**4/2 0) + x**3*(-2*a*c**2*d**4 + I*b*c**2*d**4/3) + x**2*(2*I*a*c*d**4 + 11*b*c *d**4/10) + x*(a*d**4 - 3*I*b*d**4) + (-I*b*c**4*d**4*x**5/10 - b*c**3*d** 4*x**4/2 + I*b*c**2*d**4*x**3 + b*c*d**4*x**2 - I*b*d**4*x/2)*log(I*c*x + 1) + (2*I*b*c**5*d**4*x**5 + 10*b*c**4*d**4*x**4 - 20*I*b*c**3*d**4*x**3 - 20*b*c**2*d**4*x**2 + 10*I*b*c*d**4*x - 19*b*d**4)*log(-I*c*x + 1)/(20*c)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (99) = 198\).
Time = 0.29 (sec) , antiderivative size = 264, normalized size of antiderivative = 2.11 \[ \int (d+i c d x)^4 (a+b \arctan (c x)) \, dx=\frac {1}{5} \, a c^{4} d^{4} x^{5} - i \, a c^{3} d^{4} x^{4} + \frac {1}{20} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b c^{4} d^{4} - 2 \, a c^{2} d^{4} x^{3} - \frac {1}{3} i \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b c^{3} d^{4} - {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b c^{2} d^{4} + 2 i \, a c d^{4} x^{2} + 2 i \, {\left (x^{2} \arctan \left (c x\right ) - c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )}\right )} b c d^{4} + a d^{4} x + \frac {{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{4}}{2 \, c} \]
1/5*a*c^4*d^4*x^5 - I*a*c^3*d^4*x^4 + 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*c^4*d^4 - 2*a*c^2*d^4*x^3 - 1/ 3*I*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*b*c^ 3*d^4 - (2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b*c^2*d^4 + 2*I*a*c*d^4*x^2 + 2*I*(x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*b *c*d^4 + a*d^4*x + 1/2*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*d^4/c
\[ \int (d+i c d x)^4 (a+b \arctan (c x)) \, dx=\int { {\left (i \, c d x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )} \,d x } \]
Time = 0.80 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.40 \[ \int (d+i c d x)^4 (a+b \arctan (c x)) \, dx=\frac {d^4\,\left (60\,a\,x+60\,b\,x\,\mathrm {atan}\left (c\,x\right )-b\,x\,180{}\mathrm {i}\right )}{60}+\frac {c^4\,d^4\,\left (12\,a\,x^5+12\,b\,x^5\,\mathrm {atan}\left (c\,x\right )\right )}{60}+\frac {d^4\,\left (-96\,b\,\ln \left (c^2\,x^2+1\right )+b\,\mathrm {atan}\left (c\,x\right )\,180{}\mathrm {i}\right )}{60\,c}+\frac {c\,d^4\,\left (a\,x^2\,120{}\mathrm {i}+66\,b\,x^2+b\,x^2\,\mathrm {atan}\left (c\,x\right )\,120{}\mathrm {i}\right )}{60}-\frac {c^3\,d^4\,\left (a\,x^4\,60{}\mathrm {i}+3\,b\,x^4+b\,x^4\,\mathrm {atan}\left (c\,x\right )\,60{}\mathrm {i}\right )}{60}-\frac {c^2\,d^4\,\left (120\,a\,x^3+120\,b\,x^3\,\mathrm {atan}\left (c\,x\right )-b\,x^3\,20{}\mathrm {i}\right )}{60} \]
(d^4*(60*a*x - b*x*180i + 60*b*x*atan(c*x)))/60 + (c^4*d^4*(12*a*x^5 + 12* b*x^5*atan(c*x)))/60 + (d^4*(b*atan(c*x)*180i - 96*b*log(c^2*x^2 + 1)))/(6 0*c) + (c*d^4*(a*x^2*120i + 66*b*x^2 + b*x^2*atan(c*x)*120i))/60 - (c^3*d^ 4*(a*x^4*60i + 3*b*x^4 + b*x^4*atan(c*x)*60i))/60 - (c^2*d^4*(120*a*x^3 - b*x^3*20i + 120*b*x^3*atan(c*x)))/60