Integrand size = 20, antiderivative size = 372 \[ \int \frac {(a+b \arctan (c+d x))^3}{e+f x} \, dx=-\frac {(a+b \arctan (c+d x))^3 \log \left (\frac {2}{1-i (c+d x)}\right )}{f}+\frac {(a+b \arctan (c+d x))^3 \log \left (\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{f}+\frac {3 i b (a+b \arctan (c+d x))^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-i (c+d x)}\right )}{2 f}-\frac {3 i b (a+b \arctan (c+d x))^2 \operatorname {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{2 f}-\frac {3 b^2 (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (3,1-\frac {2}{1-i (c+d x)}\right )}{2 f}+\frac {3 b^2 (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (3,1-\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{2 f}-\frac {3 i b^3 \operatorname {PolyLog}\left (4,1-\frac {2}{1-i (c+d x)}\right )}{4 f}+\frac {3 i b^3 \operatorname {PolyLog}\left (4,1-\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{4 f} \]
-(a+b*arctan(d*x+c))^3*ln(2/(1-I*(d*x+c)))/f+(a+b*arctan(d*x+c))^3*ln(2*d* (f*x+e)/(d*e+I*f-c*f)/(1-I*(d*x+c)))/f+3/2*I*b*(a+b*arctan(d*x+c))^2*polyl og(2,1-2/(1-I*(d*x+c)))/f-3/2*I*b*(a+b*arctan(d*x+c))^2*polylog(2,1-2*d*(f *x+e)/(d*e+I*f-c*f)/(1-I*(d*x+c)))/f-3/2*b^2*(a+b*arctan(d*x+c))*polylog(3 ,1-2/(1-I*(d*x+c)))/f+3/2*b^2*(a+b*arctan(d*x+c))*polylog(3,1-2*d*(f*x+e)/ (d*e+I*f-c*f)/(1-I*(d*x+c)))/f-3/4*I*b^3*polylog(4,1-2/(1-I*(d*x+c)))/f+3/ 4*I*b^3*polylog(4,1-2*d*(f*x+e)/(d*e+I*f-c*f)/(1-I*(d*x+c)))/f
\[ \int \frac {(a+b \arctan (c+d x))^3}{e+f x} \, dx=\int \frac {(a+b \arctan (c+d x))^3}{e+f x} \, dx \]
Time = 0.45 (sec) , antiderivative size = 408, normalized size of antiderivative = 1.10, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {5570, 27, 5385}
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.
| \(\displaystyle \int \frac {(a+b \arctan (c+d x))^3}{e+f x} \, dx\) |
| \(\Big \downarrow \) 5570 |
| \(\displaystyle \frac {\int \frac {d (a+b \arctan (c+d x))^3}{d \left (e-\frac {c f}{d}\right )+f (c+d x)}d(c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {(a+b \arctan (c+d x))^3}{f (c+d x)-c f+d e}d(c+d x)\) |
| \(\Big \downarrow \) 5385 |
| \(\displaystyle \frac {3 b^2 (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (3,1-\frac {2 (d e-c f+f (c+d x))}{(d e-c f+i f) (1-i (c+d x))}\right )}{2 f}-\frac {3 b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1-i (c+d x)}\right ) (a+b \arctan (c+d x))}{2 f}-\frac {3 i b (a+b \arctan (c+d x))^2 \operatorname {PolyLog}\left (2,1-\frac {2 (d e-c f+f (c+d x))}{(d e-c f+i f) (1-i (c+d x))}\right )}{2 f}+\frac {(a+b \arctan (c+d x))^3 \log \left (\frac {2 (f (c+d x)-c f+d e)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{f}+\frac {3 i b \operatorname {PolyLog}\left (2,1-\frac {2}{1-i (c+d x)}\right ) (a+b \arctan (c+d x))^2}{2 f}-\frac {\log \left (\frac {2}{1-i (c+d x)}\right ) (a+b \arctan (c+d x))^3}{f}+\frac {3 i b^3 \operatorname {PolyLog}\left (4,1-\frac {2 (d e-c f+f (c+d x))}{(d e-c f+i f) (1-i (c+d x))}\right )}{4 f}-\frac {3 i b^3 \operatorname {PolyLog}\left (4,1-\frac {2}{1-i (c+d x)}\right )}{4 f}\) |
-(((a + b*ArcTan[c + d*x])^3*Log[2/(1 - I*(c + d*x))])/f) + ((a + b*ArcTan [c + d*x])^3*Log[(2*(d*e - c*f + f*(c + d*x)))/((d*e + I*f - c*f)*(1 - I*( c + d*x)))])/f + (((3*I)/2)*b*(a + b*ArcTan[c + d*x])^2*PolyLog[2, 1 - 2/( 1 - I*(c + d*x))])/f - (((3*I)/2)*b*(a + b*ArcTan[c + d*x])^2*PolyLog[2, 1 - (2*(d*e - c*f + f*(c + d*x)))/((d*e + I*f - c*f)*(1 - I*(c + d*x)))])/f - (3*b^2*(a + b*ArcTan[c + d*x])*PolyLog[3, 1 - 2/(1 - I*(c + d*x))])/(2* f) + (3*b^2*(a + b*ArcTan[c + d*x])*PolyLog[3, 1 - (2*(d*e - c*f + f*(c + d*x)))/((d*e + I*f - c*f)*(1 - I*(c + d*x)))])/(2*f) - (((3*I)/4)*b^3*Poly Log[4, 1 - 2/(1 - I*(c + d*x))])/f + (((3*I)/4)*b^3*PolyLog[4, 1 - (2*(d*e - c*f + f*(c + d*x)))/((d*e + I*f - c*f)*(1 - I*(c + d*x)))])/f
3.1.39.3.1 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_.) + ArcTan[(c_.)*(x_)]*(b_.))^3/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^3)*(Log[2/(1 - I*c*x)]/e), x] + (Simp[(a + b*Arc Tan[c*x])^3*(Log[2*c*((d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/e), x] + Simp[3 *I*b*(a + b*ArcTan[c*x])^2*(PolyLog[2, 1 - 2/(1 - I*c*x)]/(2*e)), x] - Simp [3*I*b*(a + b*ArcTan[c*x])^2*(PolyLog[2, 1 - 2*c*((d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/(2*e)), x] - Simp[3*b^2*(a + b*ArcTan[c*x])*(PolyLog[3, 1 - 2/ (1 - I*c*x)]/(2*e)), x] + Simp[3*b^2*(a + b*ArcTan[c*x])*(PolyLog[3, 1 - 2* c*((d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/(2*e)), x] - Simp[3*I*b^3*(PolyLog [4, 1 - 2/(1 - I*c*x)]/(4*e)), x] + Simp[3*I*b^3*(PolyLog[4, 1 - 2*c*((d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/(4*e)), x]) /; FreeQ[{a, b, c, d, e}, x] & & NeQ[c^2*d^2 + e^2, 0]
Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*A rcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && I GtQ[p, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 4.95 (sec) , antiderivative size = 3817, normalized size of antiderivative = 10.26
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(3817\) |
| default | \(\text {Expression too large to display}\) | \(3817\) |
| parts | \(\text {Expression too large to display}\) | \(4072\) |
1/d*(a^3*d*ln(c*f-d*e-f*(d*x+c))/f-b^3*d*(-ln(c*f-d*e-f*(d*x+c))/f*arctan( d*x+c)^3+3/f*(1/3*arctan(d*x+c)^3*ln(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f *(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d *e)-I*d*e*arctan(d*x+c)^2*polylog(2,(c*f-d*e+I*f)*(1+I*(d*x+c))^2/(1+(d*x+ c)^2)/(d*e+I*f-c*f))/(2*I*f+2*c*f-2*d*e)-1/2*I*arctan(d*x+c)^2*polylog(2,- (1+I*(d*x+c))^2/(1+(d*x+c)^2))+1/2*arctan(d*x+c)*polylog(3,-(1+I*(d*x+c))^ 2/(1+(d*x+c)^2))+1/2*I*c*f/(c*f-d*e+I*f)*arctan(d*x+c)^2*polylog(2,(c*f-d* e+I*f)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+1/4*I*polylog(4,-(1+I* (d*x+c))^2/(1+(d*x+c)^2))-1/2*I*f/(c*f-d*e+I*f)*arctan(d*x+c)*polylog(3,(c *f-d*e+I*f)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))-1/2*f/(c*f-d*e+I* f)*arctan(d*x+c)^2*polylog(2,(c*f-d*e+I*f)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/( d*e+I*f-c*f))+1/4*f/(c*f-d*e+I*f)*polylog(4,(c*f-d*e+I*f)*(1+I*(d*x+c))^2/ (1+(d*x+c)^2)/(d*e+I*f-c*f))-1/3*c*f/(c*f-d*e+I*f)*arctan(d*x+c)^3*ln(1-(c *f-d*e+I*f)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))-1/2*c*f/(c*f-d*e+ I*f)*arctan(d*x+c)*polylog(3,(c*f-d*e+I*f)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/( d*e+I*f-c*f))-1/6*I*Pi*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I* (d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e)/(1 +(1+I*(d*x+c))^2/(1+(d*x+c)^2)))*(csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2 )+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+ c*f-d*e))*csgn(I/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))-csgn(I*(I*f*(1+I*(d...
\[ \int \frac {(a+b \arctan (c+d x))^3}{e+f x} \, dx=\int { \frac {{\left (b \arctan \left (d x + c\right ) + a\right )}^{3}}{f x + e} \,d x } \]
integral((b^3*arctan(d*x + c)^3 + 3*a*b^2*arctan(d*x + c)^2 + 3*a^2*b*arct an(d*x + c) + a^3)/(f*x + e), x)
Timed out. \[ \int \frac {(a+b \arctan (c+d x))^3}{e+f x} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \arctan (c+d x))^3}{e+f x} \, dx=\int { \frac {{\left (b \arctan \left (d x + c\right ) + a\right )}^{3}}{f x + e} \,d x } \]
a^3*log(f*x + e)/f + integrate(1/32*(28*b^3*arctan(d*x + c)^3 + 3*b^3*arct an(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2 + 96*a*b^2*arctan(d*x + c)^ 2 + 96*a^2*b*arctan(d*x + c))/(f*x + e), x)
Timed out. \[ \int \frac {(a+b \arctan (c+d x))^3}{e+f x} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \arctan (c+d x))^3}{e+f x} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^3}{e+f\,x} \,d x \]