Integrand size = 23, antiderivative size = 163 \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx=-\frac {i (a+b \arctan (c+d x))^3}{d e^2}-\frac {(a+b \arctan (c+d x))^3}{d e^2 (c+d x)}+\frac {3 b (a+b \arctan (c+d x))^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^2}-\frac {3 i b^2 (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i (c+d x)}\right )}{d e^2}+\frac {3 b^3 \operatorname {PolyLog}\left (3,-1+\frac {2}{1-i (c+d x)}\right )}{2 d e^2} \]
-I*(a+b*arctan(d*x+c))^3/d/e^2-(a+b*arctan(d*x+c))^3/d/e^2/(d*x+c)+3*b*(a+ b*arctan(d*x+c))^2*ln(2-2/(1-I*(d*x+c)))/d/e^2-3*I*b^2*(a+b*arctan(d*x+c)) *polylog(2,-1+2/(1-I*(d*x+c)))/d/e^2+3/2*b^3*polylog(3,-1+2/(1-I*(d*x+c))) /d/e^2
Time = 0.59 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.61 \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx=\frac {-\frac {2 a^3}{c+d x}-\frac {6 a^2 b \arctan (c+d x)}{c+d x}+6 a^2 b \log (c+d x)-3 a^2 b \log \left (1+c^2+2 c d x+d^2 x^2\right )+6 a b^2 \left (\arctan (c+d x) \left (\left (-i-\frac {1}{c+d x}\right ) \arctan (c+d x)+2 \log \left (1-e^{2 i \arctan (c+d x)}\right )\right )-i \operatorname {PolyLog}\left (2,e^{2 i \arctan (c+d x)}\right )\right )+2 b^3 \left (-\frac {i \pi ^3}{8}+i \arctan (c+d x)^3-\frac {\arctan (c+d x)^3}{c+d x}+3 \arctan (c+d x)^2 \log \left (1-e^{-2 i \arctan (c+d x)}\right )+3 i \arctan (c+d x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c+d x)}\right )+\frac {3}{2} \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c+d x)}\right )\right )}{2 d e^2} \]
((-2*a^3)/(c + d*x) - (6*a^2*b*ArcTan[c + d*x])/(c + d*x) + 6*a^2*b*Log[c + d*x] - 3*a^2*b*Log[1 + c^2 + 2*c*d*x + d^2*x^2] + 6*a*b^2*(ArcTan[c + d* x]*((-I - (c + d*x)^(-1))*ArcTan[c + d*x] + 2*Log[1 - E^((2*I)*ArcTan[c + d*x])]) - I*PolyLog[2, E^((2*I)*ArcTan[c + d*x])]) + 2*b^3*((-1/8*I)*Pi^3 + I*ArcTan[c + d*x]^3 - ArcTan[c + d*x]^3/(c + d*x) + 3*ArcTan[c + d*x]^2* Log[1 - E^((-2*I)*ArcTan[c + d*x])] + (3*I)*ArcTan[c + d*x]*PolyLog[2, E^( (-2*I)*ArcTan[c + d*x])] + (3*PolyLog[3, E^((-2*I)*ArcTan[c + d*x])])/2))/ (2*d*e^2)
Time = 0.73 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5566, 27, 5361, 5459, 5403, 5527, 7164}
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.
| \(\displaystyle \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx\) |
| \(\Big \downarrow \) 5566 |
| \(\displaystyle \frac {\int \frac {(a+b \arctan (c+d x))^3}{e^2 (c+d x)^2}d(c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(a+b \arctan (c+d x))^3}{(c+d x)^2}d(c+d x)}{d e^2}\) |
| \(\Big \downarrow \) 5361 |
| \(\displaystyle \frac {3 b \int \frac {(a+b \arctan (c+d x))^2}{(c+d x) \left ((c+d x)^2+1\right )}d(c+d x)-\frac {(a+b \arctan (c+d x))^3}{c+d x}}{d e^2}\) |
| \(\Big \downarrow \) 5459 |
| \(\displaystyle \frac {-\frac {(a+b \arctan (c+d x))^3}{c+d x}+3 b \left (i \int \frac {(a+b \arctan (c+d x))^2}{(c+d x) (c+d x+i)}d(c+d x)-\frac {i (a+b \arctan (c+d x))^3}{3 b}\right )}{d e^2}\) |
| \(\Big \downarrow \) 5403 |
| \(\displaystyle \frac {-\frac {(a+b \arctan (c+d x))^3}{c+d x}+3 b \left (i \left (2 i b \int \frac {(a+b \arctan (c+d x)) \log \left (2-\frac {2}{1-i (c+d x)}\right )}{(c+d x)^2+1}d(c+d x)-i \log \left (2-\frac {2}{1-i (c+d x)}\right ) (a+b \arctan (c+d x))^2\right )-\frac {i (a+b \arctan (c+d x))^3}{3 b}\right )}{d e^2}\) |
| \(\Big \downarrow \) 5527 |
| \(\displaystyle \frac {-\frac {(a+b \arctan (c+d x))^3}{c+d x}+3 b \left (i \left (2 i b \left (\frac {1}{2} i \operatorname {PolyLog}\left (2,\frac {2}{1-i (c+d x)}-1\right ) (a+b \arctan (c+d x))-\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (2,\frac {2}{1-i (c+d x)}-1\right )}{(c+d x)^2+1}d(c+d x)\right )-i \log \left (2-\frac {2}{1-i (c+d x)}\right ) (a+b \arctan (c+d x))^2\right )-\frac {i (a+b \arctan (c+d x))^3}{3 b}\right )}{d e^2}\) |
| \(\Big \downarrow \) 7164 |
| \(\displaystyle \frac {-\frac {(a+b \arctan (c+d x))^3}{c+d x}+3 b \left (i \left (2 i b \left (\frac {1}{2} i \operatorname {PolyLog}\left (2,\frac {2}{1-i (c+d x)}-1\right ) (a+b \arctan (c+d x))-\frac {1}{4} b \operatorname {PolyLog}\left (3,\frac {2}{1-i (c+d x)}-1\right )\right )-i \log \left (2-\frac {2}{1-i (c+d x)}\right ) (a+b \arctan (c+d x))^2\right )-\frac {i (a+b \arctan (c+d x))^3}{3 b}\right )}{d e^2}\) |
(-((a + b*ArcTan[c + d*x])^3/(c + d*x)) + 3*b*(((-1/3*I)*(a + b*ArcTan[c + d*x])^3)/b + I*((-I)*(a + b*ArcTan[c + d*x])^2*Log[2 - 2/(1 - I*(c + d*x) )] + (2*I)*b*((I/2)*(a + b*ArcTan[c + d*x])*PolyLog[2, -1 + 2/(1 - I*(c + d*x))] - (b*PolyLog[3, -1 + 2/(1 - I*(c + d*x))])/4))))/(d*e^2)
3.1.18.3.1 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_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] & & IntegerQ[m])) && NeQ[m, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_ Symbol] :> Simp[(a + b*ArcTan[c*x])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Si mp[b*c*(p/d) Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2 - 2/(1 + e*(x/d))]/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2* d^2 + e^2, 0]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(-I)*((a + b*ArcTan[c*x])^(p + 1)/(b*d*(p + 1))), x] + Si mp[I/d Int[(a + b*ArcTan[c*x])^p/(x*(I + c*x)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p, 0]
Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2 ), x_Symbol] :> Simp[I*(a + b*ArcTan[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)), x ] - Simp[b*p*(I/2) Int[(a + b*ArcTan[c*x])^(p - 1)*(PolyLog[2, 1 - u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2* d] && EqQ[(1 - u)^2 - (1 - 2*(I/(I + c*x)))^2, 0]
Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[(f*(x/d))^m*(a + b*ArcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0] && IGtQ[p, 0]
Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /; !FalseQ[w]] /; FreeQ[n, x]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.33 (sec) , antiderivative size = 2104, normalized size of antiderivative = 12.91
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(2104\) |
| default | \(\text {Expression too large to display}\) | \(2104\) |
| parts | \(\text {Expression too large to display}\) | \(2112\) |
1/d*(-a^3/e^2/(d*x+c)+b^3/e^2*(-1/(d*x+c)*arctan(d*x+c)^3+3*ln(d*x+c)*arct an(d*x+c)^2-3/2*arctan(d*x+c)^2*ln(1+(d*x+c)^2)+3*arctan(d*x+c)^2*ln((1+I* (d*x+c))/(1+(d*x+c)^2)^(1/2))-3*arctan(d*x+c)^2*ln((1+I*(d*x+c))^2/(1+(d*x +c)^2)-1)-I*arctan(d*x+c)^3+3/4*(2*I*Pi*csgn(I*(1+I*(d*x+c))/(1+(d*x+c)^2) ^(1/2))*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2-2*I*Pi*csgn(((1+I*(d*x+c)) ^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^2+2*I*Pi+I*Pi*csgn( I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(1+( 1+I*(d*x+c))^2/(1+(d*x+c)^2))^2)^2+2*I*Pi*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+ c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^3-I*Pi*csgn(I*(1+I*(d*x+c))/(1 +(d*x+c)^2)^(1/2))^2*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))-I*Pi*csgn(I/(1+ (1+I*(d*x+c))^2/(1+(d*x+c)^2))^2)*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))*cs gn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2)-I* Pi*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))^3-I*Pi*csgn(I*(1+I*(d*x+c))^2/(1+ (d*x+c)^2)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2)^3+2*I*Pi*csgn(I*((1+I*(d*x +c))^2/(1+(d*x+c)^2)-1))*csgn(I/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))*csgn(I* ((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))-2*I*P i*csgn(I/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))*csgn(I*((1+I*(d*x+c))^2/(1+(d* x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^2-2*I*Pi*csgn(I*((1+I*(d*x+c ))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))*csgn(((1+I*(d*x+c ))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^2+I*Pi*csgn(I*...
\[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx=\int { \frac {{\left (b \arctan \left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{2}} \,d x } \]
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)/(d^2*e^2*x^2 + 2*c*d*e^2*x + c^2*e^2), x)
\[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx=\frac {\int \frac {a^{3}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {b^{3} \operatorname {atan}^{3}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {3 a b^{2} \operatorname {atan}^{2}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {3 a^{2} b \operatorname {atan}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx}{e^{2}} \]
(Integral(a**3/(c**2 + 2*c*d*x + d**2*x**2), x) + Integral(b**3*atan(c + d *x)**3/(c**2 + 2*c*d*x + d**2*x**2), x) + Integral(3*a*b**2*atan(c + d*x)* *2/(c**2 + 2*c*d*x + d**2*x**2), x) + Integral(3*a**2*b*atan(c + d*x)/(c** 2 + 2*c*d*x + d**2*x**2), x))/e**2
\[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx=\int { \frac {{\left (b \arctan \left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{2}} \,d x } \]
-3/2*(d*(log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*e^2) - 2*log(d*x + c)/(d^2* e^2)) + 2*arctan(d*x + c)/(d^2*e^2*x + c*d*e^2))*a^2*b - a^3/(d^2*e^2*x + c*d*e^2) - 1/32*(4*b^3*arctan(d*x + c)^3 - 3*b^3*arctan(d*x + c)*log(d^2*x ^2 + 2*c*d*x + c^2 + 1)^2 - 32*(d^2*e^2*x + c*d*e^2)*integrate(1/32*(28*(b ^3*d^2*x^2 + 2*b^3*c*d*x + b^3*c^2 + b^3)*arctan(d*x + c)^3 + 12*(8*a*b^2* d^2*x^2 + 8*a*b^2*c^2 + b^3*c + 8*a*b^2 + (16*a*b^2*c + b^3)*d*x)*arctan(d *x + c)^2 - 12*(b^3*d^2*x^2 + 2*b^3*c*d*x + b^3*c^2)*arctan(d*x + c)*log(d ^2*x^2 + 2*c*d*x + c^2 + 1) - 3*(b^3*d*x + b^3*c - (b^3*d^2*x^2 + 2*b^3*c* d*x + b^3*c^2 + b^3)*arctan(d*x + c))*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2)/ (d^4*e^2*x^4 + 4*c*d^3*e^2*x^3 + (6*c^2 + 1)*d^2*e^2*x^2 + 2*(2*c^3 + c)*d *e^2*x + (c^4 + c^2)*e^2), x))/(d^2*e^2*x + c*d*e^2)
Timed out. \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^2} \,d x \]