Integrand size = 20, antiderivative size = 181 \[ \int \frac {(c+d x)^2}{a+b \tan (e+f x)} \, dx=\frac {(c+d x)^3}{3 (a+i b) d}+\frac {b (c+d x)^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f}-\frac {i b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f^2}+\frac {b d^2 \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) f^3} \]
1/3*(d*x+c)^3/(a+I*b)/d+b*(d*x+c)^2*ln(1+(a^2+b^2)*exp(2*I*(f*x+e))/(a+I*b )^2)/(a^2+b^2)/f-I*b*d*(d*x+c)*polylog(2,-(a^2+b^2)*exp(2*I*(f*x+e))/(a+I* b)^2)/(a^2+b^2)/f^2+1/2*b*d^2*polylog(3,-(a^2+b^2)*exp(2*I*(f*x+e))/(a+I*b )^2)/(a^2+b^2)/f^3
Time = 1.43 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.30 \[ \int \frac {(c+d x)^2}{a+b \tan (e+f x)} \, dx=\frac {1}{6} b \left (-\frac {4 (c+d x)^3}{(i a+b) d \left (-i b \left (-1+e^{2 i e}\right )+a \left (1+e^{2 i e}\right )\right )}+\frac {6 (c+d x)^2 \log \left (1+\frac {(a+i b) e^{-2 i (e+f x)}}{a-i b}\right )}{\left (a^2+b^2\right ) f}+\frac {3 d \left (2 i f (c+d x) \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i (e+f x)}}{a-i b}\right )+d \operatorname {PolyLog}\left (3,\frac {(-a-i b) e^{-2 i (e+f x)}}{a-i b}\right )\right )}{\left (a^2+b^2\right ) f^3}\right )+\frac {x \left (3 c^2+3 c d x+d^2 x^2\right ) \cos (e)}{3 (a \cos (e)+b \sin (e))} \]
(b*((-4*(c + d*x)^3)/((I*a + b)*d*((-I)*b*(-1 + E^((2*I)*e)) + a*(1 + E^(( 2*I)*e)))) + (6*(c + d*x)^2*Log[1 + (a + I*b)/((a - I*b)*E^((2*I)*(e + f*x )))])/((a^2 + b^2)*f) + (3*d*((2*I)*f*(c + d*x)*PolyLog[2, (-a - I*b)/((a - I*b)*E^((2*I)*(e + f*x)))] + d*PolyLog[3, (-a - I*b)/((a - I*b)*E^((2*I) *(e + f*x)))]))/((a^2 + b^2)*f^3)))/6 + (x*(3*c^2 + 3*c*d*x + d^2*x^2)*Cos [e])/(3*(a*Cos[e] + b*Sin[e]))
Time = 0.75 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 4215, 2620, 3011, 2720, 7143}
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.
| \(\displaystyle \int \frac {(c+d x)^2}{a+b \tan (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c+d x)^2}{a+b \tan (e+f x)}dx\) |
| \(\Big \downarrow \) 4215 |
| \(\displaystyle 2 i b \int \frac {e^{2 i (e+f x)} (c+d x)^2}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i (e+f x)}}dx+\frac {(c+d x)^3}{3 d (a+i b)}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle 2 i b \left (\frac {i d \int (c+d x) \log \left (\frac {e^{2 i (e+f x)} \left (a^2+b^2\right )}{(a+i b)^2}+1\right )dx}{f \left (a^2+b^2\right )}-\frac {i (c+d x)^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 f \left (a^2+b^2\right )}\right )+\frac {(c+d x)^3}{3 d (a+i b)}\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle 2 i b \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 f}-\frac {i d \int \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )dx}{2 f}\right )}{f \left (a^2+b^2\right )}-\frac {i (c+d x)^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 f \left (a^2+b^2\right )}\right )+\frac {(c+d x)^3}{3 d (a+i b)}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle 2 i b \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 f}-\frac {d \int e^{-2 i (e+f x)} \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )de^{2 i (e+f x)}}{4 f^2}\right )}{f \left (a^2+b^2\right )}-\frac {i (c+d x)^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 f \left (a^2+b^2\right )}\right )+\frac {(c+d x)^3}{3 d (a+i b)}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle 2 i b \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 f}-\frac {d \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{4 f^2}\right )}{f \left (a^2+b^2\right )}-\frac {i (c+d x)^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 f \left (a^2+b^2\right )}\right )+\frac {(c+d x)^3}{3 d (a+i b)}\) |
(c + d*x)^3/(3*(a + I*b)*d) + (2*I)*b*(((-1/2*I)*(c + d*x)^2*Log[1 + ((a^2 + b^2)*E^((2*I)*(e + f*x)))/(a + I*b)^2])/((a^2 + b^2)*f) + (I*d*(((I/2)* (c + d*x)*PolyLog[2, -(((a^2 + b^2)*E^((2*I)*(e + f*x)))/(a + I*b)^2)])/f - (d*PolyLog[3, -(((a^2 + b^2)*E^((2*I)*(e + f*x)))/(a + I*b)^2)])/(4*f^2) ))/((a^2 + b^2)*f))
3.1.55.3.1 Defintions of rubi rules used
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) *(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F])) Int[(f + g*x)^( m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e , f, g, n}, x] && GtQ[m, 0]
Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Sy mbol] :> Simp[(c + d*x)^(m + 1)/(d*(m + 1)*(a + I*b)), x] + Simp[2*I*b In t[(c + d*x)^m*(E^Simp[2*I*(e + f*x), x]/((a + I*b)^2 + (a^2 + b^2)*E^Simp[2 *I*(e + f*x), x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2 , 0] && IGtQ[m, 0]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 939 vs. \(2 (166 ) = 332\).
Time = 0.69 (sec) , antiderivative size = 940, normalized size of antiderivative = 5.19
| method | result | size |
| risch | \(-\frac {d^{2} x^{3}}{3 \left (i b -a \right )}-\frac {d c \,x^{2}}{i b -a}-\frac {c^{2} x}{i b -a}-\frac {c^{3}}{3 d \left (i b -a \right )}+\frac {4 b \,e^{3} d^{2}}{3 f^{3} \left (i a +b \right ) \left (-i b -a \right )}-\frac {2 i b c d \ln \left (1-\frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right ) x}{f \left (i a +b \right ) \left (-i b -a \right )}+\frac {2 b \,e^{2} d^{2} x}{f^{2} \left (i a +b \right ) \left (-i b -a \right )}-\frac {b \,d^{2} \operatorname {Li}_{2}\left (\frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right ) x}{f^{2} \left (i a +b \right ) \left (-i b -a \right )}-\frac {2 i b c d \ln \left (1-\frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right ) e}{f^{2} \left (i a +b \right ) \left (-i b -a \right )}-\frac {2 i b \,d^{2} e^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3} \left (i a +b \right ) \left (i b +a \right )}-\frac {2 b c d \,x^{2}}{\left (i a +b \right ) \left (-i b -a \right )}-\frac {4 b c d e x}{f \left (i a +b \right ) \left (-i b -a \right )}-\frac {2 b c d \,e^{2}}{f^{2} \left (i a +b \right ) \left (-i b -a \right )}-\frac {b c d \,\operatorname {Li}_{2}\left (\frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right )}{f^{2} \left (i a +b \right ) \left (-i b -a \right )}-\frac {i b \,d^{2} \operatorname {Li}_{3}\left (\frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right )}{2 f^{3} \left (i a +b \right ) \left (-i b -a \right )}-\frac {2 i b c d e \ln \left (i {\mathrm e}^{2 i \left (f x +e \right )} b -a \,{\mathrm e}^{2 i \left (f x +e \right )}-i b -a \right )}{f^{2} \left (i a +b \right ) \left (i b +a \right )}-\frac {2 i b \,c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f \left (i a +b \right ) \left (i b +a \right )}+\frac {i b \,c^{2} \ln \left (i {\mathrm e}^{2 i \left (f x +e \right )} b -a \,{\mathrm e}^{2 i \left (f x +e \right )}-i b -a \right )}{f \left (i a +b \right ) \left (i b +a \right )}+\frac {4 i b c d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2} \left (i a +b \right ) \left (i b +a \right )}+\frac {i b \,e^{2} d^{2} \ln \left (1-\frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right )}{f^{3} \left (i a +b \right ) \left (-i b -a \right )}-\frac {2 b \,d^{2} x^{3}}{3 \left (i a +b \right ) \left (-i b -a \right )}-\frac {i b \,d^{2} \ln \left (1-\frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right ) x^{2}}{f \left (i a +b \right ) \left (-i b -a \right )}+\frac {i b \,d^{2} e^{2} \ln \left (i {\mathrm e}^{2 i \left (f x +e \right )} b -a \,{\mathrm e}^{2 i \left (f x +e \right )}-i b -a \right )}{f^{3} \left (i a +b \right ) \left (i b +a \right )}\) | \(940\) |
-1/3*d^2/(I*b-a)*x^3-d/(I*b-a)*c*x^2-1/(I*b-a)*c^2*x-1/3/d/(I*b-a)*c^3+4/3 /f^3/(I*a+b)*b/(-I*b-a)*e^3*d^2-2*I/f/(I*a+b)*b/(-I*b-a)*c*d*ln(1-(a-I*b)* exp(2*I*(f*x+e))/(-I*b-a))*x+2/f^2/(I*a+b)*b/(-I*b-a)*e^2*d^2*x-1/f^2/(I*a +b)*b/(-I*b-a)*d^2*polylog(2,(a-I*b)*exp(2*I*(f*x+e))/(-I*b-a))*x-2*I/f^2/ (I*a+b)*b/(-I*b-a)*c*d*ln(1-(a-I*b)*exp(2*I*(f*x+e))/(-I*b-a))*e-2*I/f^3/( I*a+b)*b*d^2*e^2/(a+I*b)*ln(exp(I*(f*x+e)))-2/(I*a+b)*b/(-I*b-a)*c*d*x^2-4 /f/(I*a+b)*b/(-I*b-a)*c*d*e*x-2/f^2/(I*a+b)*b/(-I*b-a)*c*d*e^2-1/f^2/(I*a+ b)*b/(-I*b-a)*c*d*polylog(2,(a-I*b)*exp(2*I*(f*x+e))/(-I*b-a))-1/2*I/f^3/( I*a+b)*b/(-I*b-a)*d^2*polylog(3,(a-I*b)*exp(2*I*(f*x+e))/(-I*b-a))-2*I/f^2 /(I*a+b)*b*c*d*e/(a+I*b)*ln(I*exp(2*I*(f*x+e))*b-a*exp(2*I*(f*x+e))-I*b-a) -2*I/f/(I*a+b)*b*c^2/(a+I*b)*ln(exp(I*(f*x+e)))+I/f/(I*a+b)*b*c^2/(a+I*b)* ln(I*exp(2*I*(f*x+e))*b-a*exp(2*I*(f*x+e))-I*b-a)+4*I/f^2/(I*a+b)*b*c*d*e/ (a+I*b)*ln(exp(I*(f*x+e)))+I/f^3/(I*a+b)*b/(-I*b-a)*e^2*d^2*ln(1-(a-I*b)*e xp(2*I*(f*x+e))/(-I*b-a))-2/3/(I*a+b)*b/(-I*b-a)*d^2*x^3-I/f/(I*a+b)*b/(-I *b-a)*d^2*ln(1-(a-I*b)*exp(2*I*(f*x+e))/(-I*b-a))*x^2+I/f^3/(I*a+b)*b*d^2* e^2/(a+I*b)*ln(I*exp(2*I*(f*x+e))*b-a*exp(2*I*(f*x+e))-I*b-a)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 834 vs. \(2 (160) = 320\).
Time = 0.27 (sec) , antiderivative size = 834, normalized size of antiderivative = 4.61 \[ \int \frac {(c+d x)^2}{a+b \tan (e+f x)} \, dx=\frac {4 \, a d^{2} f^{3} x^{3} + 12 \, a c d f^{3} x^{2} + 12 \, a c^{2} f^{3} x + 3 \, b d^{2} {\rm polylog}\left (3, \frac {{\left (a^{2} + 2 i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} - 2 i \, a b + b^{2} - 2 \, {\left (-i \, a^{2} + 2 \, a b + i \, b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}}\right ) + 3 \, b d^{2} {\rm polylog}\left (3, \frac {{\left (a^{2} - 2 i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} + 2 i \, a b + b^{2} - 2 \, {\left (i \, a^{2} + 2 \, a b - i \, b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}}\right ) - 6 \, {\left (-i \, b d^{2} f x - i \, b c d f\right )} {\rm Li}_2\left (\frac {2 \, {\left ({\left (i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} - i \, a b + {\left (i \, a^{2} - 2 \, a b - i \, b^{2}\right )} \tan \left (f x + e\right )\right )}}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}} + 1\right ) - 6 \, {\left (i \, b d^{2} f x + i \, b c d f\right )} {\rm Li}_2\left (\frac {2 \, {\left ({\left (-i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} + i \, a b + {\left (-i \, a^{2} - 2 \, a b + i \, b^{2}\right )} \tan \left (f x + e\right )\right )}}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}} + 1\right ) + 6 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (-\frac {2 \, {\left ({\left (i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} - i \, a b + {\left (i \, a^{2} - 2 \, a b - i \, b^{2}\right )} \tan \left (f x + e\right )\right )}}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}}\right ) + 6 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (-\frac {2 \, {\left ({\left (-i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} + i \, a b + {\left (-i \, a^{2} - 2 \, a b + i \, b^{2}\right )} \tan \left (f x + e\right )\right )}}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}}\right ) + 6 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (\frac {{\left (i \, a b + b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} + i \, a b + {\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}\right ) + 6 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (\frac {{\left (i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + i \, a b + {\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}\right )}{12 \, {\left (a^{2} + b^{2}\right )} f^{3}} \]
1/12*(4*a*d^2*f^3*x^3 + 12*a*c*d*f^3*x^2 + 12*a*c^2*f^3*x + 3*b*d^2*polylo g(3, ((a^2 + 2*I*a*b - b^2)*tan(f*x + e)^2 - a^2 - 2*I*a*b + b^2 - 2*(-I*a ^2 + 2*a*b + I*b^2)*tan(f*x + e))/((a^2 + b^2)*tan(f*x + e)^2 + a^2 + b^2) ) + 3*b*d^2*polylog(3, ((a^2 - 2*I*a*b - b^2)*tan(f*x + e)^2 - a^2 + 2*I*a *b + b^2 - 2*(I*a^2 + 2*a*b - I*b^2)*tan(f*x + e))/((a^2 + b^2)*tan(f*x + e)^2 + a^2 + b^2)) - 6*(-I*b*d^2*f*x - I*b*c*d*f)*dilog(2*((I*a*b - b^2)*t an(f*x + e)^2 - a^2 - I*a*b + (I*a^2 - 2*a*b - I*b^2)*tan(f*x + e))/((a^2 + b^2)*tan(f*x + e)^2 + a^2 + b^2) + 1) - 6*(I*b*d^2*f*x + I*b*c*d*f)*dilo g(2*((-I*a*b - b^2)*tan(f*x + e)^2 - a^2 + I*a*b + (-I*a^2 - 2*a*b + I*b^2 )*tan(f*x + e))/((a^2 + b^2)*tan(f*x + e)^2 + a^2 + b^2) + 1) + 6*(b*d^2*f ^2*x^2 + 2*b*c*d*f^2*x - b*d^2*e^2 + 2*b*c*d*e*f)*log(-2*((I*a*b - b^2)*ta n(f*x + e)^2 - a^2 - I*a*b + (I*a^2 - 2*a*b - I*b^2)*tan(f*x + e))/((a^2 + b^2)*tan(f*x + e)^2 + a^2 + b^2)) + 6*(b*d^2*f^2*x^2 + 2*b*c*d*f^2*x - b* d^2*e^2 + 2*b*c*d*e*f)*log(-2*((-I*a*b - b^2)*tan(f*x + e)^2 - a^2 + I*a*b + (-I*a^2 - 2*a*b + I*b^2)*tan(f*x + e))/((a^2 + b^2)*tan(f*x + e)^2 + a^ 2 + b^2)) + 6*(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*log(((I*a*b + b^2)*tan (f*x + e)^2 - a^2 + I*a*b + (I*a^2 + I*b^2)*tan(f*x + e))/(tan(f*x + e)^2 + 1)) + 6*(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*log(((I*a*b - b^2)*tan(f*x + e)^2 + a^2 + I*a*b + (I*a^2 + I*b^2)*tan(f*x + e))/(tan(f*x + e)^2 + 1) ))/((a^2 + b^2)*f^3)
\[ \int \frac {(c+d x)^2}{a+b \tan (e+f x)} \, dx=\int \frac {\left (c + d x\right )^{2}}{a + b \tan {\left (e + f x \right )}}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 715 vs. \(2 (160) = 320\).
Time = 0.55 (sec) , antiderivative size = 715, normalized size of antiderivative = 3.95 \[ \int \frac {(c+d x)^2}{a+b \tan (e+f x)} \, dx=-\frac {6 \, c d e {\left (\frac {2 \, {\left (f x + e\right )} a}{{\left (a^{2} + b^{2}\right )} f} + \frac {2 \, b \log \left (b \tan \left (f x + e\right ) + a\right )}{{\left (a^{2} + b^{2}\right )} f} - \frac {b \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{{\left (a^{2} + b^{2}\right )} f}\right )} - 3 \, {\left (\frac {2 \, {\left (f x + e\right )} a}{a^{2} + b^{2}} + \frac {2 \, b \log \left (b \tan \left (f x + e\right ) + a\right )}{a^{2} + b^{2}} - \frac {b \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{2} + b^{2}}\right )} c^{2} - \frac {2 \, {\left (f x + e\right )}^{3} {\left (a - i \, b\right )} d^{2} + 6 \, {\left (f x + e\right )} {\left (a - i \, b\right )} d^{2} e^{2} + 6 i \, b d^{2} e^{2} \arctan \left (-b \cos \left (2 \, f x + 2 \, e\right ) + a \sin \left (2 \, f x + 2 \, e\right ) + b, a \cos \left (2 \, f x + 2 \, e\right ) + b \sin \left (2 \, f x + 2 \, e\right ) + a\right ) + 3 \, b d^{2} e^{2} \log \left ({\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a b \sin \left (2 \, f x + 2 \, e\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + a^{2} + b^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )\right ) + 3 \, b d^{2} {\rm Li}_{3}(\frac {{\left (i \, a + b\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}{-i \, a + b}) - 6 \, {\left ({\left (a - i \, b\right )} d^{2} e - {\left (a - i \, b\right )} c d f\right )} {\left (f x + e\right )}^{2} - 6 \, {\left (i \, {\left (f x + e\right )}^{2} b d^{2} + 2 \, {\left (-i \, b d^{2} e + i \, b c d f\right )} {\left (f x + e\right )}\right )} \arctan \left (\frac {2 \, a b \cos \left (2 \, f x + 2 \, e\right ) - {\left (a^{2} - b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}, \frac {2 \, a b \sin \left (2 \, f x + 2 \, e\right ) + a^{2} + b^{2} + {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) - 6 \, {\left (i \, {\left (f x + e\right )} b d^{2} - i \, b d^{2} e + i \, b c d f\right )} {\rm Li}_2\left (\frac {{\left (i \, a + b\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}{-i \, a + b}\right ) + 3 \, {\left ({\left (f x + e\right )}^{2} b d^{2} - 2 \, {\left (b d^{2} e - b c d f\right )} {\left (f x + e\right )}\right )} \log \left (\frac {{\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a b \sin \left (2 \, f x + 2 \, e\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + a^{2} + b^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right )}{{\left (a^{2} + b^{2}\right )} f^{2}}}{6 \, f} \]
-1/6*(6*c*d*e*(2*(f*x + e)*a/((a^2 + b^2)*f) + 2*b*log(b*tan(f*x + e) + a) /((a^2 + b^2)*f) - b*log(tan(f*x + e)^2 + 1)/((a^2 + b^2)*f)) - 3*(2*(f*x + e)*a/(a^2 + b^2) + 2*b*log(b*tan(f*x + e) + a)/(a^2 + b^2) - b*log(tan(f *x + e)^2 + 1)/(a^2 + b^2))*c^2 - (2*(f*x + e)^3*(a - I*b)*d^2 + 6*(f*x + e)*(a - I*b)*d^2*e^2 + 6*I*b*d^2*e^2*arctan2(-b*cos(2*f*x + 2*e) + a*sin(2 *f*x + 2*e) + b, a*cos(2*f*x + 2*e) + b*sin(2*f*x + 2*e) + a) + 3*b*d^2*e^ 2*log((a^2 + b^2)*cos(2*f*x + 2*e)^2 + 4*a*b*sin(2*f*x + 2*e) + (a^2 + b^2 )*sin(2*f*x + 2*e)^2 + a^2 + b^2 + 2*(a^2 - b^2)*cos(2*f*x + 2*e)) + 3*b*d ^2*polylog(3, (I*a + b)*e^(2*I*f*x + 2*I*e)/(-I*a + b)) - 6*((a - I*b)*d^2 *e - (a - I*b)*c*d*f)*(f*x + e)^2 - 6*(I*(f*x + e)^2*b*d^2 + 2*(-I*b*d^2*e + I*b*c*d*f)*(f*x + e))*arctan2((2*a*b*cos(2*f*x + 2*e) - (a^2 - b^2)*sin (2*f*x + 2*e))/(a^2 + b^2), (2*a*b*sin(2*f*x + 2*e) + a^2 + b^2 + (a^2 - b ^2)*cos(2*f*x + 2*e))/(a^2 + b^2)) - 6*(I*(f*x + e)*b*d^2 - I*b*d^2*e + I* b*c*d*f)*dilog((I*a + b)*e^(2*I*f*x + 2*I*e)/(-I*a + b)) + 3*((f*x + e)^2* b*d^2 - 2*(b*d^2*e - b*c*d*f)*(f*x + e))*log(((a^2 + b^2)*cos(2*f*x + 2*e) ^2 + 4*a*b*sin(2*f*x + 2*e) + (a^2 + b^2)*sin(2*f*x + 2*e)^2 + a^2 + b^2 + 2*(a^2 - b^2)*cos(2*f*x + 2*e))/(a^2 + b^2)))/((a^2 + b^2)*f^2))/f
\[ \int \frac {(c+d x)^2}{a+b \tan (e+f x)} \, dx=\int { \frac {{\left (d x + c\right )}^{2}}{b \tan \left (f x + e\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(c+d x)^2}{a+b \tan (e+f x)} \, dx=\int \frac {{\left (c+d\,x\right )}^2}{a+b\,\mathrm {tan}\left (e+f\,x\right )} \,d x \]