\(\int \frac {(c+d x)^2}{a+a \sin (e+f x)} \, dx\) [108]

Optimal result

Integrand size = 20, antiderivative size = 113 \[ \int \frac {(c+d x)^2}{a+a \sin (e+f x)} \, dx=-\frac {i (c+d x)^2}{a f}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f}+\frac {4 d (c+d x) \log \left (1-i e^{i (e+f x)}\right )}{a f^2}-\frac {4 i d^2 \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{a f^3} \] Output:

-I*(d*x+c)^2/a/f-(d*x+c)^2*cot(1/2*e+1/4*Pi+1/2*f*x)/a/f+4*d*(d*x+c)*ln(1- 
I*exp(I*(f*x+e)))/a/f^2-4*I*d^2*polylog(2,I*exp(I*(f*x+e)))/a/f^3
 

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.83 \[ \int \frac {(c+d x)^2}{a+a \sin (e+f x)} \, dx=\frac {-4 i d^2 \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )+f (c+d x) \left (-i f (c+d x)+4 d \log \left (1-i e^{i (e+f x)}\right )+f (c+d x) \tan \left (\frac {1}{4} (2 e-\pi +2 f x)\right )\right )}{a f^3} \] Input:

Integrate[(c + d*x)^2/(a + a*Sin[e + f*x]),x]
 

Output:

((-4*I)*d^2*PolyLog[2, I*E^(I*(e + f*x))] + f*(c + d*x)*((-I)*f*(c + d*x) 
+ 4*d*Log[1 - I*E^(I*(e + f*x))] + f*(c + d*x)*Tan[(2*e - Pi + 2*f*x)/4])) 
/(a*f^3)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.14, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3799, 3042, 4672, 3042, 25, 4202, 2620, 2715, 2838}

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.

\(\displaystyle \int \frac {(c+d x)^2}{a \sin (e+f x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c+d x)^2}{a \sin (e+f x)+a}dx\)

\(\Big \downarrow \) 3799

\(\displaystyle \frac {\int (c+d x)^2 \csc ^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )dx}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (c+d x)^2 \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )^2dx}{2 a}\)

\(\Big \downarrow \) 4672

\(\displaystyle \frac {\frac {4 d \int (c+d x) \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )dx}{f}-\frac {2 (c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {4 d \int -\left ((c+d x) \tan \left (\frac {e}{2}+\frac {f x}{2}+\frac {3 \pi }{4}\right )\right )dx}{f}-\frac {2 (c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}}{2 a}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {4 d \int (c+d x) \tan \left (\frac {1}{4} (2 e+3 \pi )+\frac {f x}{2}\right )dx}{f}-\frac {2 (c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}}{2 a}\)

\(\Big \downarrow \) 4202

\(\displaystyle \frac {-\frac {2 (c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}-\frac {4 d \left (\frac {i (c+d x)^2}{2 d}-2 i \int \frac {e^{\frac {1}{2} i (2 e+2 f x+3 \pi )} (c+d x)}{1+e^{\frac {1}{2} i (2 e+2 f x+3 \pi )}}dx\right )}{f}}{2 a}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {-\frac {2 (c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}-\frac {4 d \left (\frac {i (c+d x)^2}{2 d}-2 i \left (\frac {i d \int \log \left (1+e^{\frac {1}{2} i (2 e+2 f x+3 \pi )}\right )dx}{f}-\frac {i (c+d x) \log \left (1+e^{\frac {1}{2} i (2 e+2 f x+3 \pi )}\right )}{f}\right )\right )}{f}}{2 a}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {-\frac {2 (c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}-\frac {4 d \left (\frac {i (c+d x)^2}{2 d}-2 i \left (\frac {d \int e^{-\frac {1}{2} i (2 e+2 f x+3 \pi )} \log \left (1+e^{\frac {1}{2} i (2 e+2 f x+3 \pi )}\right )de^{\frac {1}{2} i (2 e+2 f x+3 \pi )}}{f^2}-\frac {i (c+d x) \log \left (1+e^{\frac {1}{2} i (2 e+2 f x+3 \pi )}\right )}{f}\right )\right )}{f}}{2 a}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {-\frac {2 (c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}-\frac {4 d \left (\frac {i (c+d x)^2}{2 d}-2 i \left (-\frac {i (c+d x) \log \left (1+e^{\frac {1}{2} i (2 e+2 f x+3 \pi )}\right )}{f}-\frac {d \operatorname {PolyLog}\left (2,-e^{\frac {1}{2} i (2 e+2 f x+3 \pi )}\right )}{f^2}\right )\right )}{f}}{2 a}\)

Input:

Int[(c + d*x)^2/(a + a*Sin[e + f*x]),x]
 

Output:

((-2*(c + d*x)^2*Cot[e/2 + Pi/4 + (f*x)/2])/f - (4*d*(((I/2)*(c + d*x)^2)/ 
d - (2*I)*(((-I)*(c + d*x)*Log[1 + E^((I/2)*(2*e + 3*Pi + 2*f*x))])/f - (d 
*PolyLog[2, -E^((I/2)*(2*e + 3*Pi + 2*f*x))])/f^2)))/f)/(2*a)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2620

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]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3799

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) 
, x_Symbol] :> Simp[(2*a)^n   Int[(c + d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) 
+ f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^ 
2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])
 

rule 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I   Int[(c + d*x)^m*(E^(2*I*( 
e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt 
Q[m, 0]
 

rule 4672

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp 
[(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1) 
*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (99 ) = 198\).

Time = 0.78 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.25

method	result	size
risch	\(-\frac {2 \left (x^{2} d^{2}+2 c d x +c^{2}\right )}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}+\frac {4 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) c d}{a \,f^{2}}-\frac {4 \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right ) c d}{a \,f^{2}}-\frac {2 i d^{2} x^{2}}{a f}-\frac {4 i d^{2} e x}{a \,f^{2}}-\frac {2 i d^{2} e^{2}}{a \,f^{3}}+\frac {4 d^{2} \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{a \,f^{2}}+\frac {4 d^{2} \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{a \,f^{3}}-\frac {4 i d^{2} \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}-\frac {4 e \,d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{a \,f^{3}}+\frac {4 e \,d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}\)	\(254\)

Input:

int((d*x+c)^2/(a+a*sin(f*x+e)),x,method=_RETURNVERBOSE)
 

Output:

-2*(d^2*x^2+2*c*d*x+c^2)/f/a/(exp(I*(f*x+e))+I)+4/a/f^2*ln(exp(I*(f*x+e))+ 
I)*c*d-4/a/f^2*ln(exp(I*(f*x+e)))*c*d-2*I/a/f*d^2*x^2-4*I/a/f^2*d^2*e*x-2* 
I/a/f^3*d^2*e^2+4/a/f^2*d^2*ln(1-I*exp(I*(f*x+e)))*x+4/a/f^3*d^2*ln(1-I*ex 
p(I*(f*x+e)))*e-4*I*d^2*polylog(2,I*exp(I*(f*x+e)))/a/f^3-4/a/f^3*e*d^2*ln 
(exp(I*(f*x+e))+I)+4/a/f^3*e*d^2*ln(exp(I*(f*x+e)))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 493 vs. \(2 (94) = 188\).

Time = 0.09 (sec) , antiderivative size = 493, normalized size of antiderivative = 4.36 \[ \int \frac {(c+d x)^2}{a+a \sin (e+f x)} \, dx=-\frac {d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2} + {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2}\right )} \cos \left (f x + e\right ) + 2 \, {\left (i \, d^{2} \cos \left (f x + e\right ) + i \, d^{2} \sin \left (f x + e\right ) + i \, d^{2}\right )} {\rm Li}_2\left (i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) + 2 \, {\left (-i \, d^{2} \cos \left (f x + e\right ) - i \, d^{2} \sin \left (f x + e\right ) - i \, d^{2}\right )} {\rm Li}_2\left (-i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) + 2 \, {\left (d^{2} e - c d f + {\left (d^{2} e - c d f\right )} \cos \left (f x + e\right ) + {\left (d^{2} e - c d f\right )} \sin \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + i\right ) - 2 \, {\left (d^{2} f x + d^{2} e + {\left (d^{2} f x + d^{2} e\right )} \cos \left (f x + e\right ) + {\left (d^{2} f x + d^{2} e\right )} \sin \left (f x + e\right )\right )} \log \left (i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (d^{2} f x + d^{2} e + {\left (d^{2} f x + d^{2} e\right )} \cos \left (f x + e\right ) + {\left (d^{2} f x + d^{2} e\right )} \sin \left (f x + e\right )\right )} \log \left (-i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (d^{2} e - c d f + {\left (d^{2} e - c d f\right )} \cos \left (f x + e\right ) + {\left (d^{2} e - c d f\right )} \sin \left (f x + e\right )\right )} \log \left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + i\right ) - {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2}\right )} \sin \left (f x + e\right )}{a f^{3} \cos \left (f x + e\right ) + a f^{3} \sin \left (f x + e\right ) + a f^{3}} \] Input:

integrate((d*x+c)^2/(a+a*sin(f*x+e)),x, algorithm="fricas")
 

Output:

-(d^2*f^2*x^2 + 2*c*d*f^2*x + c^2*f^2 + (d^2*f^2*x^2 + 2*c*d*f^2*x + c^2*f 
^2)*cos(f*x + e) + 2*(I*d^2*cos(f*x + e) + I*d^2*sin(f*x + e) + I*d^2)*dil 
og(I*cos(f*x + e) - sin(f*x + e)) + 2*(-I*d^2*cos(f*x + e) - I*d^2*sin(f*x 
 + e) - I*d^2)*dilog(-I*cos(f*x + e) - sin(f*x + e)) + 2*(d^2*e - c*d*f + 
(d^2*e - c*d*f)*cos(f*x + e) + (d^2*e - c*d*f)*sin(f*x + e))*log(cos(f*x + 
 e) + I*sin(f*x + e) + I) - 2*(d^2*f*x + d^2*e + (d^2*f*x + d^2*e)*cos(f*x 
 + e) + (d^2*f*x + d^2*e)*sin(f*x + e))*log(I*cos(f*x + e) + sin(f*x + e) 
+ 1) - 2*(d^2*f*x + d^2*e + (d^2*f*x + d^2*e)*cos(f*x + e) + (d^2*f*x + d^ 
2*e)*sin(f*x + e))*log(-I*cos(f*x + e) + sin(f*x + e) + 1) + 2*(d^2*e - c* 
d*f + (d^2*e - c*d*f)*cos(f*x + e) + (d^2*e - c*d*f)*sin(f*x + e))*log(-co 
s(f*x + e) + I*sin(f*x + e) + I) - (d^2*f^2*x^2 + 2*c*d*f^2*x + c^2*f^2)*s 
in(f*x + e))/(a*f^3*cos(f*x + e) + a*f^3*sin(f*x + e) + a*f^3)
 

Sympy [F]

\[ \int \frac {(c+d x)^2}{a+a \sin (e+f x)} \, dx=\frac {\int \frac {c^{2}}{\sin {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{2} x^{2}}{\sin {\left (e + f x \right )} + 1}\, dx + \int \frac {2 c d x}{\sin {\left (e + f x \right )} + 1}\, dx}{a} \] Input:

integrate((d*x+c)**2/(a+a*sin(f*x+e)),x)
 

Output:

(Integral(c**2/(sin(e + f*x) + 1), x) + Integral(d**2*x**2/(sin(e + f*x) + 
 1), x) + Integral(2*c*d*x/(sin(e + f*x) + 1), x))/a
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (94) = 188\).

Time = 0.14 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.73 \[ \int \frac {(c+d x)^2}{a+a \sin (e+f x)} \, dx=-\frac {2 \, {\left (-i \, c^{2} f^{2} - 2 \, {\left (c d f \cos \left (f x + e\right ) + i \, c d f \sin \left (f x + e\right ) + i \, c d f\right )} \arctan \left (\sin \left (f x + e\right ) + 1, \cos \left (f x + e\right )\right ) + 2 \, {\left (d^{2} f x \cos \left (f x + e\right ) + i \, d^{2} f x \sin \left (f x + e\right ) + i \, d^{2} f x\right )} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x\right )} \cos \left (f x + e\right ) + 2 \, {\left (d^{2} \cos \left (f x + e\right ) + i \, d^{2} \sin \left (f x + e\right ) + i \, d^{2}\right )} {\rm Li}_2\left (i \, e^{\left (i \, f x + i \, e\right )}\right ) - {\left (d^{2} f x + c d f - {\left (i \, d^{2} f x + i \, c d f\right )} \cos \left (f x + e\right ) + {\left (d^{2} f x + c d f\right )} \sin \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) + {\left (i \, d^{2} f^{2} x^{2} + 2 i \, c d f^{2} x\right )} \sin \left (f x + e\right )\right )}}{-i \, a f^{3} \cos \left (f x + e\right ) + a f^{3} \sin \left (f x + e\right ) + a f^{3}} \] Input:

integrate((d*x+c)^2/(a+a*sin(f*x+e)),x, algorithm="maxima")
 

Output:

-2*(-I*c^2*f^2 - 2*(c*d*f*cos(f*x + e) + I*c*d*f*sin(f*x + e) + I*c*d*f)*a 
rctan2(sin(f*x + e) + 1, cos(f*x + e)) + 2*(d^2*f*x*cos(f*x + e) + I*d^2*f 
*x*sin(f*x + e) + I*d^2*f*x)*arctan2(cos(f*x + e), sin(f*x + e) + 1) + (d^ 
2*f^2*x^2 + 2*c*d*f^2*x)*cos(f*x + e) + 2*(d^2*cos(f*x + e) + I*d^2*sin(f* 
x + e) + I*d^2)*dilog(I*e^(I*f*x + I*e)) - (d^2*f*x + c*d*f - (I*d^2*f*x + 
 I*c*d*f)*cos(f*x + e) + (d^2*f*x + c*d*f)*sin(f*x + e))*log(cos(f*x + e)^ 
2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) + (I*d^2*f^2*x^2 + 2*I*c*d*f^2*x) 
*sin(f*x + e))/(-I*a*f^3*cos(f*x + e) + a*f^3*sin(f*x + e) + a*f^3)
 

Giac [F]

\[ \int \frac {(c+d x)^2}{a+a \sin (e+f x)} \, dx=\int { \frac {{\left (d x + c\right )}^{2}}{a \sin \left (f x + e\right ) + a} \,d x } \] Input:

integrate((d*x+c)^2/(a+a*sin(f*x+e)),x, algorithm="giac")
 

Output:

integrate((d*x + c)^2/(a*sin(f*x + e) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^2}{a+a \sin (e+f x)} \, dx=\int \frac {{\left (c+d\,x\right )}^2}{a+a\,\sin \left (e+f\,x\right )} \,d x \] Input:

int((c + d*x)^2/(a + a*sin(e + f*x)),x)
 

Output:

int((c + d*x)^2/(a + a*sin(e + f*x)), x)
 

Reduce [F]

\[ \int \frac {(c+d x)^2}{a+a \sin (e+f x)} \, dx=\frac {-4 \left (\int \frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) x}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}d x \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d^{2} f -4 \left (\int \frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) x}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}d x \right ) d^{2} f -2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c d -2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) c d +4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c d +4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) c d +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c^{2} f +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c d f x +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d^{2} f \,x^{2}-2 c d f x}{a \,f^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \] Input:

int((d*x+c)^2/(a+a*sin(f*x+e)),x)
 

Output:

(2*( - 2*int((tan((e + f*x)/2)*x)/(tan((e + f*x)/2) + 1),x)*tan((e + f*x)/ 
2)*d**2*f - 2*int((tan((e + f*x)/2)*x)/(tan((e + f*x)/2) + 1),x)*d**2*f - 
log(tan((e + f*x)/2)**2 + 1)*tan((e + f*x)/2)*c*d - log(tan((e + f*x)/2)** 
2 + 1)*c*d + 2*log(tan((e + f*x)/2) + 1)*tan((e + f*x)/2)*c*d + 2*log(tan( 
(e + f*x)/2) + 1)*c*d + tan((e + f*x)/2)*c**2*f + tan((e + f*x)/2)*c*d*f*x 
 + tan((e + f*x)/2)*d**2*f*x**2 - c*d*f*x))/(a*f**2*(tan((e + f*x)/2) + 1) 
)