\(\int \frac {(c+d x)^2}{a-a \cos (e+f x)} \, dx\) [139]

Optimal result

Integrand size = 21, antiderivative size = 102 \[ \int \frac {(c+d x)^2}{a-a \cos (e+f x)} \, dx=-\frac {i (c+d x)^2}{a f}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}+\frac {4 d (c+d x) \log \left (1-e^{i (e+f x)}\right )}{a f^2}-\frac {4 i d^2 \operatorname {PolyLog}\left (2,e^{i (e+f x)}\right )}{a f^3} \] Output:

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

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(447\) vs. \(2(102)=204\).

Time = 6.62 (sec) , antiderivative size = 447, normalized size of antiderivative = 4.38 \[ \int \frac {(c+d x)^2}{a-a \cos (e+f x)} \, dx=\frac {2 \csc \left (\frac {e}{2}\right ) \left (c^2 \sin \left (\frac {f x}{2}\right )+2 c d x \sin \left (\frac {f x}{2}\right )+d^2 x^2 \sin \left (\frac {f x}{2}\right )\right ) \sin \left (\frac {e}{2}+\frac {f x}{2}\right )}{f (a-a \cos (e+f x))}+\frac {8 c d \csc \left (\frac {e}{2}\right ) \left (-\frac {1}{2} f x \cos \left (\frac {e}{2}\right )+\log \left (\cos \left (\frac {f x}{2}\right ) \sin \left (\frac {e}{2}\right )+\cos \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right )\right ) \sin \left (\frac {e}{2}\right )\right ) \sin ^2\left (\frac {e}{2}+\frac {f x}{2}\right )}{f^2 (a-a \cos (e+f x)) \left (\cos ^2\left (\frac {e}{2}\right )+\sin ^2\left (\frac {e}{2}\right )\right )}-\frac {8 d^2 \csc \left (\frac {e}{2}\right ) \sec \left (\frac {e}{2}\right ) \sin ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \left (\frac {1}{4} e^{i \arctan \left (\tan \left (\frac {e}{2}\right )\right )} f^2 x^2+\frac {\left (\frac {1}{2} i f x \left (-\pi +2 \arctan \left (\tan \left (\frac {e}{2}\right )\right )\right )-\pi \log \left (1+e^{-i f x}\right )-2 \left (\frac {f x}{2}+\arctan \left (\tan \left (\frac {e}{2}\right )\right )\right ) \log \left (1-e^{2 i \left (\frac {f x}{2}+\arctan \left (\tan \left (\frac {e}{2}\right )\right )\right )}\right )+\pi \log \left (\cos \left (\frac {f x}{2}\right )\right )+2 \arctan \left (\tan \left (\frac {e}{2}\right )\right ) \log \left (\sin \left (\frac {f x}{2}+\arctan \left (\tan \left (\frac {e}{2}\right )\right )\right )\right )+i \operatorname {PolyLog}\left (2,e^{2 i \left (\frac {f x}{2}+\arctan \left (\tan \left (\frac {e}{2}\right )\right )\right )}\right )\right ) \tan \left (\frac {e}{2}\right )}{\sqrt {1+\tan ^2\left (\frac {e}{2}\right )}}\right )}{f^3 (a-a \cos (e+f x)) \sqrt {\sec ^2\left (\frac {e}{2}\right ) \left (\cos ^2\left (\frac {e}{2}\right )+\sin ^2\left (\frac {e}{2}\right )\right )}} \] Input:

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

Output:

(2*Csc[e/2]*(c^2*Sin[(f*x)/2] + 2*c*d*x*Sin[(f*x)/2] + d^2*x^2*Sin[(f*x)/2 
])*Sin[e/2 + (f*x)/2])/(f*(a - a*Cos[e + f*x])) + (8*c*d*Csc[e/2]*(-1/2*(f 
*x*Cos[e/2]) + Log[Cos[(f*x)/2]*Sin[e/2] + Cos[e/2]*Sin[(f*x)/2]]*Sin[e/2] 
)*Sin[e/2 + (f*x)/2]^2)/(f^2*(a - a*Cos[e + f*x])*(Cos[e/2]^2 + Sin[e/2]^2 
)) - (8*d^2*Csc[e/2]*Sec[e/2]*Sin[e/2 + (f*x)/2]^2*((E^(I*ArcTan[Tan[e/2]] 
)*f^2*x^2)/4 + (((I/2)*f*x*(-Pi + 2*ArcTan[Tan[e/2]]) - Pi*Log[1 + E^((-I) 
*f*x)] - 2*((f*x)/2 + ArcTan[Tan[e/2]])*Log[1 - E^((2*I)*((f*x)/2 + ArcTan 
[Tan[e/2]]))] + Pi*Log[Cos[(f*x)/2]] + 2*ArcTan[Tan[e/2]]*Log[Sin[(f*x)/2 
+ ArcTan[Tan[e/2]]]] + I*PolyLog[2, E^((2*I)*((f*x)/2 + ArcTan[Tan[e/2]])) 
])*Tan[e/2])/Sqrt[1 + Tan[e/2]^2]))/(f^3*(a - a*Cos[e + f*x])*Sqrt[Sec[e/2 
]^2*(Cos[e/2]^2 + Sin[e/2]^2)])
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3799

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4672

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

((-2*(c + d*x)^2*Cot[e/2 + (f*x)/2])/f - (4*d*(((I/2)*(c + d*x)^2)/d - (2* 
I)*(((-I)*(c + d*x)*Log[1 + E^(I*(e + Pi + f*x))])/f - (d*PolyLog[2, -E^(I 
*(e + Pi + 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. 246 vs. \(2 (92 ) = 184\).

Time = 1.13 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.42

method	result	size
risch	\(-\frac {2 i \left (x^{2} d^{2}+2 c d x +c^{2}\right )}{f a \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}-\frac {4 d c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{2}}+\frac {4 d c \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{a \,f^{2}}-\frac {2 i d^{2} x^{2}}{a f}-\frac {4 i d^{2} e x}{a \,f^{2}}-\frac {2 i d^{2} e^{2}}{a \,f^{3}}+\frac {4 d^{2} \ln \left (1-{\mathrm e}^{i \left (f x +e \right )}\right ) x}{a \,f^{2}}+\frac {4 d^{2} \ln \left (1-{\mathrm e}^{i \left (f x +e \right )}\right ) e}{a \,f^{3}}-\frac {4 i d^{2} \operatorname {polylog}\left (2, {\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}+\frac {4 d^{2} e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}-\frac {4 d^{2} e \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{a \,f^{3}}\)	\(247\)

Input:

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

Output:

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

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. 283 vs. \(2 (89) = 178\).

Time = 0.08 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.77 \[ \int \frac {(c+d x)^2}{a-a \cos (e+f x)} \, dx=-\frac {d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2} + 2 i \, d^{2} {\rm Li}_2\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) \sin \left (f x + e\right ) - 2 i \, d^{2} {\rm Li}_2\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) \sin \left (f x + e\right ) + 2 \, {\left (d^{2} e - c d f\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2} i \, \sin \left (f x + e\right ) + \frac {1}{2}\right ) \sin \left (f x + e\right ) + 2 \, {\left (d^{2} e - c d f\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) - \frac {1}{2} i \, \sin \left (f x + e\right ) + \frac {1}{2}\right ) \sin \left (f x + e\right ) - 2 \, {\left (d^{2} f x + d^{2} e\right )} \log \left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 2 \, {\left (d^{2} f x + d^{2} e\right )} \log \left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2}\right )} \cos \left (f x + e\right )}{a f^{3} \sin \left (f x + e\right )} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

-(Integral(c**2/(cos(e + f*x) - 1), x) + Integral(d**2*x**2/(cos(e + f*x) 
- 1), x) + Integral(2*c*d*x/(cos(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. 310 vs. \(2 (89) = 178\).

Time = 0.13 (sec) , antiderivative size = 310, normalized size of antiderivative = 3.04 \[ \int \frac {(c+d x)^2}{a-a \cos (e+f x)} \, dx=-\frac {2 \, {\left (c^{2} f^{2} - 2 \, {\left (c d f \cos \left (f x + e\right ) + i \, c d f \sin \left (f x + e\right ) - c d f\right )} \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) - 1\right ) + 2 \, {\left (d^{2} f x \cos \left (f x + e\right ) + i \, d^{2} f x \sin \left (f x + e\right ) - d^{2} f x\right )} \arctan \left (\sin \left (f x + e\right ), -\cos \left (f x + e\right ) + 1\right ) + {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x\right )} \cos \left (f x + e\right ) + 2 \, {\left (d^{2} \cos \left (f x + e\right ) + i \, d^{2} \sin \left (f x + e\right ) - d^{2}\right )} {\rm Li}_2\left (e^{\left (i \, f x + i \, e\right )}\right ) + {\left (-i \, d^{2} f x - i \, c d f + {\left (i \, d^{2} f x + i \, c d f\right )} \cos \left (f x + e\right ) - {\left (d^{2} f x + c d f\right )} \sin \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right ) + {\left (i \, d^{2} f^{2} x^{2} + 2 i \, c d f^{2} x\right )} \sin \left (f x + e\right )\right )}}{-i \, a f^{3} \cos \left (f x + e\right ) + a f^{3} \sin \left (f x + e\right ) + i \, a f^{3}} \] Input:

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

Output:

-2*(c^2*f^2 - 2*(c*d*f*cos(f*x + e) + I*c*d*f*sin(f*x + e) - c*d*f)*arctan 
2(sin(f*x + e), cos(f*x + e) - 1) + 2*(d^2*f*x*cos(f*x + e) + I*d^2*f*x*si 
n(f*x + e) - d^2*f*x)*arctan2(sin(f*x + e), -cos(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) 
 - d^2)*dilog(e^(I*f*x + I*e)) + (-I*d^2*f*x - I*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 + s 
in(f*x + e)^2 - 2*cos(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) + I*a*f^3)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*int(x/tan((e + f*x)/2),x)*tan((e + f*x)/2)*d**2*f - 2*log(tan((e + f*x) 
/2)**2 + 1)*tan((e + f*x)/2)*c*d + 4*log(tan((e + f*x)/2))*tan((e + f*x)/2 
)*c*d - c**2*f - 2*c*d*f*x - d**2*f*x**2)/(tan((e + f*x)/2)*a*f**2)