\(\int (c+d x)^2 \sec (a+b x) \tan (a+b x) \, dx\) [249]

Optimal result

Integrand size = 20, antiderivative size = 97 \[ \int (c+d x)^2 \sec (a+b x) \tan (a+b x) \, dx=\frac {4 i d (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^2}-\frac {2 i d^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 i d^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}+\frac {(c+d x)^2 \sec (a+b x)}{b} \] Output:

4*I*d*(d*x+c)*arctan(exp(I*(b*x+a)))/b^2-2*I*d^2*polylog(2,-I*exp(I*(b*x+a 
)))/b^3+2*I*d^2*polylog(2,I*exp(I*(b*x+a)))/b^3+(d*x+c)^2*sec(b*x+a)/b
 

Mathematica [A] (warning: unable to verify)

Time = 1.15 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.79 \[ \int (c+d x)^2 \sec (a+b x) \tan (a+b x) \, dx=\frac {-4 b c d \text {arctanh}\left (\sin (a)+\cos (a) \tan \left (\frac {b x}{2}\right )\right )-4 d^2 \arctan (\cot (a)) \text {arctanh}\left (\sin (a)+\cos (a) \tan \left (\frac {b x}{2}\right )\right )+\frac {2 d^2 \csc (a) \left ((b x-\arctan (\cot (a))) \left (\log \left (1-e^{i (b x-\arctan (\cot (a)))}\right )-\log \left (1+e^{i (b x-\arctan (\cot (a)))}\right )\right )+i \operatorname {PolyLog}\left (2,-e^{i (b x-\arctan (\cot (a)))}\right )-i \operatorname {PolyLog}\left (2,e^{i (b x-\arctan (\cot (a)))}\right )\right )}{\sqrt {\csc ^2(a)}}+b^2 (c+d x)^2 \sec (a+b x)}{b^3} \] Input:

Integrate[(c + d*x)^2*Sec[a + b*x]*Tan[a + b*x],x]
 

Output:

(-4*b*c*d*ArcTanh[Sin[a] + Cos[a]*Tan[(b*x)/2]] - 4*d^2*ArcTan[Cot[a]]*Arc 
Tanh[Sin[a] + Cos[a]*Tan[(b*x)/2]] + (2*d^2*Csc[a]*((b*x - ArcTan[Cot[a]]) 
*(Log[1 - E^(I*(b*x - ArcTan[Cot[a]]))] - Log[1 + E^(I*(b*x - ArcTan[Cot[a 
]]))]) + I*PolyLog[2, -E^(I*(b*x - ArcTan[Cot[a]]))] - I*PolyLog[2, E^(I*( 
b*x - ArcTan[Cot[a]]))]))/Sqrt[Csc[a]^2] + b^2*(c + d*x)^2*Sec[a + b*x])/b 
^3
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4909, 3042, 4669, 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.

\(\displaystyle \int (c+d x)^2 \tan (a+b x) \sec (a+b x) \, dx\)

\(\Big \downarrow \) 4909

\(\displaystyle \frac {(c+d x)^2 \sec (a+b x)}{b}-\frac {2 d \int (c+d x) \sec (a+b x)dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(c+d x)^2 \sec (a+b x)}{b}-\frac {2 d \int (c+d x) \csc \left (a+b x+\frac {\pi }{2}\right )dx}{b}\)

\(\Big \downarrow \) 4669

\(\displaystyle \frac {(c+d x)^2 \sec (a+b x)}{b}-\frac {2 d \left (-\frac {d \int \log \left (1-i e^{i (a+b x)}\right )dx}{b}+\frac {d \int \log \left (1+i e^{i (a+b x)}\right )dx}{b}-\frac {2 i (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {(c+d x)^2 \sec (a+b x)}{b}-\frac {2 d \left (\frac {i d \int e^{-i (a+b x)} \log \left (1-i e^{i (a+b x)}\right )de^{i (a+b x)}}{b^2}-\frac {i d \int e^{-i (a+b x)} \log \left (1+i e^{i (a+b x)}\right )de^{i (a+b x)}}{b^2}-\frac {2 i (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {(c+d x)^2 \sec (a+b x)}{b}-\frac {2 d \left (-\frac {2 i (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {i d \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac {i d \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}\right )}{b}\)

Input:

Int[(c + d*x)^2*Sec[a + b*x]*Tan[a + b*x],x]
 

Output:

(-2*d*(((-2*I)*(c + d*x)*ArcTan[E^(I*(a + b*x))])/b + (I*d*PolyLog[2, (-I) 
*E^(I*(a + b*x))])/b^2 - (I*d*PolyLog[2, I*E^(I*(a + b*x))])/b^2))/b + ((c 
 + d*x)^2*Sec[a + b*x])/b
 

Defintions of rubi rules used

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 4669

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol 
] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Si 
mp[d*(m/f)   Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))], x], 
 x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x 
))], x], x]) /; FreeQ[{c, d, e, f}, x] && IntegerQ[2*k] && IGtQ[m, 0]
 

rule 4909

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b 
_.)*(x_)]^(p_.), x_Symbol] :> Simp[(c + d*x)^m*(Sec[a + b*x]^n/(b*n)), x] - 
 Simp[d*(m/(b*n))   Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; FreeQ[{ 
a, b, c, d, n}, x] && EqQ[p, 1] && 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. 226 vs. \(2 (86 ) = 172\).

Time = 0.52 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.34

method	result	size
risch	\(\frac {2 \,{\mathrm e}^{i \left (b x +a \right )} \left (x^{2} d^{2}+2 c d x +c^{2}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}+\frac {4 i d c \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {2 d^{2} \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{2}}+\frac {2 d^{2} \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{3}}-\frac {2 d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}-\frac {2 i d^{2} \operatorname {dilog}\left (i {\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{3}}+\frac {2 i d^{2} \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {4 i d^{2} a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}\)	\(227\)
derivativedivides	\(\frac {\frac {a^{2} d^{2}}{b^{2} \cos \left (b x +a \right )}-\frac {2 a c d}{b \cos \left (b x +a \right )}-\frac {2 a \,d^{2} \left (\frac {b x +a}{\cos \left (b x +a \right )}-\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )\right )}{b^{2}}+\frac {c^{2}}{\cos \left (b x +a \right )}+\frac {2 c d \left (\frac {b x +a}{\cos \left (b x +a \right )}-\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )\right )}{b}+\frac {d^{2} \left (\frac {\left (b x +a \right )^{2}}{\cos \left (b x +a \right )}+2 \left (b x +a \right ) \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right )-2 \left (b x +a \right ) \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )-2 i \operatorname {dilog}\left (i {\mathrm e}^{i \left (b x +a \right )}+1\right )+2 i \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )\right )}{b^{2}}}{b}\)	\(234\)
default	\(\frac {\frac {a^{2} d^{2}}{b^{2} \cos \left (b x +a \right )}-\frac {2 a c d}{b \cos \left (b x +a \right )}-\frac {2 a \,d^{2} \left (\frac {b x +a}{\cos \left (b x +a \right )}-\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )\right )}{b^{2}}+\frac {c^{2}}{\cos \left (b x +a \right )}+\frac {2 c d \left (\frac {b x +a}{\cos \left (b x +a \right )}-\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )\right )}{b}+\frac {d^{2} \left (\frac {\left (b x +a \right )^{2}}{\cos \left (b x +a \right )}+2 \left (b x +a \right ) \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right )-2 \left (b x +a \right ) \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )-2 i \operatorname {dilog}\left (i {\mathrm e}^{i \left (b x +a \right )}+1\right )+2 i \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )\right )}{b^{2}}}{b}\)	\(234\)

Input:

int((d*x+c)^2*sec(b*x+a)*tan(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

2*exp(I*(b*x+a))*(d^2*x^2+2*c*d*x+c^2)/b/(exp(2*I*(b*x+a))+1)+4*I/b^2*d*c* 
arctan(exp(I*(b*x+a)))+2/b^2*d^2*ln(I*exp(I*(b*x+a))+1)*x+2/b^3*d^2*ln(I*e 
xp(I*(b*x+a))+1)*a-2/b^2*d^2*ln(1-I*exp(I*(b*x+a)))*x-2/b^3*d^2*ln(1-I*exp 
(I*(b*x+a)))*a-2*I/b^3*d^2*dilog(I*exp(I*(b*x+a))+1)+2*I/b^3*d^2*dilog(1-I 
*exp(I*(b*x+a)))-4*I/b^3*d^2*a*arctan(exp(I*(b*x+a)))
 

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. 446 vs. \(2 (79) = 158\).

Time = 0.10 (sec) , antiderivative size = 446, normalized size of antiderivative = 4.60 \[ \int (c+d x)^2 \sec (a+b x) \tan (a+b x) \, dx=\frac {b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + i \, d^{2} \cos \left (b x + a\right ) {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + i \, d^{2} \cos \left (b x + a\right ) {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - i \, d^{2} \cos \left (b x + a\right ) {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - i \, d^{2} \cos \left (b x + a\right ) {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - {\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - {\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right )}{b^{3} \cos \left (b x + a\right )} \] Input:

integrate((d*x+c)^2*sec(b*x+a)*tan(b*x+a),x, algorithm="fricas")
 

Output:

(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 + I*d^2*cos(b*x + a)*dilog(I*cos(b*x 
+ a) + sin(b*x + a)) + I*d^2*cos(b*x + a)*dilog(I*cos(b*x + a) - sin(b*x + 
 a)) - I*d^2*cos(b*x + a)*dilog(-I*cos(b*x + a) + sin(b*x + a)) - I*d^2*co 
s(b*x + a)*dilog(-I*cos(b*x + a) - sin(b*x + a)) - (b*c*d - a*d^2)*cos(b*x 
 + a)*log(cos(b*x + a) + I*sin(b*x + a) + I) + (b*c*d - a*d^2)*cos(b*x + a 
)*log(cos(b*x + a) - I*sin(b*x + a) + I) - (b*d^2*x + a*d^2)*cos(b*x + a)* 
log(I*cos(b*x + a) + sin(b*x + a) + 1) + (b*d^2*x + a*d^2)*cos(b*x + a)*lo 
g(I*cos(b*x + a) - sin(b*x + a) + 1) - (b*d^2*x + a*d^2)*cos(b*x + a)*log( 
-I*cos(b*x + a) + sin(b*x + a) + 1) + (b*d^2*x + a*d^2)*cos(b*x + a)*log(- 
I*cos(b*x + a) - sin(b*x + a) + 1) - (b*c*d - a*d^2)*cos(b*x + a)*log(-cos 
(b*x + a) + I*sin(b*x + a) + I) + (b*c*d - a*d^2)*cos(b*x + a)*log(-cos(b* 
x + a) - I*sin(b*x + a) + I))/(b^3*cos(b*x + a))
 

Sympy [F]

\[ \int (c+d x)^2 \sec (a+b x) \tan (a+b x) \, dx=\int \left (c + d x\right )^{2} \tan {\left (a + b x \right )} \sec {\left (a + b x \right )}\, dx \] Input:

integrate((d*x+c)**2*sec(b*x+a)*tan(b*x+a),x)
 

Output:

Integral((c + d*x)**2*tan(a + b*x)*sec(a + b*x), x)
 

Maxima [F]

\[ \int (c+d x)^2 \sec (a+b x) \tan (a+b x) \, dx=\int { {\left (d x + c\right )}^{2} \sec \left (b x + a\right ) \tan \left (b x + a\right ) \,d x } \] Input:

integrate((d*x+c)^2*sec(b*x+a)*tan(b*x+a),x, algorithm="maxima")
 

Output:

(2*(b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cos(2*b*x + 2*a)*cos(b*x + a) + 2*(b*d^ 
2*x^2 + 2*b*c*d*x + b*c^2)*sin(2*b*x + 2*a)*sin(b*x + a) + 2*(b*d^2*x^2 + 
2*b*c*d*x + b*c^2)*cos(b*x + a) - 4*(b^2*d^2*cos(2*b*x + 2*a)^2 + b^2*d^2* 
sin(2*b*x + 2*a)^2 + 2*b^2*d^2*cos(2*b*x + 2*a) + b^2*d^2)*integrate((x*co 
s(2*b*x + 2*a)*cos(b*x + a) + x*sin(2*b*x + 2*a)*sin(b*x + a) + x*cos(b*x 
+ a))/(b*cos(2*b*x + 2*a)^2 + b*sin(2*b*x + 2*a)^2 + 2*b*cos(2*b*x + 2*a) 
+ b), x) - (c*d*cos(2*b*x + 2*a)^2 + c*d*sin(2*b*x + 2*a)^2 + 2*c*d*cos(2* 
b*x + 2*a) + c*d)*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*sin(b*x + a) + 1 
) + (c*d*cos(2*b*x + 2*a)^2 + c*d*sin(2*b*x + 2*a)^2 + 2*c*d*cos(2*b*x + 2 
*a) + c*d)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*sin(b*x + a) + 1))/(b^2 
*cos(2*b*x + 2*a)^2 + b^2*sin(2*b*x + 2*a)^2 + 2*b^2*cos(2*b*x + 2*a) + b^ 
2)
                                                                                    
                                                                                    
 

Giac [F]

\[ \int (c+d x)^2 \sec (a+b x) \tan (a+b x) \, dx=\int { {\left (d x + c\right )}^{2} \sec \left (b x + a\right ) \tan \left (b x + a\right ) \,d x } \] Input:

integrate((d*x+c)^2*sec(b*x+a)*tan(b*x+a),x, algorithm="giac")
 

Output:

integrate((d*x + c)^2*sec(b*x + a)*tan(b*x + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^2 \sec (a+b x) \tan (a+b x) \, dx=\int \frac {\mathrm {tan}\left (a+b\,x\right )\,{\left (c+d\,x\right )}^2}{\cos \left (a+b\,x\right )} \,d x \] Input:

int((tan(a + b*x)*(c + d*x)^2)/cos(a + b*x),x)
 

Output:

int((tan(a + b*x)*(c + d*x)^2)/cos(a + b*x), x)
 

Reduce [F]

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

int((d*x+c)^2*sec(b*x+a)*tan(b*x+a),x)
 

Output:

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