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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4494, 4266, 2317, 2438} \[ \int (c+d x)^2 \sec (a+b x) \tan (a+b x) \, dx=\frac {4 i d (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^2}-\frac {2 i d^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 i d^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}+\frac {(c+d x)^2 \sec (a+b x)}{b} \]

[In]

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

[Out]

((4*I)*d*(c + d*x)*ArcTan[E^(I*(a + b*x))])/b^2 - ((2*I)*d^2*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^3 + ((2*I)*d^
2*PolyLog[2, I*E^(I*(a + b*x))])/b^3 + ((c + d*x)^2*Sec[a + b*x])/b

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2438

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 4266

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] + (-Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[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 4494

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] - Dist[d*(m/(b*n)), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(c+d x)^2 \sec (a+b x)}{b}-\frac {(2 d) \int (c+d x) \sec (a+b x) \, dx}{b} \\ & = \frac {4 i d (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^2}+\frac {(c+d x)^2 \sec (a+b x)}{b}+\frac {\left (2 d^2\right ) \int \log \left (1-i e^{i (a+b x)}\right ) \, dx}{b^2}-\frac {\left (2 d^2\right ) \int \log \left (1+i e^{i (a+b x)}\right ) \, dx}{b^2} \\ & = \frac {4 i d (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^2}+\frac {(c+d x)^2 \sec (a+b x)}{b}-\frac {\left (2 i d^2\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}+\frac {\left (2 i d^2\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.84 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.79 \[ \int (c+d x)^2 \sec (a+b x) \tan (a+b x) \, dx=\frac {-4 b c d \text {arctanh}\left (\sin (a)+\cos (a) \tan \left (\frac {b x}{2}\right )\right )-4 d^2 \arctan (\cot (a)) \text {arctanh}\left (\sin (a)+\cos (a) \tan \left (\frac {b x}{2}\right )\right )+\frac {2 d^2 \csc (a) \left ((b x-\arctan (\cot (a))) \left (\log \left (1-e^{i (b x-\arctan (\cot (a)))}\right )-\log \left (1+e^{i (b x-\arctan (\cot (a)))}\right )\right )+i \operatorname {PolyLog}\left (2,-e^{i (b x-\arctan (\cot (a)))}\right )-i \operatorname {PolyLog}\left (2,e^{i (b x-\arctan (\cot (a)))}\right )\right )}{\sqrt {\csc ^2(a)}}+b^2 (c+d x)^2 \sec (a+b x)}{b^3} \]

[In]

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

[Out]

(-4*b*c*d*ArcTanh[Sin[a] + Cos[a]*Tan[(b*x)/2]] - 4*d^2*ArcTan[Cot[a]]*ArcTanh[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

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 = 1.23 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.34

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

[In]

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

[Out]

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(
1+I*exp(I*(b*x+a)))*x+2/b^3*d^2*ln(1+I*exp(I*(b*x+a)))*a-2/b^2*d^2*ln(1-I*exp(I*(b*x+a)))*x-2/b^3*d^2*ln(1-I*e
xp(I*(b*x+a)))*a-2*I/b^3*d^2*dilog(1+I*exp(I*(b*x+a)))+2*I/b^3*d^2*dilog(1-I*exp(I*(b*x+a)))-4*I/b^3*d^2*a*arc
tan(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 = 0.29 (sec) , antiderivative size = 446, normalized size of antiderivative = 4.60 \[ \int (c+d x)^2 \sec (a+b x) \tan (a+b x) \, dx=\frac {b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + i \, d^{2} \cos \left (b x + a\right ) {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + i \, d^{2} \cos \left (b x + a\right ) {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - i \, d^{2} \cos \left (b x + a\right ) {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - i \, d^{2} \cos \left (b x + a\right ) {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - {\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - {\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right )}{b^{3} \cos \left (b x + a\right )} \]

[In]

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

[Out]

(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*
cos(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)*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*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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

(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*cos(2*b*x + 2*a)*cos(b*x + a) + x*si
n(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]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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