Integrand size = 12, antiderivative size = 61 \[ \int (c+d x) \text {sech}(a+b x) \, dx=\frac {2 (c+d x) \arctan \left (e^{a+b x}\right )}{b}-\frac {i d \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}+\frac {i d \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b^2} \]
2*(d*x+c)*arctan(exp(b*x+a))/b-I*d*polylog(2,-I*exp(b*x+a))/b^2+I*d*polylo g(2,I*exp(b*x+a))/b^2
Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.38 \[ \int (c+d x) \text {sech}(a+b x) \, dx=\frac {c \arctan (\sinh (a+b x))}{b}+\frac {i d \left (b x \left (\log \left (1-i e^{a+b x}\right )-\log \left (1+i e^{a+b x}\right )\right )-\operatorname {PolyLog}\left (2,-i e^{a+b x}\right )+\operatorname {PolyLog}\left (2,i e^{a+b x}\right )\right )}{b^2} \]
(c*ArcTan[Sinh[a + b*x]])/b + (I*d*(b*x*(Log[1 - I*E^(a + b*x)] - Log[1 + I*E^(a + b*x)]) - PolyLog[2, (-I)*E^(a + b*x)] + PolyLog[2, I*E^(a + b*x)] ))/b^2
Time = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4668, 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.
\(\displaystyle \int (c+d x) \text {sech}(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (c+d x) \csc \left (i a+i b x+\frac {\pi }{2}\right )dx\) |
\(\Big \downarrow \) 4668 |
\(\displaystyle -\frac {i d \int \log \left (1-i e^{a+b x}\right )dx}{b}+\frac {i d \int \log \left (1+i e^{a+b x}\right )dx}{b}+\frac {2 (c+d x) \arctan \left (e^{a+b x}\right )}{b}\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle -\frac {i d \int e^{-a-b x} \log \left (1-i e^{a+b x}\right )de^{a+b x}}{b^2}+\frac {i d \int e^{-a-b x} \log \left (1+i e^{a+b x}\right )de^{a+b x}}{b^2}+\frac {2 (c+d x) \arctan \left (e^{a+b x}\right )}{b}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {2 (c+d x) \arctan \left (e^{a+b x}\right )}{b}-\frac {i d \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}+\frac {i d \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b^2}\) |
(2*(c + d*x)*ArcTan[E^(a + b*x)])/b - (I*d*PolyLog[2, (-I)*E^(a + b*x)])/b ^2 + (I*d*PolyLog[2, I*E^(a + b*x)])/b^2
3.1.3.3.1 Defintions of rubi rules used
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]
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]
Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_ ))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^( I*k*Pi)]/(f*fz*I)), x] + (-Simp[d*(m/(f*fz*I)) Int[(c + d*x)^(m - 1)*Log[ 1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Simp[d*(m/(f*fz*I)) Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c , d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]
Time = 0.31 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.66
method | result | size |
derivativedivides | \(\frac {\frac {d \left (i \left (b x +a \right ) \left (\ln \left (1-i {\mathrm e}^{b x +a}\right )-\ln \left (1+i {\mathrm e}^{b x +a}\right )\right )-i \operatorname {dilog}\left (1+i {\mathrm e}^{b x +a}\right )+i \operatorname {dilog}\left (1-i {\mathrm e}^{b x +a}\right )\right )}{b}-\frac {2 d a \arctan \left ({\mathrm e}^{b x +a}\right )}{b}+2 c \arctan \left ({\mathrm e}^{b x +a}\right )}{b}\) | \(101\) |
default | \(\frac {\frac {d \left (i \left (b x +a \right ) \left (\ln \left (1-i {\mathrm e}^{b x +a}\right )-\ln \left (1+i {\mathrm e}^{b x +a}\right )\right )-i \operatorname {dilog}\left (1+i {\mathrm e}^{b x +a}\right )+i \operatorname {dilog}\left (1-i {\mathrm e}^{b x +a}\right )\right )}{b}-\frac {2 d a \arctan \left ({\mathrm e}^{b x +a}\right )}{b}+2 c \arctan \left ({\mathrm e}^{b x +a}\right )}{b}\) | \(101\) |
parts | \(\frac {\arctan \left (\sinh \left (b x +a \right )\right ) d x}{b}+\frac {\arctan \left (\sinh \left (b x +a \right )\right ) c}{b}-\frac {d \left (x \arctan \left (\sinh \left (b x +a \right )\right )-\frac {i \left (b x +a \right ) \left (\ln \left (1-i {\mathrm e}^{b x +a}\right )-\ln \left (1+i {\mathrm e}^{b x +a}\right )\right )-i \operatorname {dilog}\left (1+i {\mathrm e}^{b x +a}\right )+i \operatorname {dilog}\left (1-i {\mathrm e}^{b x +a}\right )-2 a \arctan \left ({\mathrm e}^{b x +a}\right )}{b}\right )}{b}\) | \(124\) |
risch | \(\frac {2 c \arctan \left ({\mathrm e}^{b x +a}\right )}{b}-\frac {i d \ln \left (1+i {\mathrm e}^{b x +a}\right ) x}{b}-\frac {i d \ln \left (1+i {\mathrm e}^{b x +a}\right ) a}{b^{2}}+\frac {i d \ln \left (1-i {\mathrm e}^{b x +a}\right ) x}{b}+\frac {i d \ln \left (1-i {\mathrm e}^{b x +a}\right ) a}{b^{2}}-\frac {i d \operatorname {dilog}\left (1+i {\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {i d \operatorname {dilog}\left (1-i {\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {2 d a \arctan \left ({\mathrm e}^{b x +a}\right )}{b^{2}}\) | \(147\) |
1/b*(d/b*(I*(b*x+a)*(ln(1-I*exp(b*x+a))-ln(1+I*exp(b*x+a)))-I*dilog(1+I*ex p(b*x+a))+I*dilog(1-I*exp(b*x+a)))-2*d/b*a*arctan(exp(b*x+a))+2*c*arctan(e xp(b*x+a)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 157 vs. \(2 (48) = 96\).
Time = 0.26 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.57 \[ \int (c+d x) \text {sech}(a+b x) \, dx=\frac {i \, d {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - i \, d {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + {\left (i \, b c - i \, a d\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + {\left (-i \, b c + i \, a d\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) + {\left (-i \, b d x - i \, a d\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) + {\left (i \, b d x + i \, a d\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right )}{b^{2}} \]
(I*d*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - I*d*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + (I*b*c - I*a*d)*log(cosh(b*x + a) + sinh(b*x + a) + I) + (-I*b*c + I*a*d)*log(cosh(b*x + a) + sinh(b*x + a) - I) + (-I*b*d*x - I*a*d)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) + (I*b*d*x + I*a*d)*log (-I*cosh(b*x + a) - I*sinh(b*x + a) + 1))/b^2
\[ \int (c+d x) \text {sech}(a+b x) \, dx=\int \left (c + d x\right ) \operatorname {sech}{\left (a + b x \right )}\, dx \]
\[ \int (c+d x) \text {sech}(a+b x) \, dx=\int { {\left (d x + c\right )} \operatorname {sech}\left (b x + a\right ) \,d x } \]
\[ \int (c+d x) \text {sech}(a+b x) \, dx=\int { {\left (d x + c\right )} \operatorname {sech}\left (b x + a\right ) \,d x } \]
Timed out. \[ \int (c+d x) \text {sech}(a+b x) \, dx=\int \frac {c+d\,x}{\mathrm {cosh}\left (a+b\,x\right )} \,d x \]