Integrand size = 18, antiderivative size = 108 \[ \int \frac {c+d x}{a+b \coth (e+f x)} \, dx=\frac {(c+d x)^2}{2 (a+b) d}-\frac {b (c+d x) \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac {b d \operatorname {PolyLog}\left (2,\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 \left (a^2-b^2\right ) f^2} \]
1/2*(d*x+c)^2/(a+b)/d-b*(d*x+c)*ln(1+(-a+b)/(a+b)/exp(2*f*x+2*e))/(a^2-b^2 )/f+1/2*b*d*polylog(2,(a-b)/(a+b)/exp(2*f*x+2*e))/(a^2-b^2)/f^2
Time = 1.04 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.41 \[ \int \frac {c+d x}{a+b \coth (e+f x)} \, dx=\frac {1}{2} \left (\frac {2 b (c+d x)^2}{(a+b) d \left (a \left (-1+e^{2 e}\right )+b \left (1+e^{2 e}\right )\right )}-\frac {2 b (c+d x) \log \left (1+\frac {(-a+b) e^{-2 (e+f x)}}{a+b}\right )}{(a-b) (a+b) f}+\frac {b d \operatorname {PolyLog}\left (2,\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{(a-b) (a+b) f^2}+\frac {x (2 c+d x) \sinh (e)}{b \cosh (e)+a \sinh (e)}\right ) \]
((2*b*(c + d*x)^2)/((a + b)*d*(a*(-1 + E^(2*e)) + b*(1 + E^(2*e)))) - (2*b *(c + d*x)*Log[1 + (-a + b)/((a + b)*E^(2*(e + f*x)))])/((a - b)*(a + b)*f ) + (b*d*PolyLog[2, (a - b)/((a + b)*E^(2*(e + f*x)))])/((a - b)*(a + b)*f ^2) + (x*(2*c + d*x)*Sinh[e])/(b*Cosh[e] + a*Sinh[e]))/2
Time = 0.50 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4214, 25, 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.
| \(\displaystyle \int \frac {c+d x}{a+b \coth (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {c+d x}{a-i b \tan \left (i e+i f x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4214 |
| \(\displaystyle 2 b \int -\frac {e^{-2 (e+f x)} (c+d x)}{(a+b)^2-\left (a^2-b^2\right ) e^{-2 (e+f x)}}dx+\frac {(c+d x)^2}{2 d (a+b)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(c+d x)^2}{2 d (a+b)}-2 b \int \frac {e^{-2 (e+f x)} (c+d x)}{(a+b)^2-\left (a^2-b^2\right ) e^{-2 (e+f x)}}dx\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {(c+d x)^2}{2 d (a+b)}-2 b \left (\frac {(c+d x) \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 f \left (a^2-b^2\right )}-\frac {d \int \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )dx}{2 f \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle \frac {(c+d x)^2}{2 d (a+b)}-2 b \left (\frac {d \int e^{2 (e+f x)} \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )de^{-2 (e+f x)}}{4 f^2 \left (a^2-b^2\right )}+\frac {(c+d x) \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 f \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle \frac {(c+d x)^2}{2 d (a+b)}-2 b \left (\frac {(c+d x) \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 f \left (a^2-b^2\right )}-\frac {d \operatorname {PolyLog}\left (2,\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{4 f^2 \left (a^2-b^2\right )}\right )\) |
(c + d*x)^2/(2*(a + b)*d) - 2*b*(((c + d*x)*Log[1 - (a - b)/((a + b)*E^(2* (e + f*x)))])/(2*(a^2 - b^2)*f) - (d*PolyLog[2, (a - b)/((a + b)*E^(2*(e + f*x)))])/(4*(a^2 - b^2)*f^2))
3.1.54.3.1 Defintions of rubi rules used
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]
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[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + Pi*(k_.) + (f_.)*( x_)]), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(d*(m + 1)*(a + I*b)), x] + Simp [2*I*b Int[(c + d*x)^m*E^(2*I*k*Pi)*(E^Simp[2*I*(e + f*x), x]/((a + I*b)^ 2 + (a^2 + b^2)*E^(2*I*k*Pi)*E^Simp[2*I*(e + f*x), x])), x], x] /; FreeQ[{a , b, c, d, e, f}, x] && IntegerQ[4*k] && NeQ[a^2 + b^2, 0] && IGtQ[m, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(356\) vs. \(2(107)=214\).
Time = 0.36 (sec) , antiderivative size = 357, normalized size of antiderivative = 3.31
| method | result | size |
| risch | \(\frac {d \,x^{2}}{2 a +2 b}+\frac {c x}{a +b}+\frac {2 b c \ln \left ({\mathrm e}^{f x +e}\right )}{f \left (a +b \right ) \left (a -b \right )}-\frac {b c \ln \left ({\mathrm e}^{2 f x +2 e} a +b \,{\mathrm e}^{2 f x +2 e}-a +b \right )}{f \left (a +b \right ) \left (a -b \right )}+\frac {b d \,x^{2}}{\left (a +b \right ) \left (a -b \right )}-\frac {b d \ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right ) x}{f \left (a +b \right ) \left (a -b \right )}+\frac {2 b d e x}{f \left (a +b \right ) \left (a -b \right )}-\frac {b d \ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right ) e}{f^{2} \left (a +b \right ) \left (a -b \right )}+\frac {b d \,e^{2}}{f^{2} \left (a +b \right ) \left (a -b \right )}-\frac {b d \operatorname {polylog}\left (2, \frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right )}{2 f^{2} \left (a +b \right ) \left (a -b \right )}-\frac {2 b d e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2} \left (a +b \right ) \left (a -b \right )}+\frac {b d e \ln \left ({\mathrm e}^{2 f x +2 e} a +b \,{\mathrm e}^{2 f x +2 e}-a +b \right )}{f^{2} \left (a +b \right ) \left (a -b \right )}\) | \(357\) |
1/2/(a+b)*d*x^2+1/(a+b)*c*x+2/f*b/(a+b)*c/(a-b)*ln(exp(f*x+e))-1/f*b/(a+b) *c/(a-b)*ln(exp(2*f*x+2*e)*a+b*exp(2*f*x+2*e)-a+b)+b/(a+b)/(a-b)*d*x^2-1/f *b/(a+b)/(a-b)*d*ln(1-(a+b)*exp(2*f*x+2*e)/(a-b))*x+2/f*b/(a+b)/(a-b)*d*e* x-1/f^2*b/(a+b)/(a-b)*d*ln(1-(a+b)*exp(2*f*x+2*e)/(a-b))*e+1/f^2*b/(a+b)/( a-b)*d*e^2-1/2/f^2*b/(a+b)/(a-b)*d*polylog(2,(a+b)*exp(2*f*x+2*e)/(a-b))-2 /f^2*b/(a+b)*d*e/(a-b)*ln(exp(f*x+e))+1/f^2*b/(a+b)*d*e/(a-b)*ln(exp(2*f*x +2*e)*a+b*exp(2*f*x+2*e)-a+b)
Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (103) = 206\).
Time = 0.28 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.78 \[ \int \frac {c+d x}{a+b \coth (e+f x)} \, dx=\frac {{\left (a + b\right )} d f^{2} x^{2} + 2 \, {\left (a + b\right )} c f^{2} x - 2 \, b d {\rm Li}_2\left (\sqrt {\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right ) - 2 \, b d {\rm Li}_2\left (-\sqrt {\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right ) + 2 \, {\left (b d e - b c f\right )} \log \left (2 \, {\left (a + b\right )} \cosh \left (f x + e\right ) + 2 \, {\left (a + b\right )} \sinh \left (f x + e\right ) + 2 \, {\left (a - b\right )} \sqrt {\frac {a + b}{a - b}}\right ) + 2 \, {\left (b d e - b c f\right )} \log \left (2 \, {\left (a + b\right )} \cosh \left (f x + e\right ) + 2 \, {\left (a + b\right )} \sinh \left (f x + e\right ) - 2 \, {\left (a - b\right )} \sqrt {\frac {a + b}{a - b}}\right ) - 2 \, {\left (b d f x + b d e\right )} \log \left (\sqrt {\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )} + 1\right ) - 2 \, {\left (b d f x + b d e\right )} \log \left (-\sqrt {\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )} + 1\right )}{2 \, {\left (a^{2} - b^{2}\right )} f^{2}} \]
1/2*((a + b)*d*f^2*x^2 + 2*(a + b)*c*f^2*x - 2*b*d*dilog(sqrt((a + b)/(a - b))*(cosh(f*x + e) + sinh(f*x + e))) - 2*b*d*dilog(-sqrt((a + b)/(a - b)) *(cosh(f*x + e) + sinh(f*x + e))) + 2*(b*d*e - b*c*f)*log(2*(a + b)*cosh(f *x + e) + 2*(a + b)*sinh(f*x + e) + 2*(a - b)*sqrt((a + b)/(a - b))) + 2*( b*d*e - b*c*f)*log(2*(a + b)*cosh(f*x + e) + 2*(a + b)*sinh(f*x + e) - 2*( a - b)*sqrt((a + b)/(a - b))) - 2*(b*d*f*x + b*d*e)*log(sqrt((a + b)/(a - b))*(cosh(f*x + e) + sinh(f*x + e)) + 1) - 2*(b*d*f*x + b*d*e)*log(-sqrt(( a + b)/(a - b))*(cosh(f*x + e) + sinh(f*x + e)) + 1))/((a^2 - b^2)*f^2)
\[ \int \frac {c+d x}{a+b \coth (e+f x)} \, dx=\int \frac {c + d x}{a + b \coth {\left (e + f x \right )}}\, dx \]
\[ \int \frac {c+d x}{a+b \coth (e+f x)} \, dx=\int { \frac {d x + c}{b \coth \left (f x + e\right ) + a} \,d x } \]
-1/2*(4*b*integrate(-x/(a^2 - b^2 - (a^2*e^(2*e) + 2*a*b*e^(2*e) + b^2*e^( 2*e))*e^(2*f*x)), x) - x^2/(a + b))*d - c*(b*log(-(a - b)*e^(-2*f*x - 2*e) + a + b)/((a^2 - b^2)*f) - (f*x + e)/((a + b)*f))
\[ \int \frac {c+d x}{a+b \coth (e+f x)} \, dx=\int { \frac {d x + c}{b \coth \left (f x + e\right ) + a} \,d x } \]
Timed out. \[ \int \frac {c+d x}{a+b \coth (e+f x)} \, dx=\int \frac {c+d\,x}{a+b\,\mathrm {coth}\left (e+f\,x\right )} \,d x \]