Integrand size = 23, antiderivative size = 32 \[ \int \frac {\text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {\arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} d} \] Output:
arctan(b^(1/2)*tanh(d*x+c)/a^(1/2))/a^(1/2)/b^(1/2)/d
Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {\text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {\arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} d} \] Input:
Integrate[Sech[c + d*x]^2/(a + b*Tanh[c + d*x]^2),x]
Output:
ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]]/(Sqrt[a]*Sqrt[b]*d)
Time = 0.41 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 4158, 218}
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.
\(\displaystyle \int \frac {\text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (i c+i d x)^2}{a-b \tan (i c+i d x)^2}dx\) |
\(\Big \downarrow \) 4158 |
\(\displaystyle \frac {\int \frac {1}{b \tanh ^2(c+d x)+a}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} d}\) |
Input:
Int[Sech[c + d*x]^2/(a + b*Tanh[c + d*x]^2),x]
Output:
ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]]/(Sqrt[a]*Sqrt[b]*d)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/(c^(m - 1)*f) Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)^n)^ p, x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && I ntegerQ[m/2] && (IntegersQ[n, p] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Leaf count of result is larger than twice the leaf count of optimal. \(113\) vs. \(2(24)=48\).
Time = 6.20 (sec) , antiderivative size = 114, normalized size of antiderivative = 3.56
method | result | size |
risch | \(-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {-a b}-b \sqrt {-a b}-2 a b}{\left (a +b \right ) \sqrt {-a b}}\right )}{2 \sqrt {-a b}\, d}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {-a b}-b \sqrt {-a b}+2 a b}{\left (a +b \right ) \sqrt {-a b}}\right )}{2 \sqrt {-a b}\, d}\) | \(114\) |
derivativedivides | \(\frac {2 a \left (\frac {\left (-a -\sqrt {\left (a +b \right ) b}-b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (a -\sqrt {\left (a +b \right ) b}+b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{d}\) | \(160\) |
default | \(\frac {2 a \left (\frac {\left (-a -\sqrt {\left (a +b \right ) b}-b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (a -\sqrt {\left (a +b \right ) b}+b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{d}\) | \(160\) |
Input:
int(sech(d*x+c)^2/(a+tanh(d*x+c)^2*b),x,method=_RETURNVERBOSE)
Output:
-1/2/(-a*b)^(1/2)/d*ln(exp(2*d*x+2*c)+(a*(-a*b)^(1/2)-b*(-a*b)^(1/2)-2*a*b )/(a+b)/(-a*b)^(1/2))+1/2/(-a*b)^(1/2)/d*ln(exp(2*d*x+2*c)+(a*(-a*b)^(1/2) -b*(-a*b)^(1/2)+2*a*b)/(a+b)/(-a*b)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (24) = 48\).
Time = 0.11 (sec) , antiderivative size = 455, normalized size of antiderivative = 14.22 \[ \int \frac {\text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\left [-\frac {\sqrt {-a b} \log \left (\frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{4} + 4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{4} + 2 \, {\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} + a^{2} - b^{2}\right )} \sinh \left (d x + c\right )^{2} + a^{2} - 6 \, a b + b^{2} + 4 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{3} + {\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 4 \, {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a + b\right )} \sinh \left (d x + c\right )^{2} + a - b\right )} \sqrt {-a b}}{{\left (a + b\right )} \cosh \left (d x + c\right )^{4} + 4 \, {\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (a + b\right )} \sinh \left (d x + c\right )^{4} + 2 \, {\left (a - b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (a + b\right )} \cosh \left (d x + c\right )^{2} + a - b\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{3} + {\left (a - b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a + b}\right )}{2 \, a b d}, \frac {\sqrt {a b} \arctan \left (\frac {{\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a + b\right )} \sinh \left (d x + c\right )^{2} + a - b\right )} \sqrt {a b}}{2 \, a b}\right )}{a b d}\right ] \] Input:
integrate(sech(d*x+c)^2/(a+b*tanh(d*x+c)^2),x, algorithm="fricas")
Output:
[-1/2*sqrt(-a*b)*log(((a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^ 4 + 2*(a^2 - b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(d*x + c) ^2 + a^2 - b^2)*sinh(d*x + c)^2 + a^2 - 6*a*b + b^2 + 4*((a^2 + 2*a*b + b^ 2)*cosh(d*x + c)^3 + (a^2 - b^2)*cosh(d*x + c))*sinh(d*x + c) - 4*((a + b) *cosh(d*x + c)^2 + 2*(a + b)*cosh(d*x + c)*sinh(d*x + c) + (a + b)*sinh(d* x + c)^2 + a - b)*sqrt(-a*b))/((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d* x + c)*sinh(d*x + c)^3 + (a + b)*sinh(d*x + c)^4 + 2*(a - b)*cosh(d*x + c) ^2 + 2*(3*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^2 + 4*((a + b)*co sh(d*x + c)^3 + (a - b)*cosh(d*x + c))*sinh(d*x + c) + a + b))/(a*b*d), sq rt(a*b)*arctan(1/2*((a + b)*cosh(d*x + c)^2 + 2*(a + b)*cosh(d*x + c)*sinh (d*x + c) + (a + b)*sinh(d*x + c)^2 + a - b)*sqrt(a*b)/(a*b))/(a*b*d)]
\[ \int \frac {\text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\int \frac {\operatorname {sech}^{2}{\left (c + d x \right )}}{a + b \tanh ^{2}{\left (c + d x \right )}}\, dx \] Input:
integrate(sech(d*x+c)**2/(a+b*tanh(d*x+c)**2),x)
Output:
Integral(sech(c + d*x)**2/(a + b*tanh(c + d*x)**2), x)
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {\text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=-\frac {\arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{\sqrt {a b} d} \] Input:
integrate(sech(d*x+c)^2/(a+b*tanh(d*x+c)^2),x, algorithm="maxima")
Output:
-arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b))/(sqrt(a*b)*d)
Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {\text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {\arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{\sqrt {a b} d} \] Input:
integrate(sech(d*x+c)^2/(a+b*tanh(d*x+c)^2),x, algorithm="giac")
Output:
arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/(sqr t(a*b)*d)
Time = 2.57 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.53 \[ \int \frac {\text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {\mathrm {atan}\left (\frac {a\,\sqrt {a\,b\,d^2}-b\,\sqrt {a\,b\,d^2}+a\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\sqrt {a\,b\,d^2}+b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\sqrt {a\,b\,d^2}}{2\,a\,b\,d}\right )}{\sqrt {a\,b\,d^2}} \] Input:
int(1/(cosh(c + d*x)^2*(a + b*tanh(c + d*x)^2)),x)
Output:
atan((a*(a*b*d^2)^(1/2) - b*(a*b*d^2)^(1/2) + a*exp(2*c)*exp(2*d*x)*(a*b*d ^2)^(1/2) + b*exp(2*c)*exp(2*d*x)*(a*b*d^2)^(1/2))/(2*a*b*d))/(a*b*d^2)^(1 /2)
Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.91 \[ \int \frac {\text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {\sqrt {b}\, \sqrt {a}\, \left (\mathit {atan} \left (\frac {e^{d x +c} \sqrt {a +b}-\sqrt {b}}{\sqrt {a}}\right )-\mathit {atan} \left (\frac {e^{d x +c} \sqrt {a +b}+\sqrt {b}}{\sqrt {a}}\right )\right )}{a b d} \] Input:
int(sech(d*x+c)^2/(a+b*tanh(d*x+c)^2),x)
Output:
(sqrt(b)*sqrt(a)*(atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a)) - ata n((e**(c + d*x)*sqrt(a + b) + sqrt(b))/sqrt(a))))/(a*b*d)