Integrand size = 23, antiderivative size = 78 \[ \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=-\frac {(a-b) x}{2 (a+b)^2}-\frac {\sqrt {a} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{(a+b)^2 d}+\frac {\cosh (c+d x) \sinh (c+d x)}{2 (a+b) d} \] Output:
-1/2*(a-b)*x/(a+b)^2-a^(1/2)*b^(1/2)*arctan(b^(1/2)*tanh(d*x+c)/a^(1/2))/( a+b)^2/d+1/2*cosh(d*x+c)*sinh(d*x+c)/(a+b)/d
Time = 0.40 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.86 \[ \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {-2 (a-b) (c+d x)-4 \sqrt {a} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )+(a+b) \sinh (2 (c+d x))}{4 (a+b)^2 d} \] Input:
Integrate[Sinh[c + d*x]^2/(a + b*Tanh[c + d*x]^2),x]
Output:
(-2*(a - b)*(c + d*x) - 4*Sqrt[a]*Sqrt[b]*ArcTan[(Sqrt[b]*Tanh[c + d*x])/S qrt[a]] + (a + b)*Sinh[2*(c + d*x)])/(4*(a + b)^2*d)
Time = 0.52 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.24, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 25, 4146, 373, 397, 218, 219}
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.
\(\displaystyle \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin (i c+i d x)^2}{a-b \tan (i c+i d x)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\sin (i c+i d x)^2}{a-b \tan (i c+i d x)^2}dx\) |
\(\Big \downarrow \) 4146 |
\(\displaystyle \frac {\int \frac {\tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right )^2 \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 373 |
\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right )}-\frac {\int \frac {a-b \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{2 (a+b)}}{d}\) |
\(\Big \downarrow \) 397 |
\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {(a-b) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a+b}+\frac {2 a b \int \frac {1}{b \tanh ^2(c+d x)+a}d\tanh (c+d x)}{a+b}}{2 (a+b)}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {(a-b) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a+b}+\frac {2 \sqrt {a} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a+b}}{2 (a+b)}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {2 \sqrt {a} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a+b}+\frac {(a-b) \text {arctanh}(\tanh (c+d x))}{a+b}}{2 (a+b)}}{d}\) |
Input:
Int[Sinh[c + d*x]^2/(a + b*Tanh[c + d*x]^2),x]
Output:
(-1/2*((2*Sqrt[a]*Sqrt[b]*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(a + b) + ((a - b)*ArcTanh[Tanh[c + d*x]])/(a + b))/(a + b) + Tanh[c + d*x]/(2*(a + b)*(1 - Tanh[c + d*x]^2)))/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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1)) Int[(e *x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[sin[(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[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
Leaf count of result is larger than twice the leaf count of optimal. \(150\) vs. \(2(66)=132\).
Time = 2.62 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.94
method | result | size |
risch | \(-\frac {a x}{2 \left (a +b \right )^{2}}+\frac {x b}{2 \left (a +b \right )^{2}}+\frac {{\mathrm e}^{2 d x +2 c}}{8 \left (a +b \right ) d}-\frac {{\mathrm e}^{-2 d x -2 c}}{8 \left (a +b \right ) d}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}-a +b}{a +b}\right )}{2 \left (a +b \right )^{2} d}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}+a -b}{a +b}\right )}{2 \left (a +b \right )^{2} d}\) | \(151\) |
derivativedivides | \(\frac {\frac {4}{\left (8 a +8 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (a -b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 \left (a +b \right )^{2}}-\frac {4}{\left (8 a +8 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-a +b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 \left (a +b \right )^{2}}+\frac {2 a^{2} b \left (\frac {\left (a +\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (-a +\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{\left (a +b \right )^{2}}}{d}\) | \(310\) |
default | \(\frac {\frac {4}{\left (8 a +8 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (a -b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 \left (a +b \right )^{2}}-\frac {4}{\left (8 a +8 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-a +b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 \left (a +b \right )^{2}}+\frac {2 a^{2} b \left (\frac {\left (a +\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (-a +\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{\left (a +b \right )^{2}}}{d}\) | \(310\) |
Input:
int(sinh(d*x+c)^2/(a+tanh(d*x+c)^2*b),x,method=_RETURNVERBOSE)
Output:
-1/2*a*x/(a+b)^2+1/2*x/(a+b)^2*b+1/8/(a+b)/d*exp(2*d*x+2*c)-1/8/(a+b)/d*ex p(-2*d*x-2*c)+1/2*(-a*b)^(1/2)/(a+b)^2/d*ln(exp(2*d*x+2*c)-(2*(-a*b)^(1/2) -a+b)/(a+b))-1/2*(-a*b)^(1/2)/(a+b)^2/d*ln(exp(2*d*x+2*c)+(2*(-a*b)^(1/2)+ a-b)/(a+b))
Leaf count of result is larger than twice the leaf count of optimal. 306 vs. \(2 (66) = 132\).
Time = 0.14 (sec) , antiderivative size = 916, normalized size of antiderivative = 11.74 \[ \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(sinh(d*x+c)^2/(a+b*tanh(d*x+c)^2),x, algorithm="fricas")
Output:
[-1/8*(4*(a - b)*d*x*cosh(d*x + c)^2 - (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*(2*(a - b)*d* x - 3*(a + b)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 4*sqrt(-a*b)*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)*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)*co sh(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)*si nh(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)*cosh(d*x + c)^3 + (a - b)*cosh(d*x + c))*sinh(d*x + c) + a + b)) + 4*(2*(a - b)*d*x*cosh(d*x + c) - (a + b)*cos h(d*x + c)^3)*sinh(d*x + c) + a + b)/((a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^ 2 + 2*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)*sinh(d*x + c) + (a^2 + 2*a*b + b ^2)*d*sinh(d*x + c)^2), -1/8*(4*(a - b)*d*x*cosh(d*x + c)^2 - (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*(2*(a - b)*d*x - 3*(a + b)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 8* sqrt(a*b)*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x +...
\[ \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\int \frac {\sinh ^{2}{\left (c + d x \right )}}{a + b \tanh ^{2}{\left (c + d x \right )}}\, dx \] Input:
integrate(sinh(d*x+c)**2/(a+b*tanh(d*x+c)**2),x)
Output:
Integral(sinh(c + d*x)**2/(a + b*tanh(c + d*x)**2), x)
Leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (66) = 132\).
Time = 0.14 (sec) , antiderivative size = 316, normalized size of antiderivative = 4.05 \[ \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {b \log \left ({\left (a + b\right )} e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a - b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d} - \frac {b \log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d} - \frac {{\left (a b - b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b} d} + \frac {{\left (a b - b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b} d} + \frac {b \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{2 \, \sqrt {a b} {\left (a + b\right )} d} - \frac {d x + c}{2 \, {\left (a + b\right )} d} + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{8 \, {\left (a + b\right )} d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, {\left (a + b\right )} d} \] Input:
integrate(sinh(d*x+c)^2/(a+b*tanh(d*x+c)^2),x, algorithm="maxima")
Output:
1/4*b*log((a + b)*e^(4*d*x + 4*c) + 2*(a - b)*e^(2*d*x + 2*c) + a + b)/((a ^2 + 2*a*b + b^2)*d) - 1/4*b*log(2*(a - b)*e^(-2*d*x - 2*c) + (a + b)*e^(- 4*d*x - 4*c) + a + b)/((a^2 + 2*a*b + b^2)*d) - 1/4*(a*b - b^2)*arctan(1/2 *((a + b)*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^2 + 2*a*b + b^2)*sqrt(a* b)*d) + 1/4*(a*b - b^2)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt (a*b))/((a^2 + 2*a*b + b^2)*sqrt(a*b)*d) + 1/2*b*arctan(1/2*((a + b)*e^(-2 *d*x - 2*c) + a - b)/sqrt(a*b))/(sqrt(a*b)*(a + b)*d) - 1/2*(d*x + c)/((a + b)*d) + 1/8*e^(2*d*x + 2*c)/((a + b)*d) - 1/8*e^(-2*d*x - 2*c)/((a + b)* d)
Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (66) = 132\).
Time = 0.61 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.04 \[ \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=-\frac {\frac {8 \, a b \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b}} + \frac {4 \, {\left (d x + c\right )} {\left (a - b\right )}}{a^{2} + 2 \, a b + b^{2}} - \frac {{\left (2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} - a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{a^{2} + 2 \, a b + b^{2}} - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{a + b}}{8 \, d} \] Input:
integrate(sinh(d*x+c)^2/(a+b*tanh(d*x+c)^2),x, algorithm="giac")
Output:
-1/8*(8*a*b*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sqr t(a*b))/((a^2 + 2*a*b + b^2)*sqrt(a*b)) + 4*(d*x + c)*(a - b)/(a^2 + 2*a*b + b^2) - (2*a*e^(2*d*x + 2*c) - 2*b*e^(2*d*x + 2*c) - a - b)*e^(-2*d*x - 2*c)/(a^2 + 2*a*b + b^2) - e^(2*d*x + 2*c)/(a + b))/d
Time = 1.58 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.54 \[ \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d\,\left (a+b\right )}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d\,\left (a+b\right )}-\frac {x\,\left (a-b\right )}{2\,{\left (a+b\right )}^2}-\frac {\sqrt {-a}\,\sqrt {b}\,\ln \left (\sqrt {-a}\,b^{3/2}\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )+{\left (-a\right )}^{3/2}\,\sqrt {b}\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )-2\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{2\,d\,{\left (a+b\right )}^2}+\frac {\sqrt {-a}\,\sqrt {b}\,\ln \left (\sqrt {-a}\,b^{3/2}\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )+{\left (-a\right )}^{3/2}\,\sqrt {b}\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )+2\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{2\,d\,{\left (a+b\right )}^2} \] Input:
int(sinh(c + d*x)^2/(a + b*tanh(c + d*x)^2),x)
Output:
exp(2*c + 2*d*x)/(8*d*(a + b)) - exp(- 2*c - 2*d*x)/(8*d*(a + b)) - (x*(a - b))/(2*(a + b)^2) - ((-a)^(1/2)*b^(1/2)*log((-a)^(1/2)*b^(3/2)*(exp(2*c + 2*d*x) - 1) + (-a)^(3/2)*b^(1/2)*(exp(2*c + 2*d*x) + 1) - 2*a*b*exp(2*c + 2*d*x)))/(2*d*(a + b)^2) + ((-a)^(1/2)*b^(1/2)*log((-a)^(1/2)*b^(3/2)*(e xp(2*c + 2*d*x) - 1) + (-a)^(3/2)*b^(1/2)*(exp(2*c + 2*d*x) + 1) + 2*a*b*e xp(2*c + 2*d*x)))/(2*d*(a + b)^2)
Time = 0.23 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.14 \[ \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {-8 e^{2 d x +2 c} \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {e^{d x +c} \sqrt {a +b}-\sqrt {b}}{\sqrt {a}}\right )+8 e^{2 d x +2 c} \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {e^{d x +c} \sqrt {a +b}+\sqrt {b}}{\sqrt {a}}\right )+e^{4 d x +4 c} a +e^{4 d x +4 c} b -4 e^{2 d x +2 c} a d x +4 e^{2 d x +2 c} b d x -a -b}{8 e^{2 d x +2 c} d \left (a^{2}+2 a b +b^{2}\right )} \] Input:
int(sinh(d*x+c)^2/(a+b*tanh(d*x+c)^2),x)
Output:
( - 8*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sq rt(b))/sqrt(a)) + 8*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sq rt(a + b) + sqrt(b))/sqrt(a)) + e**(4*c + 4*d*x)*a + e**(4*c + 4*d*x)*b - 4*e**(2*c + 2*d*x)*a*d*x + 4*e**(2*c + 2*d*x)*b*d*x - a - b)/(8*e**(2*c + 2*d*x)*d*(a**2 + 2*a*b + b**2))