\(\int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx\) [27]

Optimal result
Mathematica [B] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [B] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 23, antiderivative size = 75 \[ \int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=-\frac {(a+2 b) x}{2 a^2}+\frac {\sqrt {b} \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a^2 d}+\frac {\cosh (c+d x) \sinh (c+d x)}{2 a d} \] Output:

-1/2*(a+2*b)*x/a^2+b^(1/2)*(a+b)^(1/2)*arctanh(b^(1/2)*tanh(d*x+c)/(a+b)^( 
1/2))/a^2/d+1/2*cosh(d*x+c)*sinh(d*x+c)/a/d
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(236\) vs. \(2(75)=150\).

Time = 0.61 (sec) , antiderivative size = 236, normalized size of antiderivative = 3.15 \[ \int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x))) \text {sech}^2(c+d x) \left (-\frac {\text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{\sqrt {b} \sqrt {a+b} d}+\frac {-4 (a+2 b) x+\frac {\left (a^2+8 a b+8 b^2\right ) \text {arctanh}\left (\frac {\text {sech}(d x) (\cosh (2 c)-\sinh (2 c)) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right ) (\cosh (2 c)-\sinh (2 c))}{\sqrt {a+b} d \sqrt {b (\cosh (c)-\sinh (c))^4}}+\frac {2 a \cosh (2 d x) \sinh (2 c)}{d}+\frac {2 a \cosh (2 c) \sinh (2 d x)}{d}}{a^2}\right )}{16 \left (a+b \text {sech}^2(c+d x)\right )} \] Input:

Integrate[Sinh[c + d*x]^2/(a + b*Sech[c + d*x]^2),x]
 

Output:

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(-(ArcTanh[(Sqrt[b]*Tanh[ 
c + d*x])/Sqrt[a + b]]/(Sqrt[b]*Sqrt[a + b]*d)) + (-4*(a + 2*b)*x + ((a^2 
+ 8*a*b + 8*b^2)*ArcTanh[(Sech[d*x]*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*Sin 
h[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4]) 
]*(Cosh[2*c] - Sinh[2*c]))/(Sqrt[a + b]*d*Sqrt[b*(Cosh[c] - Sinh[c])^4]) + 
 (2*a*Cosh[2*d*x]*Sinh[2*c])/d + (2*a*Cosh[2*c]*Sinh[2*d*x])/d)/a^2))/(16* 
(a + b*Sech[c + d*x]^2))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 93, 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, 4620, 373, 397, 219, 221}

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.

\(\displaystyle \int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {\sin (i c+i d x)^2}{a+b \sec (i c+i d x)^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\sin (i c+i d x)^2}{b \sec (i c+i d x)^2+a}dx\)

\(\Big \downarrow \) 4620

\(\displaystyle \frac {\int \frac {\tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right )^2 \left (-b \tanh ^2(c+d x)+a+b\right )}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 373

\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 a \left (1-\tanh ^2(c+d x)\right )}-\frac {\int \frac {b \tanh ^2(c+d x)+a+b}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )}d\tanh (c+d x)}{2 a}}{d}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 a \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {(a+2 b) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a}-\frac {2 b (a+b) \int \frac {1}{-b \tanh ^2(c+d x)+a+b}d\tanh (c+d x)}{a}}{2 a}}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 a \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {(a+2 b) \text {arctanh}(\tanh (c+d x))}{a}-\frac {2 b (a+b) \int \frac {1}{-b \tanh ^2(c+d x)+a+b}d\tanh (c+d x)}{a}}{2 a}}{d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 a \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {(a+2 b) \text {arctanh}(\tanh (c+d x))}{a}-\frac {2 \sqrt {b} \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a}}{2 a}}{d}\)

Input:

Int[Sinh[c + d*x]^2/(a + b*Sech[c + d*x]^2),x]
 

Output:

(-1/2*(((a + 2*b)*ArcTanh[Tanh[c + d*x]])/a - (2*Sqrt[b]*Sqrt[a + b]*ArcTa 
nh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/a)/a + Tanh[c + d*x]/(2*a*(1 - Ta 
nh[c + d*x]^2)))/d
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 219
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])
 

rule 221
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 373
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]
 

rule 397
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]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4620
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ 
)]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m 
+ 1)/f   Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f 
f^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, 
 x] && IntegerQ[m/2] && IntegerQ[n/2]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(147\) vs. \(2(63)=126\).

Time = 9.20 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.97

method result size
risch \(-\frac {x}{2 a}-\frac {x b}{a^{2}}+\frac {{\mathrm e}^{2 d x +2 c}}{8 a d}-\frac {{\mathrm e}^{-2 d x -2 c}}{8 a d}+\frac {\sqrt {a b +b^{2}}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-a +2 \sqrt {a b +b^{2}}-2 b}{a}\right )}{2 d \,a^{2}}-\frac {\sqrt {a b +b^{2}}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a +2 \sqrt {a b +b^{2}}+2 b}{a}\right )}{2 d \,a^{2}}\) \(148\)
derivativedivides \(\frac {-\frac {1}{2 a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-a -2 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (a +2 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{2}}-\frac {2 b \left (a +b \right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{a^{2}}}{d}\) \(230\)
default \(\frac {-\frac {1}{2 a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-a -2 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (a +2 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{2}}-\frac {2 b \left (a +b \right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{a^{2}}}{d}\) \(230\)

Input:

int(sinh(d*x+c)^2/(a+b*sech(d*x+c)^2),x,method=_RETURNVERBOSE)
 

Output:

-1/2*x/a-x*b/a^2+1/8/a/d*exp(2*d*x+2*c)-1/8/a/d*exp(-2*d*x-2*c)+1/2*(a*b+b 
^2)^(1/2)/d/a^2*ln(exp(2*d*x+2*c)-(-a+2*(a*b+b^2)^(1/2)-2*b)/a)-1/2*(a*b+b 
^2)^(1/2)/d/a^2*ln(exp(2*d*x+2*c)+(a+2*(a*b+b^2)^(1/2)+2*b)/a)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (63) = 126\).

Time = 0.17 (sec) , antiderivative size = 805, normalized size of antiderivative = 10.73 \[ \int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(sinh(d*x+c)^2/(a+b*sech(d*x+c)^2),x, algorithm="fricas")
 

Output:

[-1/8*(4*(a + 2*b)*d*x*cosh(d*x + c)^2 - a*cosh(d*x + c)^4 - 4*a*cosh(d*x 
+ c)*sinh(d*x + c)^3 - a*sinh(d*x + c)^4 + 2*(2*(a + 2*b)*d*x - 3*a*cosh(d 
*x + c)^2)*sinh(d*x + c)^2 - 4*sqrt(a*b + b^2)*(cosh(d*x + c)^2 + 2*cosh(d 
*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)*log((a^2*cosh(d*x + c)^4 + 4*a^2* 
cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b)*cosh 
(d*x + c)^2 + 2*(3*a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^2 + a^ 
2 + 8*a*b + 8*b^2 + 4*(a^2*cosh(d*x + c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))* 
sinh(d*x + c) - 4*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a 
*sinh(d*x + c)^2 + a + 2*b)*sqrt(a*b + b^2))/(a*cosh(d*x + c)^4 + 4*a*cosh 
(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^ 
2 + 2*(3*a*cosh(d*x + c)^2 + a + 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c) 
^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a)) + 4*(2*(a + 2*b)*d*x*cos 
h(d*x + c) - a*cosh(d*x + c)^3)*sinh(d*x + c) + a)/(a^2*d*cosh(d*x + c)^2 
+ 2*a^2*d*cosh(d*x + c)*sinh(d*x + c) + a^2*d*sinh(d*x + c)^2), -1/8*(4*(a 
 + 2*b)*d*x*cosh(d*x + c)^2 - a*cosh(d*x + c)^4 - 4*a*cosh(d*x + c)*sinh(d 
*x + c)^3 - a*sinh(d*x + c)^4 + 2*(2*(a + 2*b)*d*x - 3*a*cosh(d*x + c)^2)* 
sinh(d*x + c)^2 - 8*sqrt(-a*b - b^2)*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*si 
nh(d*x + c) + sinh(d*x + c)^2)*arctan(1/2*(a*cosh(d*x + c)^2 + 2*a*cosh(d* 
x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(-a*b - b^2)/(a*b 
+ b^2)) + 4*(2*(a + 2*b)*d*x*cosh(d*x + c) - a*cosh(d*x + c)^3)*sinh(d*...
 

Sympy [F]

\[ \int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\int \frac {\sinh ^{2}{\left (c + d x \right )}}{a + b \operatorname {sech}^{2}{\left (c + d x \right )}}\, dx \] Input:

integrate(sinh(d*x+c)**2/(a+b*sech(d*x+c)**2),x)
 

Output:

Integral(sinh(c + d*x)**2/(a + b*sech(c + d*x)**2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (63) = 126\).

Time = 0.15 (sec) , antiderivative size = 352, normalized size of antiderivative = 4.69 \[ \int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=-\frac {b \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{4 \, \sqrt {{\left (a + b\right )} b} a d} - \frac {d x + c}{2 \, a d} + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{8 \, a d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, a d} - \frac {b \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a + 2 \, b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a\right )}{4 \, a^{2} d} + \frac {b \log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{4 \, a^{2} d} + \frac {{\left (a b + 2 \, b^{2}\right )} \log \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{8 \, \sqrt {{\left (a + b\right )} b} a^{2} d} - \frac {{\left (a b + 2 \, b^{2}\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{8 \, \sqrt {{\left (a + b\right )} b} a^{2} d} \] Input:

integrate(sinh(d*x+c)^2/(a+b*sech(d*x+c)^2),x, algorithm="maxima")
 

Output:

-1/4*b*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(-2*d*x 
 - 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/(sqrt((a + b)*b)*a*d) - 1/2*(d*x + 
 c)/(a*d) + 1/8*e^(2*d*x + 2*c)/(a*d) - 1/8*e^(-2*d*x - 2*c)/(a*d) - 1/4*b 
*log(a*e^(4*d*x + 4*c) + 2*(a + 2*b)*e^(2*d*x + 2*c) + a)/(a^2*d) + 1/4*b* 
log(2*(a + 2*b)*e^(-2*d*x - 2*c) + a*e^(-4*d*x - 4*c) + a)/(a^2*d) + 1/8*( 
a*b + 2*b^2)*log((a*e^(2*d*x + 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(2 
*d*x + 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/(sqrt((a + b)*b)*a^2*d) - 1/8* 
(a*b + 2*b^2)*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^ 
(-2*d*x - 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/(sqrt((a + b)*b)*a^2*d)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 132 vs. \(2 (63) = 126\).

Time = 0.35 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.76 \[ \int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=-\frac {\frac {4 \, {\left (d x + c\right )} {\left (a + 2 \, b\right )}}{a^{2}} - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{a} - \frac {{\left (2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} - a\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{a^{2}} - \frac {8 \, {\left (a b + b^{2}\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a^{2}}}{8 \, d} \] Input:

integrate(sinh(d*x+c)^2/(a+b*sech(d*x+c)^2),x, algorithm="giac")
 

Output:

-1/8*(4*(d*x + c)*(a + 2*b)/a^2 - e^(2*d*x + 2*c)/a - (2*a*e^(2*d*x + 2*c) 
 + 4*b*e^(2*d*x + 2*c) - a)*e^(-2*d*x - 2*c)/a^2 - 8*(a*b + b^2)*arctan(1/ 
2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2))/(sqrt(-a*b - b^2)*a^2))/ 
d
 

Mupad [B] (verification not implemented)

Time = 2.91 (sec) , antiderivative size = 276, normalized size of antiderivative = 3.68 \[ \int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,a\,d}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,a\,d}-\frac {x\,\left (a+2\,b\right )}{2\,a^2}-\frac {\sqrt {b}\,\ln \left (2\,a\,b+a^2+a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}-2\,a\,\sqrt {b}\,\sqrt {a+b}-8\,b^{3/2}\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\sqrt {a+b}+8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,a\,\sqrt {b}\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\sqrt {a+b}\right )\,\sqrt {a+b}}{2\,a^2\,d}+\frac {\sqrt {b}\,\ln \left (2\,a\,b+a^2+a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+2\,a\,\sqrt {b}\,\sqrt {a+b}+8\,b^{3/2}\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\sqrt {a+b}+8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}+4\,a\,\sqrt {b}\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\sqrt {a+b}\right )\,\sqrt {a+b}}{2\,a^2\,d} \] Input:

int(sinh(c + d*x)^2/(a + b/cosh(c + d*x)^2),x)
 

Output:

exp(2*c + 2*d*x)/(8*a*d) - exp(- 2*c - 2*d*x)/(8*a*d) - (x*(a + 2*b))/(2*a 
^2) - (b^(1/2)*log(2*a*b + a^2 + a^2*exp(2*c + 2*d*x) + 8*b^2*exp(2*c + 2* 
d*x) - 2*a*b^(1/2)*(a + b)^(1/2) - 8*b^(3/2)*exp(2*c + 2*d*x)*(a + b)^(1/2 
) + 8*a*b*exp(2*c + 2*d*x) - 4*a*b^(1/2)*exp(2*c + 2*d*x)*(a + b)^(1/2))*( 
a + b)^(1/2))/(2*a^2*d) + (b^(1/2)*log(2*a*b + a^2 + a^2*exp(2*c + 2*d*x) 
+ 8*b^2*exp(2*c + 2*d*x) + 2*a*b^(1/2)*(a + b)^(1/2) + 8*b^(3/2)*exp(2*c + 
 2*d*x)*(a + b)^(1/2) + 8*a*b*exp(2*c + 2*d*x) + 4*a*b^(1/2)*exp(2*c + 2*d 
*x)*(a + b)^(1/2))*(a + b)^(1/2))/(2*a^2*d)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.72 \[ \int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {4 e^{2 d x +2 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (-\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+e^{d x +c} \sqrt {a}\right )+4 e^{2 d x +2 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+e^{d x +c} \sqrt {a}\right )-4 e^{2 d x +2 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (2 \sqrt {b}\, \sqrt {a +b}+e^{2 d x +2 c} a +a +2 b \right )+e^{4 d x +4 c} a -4 e^{2 d x +2 c} a d x -8 e^{2 d x +2 c} b d x -a}{8 e^{2 d x +2 c} a^{2} d} \] Input:

int(sinh(d*x+c)^2/(a+b*sech(d*x+c)^2),x)
 

Output:

(4*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) 
- a - 2*b) + e**(c + d*x)*sqrt(a)) + 4*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a + b 
)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a)) - 4*e* 
*(2*c + 2*d*x)*sqrt(b)*sqrt(a + b)*log(2*sqrt(b)*sqrt(a + b) + e**(2*c + 2 
*d*x)*a + a + 2*b) + e**(4*c + 4*d*x)*a - 4*e**(2*c + 2*d*x)*a*d*x - 8*e** 
(2*c + 2*d*x)*b*d*x - a)/(8*e**(2*c + 2*d*x)*a**2*d)