Integrand size = 23, antiderivative size = 75 \[ \int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=-\frac {(a+2 b) x}{2 a^2}+\frac {\sqrt {b} \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a^2 d}+\frac {\cosh (c+d x) \sinh (c+d x)}{2 a d} \]
-1/2*(a+2*b)*x/a^2+1/2*cosh(d*x+c)*sinh(d*x+c)/a/d+arctanh(b^(1/2)*tanh(d* x+c)/(a+b)^(1/2))*b^(1/2)*(a+b)^(1/2)/a^2/d
Leaf count is larger than twice the leaf count of optimal. \(236\) vs. \(2(75)=150\).
Time = 0.97 (sec) , antiderivative size = 236, normalized size of antiderivative = 3.15 \[ \int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x))) \text {sech}^2(c+d x) \left (-\frac {\text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{\sqrt {b} \sqrt {a+b} d}+\frac {-4 (a+2 b) x+\frac {\left (a^2+8 a b+8 b^2\right ) \text {arctanh}\left (\frac {\text {sech}(d x) (\cosh (2 c)-\sinh (2 c)) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right ) (\cosh (2 c)-\sinh (2 c))}{\sqrt {a+b} d \sqrt {b (\cosh (c)-\sinh (c))^4}}+\frac {2 a \cosh (2 d x) \sinh (2 c)}{d}+\frac {2 a \cosh (2 c) \sinh (2 d x)}{d}}{a^2}\right )}{16 \left (a+b \text {sech}^2(c+d x)\right )} \]
((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))
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 |
| \(\displaystyle \frac {\frac {\tanh (c+d x)}{2 a \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {(a+2 b) \text {arctanh}(\tanh (c+d x))}{a}-\frac {2 \sqrt {b} \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a}}{2 a}}{d}\) |
(-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
3.1.27.3.1 Defintions of rubi rules used
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[((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]
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[((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]
Leaf count of result is larger than twice the leaf count of optimal. \(147\) vs. \(2(63)=126\).
Time = 8.79 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.97
| method | result | size |
| risch | \(-\frac {x}{2 a}-\frac {b x}{a^{2}}+\frac {{\mathrm e}^{2 d x +2 c}}{8 a d}-\frac {{\mathrm e}^{-2 d x -2 c}}{8 a d}+\frac {\sqrt {a b +b^{2}}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-a +2 \sqrt {a b +b^{2}}-2 b}{a}\right )}{2 d \,a^{2}}-\frac {\sqrt {a b +b^{2}}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a +2 \sqrt {a b +b^{2}}+2 b}{a}\right )}{2 d \,a^{2}}\) | \(148\) |
| derivativedivides | \(\frac {-\frac {1}{2 a \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {1}{2 a \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\left (-a -2 b \right ) \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2}}-\frac {2 b \left (a +b \right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{a^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (a +2 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{2}}}{d}\) | \(230\) |
| default | \(\frac {-\frac {1}{2 a \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {1}{2 a \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\left (-a -2 b \right ) \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2}}-\frac {2 b \left (a +b \right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{a^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (a +2 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{2}}}{d}\) | \(230\) |
-1/2*x/a-b*x/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)
Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (63) = 126\).
Time = 0.28 (sec) , antiderivative size = 805, normalized size of antiderivative = 10.73 \[ \int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\text {Too large to display} \]
[-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*...
\[ \int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\int \frac {\sinh ^{2}{\left (c + d x \right )}}{a + b \operatorname {sech}^{2}{\left (c + d x \right )}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (63) = 126\).
Time = 0.28 (sec) , antiderivative size = 352, normalized size of antiderivative = 4.69 \[ \int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=-\frac {b \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{4 \, \sqrt {{\left (a + b\right )} b} a d} - \frac {d x + c}{2 \, a d} + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{8 \, a d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, a d} - \frac {b \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a + 2 \, b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a\right )}{4 \, a^{2} d} + \frac {b \log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{4 \, a^{2} d} + \frac {{\left (a b + 2 \, b^{2}\right )} \log \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{8 \, \sqrt {{\left (a + b\right )} b} a^{2} d} - \frac {{\left (a b + 2 \, b^{2}\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{8 \, \sqrt {{\left (a + b\right )} b} a^{2} d} \]
-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)
\[ \int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\int { \frac {\sinh \left (d x + c\right )^{2}}{b \operatorname {sech}\left (d x + c\right )^{2} + a} \,d x } \]
Time = 2.64 (sec) , antiderivative size = 276, normalized size of antiderivative = 3.68 \[ \int \frac {\sinh ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,a\,d}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,a\,d}-\frac {x\,\left (a+2\,b\right )}{2\,a^2}-\frac {\sqrt {b}\,\ln \left (2\,a\,b+a^2+a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}-2\,a\,\sqrt {b}\,\sqrt {a+b}-8\,b^{3/2}\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\sqrt {a+b}+8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,a\,\sqrt {b}\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\sqrt {a+b}\right )\,\sqrt {a+b}}{2\,a^2\,d}+\frac {\sqrt {b}\,\ln \left (2\,a\,b+a^2+a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+2\,a\,\sqrt {b}\,\sqrt {a+b}+8\,b^{3/2}\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\sqrt {a+b}+8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}+4\,a\,\sqrt {b}\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\sqrt {a+b}\right )\,\sqrt {a+b}}{2\,a^2\,d} \]
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)