Integrand size = 23, antiderivative size = 75 \[ \int \frac {\text {csch}^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=-\frac {a \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{(a+b)^{5/2} d}+\frac {a \coth (c+d x)}{(a+b)^2 d}-\frac {\coth ^3(c+d x)}{3 (a+b) d} \]
a*coth(d*x+c)/(a+b)^2/d-1/3*coth(d*x+c)^3/(a+b)/d-a*arctanh(b^(1/2)*tanh(d *x+c)/(a+b)^(1/2))*b^(1/2)/(a+b)^(5/2)/d
Leaf count is larger than twice the leaf count of optimal. \(216\) vs. \(2(75)=150\).
Time = 2.93 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.88 \[ \int \frac {\text {csch}^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x))) \text {sech}^2(c+d x) \left (3 a b \text {arctanh}\left (\frac {\text {sech}(d x) (\cosh (2 c)-\sinh (2 c)) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right ) (-\cosh (2 c)+\sinh (2 c))+\frac {1}{4} \sqrt {a+b} \text {csch}(c) \text {csch}^3(c+d x) \sqrt {b (\cosh (c)-\sinh (c))^4} (6 a \sinh (d x)-3 b \sinh (2 c+d x)+(-2 a+b) \sinh (2 c+3 d x))\right )}{6 (a+b)^{5/2} d \left (a+b \text {sech}^2(c+d x)\right ) \sqrt {b (\cosh (c)-\sinh (c))^4}} \]
((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(3*a*b*ArcTanh[(Sech[d*x] *(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqr t[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4])]*(-Cosh[2*c] + Sinh[2*c]) + (Sqrt[ a + b]*Csch[c]*Csch[c + d*x]^3*Sqrt[b*(Cosh[c] - Sinh[c])^4]*(6*a*Sinh[d*x ] - 3*b*Sinh[2*c + d*x] + (-2*a + b)*Sinh[2*c + 3*d*x]))/4))/(6*(a + b)^(5 /2)*d*(a + b*Sech[c + d*x]^2)*Sqrt[b*(Cosh[c] - Sinh[c])^4])
Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4620, 359, 264, 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.
\(\displaystyle \int \frac {\text {csch}^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (i c+i d x)^4 \left (a+b \sec (i c+i d x)^2\right )}dx\) |
\(\Big \downarrow \) 4620 |
\(\displaystyle \frac {\int \frac {\coth ^4(c+d x) \left (1-\tanh ^2(c+d x)\right )}{-b \tanh ^2(c+d x)+a+b}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {-\frac {a \int \frac {\coth ^2(c+d x)}{-b \tanh ^2(c+d x)+a+b}d\tanh (c+d x)}{a+b}-\frac {\coth ^3(c+d x)}{3 (a+b)}}{d}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {-\frac {a \left (\frac {b \int \frac {1}{-b \tanh ^2(c+d x)+a+b}d\tanh (c+d x)}{a+b}-\frac {\coth (c+d x)}{a+b}\right )}{a+b}-\frac {\coth ^3(c+d x)}{3 (a+b)}}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {-\frac {a \left (\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {\coth (c+d x)}{a+b}\right )}{a+b}-\frac {\coth ^3(c+d x)}{3 (a+b)}}{d}\) |
(-1/3*Coth[c + d*x]^3/(a + b) - (a*((Sqrt[b]*ArcTanh[(Sqrt[b]*Tanh[c + d*x ])/Sqrt[a + b]])/(a + b)^(3/2) - Coth[c + d*x]/(a + b)))/(a + b))/d
3.1.32.3.1 Defintions of rubi rules used
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
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. \(150\) vs. \(2(65)=130\).
Time = 2.29 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.01
method | result | size |
risch | \(-\frac {2 \left (3 b \,{\mathrm e}^{4 d x +4 c}+6 \,{\mathrm e}^{2 d x +2 c} a -2 a +b \right )}{3 d \left (a +b \right )^{2} \left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}+\frac {\sqrt {\left (a +b \right ) b}\, a \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {\left (a +b \right ) b}+a +2 b}{a}\right )}{2 \left (a +b \right )^{3} d}-\frac {\sqrt {\left (a +b \right ) b}\, a \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {\left (a +b \right ) b}-a -2 b}{a}\right )}{2 \left (a +b \right )^{3} d}\) | \(151\) |
derivativedivides | \(\frac {-\frac {\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a}{3}+\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b}{3}-3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 \left (a +b \right )^{2}}+\frac {2 a b \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{\left (a +b \right )^{2}}-\frac {1}{24 \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {-3 a +b}{8 \left (a +b \right )^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\) | \(212\) |
default | \(\frac {-\frac {\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a}{3}+\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b}{3}-3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 \left (a +b \right )^{2}}+\frac {2 a b \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{\left (a +b \right )^{2}}-\frac {1}{24 \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {-3 a +b}{8 \left (a +b \right )^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\) | \(212\) |
-2/3*(3*b*exp(4*d*x+4*c)+6*exp(2*d*x+2*c)*a-2*a+b)/d/(a+b)^2/(exp(2*d*x+2* c)-1)^3+1/2*((a+b)*b)^(1/2)/(a+b)^3*a/d*ln(exp(2*d*x+2*c)+(2*((a+b)*b)^(1/ 2)+a+2*b)/a)-1/2*((a+b)*b)^(1/2)/(a+b)^3*a/d*ln(exp(2*d*x+2*c)-(2*((a+b)*b )^(1/2)-a-2*b)/a)
Leaf count of result is larger than twice the leaf count of optimal. 738 vs. \(2 (65) = 130\).
Time = 0.28 (sec) , antiderivative size = 1753, normalized size of antiderivative = 23.37 \[ \int \frac {\text {csch}^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\text {Too large to display} \]
[-1/6*(12*b*cosh(d*x + c)^4 + 48*b*cosh(d*x + c)*sinh(d*x + c)^3 + 12*b*si nh(d*x + c)^4 + 24*a*cosh(d*x + c)^2 + 24*(3*b*cosh(d*x + c)^2 + a)*sinh(d *x + c)^2 - 3*(a*cosh(d*x + c)^6 + 6*a*cosh(d*x + c)*sinh(d*x + c)^5 + a*s inh(d*x + c)^6 - 3*a*cosh(d*x + c)^4 + 3*(5*a*cosh(d*x + c)^2 - a)*sinh(d* x + c)^4 + 4*(5*a*cosh(d*x + c)^3 - 3*a*cosh(d*x + c))*sinh(d*x + c)^3 + 3 *a*cosh(d*x + c)^2 + 3*(5*a*cosh(d*x + c)^4 - 6*a*cosh(d*x + c)^2 + a)*sin h(d*x + c)^2 + 6*(a*cosh(d*x + c)^5 - 2*a*cosh(d*x + c)^3 + a*cosh(d*x + c ))*sinh(d*x + c) - a)*sqrt(b/(a + b))*log((a^2*cosh(d*x + c)^4 + 4*a^2*cos h(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))*sin h(d*x + c) + 4*((a^2 + a*b)*cosh(d*x + c)^2 + 2*(a^2 + a*b)*cosh(d*x + c)* sinh(d*x + c) + (a^2 + a*b)*sinh(d*x + c)^2 + a^2 + 3*a*b + 2*b^2)*sqrt(b/ (a + b)))/(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)) + 48*(b*cosh(d*x + c)^3 + a*cosh(d*x + c))*sinh(d*x + c) - 8*a + 4*b)/((a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^6 + 6*(a^2 + 2*a*b + b^2)* d*cosh(d*x + c)*sinh(d*x + c)^5 + (a^2 + 2*a*b + b^2)*d*sinh(d*x + c)^6 - 3*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^4 + 3*(5*(a^2 + 2*a*b + b^2)*d*co...
\[ \int \frac {\text {csch}^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\int \frac {\operatorname {csch}^{4}{\left (c + d x \right )}}{a + b \operatorname {sech}^{2}{\left (c + d x \right )}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (65) = 130\).
Time = 0.30 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.60 \[ \int \frac {\text {csch}^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {a b \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {{\left (a + b\right )} b} d} - \frac {2 \, {\left (6 \, a e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, b e^{\left (-4 \, d x - 4 \, c\right )} - 2 \, a + b\right )}}{3 \, {\left (a^{2} + 2 \, a b + b^{2} - 3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - {\left (a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )} d} \]
1/2*a*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)))/((a^2 + 2*a*b + b^2)*sqrt((a + b) *b)*d) - 2/3*(6*a*e^(-2*d*x - 2*c) + 3*b*e^(-4*d*x - 4*c) - 2*a + b)/((a^2 + 2*a*b + b^2 - 3*(a^2 + 2*a*b + b^2)*e^(-2*d*x - 2*c) + 3*(a^2 + 2*a*b + b^2)*e^(-4*d*x - 4*c) - (a^2 + 2*a*b + b^2)*e^(-6*d*x - 6*c))*d)
\[ \int \frac {\text {csch}^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\int { \frac {\operatorname {csch}\left (d x + c\right )^{4}}{b \operatorname {sech}\left (d x + c\right )^{2} + a} \,d x } \]
Time = 2.84 (sec) , antiderivative size = 248, normalized size of antiderivative = 3.31 \[ \int \frac {\text {csch}^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {a\,\sqrt {b}\,\ln \left (\frac {4\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{{\left (a+b\right )}^2}-\frac {2\,\sqrt {b}\,\left (a+a\,{\mathrm {e}}^{2\,c+2\,d\,x}+2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{{\left (a+b\right )}^{5/2}}\right )}{2\,d\,{\left (a+b\right )}^{5/2}}-\frac {8}{3\,\left (a\,d+b\,d\right )\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1\right )}-\frac {2\,b}{\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )\,\left (a+b\right )\,\left (a\,d+b\,d\right )}-\frac {4}{\left (a\,d+b\,d\right )\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {a\,\sqrt {b}\,\ln \left (\frac {4\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{{\left (a+b\right )}^2}+\frac {2\,\sqrt {b}\,\left (a+a\,{\mathrm {e}}^{2\,c+2\,d\,x}+2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{{\left (a+b\right )}^{5/2}}\right )}{2\,d\,{\left (a+b\right )}^{5/2}} \]
(a*b^(1/2)*log((4*b*exp(2*c + 2*d*x))/(a + b)^2 - (2*b^(1/2)*(a + a*exp(2* c + 2*d*x) + 2*b*exp(2*c + 2*d*x)))/(a + b)^(5/2)))/(2*d*(a + b)^(5/2)) - 8/(3*(a*d + b*d)*(3*exp(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d* x) - 1)) - (2*b)/((exp(2*c + 2*d*x) - 1)*(a + b)*(a*d + b*d)) - 4/((a*d + b*d)*(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1)) - (a*b^(1/2)*log((4*b*ex p(2*c + 2*d*x))/(a + b)^2 + (2*b^(1/2)*(a + a*exp(2*c + 2*d*x) + 2*b*exp(2 *c + 2*d*x)))/(a + b)^(5/2)))/(2*d*(a + b)^(5/2))