Integrand size = 17, antiderivative size = 70 \[ \int \text {csch}^5(a+b x) \text {sech}^2(a+b x) \, dx=-\frac {15 \text {arctanh}(\cosh (a+b x))}{8 b}+\frac {15 \text {sech}(a+b x)}{8 b}+\frac {5 \text {csch}^2(a+b x) \text {sech}(a+b x)}{8 b}-\frac {\text {csch}^4(a+b x) \text {sech}(a+b x)}{4 b} \]
-15/8*arctanh(cosh(b*x+a))/b+15/8*sech(b*x+a)/b+5/8*csch(b*x+a)^2*sech(b*x +a)/b-1/4*csch(b*x+a)^4*sech(b*x+a)/b
Time = 0.13 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.76 \[ \int \text {csch}^5(a+b x) \text {sech}^2(a+b x) \, dx=\frac {7 \text {csch}^2\left (\frac {1}{2} (a+b x)\right )}{32 b}-\frac {\text {csch}^4\left (\frac {1}{2} (a+b x)\right )}{64 b}-\frac {15 \log \left (\cosh \left (\frac {1}{2} (a+b x)\right )\right )}{8 b}+\frac {15 \log \left (\sinh \left (\frac {1}{2} (a+b x)\right )\right )}{8 b}+\frac {7 \text {sech}^2\left (\frac {1}{2} (a+b x)\right )}{32 b}+\frac {\text {sech}^4\left (\frac {1}{2} (a+b x)\right )}{64 b}+\frac {\text {sech}(a+b x)}{b} \]
(7*Csch[(a + b*x)/2]^2)/(32*b) - Csch[(a + b*x)/2]^4/(64*b) - (15*Log[Cosh [(a + b*x)/2]])/(8*b) + (15*Log[Sinh[(a + b*x)/2]])/(8*b) + (7*Sech[(a + b *x)/2]^2)/(32*b) + Sech[(a + b*x)/2]^4/(64*b) + Sech[a + b*x]/b
Time = 0.23 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.17, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {3042, 26, 3102, 25, 252, 252, 262, 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.
\(\displaystyle \int \text {csch}^5(a+b x) \text {sech}^2(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int i \csc (i a+i b x)^5 \sec (i a+i b x)^2dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \csc (i a+i b x)^5 \sec (i a+i b x)^2dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {\int -\frac {\text {sech}^6(a+b x)}{\left (1-\text {sech}^2(a+b x)\right )^3}d\text {sech}(a+b x)}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\text {sech}^6(a+b x)}{\left (1-\text {sech}^2(a+b x)\right )^3}d\text {sech}(a+b x)}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {5}{4} \int \frac {\text {sech}^4(a+b x)}{\left (1-\text {sech}^2(a+b x)\right )^2}d\text {sech}(a+b x)-\frac {\text {sech}^5(a+b x)}{4 \left (1-\text {sech}^2(a+b x)\right )^2}}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {5}{4} \left (\frac {\text {sech}^3(a+b x)}{2 \left (1-\text {sech}^2(a+b x)\right )}-\frac {3}{2} \int \frac {\text {sech}^2(a+b x)}{1-\text {sech}^2(a+b x)}d\text {sech}(a+b x)\right )-\frac {\text {sech}^5(a+b x)}{4 \left (1-\text {sech}^2(a+b x)\right )^2}}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {5}{4} \left (\frac {\text {sech}^3(a+b x)}{2 \left (1-\text {sech}^2(a+b x)\right )}-\frac {3}{2} \left (\int \frac {1}{1-\text {sech}^2(a+b x)}d\text {sech}(a+b x)-\text {sech}(a+b x)\right )\right )-\frac {\text {sech}^5(a+b x)}{4 \left (1-\text {sech}^2(a+b x)\right )^2}}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {5}{4} \left (\frac {\text {sech}^3(a+b x)}{2 \left (1-\text {sech}^2(a+b x)\right )}-\frac {3}{2} (\text {arctanh}(\text {sech}(a+b x))-\text {sech}(a+b x))\right )-\frac {\text {sech}^5(a+b x)}{4 \left (1-\text {sech}^2(a+b x)\right )^2}}{b}\) |
(-1/4*Sech[a + b*x]^5/(1 - Sech[a + b*x]^2)^2 + (5*((-3*(ArcTanh[Sech[a + b*x]] - Sech[a + b*x]))/2 + Sech[a + b*x]^3/(2*(1 - Sech[a + b*x]^2))))/4) /b
3.1.45.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_S ymbol] :> Simp[1/(f*a^n) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/ 2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1 )/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Time = 15.40 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.87
method | result | size |
derivativedivides | \(\frac {-\frac {1}{4 \sinh \left (b x +a \right )^{4} \cosh \left (b x +a \right )}+\frac {5}{8 \sinh \left (b x +a \right )^{2} \cosh \left (b x +a \right )}+\frac {15}{8 \cosh \left (b x +a \right )}-\frac {15 \,\operatorname {arctanh}\left ({\mathrm e}^{b x +a}\right )}{4}}{b}\) | \(61\) |
default | \(\frac {-\frac {1}{4 \sinh \left (b x +a \right )^{4} \cosh \left (b x +a \right )}+\frac {5}{8 \sinh \left (b x +a \right )^{2} \cosh \left (b x +a \right )}+\frac {15}{8 \cosh \left (b x +a \right )}-\frac {15 \,\operatorname {arctanh}\left ({\mathrm e}^{b x +a}\right )}{4}}{b}\) | \(61\) |
risch | \(\frac {{\mathrm e}^{b x +a} \left (15 \,{\mathrm e}^{8 b x +8 a}-40 \,{\mathrm e}^{6 b x +6 a}+18 \,{\mathrm e}^{4 b x +4 a}-40 \,{\mathrm e}^{2 b x +2 a}+15\right )}{4 b \left ({\mathrm e}^{2 b x +2 a}-1\right )^{4} \left (1+{\mathrm e}^{2 b x +2 a}\right )}-\frac {15 \ln \left ({\mathrm e}^{b x +a}+1\right )}{8 b}+\frac {15 \ln \left ({\mathrm e}^{b x +a}-1\right )}{8 b}\) | \(113\) |
1/b*(-1/4/sinh(b*x+a)^4/cosh(b*x+a)+5/8/sinh(b*x+a)^2/cosh(b*x+a)+15/8/cos h(b*x+a)-15/4*arctanh(exp(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 1591 vs. \(2 (62) = 124\).
Time = 0.26 (sec) , antiderivative size = 1591, normalized size of antiderivative = 22.73 \[ \int \text {csch}^5(a+b x) \text {sech}^2(a+b x) \, dx=\text {Too large to display} \]
1/8*(30*cosh(b*x + a)^9 + 270*cosh(b*x + a)*sinh(b*x + a)^8 + 30*sinh(b*x + a)^9 + 40*(27*cosh(b*x + a)^2 - 2)*sinh(b*x + a)^7 - 80*cosh(b*x + a)^7 + 280*(9*cosh(b*x + a)^3 - 2*cosh(b*x + a))*sinh(b*x + a)^6 + 12*(315*cosh (b*x + a)^4 - 140*cosh(b*x + a)^2 + 3)*sinh(b*x + a)^5 + 36*cosh(b*x + a)^ 5 + 20*(189*cosh(b*x + a)^5 - 140*cosh(b*x + a)^3 + 9*cosh(b*x + a))*sinh( b*x + a)^4 + 40*(63*cosh(b*x + a)^6 - 70*cosh(b*x + a)^4 + 9*cosh(b*x + a) ^2 - 2)*sinh(b*x + a)^3 - 80*cosh(b*x + a)^3 + 120*(9*cosh(b*x + a)^7 - 14 *cosh(b*x + a)^5 + 3*cosh(b*x + a)^3 - 2*cosh(b*x + a))*sinh(b*x + a)^2 - 15*(cosh(b*x + a)^10 + 10*cosh(b*x + a)*sinh(b*x + a)^9 + sinh(b*x + a)^10 + 3*(15*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^8 - 3*cosh(b*x + a)^8 + 24*(5* cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a)^7 + 2*(105*cosh(b*x + a)^4 - 42*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^6 + 2*cosh(b*x + a)^6 + 12*(21*cos h(b*x + a)^5 - 14*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)^5 + 2*(10 5*cosh(b*x + a)^6 - 105*cosh(b*x + a)^4 + 15*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^4 + 2*cosh(b*x + a)^4 + 8*(15*cosh(b*x + a)^7 - 21*cosh(b*x + a)^5 + 5*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)^3 + 3*(15*cosh(b*x + a)^ 8 - 28*cosh(b*x + a)^6 + 10*cosh(b*x + a)^4 + 4*cosh(b*x + a)^2 - 1)*sinh( b*x + a)^2 - 3*cosh(b*x + a)^2 + 2*(5*cosh(b*x + a)^9 - 12*cosh(b*x + a)^7 + 6*cosh(b*x + a)^5 + 4*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 15*(cosh(b*x + a)^10 + 10...
\[ \int \text {csch}^5(a+b x) \text {sech}^2(a+b x) \, dx=\int \operatorname {csch}^{5}{\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (62) = 124\).
Time = 0.18 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.21 \[ \int \text {csch}^5(a+b x) \text {sech}^2(a+b x) \, dx=-\frac {15 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{8 \, b} + \frac {15 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{8 \, b} - \frac {15 \, e^{\left (-b x - a\right )} - 40 \, e^{\left (-3 \, b x - 3 \, a\right )} + 18 \, e^{\left (-5 \, b x - 5 \, a\right )} - 40 \, e^{\left (-7 \, b x - 7 \, a\right )} + 15 \, e^{\left (-9 \, b x - 9 \, a\right )}}{4 \, b {\left (3 \, e^{\left (-2 \, b x - 2 \, a\right )} - 2 \, e^{\left (-4 \, b x - 4 \, a\right )} - 2 \, e^{\left (-6 \, b x - 6 \, a\right )} + 3 \, e^{\left (-8 \, b x - 8 \, a\right )} - e^{\left (-10 \, b x - 10 \, a\right )} - 1\right )}} \]
-15/8*log(e^(-b*x - a) + 1)/b + 15/8*log(e^(-b*x - a) - 1)/b - 1/4*(15*e^( -b*x - a) - 40*e^(-3*b*x - 3*a) + 18*e^(-5*b*x - 5*a) - 40*e^(-7*b*x - 7*a ) + 15*e^(-9*b*x - 9*a))/(b*(3*e^(-2*b*x - 2*a) - 2*e^(-4*b*x - 4*a) - 2*e ^(-6*b*x - 6*a) + 3*e^(-8*b*x - 8*a) - e^(-10*b*x - 10*a) - 1))
Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (62) = 124\).
Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.86 \[ \int \text {csch}^5(a+b x) \text {sech}^2(a+b x) \, dx=\frac {\frac {4 \, {\left (7 \, {\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}^{3} - 36 \, e^{\left (b x + a\right )} - 36 \, e^{\left (-b x - a\right )}\right )}}{{\left ({\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}^{2} - 4\right )}^{2}} + \frac {32}{e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}} - 15 \, \log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} + 2\right ) + 15 \, \log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} - 2\right )}{16 \, b} \]
1/16*(4*(7*(e^(b*x + a) + e^(-b*x - a))^3 - 36*e^(b*x + a) - 36*e^(-b*x - a))/((e^(b*x + a) + e^(-b*x - a))^2 - 4)^2 + 32/(e^(b*x + a) + e^(-b*x - a )) - 15*log(e^(b*x + a) + e^(-b*x - a) + 2) + 15*log(e^(b*x + a) + e^(-b*x - a) - 2))/b
Time = 2.13 (sec) , antiderivative size = 214, normalized size of antiderivative = 3.06 \[ \int \text {csch}^5(a+b x) \text {sech}^2(a+b x) \, dx=\frac {3\,{\mathrm {e}}^{a+b\,x}}{2\,b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {15\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{4\,\sqrt {-b^2}}-\frac {6\,{\mathrm {e}}^{a+b\,x}}{b\,\left (3\,{\mathrm {e}}^{2\,a+2\,b\,x}-3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}-1\right )}-\frac {4\,{\mathrm {e}}^{a+b\,x}}{b\,\left (6\,{\mathrm {e}}^{4\,a+4\,b\,x}-4\,{\mathrm {e}}^{2\,a+2\,b\,x}-4\,{\mathrm {e}}^{6\,a+6\,b\,x}+{\mathrm {e}}^{8\,a+8\,b\,x}+1\right )}+\frac {7\,{\mathrm {e}}^{a+b\,x}}{4\,b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )}+\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]
(3*exp(a + b*x))/(2*b*(exp(4*a + 4*b*x) - 2*exp(2*a + 2*b*x) + 1)) - (15*a tan((exp(b*x)*exp(a)*(-b^2)^(1/2))/b))/(4*(-b^2)^(1/2)) - (6*exp(a + b*x)) /(b*(3*exp(2*a + 2*b*x) - 3*exp(4*a + 4*b*x) + exp(6*a + 6*b*x) - 1)) - (4 *exp(a + b*x))/(b*(6*exp(4*a + 4*b*x) - 4*exp(2*a + 2*b*x) - 4*exp(6*a + 6 *b*x) + exp(8*a + 8*b*x) + 1)) + (7*exp(a + b*x))/(4*b*(exp(2*a + 2*b*x) - 1)) + (2*exp(a + b*x))/(b*(exp(2*a + 2*b*x) + 1))