Integrand size = 21, antiderivative size = 45 \[ \int \text {csch}^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx=\frac {(a+2 b) \coth (c+d x)}{d}-\frac {(a+b) \coth ^3(c+d x)}{3 d}+\frac {b \tanh (c+d x)}{d} \]
Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.87 \[ \int \text {csch}^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx=\frac {2 a \coth (c+d x)}{3 d}+\frac {5 b \coth (c+d x)}{3 d}-\frac {a \coth (c+d x) \text {csch}^2(c+d x)}{3 d}-\frac {b \coth (c+d x) \text {csch}^2(c+d x)}{3 d}+\frac {b \tanh (c+d x)}{d} \]
(2*a*Coth[c + d*x])/(3*d) + (5*b*Coth[c + d*x])/(3*d) - (a*Coth[c + d*x]*C sch[c + d*x]^2)/(3*d) - (b*Coth[c + d*x]*Csch[c + d*x]^2)/(3*d) + (b*Tanh[ c + d*x])/d
Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 4620, 355, 2009}
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.
\(\displaystyle \int \text {csch}^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \sec (i c+i d x)^2}{\sin (i c+i d x)^4}dx\) |
\(\Big \downarrow \) 4620 |
\(\displaystyle \frac {\int \coth ^4(c+d x) \left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \frac {\int \left ((a+b) \coth ^4(c+d x)+(-a-2 b) \coth ^2(c+d x)+b\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{3} (a+b) \coth ^3(c+d x)+(a+2 b) \coth (c+d x)+b \tanh (c+d x)}{d}\) |
3.1.8.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] & & IGtQ[q, 0]
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]
Time = 12.00 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.62
method | result | size |
derivativedivides | \(\frac {a \left (\frac {2}{3}-\frac {\operatorname {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )+b \left (-\frac {1}{3 \sinh \left (d x +c \right )^{3} \cosh \left (d x +c \right )}+\frac {4}{3 \sinh \left (d x +c \right ) \cosh \left (d x +c \right )}+\frac {8 \tanh \left (d x +c \right )}{3}\right )}{d}\) | \(73\) |
default | \(\frac {a \left (\frac {2}{3}-\frac {\operatorname {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )+b \left (-\frac {1}{3 \sinh \left (d x +c \right )^{3} \cosh \left (d x +c \right )}+\frac {4}{3 \sinh \left (d x +c \right ) \cosh \left (d x +c \right )}+\frac {8 \tanh \left (d x +c \right )}{3}\right )}{d}\) | \(73\) |
risch | \(-\frac {4 \left (3 a \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a +8 b \,{\mathrm e}^{2 d x +2 c}-a -4 b \right )}{3 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{3} \left ({\mathrm e}^{2 d x +2 c}+1\right )}\) | \(75\) |
1/d*(a*(2/3-1/3*csch(d*x+c)^2)*coth(d*x+c)+b*(-1/3/sinh(d*x+c)^3/cosh(d*x+ c)+4/3/sinh(d*x+c)/cosh(d*x+c)+8/3*tanh(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (43) = 86\).
Time = 0.24 (sec) , antiderivative size = 246, normalized size of antiderivative = 5.47 \[ \int \text {csch}^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx=-\frac {8 \, {\left ({\left (a - 2 \, b\right )} \cosh \left (d x + c\right )^{2} + 4 \, {\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a - 2 \, b\right )} \sinh \left (d x + c\right )^{2} + a + 4 \, b\right )}}{3 \, {\left (d \cosh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + d \sinh \left (d x + c\right )^{6} - 2 \, d \cosh \left (d x + c\right )^{4} + {\left (15 \, d \cosh \left (d x + c\right )^{2} - 2 \, d\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (5 \, d \cosh \left (d x + c\right )^{3} - 2 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} - d \cosh \left (d x + c\right )^{2} + {\left (15 \, d \cosh \left (d x + c\right )^{4} - 12 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{2} + 2 \, {\left (3 \, d \cosh \left (d x + c\right )^{5} - 4 \, d \cosh \left (d x + c\right )^{3} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 2 \, d\right )}} \]
-8/3*((a - 2*b)*cosh(d*x + c)^2 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c) + (a - 2*b)*sinh(d*x + c)^2 + a + 4*b)/(d*cosh(d*x + c)^6 + 6*d*cosh(d*x + c )*sinh(d*x + c)^5 + d*sinh(d*x + c)^6 - 2*d*cosh(d*x + c)^4 + (15*d*cosh(d *x + c)^2 - 2*d)*sinh(d*x + c)^4 + 4*(5*d*cosh(d*x + c)^3 - 2*d*cosh(d*x + c))*sinh(d*x + c)^3 - d*cosh(d*x + c)^2 + (15*d*cosh(d*x + c)^4 - 12*d*co sh(d*x + c)^2 - d)*sinh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^5 - 4*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c) + 2*d)
\[ \int \text {csch}^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx=\int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right ) \operatorname {csch}^{4}{\left (c + d x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (43) = 86\).
Time = 0.20 (sec) , antiderivative size = 187, normalized size of antiderivative = 4.16 \[ \int \text {csch}^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx=\frac {4}{3} \, a {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} - \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + \frac {16}{3} \, b {\left (\frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}} - \frac {1}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} \]
4/3*a*(3*e^(-2*d*x - 2*c)/(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^ (-6*d*x - 6*c) - 1)) - 1/(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^( -6*d*x - 6*c) - 1))) + 16/3*b*(2*e^(-2*d*x - 2*c)/(d*(2*e^(-2*d*x - 2*c) - 2*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c) - 1)) - 1/(d*(2*e^(-2*d*x - 2*c) - 2*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c) - 1)))
Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.78 \[ \int \text {csch}^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (\frac {3 \, b}{e^{\left (2 \, d x + 2 \, c\right )} + 1} - \frac {3 \, b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a e^{\left (2 \, d x + 2 \, c\right )} - 12 \, b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a + 5 \, b}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}\right )}}{3 \, d} \]
-2/3*(3*b/(e^(2*d*x + 2*c) + 1) - (3*b*e^(4*d*x + 4*c) - 6*a*e^(2*d*x + 2* c) - 12*b*e^(2*d*x + 2*c) + 2*a + 5*b)/(e^(2*d*x + 2*c) - 1)^3)/d
Time = 2.07 (sec) , antiderivative size = 172, normalized size of antiderivative = 3.82 \[ \int \text {csch}^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx=\frac {\frac {2\,b}{3\,d}+\frac {2\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{3\,d}-\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a+3\,b\right )}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {\frac {2\,\left (2\,a+3\,b\right )}{3\,d}-\frac {2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{3\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}+\frac {2\,b}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {2\,b}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]
((2*b)/(3*d) + (2*b*exp(4*c + 4*d*x))/(3*d) - (4*exp(2*c + 2*d*x)*(2*a + 3 *b))/(3*d))/(3*exp(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1) - ((2*(2*a + 3*b))/(3*d) - (2*b*exp(2*c + 2*d*x))/(3*d))/(exp(4*c + 4*d *x) - 2*exp(2*c + 2*d*x) + 1) + (2*b)/(3*d*(exp(2*c + 2*d*x) - 1)) - (2*b) /(d*(exp(2*c + 2*d*x) + 1))