Integrand size = 21, antiderivative size = 34 \[ \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx=a x-\frac {a \coth (c+d x)}{d}-\frac {(a+b) \coth ^3(c+d x)}{3 d} \] Output:
a*x-a*coth(d*x+c)/d-1/3*(a+b)*coth(d*x+c)^3/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.44 \[ \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx=-\frac {b \coth ^3(c+d x)}{3 d}-\frac {a \coth ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\tanh ^2(c+d x)\right )}{3 d} \] Input:
Integrate[Coth[c + d*x]^4*(a + b*Sech[c + d*x]^2),x]
Output:
-1/3*(b*Coth[c + d*x]^3)/d - (a*Coth[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, Tanh[c + d*x]^2])/(3*d)
Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4629, 2075, 359, 264, 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.
\(\displaystyle \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \sec (i c+i d x)^2}{\tan (i c+i d x)^4}dx\) |
\(\Big \downarrow \) 4629 |
\(\displaystyle \frac {\int \frac {\coth ^4(c+d x) \left (a+b \left (1-\tanh ^2(c+d x)\right )\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \frac {\int \frac {\coth ^4(c+d x) \left (-b \tanh ^2(c+d x)+a+b\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {a \int \frac {\coth ^2(c+d x)}{1-\tanh ^2(c+d x)}d\tanh (c+d x)-\frac {1}{3} (a+b) \coth ^3(c+d x)}{d}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {a \left (\int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)-\coth (c+d x)\right )-\frac {1}{3} (a+b) \coth ^3(c+d x)}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {a (\text {arctanh}(\tanh (c+d x))-\coth (c+d x))-\frac {1}{3} (a+b) \coth ^3(c+d x)}{d}\) |
Input:
Int[Coth[c + d*x]^4*(a + b*Sech[c + d*x]^2),x]
Output:
(a*(ArcTanh[Tanh[c + d*x]] - Coth[c + d*x]) - ((a + b)*Coth[c + d*x]^3)/3) /d
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*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[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && ! BinomialMatchQ[{u, v}, x]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f _.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/f Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 )), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
Time = 2.91 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.88
method | result | size |
risch | \(x a -\frac {2 \left (6 \,{\mathrm e}^{4 d x +4 c} a +3 \,{\mathrm e}^{4 d x +4 c} b -6 a \,{\mathrm e}^{2 d x +2 c}+4 a +b \right )}{3 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\) | \(64\) |
derivativedivides | \(\frac {a \left (d x +c -\coth \left (d x +c \right )-\frac {\coth \left (d x +c \right )^{3}}{3}\right )+b \left (-\frac {\cosh \left (d x +c \right )}{2 \sinh \left (d x +c \right )^{3}}-\frac {\left (\frac {2}{3}-\frac {\operatorname {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )}{2}\right )}{d}\) | \(70\) |
default | \(\frac {a \left (d x +c -\coth \left (d x +c \right )-\frac {\coth \left (d x +c \right )^{3}}{3}\right )+b \left (-\frac {\cosh \left (d x +c \right )}{2 \sinh \left (d x +c \right )^{3}}-\frac {\left (\frac {2}{3}-\frac {\operatorname {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )}{2}\right )}{d}\) | \(70\) |
Input:
int(coth(d*x+c)^4*(a+b*sech(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
x*a-2/3*(6*exp(4*d*x+4*c)*a+3*exp(4*d*x+4*c)*b-6*a*exp(2*d*x+2*c)+4*a+b)/d /(exp(2*d*x+2*c)-1)^3
Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (32) = 64\).
Time = 0.20 (sec) , antiderivative size = 140, normalized size of antiderivative = 4.12 \[ \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx=-\frac {{\left (4 \, a + b\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (4 \, a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} - {\left (3 \, a d x + 4 \, a + b\right )} \sinh \left (d x + c\right )^{3} + 3 \, b \cosh \left (d x + c\right ) + 3 \, {\left (3 \, a d x - {\left (3 \, a d x + 4 \, a + b\right )} \cosh \left (d x + c\right )^{2} + 4 \, a + b\right )} \sinh \left (d x + c\right )}{3 \, {\left (d \sinh \left (d x + c\right )^{3} + 3 \, {\left (d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )\right )}} \] Input:
integrate(coth(d*x+c)^4*(a+b*sech(d*x+c)^2),x, algorithm="fricas")
Output:
-1/3*((4*a + b)*cosh(d*x + c)^3 + 3*(4*a + b)*cosh(d*x + c)*sinh(d*x + c)^ 2 - (3*a*d*x + 4*a + b)*sinh(d*x + c)^3 + 3*b*cosh(d*x + c) + 3*(3*a*d*x - (3*a*d*x + 4*a + b)*cosh(d*x + c)^2 + 4*a + b)*sinh(d*x + c))/(d*sinh(d*x + c)^3 + 3*(d*cosh(d*x + c)^2 - d)*sinh(d*x + c))
\[ \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx=\int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right ) \coth ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(coth(d*x+c)**4*(a+b*sech(d*x+c)**2),x)
Output:
Integral((a + b*sech(c + d*x)**2)*coth(c + d*x)**4, x)
Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (32) = 64\).
Time = 0.05 (sec) , antiderivative size = 170, normalized size of antiderivative = 5.00 \[ \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx=\frac {1}{3} \, a {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + \frac {2}{3} \, b {\left (\frac {3 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} \] Input:
integrate(coth(d*x+c)^4*(a+b*sech(d*x+c)^2),x, algorithm="maxima")
Output:
1/3*a*(3*x + 3*c/d - 4*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) - 2)/(d*(3 *e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) - 1))) + 2/3*b*( 3*e^(-4*d*x - 4*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)))
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (32) = 64\).
Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.06 \[ \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx=\frac {3 \, {\left (d x + c\right )} a - \frac {2 \, {\left (6 \, a e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a + b\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{3 \, d} \] Input:
integrate(coth(d*x+c)^4*(a+b*sech(d*x+c)^2),x, algorithm="giac")
Output:
1/3*(3*(d*x + c)*a - 2*(6*a*e^(4*d*x + 4*c) + 3*b*e^(4*d*x + 4*c) - 6*a*e^ (2*d*x + 2*c) + 4*a + b)/(e^(2*d*x + 2*c) - 1)^3)/d
Time = 2.31 (sec) , antiderivative size = 161, normalized size of antiderivative = 4.74 \[ \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx=a\,x-\frac {\frac {2\,b}{3\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a+b\right )}{3\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {\frac {2\,\left (2\,a+b\right )}{3\,d}+\frac {4\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{3\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (2\,a+b\right )}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {2\,\left (2\,a+b\right )}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \] Input:
int(coth(c + d*x)^4*(a + b/cosh(c + d*x)^2),x)
Output:
a*x - ((2*b)/(3*d) + (2*exp(2*c + 2*d*x)*(2*a + b))/(3*d))/(exp(4*c + 4*d* x) - 2*exp(2*c + 2*d*x) + 1) - ((2*(2*a + b))/(3*d) + (4*b*exp(2*c + 2*d*x ))/(3*d) + (2*exp(4*c + 4*d*x)*(2*a + b))/(3*d))/(3*exp(2*c + 2*d*x) - 3*e xp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1) - (2*(2*a + b))/(3*d*(exp(2*c + 2* d*x) - 1))
Time = 0.20 (sec) , antiderivative size = 136, normalized size of antiderivative = 4.00 \[ \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx=\frac {3 e^{6 d x +6 c} a d x -4 e^{6 d x +6 c} a -2 e^{6 d x +6 c} b -9 e^{4 d x +4 c} a d x +9 e^{2 d x +2 c} a d x -6 e^{2 d x +2 c} b -3 a d x -4 a}{3 d \left (e^{6 d x +6 c}-3 e^{4 d x +4 c}+3 e^{2 d x +2 c}-1\right )} \] Input:
int(coth(d*x+c)^4*(a+b*sech(d*x+c)^2),x)
Output:
(3*e**(6*c + 6*d*x)*a*d*x - 4*e**(6*c + 6*d*x)*a - 2*e**(6*c + 6*d*x)*b - 9*e**(4*c + 4*d*x)*a*d*x + 9*e**(2*c + 2*d*x)*a*d*x - 6*e**(2*c + 2*d*x)*b - 3*a*d*x - 4*a)/(3*d*(e**(6*c + 6*d*x) - 3*e**(4*c + 4*d*x) + 3*e**(2*c + 2*d*x) - 1))