Integrand size = 23, antiderivative size = 60 \[ \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=a^3 x-\frac {(a-2 b) (a+b)^2 \coth (c+d x)}{d}-\frac {(a+b)^3 \coth ^3(c+d x)}{3 d}+\frac {b^3 \tanh (c+d x)}{d} \]
Leaf count is larger than twice the leaf count of optimal. \(343\) vs. \(2(60)=120\).
Time = 3.65 (sec) , antiderivative size = 343, normalized size of antiderivative = 5.72 \[ \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {\text {csch}(c) \text {csch}^3(c+d x) \text {sech}(c) \text {sech}(c+d x) \left (6 a^3 d x \cosh (2 d x)-3 a^3 d x \cosh (2 (c+2 d x))-6 a^3 d x \cosh (4 c+2 d x)+3 a^3 d x \cosh (6 c+4 d x)-18 a^2 b \sinh (2 c)-36 a b^2 \sinh (2 c)-4 a^3 \sinh (2 d x)+6 a^2 b \sinh (2 d x)+24 a b^2 \sinh (2 d x)+32 b^3 \sinh (2 d x)-16 a^3 \sinh (2 (c+d x))-12 a^2 b \sinh (2 (c+d x))+24 a b^2 \sinh (2 (c+d x))+8 b^3 \sinh (2 (c+d x))+8 a^3 \sinh (4 (c+d x))+6 a^2 b \sinh (4 (c+d x))-12 a b^2 \sinh (4 (c+d x))-4 b^3 \sinh (4 (c+d x))+8 a^3 \sinh (2 (c+2 d x))+6 a^2 b \sinh (2 (c+2 d x))-12 a b^2 \sinh (2 (c+2 d x))-16 b^3 \sinh (2 (c+2 d x))-12 a^3 \sinh (4 c+2 d x)-18 a^2 b \sinh (4 c+2 d x)\right )}{96 d} \]
(Csch[c]*Csch[c + d*x]^3*Sech[c]*Sech[c + d*x]*(6*a^3*d*x*Cosh[2*d*x] - 3* a^3*d*x*Cosh[2*(c + 2*d*x)] - 6*a^3*d*x*Cosh[4*c + 2*d*x] + 3*a^3*d*x*Cosh [6*c + 4*d*x] - 18*a^2*b*Sinh[2*c] - 36*a*b^2*Sinh[2*c] - 4*a^3*Sinh[2*d*x ] + 6*a^2*b*Sinh[2*d*x] + 24*a*b^2*Sinh[2*d*x] + 32*b^3*Sinh[2*d*x] - 16*a ^3*Sinh[2*(c + d*x)] - 12*a^2*b*Sinh[2*(c + d*x)] + 24*a*b^2*Sinh[2*(c + d *x)] + 8*b^3*Sinh[2*(c + d*x)] + 8*a^3*Sinh[4*(c + d*x)] + 6*a^2*b*Sinh[4* (c + d*x)] - 12*a*b^2*Sinh[4*(c + d*x)] - 4*b^3*Sinh[4*(c + d*x)] + 8*a^3* Sinh[2*(c + 2*d*x)] + 6*a^2*b*Sinh[2*(c + 2*d*x)] - 12*a*b^2*Sinh[2*(c + 2 *d*x)] - 16*b^3*Sinh[2*(c + 2*d*x)] - 12*a^3*Sinh[4*c + 2*d*x] - 18*a^2*b* Sinh[4*c + 2*d*x]))/(96*d)
Time = 0.34 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4629, 2075, 364, 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.
\(\displaystyle \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sec (i c+i d x)^2\right )^3}{\tan (i c+i d x)^4}dx\) |
\(\Big \downarrow \) 4629 |
\(\displaystyle \frac {\int \frac {\coth ^4(c+d x) \left (a+b \left (1-\tanh ^2(c+d x)\right )\right )^3}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \frac {\int \frac {\coth ^4(c+d x) \left (-b \tanh ^2(c+d x)+a+b\right )^3}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 364 |
\(\displaystyle \frac {\int \left ((a+b)^3 \coth ^4(c+d x)+(a-2 b) (a+b)^2 \coth ^2(c+d x)+b^3-\frac {a^3}{\tanh ^2(c+d x)-1}\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 \text {arctanh}(\tanh (c+d x))-\frac {1}{3} (a+b)^3 \coth ^3(c+d x)-(a-2 b) (a+b)^2 \coth (c+d x)+b^3 \tanh (c+d x)}{d}\) |
(a^3*ArcTanh[Tanh[c + d*x]] - (a - 2*b)*(a + b)^2*Coth[c + d*x] - ((a + b) ^3*Coth[c + d*x]^3)/3 + b^3*Tanh[c + d*x])/d
3.2.32.3.1 Defintions of rubi rules used
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
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])
Leaf count of result is larger than twice the leaf count of optimal. \(148\) vs. \(2(58)=116\).
Time = 67.92 (sec) , antiderivative size = 149, normalized size of antiderivative = 2.48
method | result | size |
derivativedivides | \(\frac {a^{3} \left (d x +c -\coth \left (d x +c \right )-\frac {\coth \left (d x +c \right )^{3}}{3}\right )+3 a^{2} b \left (-\frac {\cosh \left (d x +c \right )}{2 \sinh \left (d x +c \right )^{3}}-\frac {\left (\frac {2}{3}-\frac {\operatorname {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )}{2}\right )+3 a \,b^{2} \left (\frac {2}{3}-\frac {\operatorname {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )+b^{3} \left (-\frac {1}{3 \sinh \left (d x +c \right )^{3} \cosh \left (d x +c \right )}+\frac {4}{3 \sinh \left (d x +c \right ) \cosh \left (d x +c \right )}+\frac {8 \tanh \left (d x +c \right )}{3}\right )}{d}\) | \(149\) |
default | \(\frac {a^{3} \left (d x +c -\coth \left (d x +c \right )-\frac {\coth \left (d x +c \right )^{3}}{3}\right )+3 a^{2} b \left (-\frac {\cosh \left (d x +c \right )}{2 \sinh \left (d x +c \right )^{3}}-\frac {\left (\frac {2}{3}-\frac {\operatorname {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )}{2}\right )+3 a \,b^{2} \left (\frac {2}{3}-\frac {\operatorname {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )+b^{3} \left (-\frac {1}{3 \sinh \left (d x +c \right )^{3} \cosh \left (d x +c \right )}+\frac {4}{3 \sinh \left (d x +c \right ) \cosh \left (d x +c \right )}+\frac {8 \tanh \left (d x +c \right )}{3}\right )}{d}\) | \(149\) |
risch | \(a^{3} x -\frac {2 \left (6 a^{3} {\mathrm e}^{6 d x +6 c}+9 a^{2} b \,{\mathrm e}^{6 d x +6 c}+9 a^{2} b \,{\mathrm e}^{4 d x +4 c}+18 a \,b^{2} {\mathrm e}^{4 d x +4 c}-2 a^{3} {\mathrm e}^{2 d x +2 c}+3 a^{2} b \,{\mathrm e}^{2 d x +2 c}+12 \,{\mathrm e}^{2 d x +2 c} a \,b^{2}+16 \,{\mathrm e}^{2 d x +2 c} b^{3}+4 a^{3}+3 a^{2} b -6 a \,b^{2}-8 b^{3}\right )}{3 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{3} \left ({\mathrm e}^{2 d x +2 c}+1\right )}\) | \(178\) |
1/d*(a^3*(d*x+c-coth(d*x+c)-1/3*coth(d*x+c)^3)+3*a^2*b*(-1/2/sinh(d*x+c)^3 *cosh(d*x+c)-1/2*(2/3-1/3*csch(d*x+c)^2)*coth(d*x+c))+3*a*b^2*(2/3-1/3*csc h(d*x+c)^2)*coth(d*x+c)+b^3*(-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. 354 vs. \(2 (58) = 116\).
Time = 0.26 (sec) , antiderivative size = 354, normalized size of antiderivative = 5.90 \[ \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=-\frac {{\left (4 \, a^{3} + 3 \, a^{2} b - 6 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} - 4 \, {\left (3 \, a^{3} d x + 4 \, a^{3} + 3 \, a^{2} b - 6 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (4 \, a^{3} + 3 \, a^{2} b - 6 \, a b^{2} - 8 \, b^{3}\right )} \sinh \left (d x + c\right )^{4} + 9 \, a^{2} b + 18 \, a b^{2} + 4 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + 4 \, b^{3}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{3} + 6 \, a^{2} b + 6 \, a b^{2} + 8 \, b^{3} + 3 \, {\left (4 \, a^{3} + 3 \, a^{2} b - 6 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2} - 4 \, {\left ({\left (3 \, a^{3} d x + 4 \, a^{3} + 3 \, a^{2} b - 6 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} - {\left (3 \, a^{3} d x + 4 \, a^{3} + 3 \, a^{2} b - 6 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{12 \, {\left (d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (d \cosh \left (d x + c\right )^{3} - d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}} \]
-1/12*((4*a^3 + 3*a^2*b - 6*a*b^2 - 8*b^3)*cosh(d*x + c)^4 - 4*(3*a^3*d*x + 4*a^3 + 3*a^2*b - 6*a*b^2 - 8*b^3)*cosh(d*x + c)*sinh(d*x + c)^3 + (4*a^ 3 + 3*a^2*b - 6*a*b^2 - 8*b^3)*sinh(d*x + c)^4 + 9*a^2*b + 18*a*b^2 + 4*(a ^3 + 3*a^2*b + 3*a*b^2 + 4*b^3)*cosh(d*x + c)^2 + 2*(2*a^3 + 6*a^2*b + 6*a *b^2 + 8*b^3 + 3*(4*a^3 + 3*a^2*b - 6*a*b^2 - 8*b^3)*cosh(d*x + c)^2)*sinh (d*x + c)^2 - 4*((3*a^3*d*x + 4*a^3 + 3*a^2*b - 6*a*b^2 - 8*b^3)*cosh(d*x + c)^3 - (3*a^3*d*x + 4*a^3 + 3*a^2*b - 6*a*b^2 - 8*b^3)*cosh(d*x + c))*si nh(d*x + c))/(d*cosh(d*x + c)*sinh(d*x + c)^3 + (d*cosh(d*x + c)^3 - d*cos h(d*x + c))*sinh(d*x + c))
Timed out. \[ \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 366 vs. \(2 (58) = 116\).
Time = 0.20 (sec) , antiderivative size = 366, normalized size of antiderivative = 6.10 \[ \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {1}{3} \, a^{3} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + 4 \, a b^{2} {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} - \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + \frac {16}{3} \, b^{3} {\left (\frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}} - \frac {1}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} + 2 \, a^{2} b {\left (\frac {3 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} \]
1/3*a^3*(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))) + 4*a*b ^2*(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^3*(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))) + 2*a^2*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. 158 vs. \(2 (58) = 116\).
Time = 0.38 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.63 \[ \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {3 \, {\left (d x + c\right )} a^{3} - \frac {6 \, b^{3}}{e^{\left (2 \, d x + 2 \, c\right )} + 1} - \frac {2 \, {\left (6 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 9 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 18 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a^{3} + 3 \, a^{2} b - 6 \, a b^{2} - 5 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{3 \, d} \]
1/3*(3*(d*x + c)*a^3 - 6*b^3/(e^(2*d*x + 2*c) + 1) - 2*(6*a^3*e^(4*d*x + 4 *c) + 9*a^2*b*e^(4*d*x + 4*c) - 3*b^3*e^(4*d*x + 4*c) - 6*a^3*e^(2*d*x + 2 *c) + 18*a*b^2*e^(2*d*x + 2*c) + 12*b^3*e^(2*d*x + 2*c) + 4*a^3 + 3*a^2*b - 6*a*b^2 - 5*b^3)/(e^(2*d*x + 2*c) - 1)^3)/d
Time = 2.17 (sec) , antiderivative size = 260, normalized size of antiderivative = 4.33 \[ \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=a^3\,x-\frac {\frac {2\,\left (a^2\,b+2\,a\,b^2+b^3\right )}{d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a^3+3\,a^2\,b-b^3\right )}{3\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {\frac {2\,\left (2\,a^3+3\,a^2\,b-b^3\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (2\,a^3+3\,a^2\,b-b^3\right )}{3\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b+2\,a\,b^2+b^3\right )}{d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {2\,b^3}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {2\,\left (2\,a^3+3\,a^2\,b-b^3\right )}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]
a^3*x - ((2*(2*a*b^2 + a^2*b + b^3))/d + (2*exp(2*c + 2*d*x)*(3*a^2*b + 2* a^3 - b^3))/(3*d))/(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1) - ((2*(3*a^ 2*b + 2*a^3 - b^3))/(3*d) + (2*exp(4*c + 4*d*x)*(3*a^2*b + 2*a^3 - b^3))/( 3*d) + (4*exp(2*c + 2*d*x)*(2*a*b^2 + a^2*b + 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*b^3)/(d*(exp(2*c + 2*d* x) + 1)) - (2*(3*a^2*b + 2*a^3 - b^3))/(3*d*(exp(2*c + 2*d*x) - 1))