Integrand size = 23, antiderivative size = 66 \[ \int \frac {\tanh ^5(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {\log (\cosh (c+d x))}{(a+b) d}+\frac {a^2 \log \left (a+b \tanh ^2(c+d x)\right )}{2 b^2 (a+b) d}-\frac {\tanh ^2(c+d x)}{2 b d} \]
Time = 0.21 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.91 \[ \int \frac {\tanh ^5(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=-\frac {-\frac {2 \log (\cosh (c+d x))}{a+b}-\frac {a^2 \log \left (a+b \tanh ^2(c+d x)\right )}{b^2 (a+b)}+\frac {\tanh ^2(c+d x)}{b}}{2 d} \]
-1/2*((-2*Log[Cosh[c + d*x]])/(a + b) - (a^2*Log[a + b*Tanh[c + d*x]^2])/( b^2*(a + b)) + Tanh[c + d*x]^2/b)/d
Time = 0.31 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 26, 4153, 26, 354, 93, 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.
| \(\displaystyle \int \frac {\tanh ^5(c+d x)}{a+b \tanh ^2(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {i \tan (i c+i d x)^5}{a-b \tan (i c+i d x)^2}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle -i \int \frac {\tan (i c+i d x)^5}{a-b \tan (i c+i d x)^2}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle -\frac {i \int \frac {i \tanh ^5(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {\int \frac {\tanh ^5(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {\int \frac {\tanh ^4(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 93 |
| \(\displaystyle \frac {\int \left (\frac {a^2}{b (a+b) \left (b \tanh ^2(c+d x)+a\right )}-\frac {1}{b}-\frac {1}{(a+b) \left (\tanh ^2(c+d x)-1\right )}\right )d\tanh ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^2 \log \left (a+b \tanh ^2(c+d x)\right )}{b^2 (a+b)}-\frac {\log \left (1-\tanh ^2(c+d x)\right )}{a+b}-\frac {\tanh ^2(c+d x)}{b}}{2 d}\) |
(-(Log[1 - Tanh[c + d*x]^2]/(a + b)) + (a^2*Log[a + b*Tanh[c + d*x]^2])/(b ^2*(a + b)) - Tanh[c + d*x]^2/b)/(2*d)
3.2.70.3.1 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[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Int[ExpandIntegrand[(e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; Fre eQ[{a, b, c, d, e, f}, x] && IntegerQ[p]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.20
| method | result | size |
| parallelrisch | \(-\frac {2 b^{2} d x +\tanh \left (d x +c \right )^{2} a b +b^{2} \tanh \left (d x +c \right )^{2}+2 \ln \left (1-\tanh \left (d x +c \right )\right ) b^{2}-a^{2} \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{2 b^{2} d \left (a +b \right )}\) | \(79\) |
| derivativedivides | \(\frac {-\frac {\tanh \left (d x +c \right )^{2}}{2 b}+\frac {a^{2} \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{2 \left (a +b \right ) b^{2}}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 a +2 b}}{d}\) | \(85\) |
| default | \(\frac {-\frac {\tanh \left (d x +c \right )^{2}}{2 b}+\frac {a^{2} \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{2 \left (a +b \right ) b^{2}}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 a +2 b}}{d}\) | \(85\) |
| risch | \(\frac {x}{a +b}-\frac {2 a^{2} x}{b^{2} \left (a +b \right )}-\frac {2 a^{2} c}{b^{2} d \left (a +b \right )}+\frac {2 a x}{b^{2}}+\frac {2 a c}{b^{2} d}-\frac {2 x}{b}-\frac {2 c}{b d}+\frac {2 \,{\mathrm e}^{2 d x +2 c}}{b d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}+\frac {a^{2} \ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{2 b^{2} d \left (a +b \right )}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+1\right ) a}{b^{2} d}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{b d}\) | \(191\) |
-1/2*(2*b^2*d*x+tanh(d*x+c)^2*a*b+b^2*tanh(d*x+c)^2+2*ln(1-tanh(d*x+c))*b^ 2-a^2*ln(a+b*tanh(d*x+c)^2))/b^2/d/(a+b)
Leaf count of result is larger than twice the leaf count of optimal. 742 vs. \(2 (62) = 124\).
Time = 0.30 (sec) , antiderivative size = 742, normalized size of antiderivative = 11.24 \[ \int \frac {\tanh ^5(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=-\frac {2 \, b^{2} d x \cosh \left (d x + c\right )^{4} + 8 \, b^{2} d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 2 \, b^{2} d x \sinh \left (d x + c\right )^{4} + 2 \, b^{2} d x + 4 \, {\left (b^{2} d x - a b - b^{2}\right )} \cosh \left (d x + c\right )^{2} + 4 \, {\left (3 \, b^{2} d x \cosh \left (d x + c\right )^{2} + b^{2} d x - a b - b^{2}\right )} \sinh \left (d x + c\right )^{2} - {\left (a^{2} \cosh \left (d x + c\right )^{4} + 4 \, a^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a^{2} \sinh \left (d x + c\right )^{4} + 2 \, a^{2} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{2} \cosh \left (d x + c\right )^{2} + a^{2}\right )} \sinh \left (d x + c\right )^{2} + a^{2} + 4 \, {\left (a^{2} \cosh \left (d x + c\right )^{3} + a^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \log \left (\frac {2 \, {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{2} + {\left (a + b\right )} \sinh \left (d x + c\right )^{2} + a - b\right )}}{\cosh \left (d x + c\right )^{2} - 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2}}\right ) + 2 \, {\left ({\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right )^{4} + 4 \, {\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (a^{2} - b^{2}\right )} \sinh \left (d x + c\right )^{4} + 2 \, {\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right )^{2} + a^{2} - b^{2}\right )} \sinh \left (d x + c\right )^{2} + a^{2} - b^{2} + 4 \, {\left ({\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right )^{3} + {\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \log \left (\frac {2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 8 \, {\left (b^{2} d x \cosh \left (d x + c\right )^{3} + {\left (b^{2} d x - a b - b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{2 \, {\left ({\left (a b^{2} + b^{3}\right )} d \cosh \left (d x + c\right )^{4} + 4 \, {\left (a b^{2} + b^{3}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (a b^{2} + b^{3}\right )} d \sinh \left (d x + c\right )^{4} + 2 \, {\left (a b^{2} + b^{3}\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (a b^{2} + b^{3}\right )} d \cosh \left (d x + c\right )^{2} + {\left (a b^{2} + b^{3}\right )} d\right )} \sinh \left (d x + c\right )^{2} + {\left (a b^{2} + b^{3}\right )} d + 4 \, {\left ({\left (a b^{2} + b^{3}\right )} d \cosh \left (d x + c\right )^{3} + {\left (a b^{2} + b^{3}\right )} d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}} \]
-1/2*(2*b^2*d*x*cosh(d*x + c)^4 + 8*b^2*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*b^2*d*x*sinh(d*x + c)^4 + 2*b^2*d*x + 4*(b^2*d*x - a*b - b^2)*cosh(d*x + c)^2 + 4*(3*b^2*d*x*cosh(d*x + c)^2 + b^2*d*x - a*b - b^2)*sinh(d*x + c )^2 - (a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sin h(d*x + c)^4 + 2*a^2*cosh(d*x + c)^2 + 2*(3*a^2*cosh(d*x + c)^2 + a^2)*sin h(d*x + c)^2 + a^2 + 4*(a^2*cosh(d*x + c)^3 + a^2*cosh(d*x + c))*sinh(d*x + c))*log(2*((a + b)*cosh(d*x + c)^2 + (a + b)*sinh(d*x + c)^2 + a - b)/(c osh(d*x + c)^2 - 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)) + 2*((a ^2 - b^2)*cosh(d*x + c)^4 + 4*(a^2 - b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 - b^2)*sinh(d*x + c)^4 + 2*(a^2 - b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 - b^2)*cosh(d*x + c)^2 + a^2 - b^2)*sinh(d*x + c)^2 + a^2 - b^2 + 4*((a^2 - b^2)*cosh(d*x + c)^3 + (a^2 - b^2)*cosh(d*x + c))*sinh(d*x + c))*log(2*cos h(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 8*(b^2*d*x*cosh(d*x + c)^3 + (b^2*d*x - a*b - b^2)*cosh(d*x + c))*sinh(d*x + c))/((a*b^2 + b^3)*d*cosh (d*x + c)^4 + 4*(a*b^2 + b^3)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a*b^2 + b ^3)*d*sinh(d*x + c)^4 + 2*(a*b^2 + b^3)*d*cosh(d*x + c)^2 + 2*(3*(a*b^2 + b^3)*d*cosh(d*x + c)^2 + (a*b^2 + b^3)*d)*sinh(d*x + c)^2 + (a*b^2 + b^3)* d + 4*((a*b^2 + b^3)*d*cosh(d*x + c)^3 + (a*b^2 + b^3)*d*cosh(d*x + c))*si nh(d*x + c))
Leaf count of result is larger than twice the leaf count of optimal. 415 vs. \(2 (53) = 106\).
Time = 10.07 (sec) , antiderivative size = 415, normalized size of antiderivative = 6.29 \[ \int \frac {\tanh ^5(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\begin {cases} \tilde {\infty } x \tanh ^{3}{\left (c \right )} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {x - \frac {\log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {\tanh ^{4}{\left (c + d x \right )}}{4 d} - \frac {\tanh ^{2}{\left (c + d x \right )}}{2 d}}{a} & \text {for}\: b = 0 \\\frac {4 d x \tanh ^{2}{\left (c + d x \right )}}{2 b d \tanh ^{2}{\left (c + d x \right )} - 2 b d} - \frac {4 d x}{2 b d \tanh ^{2}{\left (c + d x \right )} - 2 b d} - \frac {4 \log {\left (\tanh {\left (c + d x \right )} + 1 \right )} \tanh ^{2}{\left (c + d x \right )}}{2 b d \tanh ^{2}{\left (c + d x \right )} - 2 b d} + \frac {4 \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{2 b d \tanh ^{2}{\left (c + d x \right )} - 2 b d} - \frac {\tanh ^{4}{\left (c + d x \right )}}{2 b d \tanh ^{2}{\left (c + d x \right )} - 2 b d} + \frac {2}{2 b d \tanh ^{2}{\left (c + d x \right )} - 2 b d} & \text {for}\: a = - b \\\frac {x \tanh ^{5}{\left (c \right )}}{a + b \tanh ^{2}{\left (c \right )}} & \text {for}\: d = 0 \\\frac {a^{2} \log {\left (- \sqrt {- \frac {a}{b}} + \tanh {\left (c + d x \right )} \right )}}{2 a b^{2} d + 2 b^{3} d} + \frac {a^{2} \log {\left (\sqrt {- \frac {a}{b}} + \tanh {\left (c + d x \right )} \right )}}{2 a b^{2} d + 2 b^{3} d} - \frac {a b \tanh ^{2}{\left (c + d x \right )}}{2 a b^{2} d + 2 b^{3} d} + \frac {2 b^{2} d x}{2 a b^{2} d + 2 b^{3} d} - \frac {2 b^{2} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{2 a b^{2} d + 2 b^{3} d} - \frac {b^{2} \tanh ^{2}{\left (c + d x \right )}}{2 a b^{2} d + 2 b^{3} d} & \text {otherwise} \end {cases} \]
Piecewise((zoo*x*tanh(c)**3, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((x - log(ta nh(c + d*x) + 1)/d - tanh(c + d*x)**4/(4*d) - tanh(c + d*x)**2/(2*d))/a, E q(b, 0)), (4*d*x*tanh(c + d*x)**2/(2*b*d*tanh(c + d*x)**2 - 2*b*d) - 4*d*x /(2*b*d*tanh(c + d*x)**2 - 2*b*d) - 4*log(tanh(c + d*x) + 1)*tanh(c + d*x) **2/(2*b*d*tanh(c + d*x)**2 - 2*b*d) + 4*log(tanh(c + d*x) + 1)/(2*b*d*tan h(c + d*x)**2 - 2*b*d) - tanh(c + d*x)**4/(2*b*d*tanh(c + d*x)**2 - 2*b*d) + 2/(2*b*d*tanh(c + d*x)**2 - 2*b*d), Eq(a, -b)), (x*tanh(c)**5/(a + b*ta nh(c)**2), Eq(d, 0)), (a**2*log(-sqrt(-a/b) + tanh(c + d*x))/(2*a*b**2*d + 2*b**3*d) + a**2*log(sqrt(-a/b) + tanh(c + d*x))/(2*a*b**2*d + 2*b**3*d) - a*b*tanh(c + d*x)**2/(2*a*b**2*d + 2*b**3*d) + 2*b**2*d*x/(2*a*b**2*d + 2*b**3*d) - 2*b**2*log(tanh(c + d*x) + 1)/(2*a*b**2*d + 2*b**3*d) - b**2*t anh(c + d*x)**2/(2*a*b**2*d + 2*b**3*d), True))
Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (62) = 124\).
Time = 0.28 (sec) , antiderivative size = 133, normalized size of antiderivative = 2.02 \[ \int \frac {\tanh ^5(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {a^{2} \log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \, {\left (a b^{2} + b^{3}\right )} d} + \frac {d x + c}{{\left (a + b\right )} d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (2 \, b e^{\left (-2 \, d x - 2 \, c\right )} + b e^{\left (-4 \, d x - 4 \, c\right )} + b\right )} d} - \frac {{\left (a - b\right )} \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{b^{2} d} \]
1/2*a^2*log(2*(a - b)*e^(-2*d*x - 2*c) + (a + b)*e^(-4*d*x - 4*c) + a + b) /((a*b^2 + b^3)*d) + (d*x + c)/((a + b)*d) + 2*e^(-2*d*x - 2*c)/((2*b*e^(- 2*d*x - 2*c) + b*e^(-4*d*x - 4*c) + b)*d) - (a - b)*log(e^(-2*d*x - 2*c) + 1)/(b^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 132 vs. \(2 (62) = 124\).
Time = 0.35 (sec) , antiderivative size = 132, normalized size of antiderivative = 2.00 \[ \int \frac {\tanh ^5(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {\frac {a^{2} \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}{a b^{2} + b^{3}} - \frac {2 \, {\left (d x + c\right )}}{a + b} - \frac {2 \, {\left (a - b\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{b^{2}} + \frac {4 \, e^{\left (2 \, d x + 2 \, c\right )}}{b {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}}}{2 \, d} \]
1/2*(a^2*log(a*e^(4*d*x + 4*c) + b*e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c) - 2*b*e^(2*d*x + 2*c) + a + b)/(a*b^2 + b^3) - 2*(d*x + c)/(a + b) - 2*(a - b)*log(e^(2*d*x + 2*c) + 1)/b^2 + 4*e^(2*d*x + 2*c)/(b*(e^(2*d*x + 2*c) + 1)^2))/d
Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.09 \[ \int \frac {\tanh ^5(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=-\frac {b^2\,\left (\ln \left (\mathrm {tanh}\left (c+d\,x\right )+1\right )-d\,x+\frac {{\mathrm {tanh}\left (c+d\,x\right )}^2}{2}\right )-\frac {a^2\,\ln \left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}{2}+\frac {a\,b\,{\mathrm {tanh}\left (c+d\,x\right )}^2}{2}}{b^2\,d\,\left (a+b\right )} \]