Integrand size = 23, antiderivative size = 83 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\log (\cosh (c+d x))}{(a+b)^2 d}-\frac {a (a+2 b) \log \left (a+b \tanh ^2(c+d x)\right )}{2 b^2 (a+b)^2 d}-\frac {a^2}{2 b^2 (a+b) d \left (a+b \tanh ^2(c+d x)\right )} \]
ln(cosh(d*x+c))/(a+b)^2/d-1/2*a*(a+2*b)*ln(a+b*tanh(d*x+c)^2)/b^2/(a+b)^2/ d-1/2*a^2/b^2/(a+b)/d/(a+b*tanh(d*x+c)^2)
Time = 0.56 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.83 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {-2 \log (\cosh (c+d x))+\frac {a (a+2 b) \log \left (a+b \tanh ^2(c+d x)\right )}{b^2}+\frac {a^2 (a+b)}{b^2 \left (a+b \tanh ^2(c+d x)\right )}}{2 (a+b)^2 d} \]
-1/2*(-2*Log[Cosh[c + d*x]] + (a*(a + 2*b)*Log[a + b*Tanh[c + d*x]^2])/b^2 + (a^2*(a + b))/(b^2*(a + b*Tanh[c + d*x]^2)))/((a + b)^2*d)
Time = 0.33 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.01, 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, 99, 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.
\(\displaystyle \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \tan (i c+i d x)^5}{\left (a-b \tan (i c+i d x)^2\right )^2}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\tan (i c+i d x)^5}{\left (a-b \tan (i c+i d x)^2\right )^2}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle -\frac {i \int \frac {i \tanh ^5(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\int \frac {\tanh ^5(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {\int \frac {\tanh ^4(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {a^2}{b (a+b) \left (b \tanh ^2(c+d x)+a\right )^2}-\frac {(a+2 b) a}{b (a+b)^2 \left (b \tanh ^2(c+d x)+a\right )}-\frac {1}{(a+b)^2 \left (\tanh ^2(c+d x)-1\right )}\right )d\tanh ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {a^2}{b^2 (a+b) \left (a+b \tanh ^2(c+d x)\right )}-\frac {a (a+2 b) \log \left (a+b \tanh ^2(c+d x)\right )}{b^2 (a+b)^2}-\frac {\log \left (1-\tanh ^2(c+d x)\right )}{(a+b)^2}}{2 d}\) |
(-(Log[1 - Tanh[c + d*x]^2]/(a + b)^2) - (a*(a + 2*b)*Log[a + b*Tanh[c + d *x]^2])/(b^2*(a + b)^2) - a^2/(b^2*(a + b)*(a + b*Tanh[c + d*x]^2)))/(2*d)
3.2.80.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 = 91, normalized size of antiderivative = 1.10
method | result | size |
derivativedivides | \(\frac {-\frac {a \left (\frac {\left (a +2 b \right ) \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{b^{2}}+\frac {a \left (a +b \right )}{b^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right )}\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}}{d}\) | \(91\) |
default | \(\frac {-\frac {a \left (\frac {\left (a +2 b \right ) \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{b^{2}}+\frac {a \left (a +b \right )}{b^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right )}\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}}{d}\) | \(91\) |
parallelrisch | \(-\frac {2 x \tanh \left (d x +c \right )^{2} b^{3} d +2 \ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) \tanh \left (d x +c \right )^{2} a \,b^{2}+\ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) a^{3}+a^{3}+a^{2} b +2 a \,b^{2} d x +2 \ln \left (1-\tanh \left (d x +c \right )\right ) a \,b^{2}+\ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) \tanh \left (d x +c \right )^{2} a^{2} b +2 \ln \left (1-\tanh \left (d x +c \right )\right ) \tanh \left (d x +c \right )^{2} b^{3}+2 \ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) a^{2} b}{2 \left (a +b \tanh \left (d x +c \right )^{2}\right ) d \left (a +b \right )^{2} b^{2}}\) | \(190\) |
risch | \(\frac {x}{a^{2}+2 a b +b^{2}}-\frac {2 x}{b^{2}}-\frac {2 c}{b^{2} d}+\frac {2 a^{2} x}{b^{2} \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 a^{2} c}{b^{2} d \left (a^{2}+2 a b +b^{2}\right )}+\frac {4 a x}{b \left (a^{2}+2 a b +b^{2}\right )}+\frac {4 a c}{b d \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 a^{2} {\mathrm e}^{2 d x +2 c}}{b d \left (a +b \right )^{2} \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{b^{2} d}-\frac {a^{2} \ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{2 b^{2} d \left (a^{2}+2 a b +b^{2}\right )}-\frac {a \ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{b d \left (a^{2}+2 a b +b^{2}\right )}\) | \(329\) |
1/d*(-1/2*a/(a+b)^2*((a+2*b)/b^2*ln(a+b*tanh(d*x+c)^2)+a*(a+b)/b^2/(a+b*ta nh(d*x+c)^2))-1/2/(a+b)^2*ln(tanh(d*x+c)+1)-1/2/(a+b)^2*ln(tanh(d*x+c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 1141 vs. \(2 (79) = 158\).
Time = 0.33 (sec) , antiderivative size = 1141, normalized size of antiderivative = 13.75 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]
-1/2*(2*(a*b^2 + b^3)*d*x*cosh(d*x + c)^4 + 8*(a*b^2 + b^3)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(a*b^2 + b^3)*d*x*sinh(d*x + c)^4 + 2*(a*b^2 + b^3 )*d*x + 4*(a^2*b + (a*b^2 - b^3)*d*x)*cosh(d*x + c)^2 + 4*(3*(a*b^2 + b^3) *d*x*cosh(d*x + c)^2 + a^2*b + (a*b^2 - b^3)*d*x)*sinh(d*x + c)^2 + ((a^3 + 3*a^2*b + 2*a*b^2)*cosh(d*x + c)^4 + 4*(a^3 + 3*a^2*b + 2*a*b^2)*cosh(d* x + c)*sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 2*a*b^2)*sinh(d*x + c)^4 + a^3 + 3*a^2*b + 2*a*b^2 + 2*(a^3 + a^2*b - 2*a*b^2)*cosh(d*x + c)^2 + 2*(a^3 + a^2*b - 2*a*b^2 + 3*(a^3 + 3*a^2*b + 2*a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a^3 + 3*a^2*b + 2*a*b^2)*cosh(d*x + c)^3 + (a^3 + a^2*b - 2*a*b ^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)/(cosh(d*x + c)^2 - 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)) - 2*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 4*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 + 2*(a^3 + a^2*b - a*b^2 - b^3)*cosh(d*x + c)^2 + 2*(a^3 + a^2*b - a*b^2 - b^3 + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^3 + (a^3 + a^2*b - a*b^ 2 - b^3)*cosh(d*x + c))*sinh(d*x + c))*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 8*((a*b^2 + b^3)*d*x*cosh(d*x + c)^3 + (a^2*b + (a*b^2 - b^3)*d*x)*cosh(d*x + c))*sinh(d*x + c))/((a^3*b^2 + 3*a^2*b^3 + 3*a*...
Time = 74.91 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.23 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {a^{2} \left (\begin {cases} \frac {\tanh ^{2}{\left (c + d x \right )}}{a^{2}} & \text {for}\: b = 0 \\- \frac {1}{b \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )} & \text {otherwise} \end {cases}\right )}{2 b d \left (a + b\right )} - \frac {a \left (a + 2 b\right ) \left (\begin {cases} \frac {\tanh ^{2}{\left (c + d x \right )}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b \tanh ^{2}{\left (c + d x \right )} \right )}}{b} & \text {otherwise} \end {cases}\right )}{2 b d \left (a + b\right )^{2}} - \frac {\log {\left (\tanh ^{2}{\left (c + d x \right )} - 1 \right )}}{2 d \left (a + b\right )^{2}} \]
a**2*Piecewise((tanh(c + d*x)**2/a**2, Eq(b, 0)), (-1/(b*(a + b*tanh(c + d *x)**2)), True))/(2*b*d*(a + b)) - a*(a + 2*b)*Piecewise((tanh(c + d*x)**2 /a, Eq(b, 0)), (log(a + b*tanh(c + d*x)**2)/b, True))/(2*b*d*(a + b)**2) - log(tanh(c + d*x)**2 - 1)/(2*d*(a + b)**2)
Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (79) = 158\).
Time = 0.30 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.61 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {2 \, a^{2} e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4} + 2 \, {\left (a^{3} b + a^{2} b^{2} - a b^{3} - b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} - \frac {{\left (a^{2} + 2 \, a b\right )} \log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} d} + \frac {d x + c}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{b^{2} d} \]
-2*a^2*e^(-2*d*x - 2*c)/((a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4 + 2*(a^3*b + a ^2*b^2 - a*b^3 - b^4)*e^(-2*d*x - 2*c) + (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^ 4)*e^(-4*d*x - 4*c))*d) - 1/2*(a^2 + 2*a*b)*log(2*(a - b)*e^(-2*d*x - 2*c) + (a + b)*e^(-4*d*x - 4*c) + a + b)/((a^2*b^2 + 2*a*b^3 + b^4)*d) + (d*x + c)/((a^2 + 2*a*b + b^2)*d) + log(e^(-2*d*x - 2*c) + 1)/(b^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (79) = 158\).
Time = 0.43 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.34 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {\frac {{\left (a^{2} + 2 \, a b\right )} \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}{a^{2} b^{2} + 2 \, a b^{3} + b^{4}} + \frac {2 \, {\left (d x + c\right )}}{a^{2} + 2 \, a b + b^{2}} + \frac {4 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )}}{{\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )} {\left (a + b\right )}^{2} b} - \frac {2 \, \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{b^{2}}}{2 \, d} \]
-1/2*((a^2 + 2*a*b)*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^2*b^2 + 2*a*b^3 + b^4) + 2*(d* x + c)/(a^2 + 2*a*b + b^2) + 4*a^2*e^(2*d*x + 2*c)/((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) - 2*log(e^(2*d*x + 2*c) + 1)/b^2)/d
Time = 2.32 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.05 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {a^2}{2\,\left (d\,a^2\,b^2+d\,a\,b^3\,{\mathrm {tanh}\left (c+d\,x\right )}^2+d\,a\,b^3+d\,b^4\,{\mathrm {tanh}\left (c+d\,x\right )}^2\right )}-\frac {\ln \left ({\mathrm {tanh}\left (c+d\,x\right )}^2-1\right )}{2\,\left (d\,a^2+2\,d\,a\,b+d\,b^2\right )}-\frac {a^2\,\ln \left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}{2\,\left (d\,a^2\,b^2+2\,d\,a\,b^3+d\,b^4\right )}-\frac {a\,b\,\ln \left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}{d\,a^2\,b^2+2\,d\,a\,b^3+d\,b^4} \]