Integrand size = 13, antiderivative size = 64 \[ \int \frac {\tanh ^3(x)}{a+b \tanh (x)} \, dx=-\frac {b x}{a^2-b^2}+\frac {a \log (\cosh (x))}{a^2-b^2}+\frac {a^3 \log (a+b \tanh (x))}{b^2 \left (a^2-b^2\right )}-\frac {\tanh (x)}{b} \]
Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.02 \[ \int \frac {\tanh ^3(x)}{a+b \tanh (x)} \, dx=-\frac {\log (1-\tanh (x))}{2 (a+b)}-\frac {\log (1+\tanh (x))}{2 (a-b)}+\frac {a^3 \log (a+b \tanh (x))}{b^2 \left (a^2-b^2\right )}-\frac {\tanh (x)}{b} \]
-1/2*Log[1 - Tanh[x]]/(a + b) - Log[1 + Tanh[x]]/(2*(a - b)) + (a^3*Log[a + b*Tanh[x]])/(b^2*(a^2 - b^2)) - Tanh[x]/b
Result contains complex when optimal does not.
Time = 0.55 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.27, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.923, Rules used = {3042, 26, 4049, 25, 3042, 4109, 26, 3042, 26, 3956, 4100, 16}
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.
\(\displaystyle \int \frac {\tanh ^3(x)}{a+b \tanh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i \tan (i x)^3}{a-i b \tan (i x)}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {\tan (i x)^3}{a-i b \tan (i x)}dx\) |
\(\Big \downarrow \) 4049 |
\(\displaystyle i \left (\frac {i \int -\frac {-a \tanh ^2(x)+b \tanh (x)+a}{a+b \tanh (x)}dx}{b}+\frac {i \tanh (x)}{b}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle i \left (\frac {i \tanh (x)}{b}-\frac {i \int \frac {-a \tanh ^2(x)+b \tanh (x)+a}{a+b \tanh (x)}dx}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle i \left (\frac {i \tanh (x)}{b}-\frac {i \int \frac {a \tan (i x)^2-i b \tan (i x)+a}{a-i b \tan (i x)}dx}{b}\right )\) |
\(\Big \downarrow \) 4109 |
\(\displaystyle i \left (\frac {i \tanh (x)}{b}-\frac {i \left (-\frac {i a b \int i \tanh (x)dx}{a^2-b^2}+\frac {a^3 \int \frac {1-\tanh ^2(x)}{a+b \tanh (x)}dx}{a^2-b^2}-\frac {b^2 x}{a^2-b^2}\right )}{b}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \left (\frac {i \tanh (x)}{b}-\frac {i \left (\frac {a b \int \tanh (x)dx}{a^2-b^2}+\frac {a^3 \int \frac {1-\tanh ^2(x)}{a+b \tanh (x)}dx}{a^2-b^2}-\frac {b^2 x}{a^2-b^2}\right )}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle i \left (\frac {i \tanh (x)}{b}-\frac {i \left (\frac {a b \int -i \tan (i x)dx}{a^2-b^2}+\frac {a^3 \int \frac {\tan (i x)^2+1}{a-i b \tan (i x)}dx}{a^2-b^2}-\frac {b^2 x}{a^2-b^2}\right )}{b}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \left (\frac {i \tanh (x)}{b}-\frac {i \left (-\frac {i a b \int \tan (i x)dx}{a^2-b^2}+\frac {a^3 \int \frac {\tan (i x)^2+1}{a-i b \tan (i x)}dx}{a^2-b^2}-\frac {b^2 x}{a^2-b^2}\right )}{b}\right )\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle i \left (\frac {i \tanh (x)}{b}-\frac {i \left (\frac {a^3 \int \frac {\tan (i x)^2+1}{a-i b \tan (i x)}dx}{a^2-b^2}-\frac {b^2 x}{a^2-b^2}+\frac {a b \log (\cosh (x))}{a^2-b^2}\right )}{b}\right )\) |
\(\Big \downarrow \) 4100 |
\(\displaystyle i \left (\frac {i \tanh (x)}{b}-\frac {i \left (\frac {a^3 \int \frac {1}{a+b \tanh (x)}d(b \tanh (x))}{b \left (a^2-b^2\right )}-\frac {b^2 x}{a^2-b^2}+\frac {a b \log (\cosh (x))}{a^2-b^2}\right )}{b}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle i \left (\frac {i \tanh (x)}{b}-\frac {i \left (-\frac {b^2 x}{a^2-b^2}+\frac {a b \log (\cosh (x))}{a^2-b^2}+\frac {a^3 \log (a+b \tanh (x))}{b \left (a^2-b^2\right )}\right )}{b}\right )\) |
I*(((-I)*(-((b^2*x)/(a^2 - b^2)) + (a*b*Log[Cosh[x]])/(a^2 - b^2) + (a^3*L og[a + b*Tanh[x]])/(b*(a^2 - b^2))))/b + (I*Tanh[x])/b)
3.2.35.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
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[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 , 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I ntegerQ[m]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) )
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A/(b*f) Subst[Int[(a + x)^m, x], x, b* Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A, C]
Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2 )/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*A + b*B - a *C)*(x/(a^2 + b^2)), x] + (Simp[(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2) Int[( 1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x]), x], x] - Simp[(A*b - a*B - b*C)/( a^2 + b^2) Int[Tan[e + f*x], x], x]) /; FreeQ[{a, b, e, f, A, B, C}, x] & & NeQ[A*b^2 - a*b*B + a^2*C, 0] && NeQ[a^2 + b^2, 0] && NeQ[A*b - a*B - b*C , 0]
Time = 0.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.05
method | result | size |
derivativedivides | \(-\frac {\tanh \left (x \right )}{b}-\frac {\ln \left (1+\tanh \left (x \right )\right )}{2 a -2 b}+\frac {a^{3} \ln \left (a +b \tanh \left (x \right )\right )}{b^{2} \left (a +b \right ) \left (a -b \right )}-\frac {\ln \left (\tanh \left (x \right )-1\right )}{2 a +2 b}\) | \(67\) |
default | \(-\frac {\tanh \left (x \right )}{b}-\frac {\ln \left (1+\tanh \left (x \right )\right )}{2 a -2 b}+\frac {a^{3} \ln \left (a +b \tanh \left (x \right )\right )}{b^{2} \left (a +b \right ) \left (a -b \right )}-\frac {\ln \left (\tanh \left (x \right )-1\right )}{2 a +2 b}\) | \(67\) |
parallelrisch | \(-\frac {\ln \left (1-\tanh \left (x \right )\right ) a \,b^{2}-a^{3} \ln \left (a +b \tanh \left (x \right )\right )+a \,b^{2} x +b^{3} x +\tanh \left (x \right ) a^{2} b -\tanh \left (x \right ) b^{3}}{b^{2} \left (a^{2}-b^{2}\right )}\) | \(67\) |
risch | \(\frac {x}{a +b}-\frac {2 a^{3} x}{b^{2} \left (a^{2}-b^{2}\right )}+\frac {2 a x}{b^{2}}+\frac {2}{b \left (1+{\mathrm e}^{2 x}\right )}+\frac {a^{3} \ln \left ({\mathrm e}^{2 x}+\frac {a -b}{a +b}\right )}{b^{2} \left (a^{2}-b^{2}\right )}-\frac {a \ln \left (1+{\mathrm e}^{2 x}\right )}{b^{2}}\) | \(97\) |
-tanh(x)/b-1/(2*a-2*b)*ln(1+tanh(x))+1/b^2*a^3/(a+b)/(a-b)*ln(a+b*tanh(x)) -1/(2*a+2*b)*ln(tanh(x)-1)
Leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (64) = 128\).
Time = 0.26 (sec) , antiderivative size = 264, normalized size of antiderivative = 4.12 \[ \int \frac {\tanh ^3(x)}{a+b \tanh (x)} \, dx=-\frac {{\left (a b^{2} + b^{3}\right )} x \cosh \left (x\right )^{2} + 2 \, {\left (a b^{2} + b^{3}\right )} x \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a b^{2} + b^{3}\right )} x \sinh \left (x\right )^{2} - 2 \, a^{2} b + 2 \, b^{3} + {\left (a b^{2} + b^{3}\right )} x - {\left (a^{3} \cosh \left (x\right )^{2} + 2 \, a^{3} \cosh \left (x\right ) \sinh \left (x\right ) + a^{3} \sinh \left (x\right )^{2} + a^{3}\right )} \log \left (\frac {2 \, {\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + {\left (a^{3} - a b^{2} + {\left (a^{3} - a b^{2}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{3} - a b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a^{3} - a b^{2}\right )} \sinh \left (x\right )^{2}\right )} \log \left (\frac {2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} b^{2} - b^{4} + {\left (a^{2} b^{2} - b^{4}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{2} b^{2} - b^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a^{2} b^{2} - b^{4}\right )} \sinh \left (x\right )^{2}} \]
-((a*b^2 + b^3)*x*cosh(x)^2 + 2*(a*b^2 + b^3)*x*cosh(x)*sinh(x) + (a*b^2 + b^3)*x*sinh(x)^2 - 2*a^2*b + 2*b^3 + (a*b^2 + b^3)*x - (a^3*cosh(x)^2 + 2 *a^3*cosh(x)*sinh(x) + a^3*sinh(x)^2 + a^3)*log(2*(a*cosh(x) + b*sinh(x))/ (cosh(x) - sinh(x))) + (a^3 - a*b^2 + (a^3 - a*b^2)*cosh(x)^2 + 2*(a^3 - a *b^2)*cosh(x)*sinh(x) + (a^3 - a*b^2)*sinh(x)^2)*log(2*cosh(x)/(cosh(x) - sinh(x))))/(a^2*b^2 - b^4 + (a^2*b^2 - b^4)*cosh(x)^2 + 2*(a^2*b^2 - b^4)* cosh(x)*sinh(x) + (a^2*b^2 - b^4)*sinh(x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (49) = 98\).
Time = 0.31 (sec) , antiderivative size = 330, normalized size of antiderivative = 5.16 \[ \int \frac {\tanh ^3(x)}{a+b \tanh (x)} \, dx=\begin {cases} \tilde {\infty } \left (x - \tanh {\left (x \right )}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {x - \log {\left (\tanh {\left (x \right )} + 1 \right )} - \frac {\tanh ^{2}{\left (x \right )}}{2}}{a} & \text {for}\: b = 0 \\\frac {5 x \tanh {\left (x \right )}}{2 b \tanh {\left (x \right )} - 2 b} - \frac {5 x}{2 b \tanh {\left (x \right )} - 2 b} - \frac {2 \log {\left (\tanh {\left (x \right )} + 1 \right )} \tanh {\left (x \right )}}{2 b \tanh {\left (x \right )} - 2 b} + \frac {2 \log {\left (\tanh {\left (x \right )} + 1 \right )}}{2 b \tanh {\left (x \right )} - 2 b} - \frac {2 \tanh ^{2}{\left (x \right )}}{2 b \tanh {\left (x \right )} - 2 b} + \frac {3}{2 b \tanh {\left (x \right )} - 2 b} & \text {for}\: a = - b \\\frac {x \tanh {\left (x \right )}}{2 b \tanh {\left (x \right )} + 2 b} + \frac {x}{2 b \tanh {\left (x \right )} + 2 b} + \frac {2 \log {\left (\tanh {\left (x \right )} + 1 \right )} \tanh {\left (x \right )}}{2 b \tanh {\left (x \right )} + 2 b} + \frac {2 \log {\left (\tanh {\left (x \right )} + 1 \right )}}{2 b \tanh {\left (x \right )} + 2 b} - \frac {2 \tanh ^{2}{\left (x \right )}}{2 b \tanh {\left (x \right )} + 2 b} + \frac {3}{2 b \tanh {\left (x \right )} + 2 b} & \text {for}\: a = b \\\frac {a^{3} \log {\left (\frac {a}{b} + \tanh {\left (x \right )} \right )}}{a^{2} b^{2} - b^{4}} - \frac {a^{2} b \tanh {\left (x \right )}}{a^{2} b^{2} - b^{4}} + \frac {a b^{2} x}{a^{2} b^{2} - b^{4}} - \frac {a b^{2} \log {\left (\tanh {\left (x \right )} + 1 \right )}}{a^{2} b^{2} - b^{4}} - \frac {b^{3} x}{a^{2} b^{2} - b^{4}} + \frac {b^{3} \tanh {\left (x \right )}}{a^{2} b^{2} - b^{4}} & \text {otherwise} \end {cases} \]
Piecewise((zoo*(x - tanh(x)), Eq(a, 0) & Eq(b, 0)), ((x - log(tanh(x) + 1) - tanh(x)**2/2)/a, Eq(b, 0)), (5*x*tanh(x)/(2*b*tanh(x) - 2*b) - 5*x/(2*b *tanh(x) - 2*b) - 2*log(tanh(x) + 1)*tanh(x)/(2*b*tanh(x) - 2*b) + 2*log(t anh(x) + 1)/(2*b*tanh(x) - 2*b) - 2*tanh(x)**2/(2*b*tanh(x) - 2*b) + 3/(2* b*tanh(x) - 2*b), Eq(a, -b)), (x*tanh(x)/(2*b*tanh(x) + 2*b) + x/(2*b*tanh (x) + 2*b) + 2*log(tanh(x) + 1)*tanh(x)/(2*b*tanh(x) + 2*b) + 2*log(tanh(x ) + 1)/(2*b*tanh(x) + 2*b) - 2*tanh(x)**2/(2*b*tanh(x) + 2*b) + 3/(2*b*tan h(x) + 2*b), Eq(a, b)), (a**3*log(a/b + tanh(x))/(a**2*b**2 - b**4) - a**2 *b*tanh(x)/(a**2*b**2 - b**4) + a*b**2*x/(a**2*b**2 - b**4) - a*b**2*log(t anh(x) + 1)/(a**2*b**2 - b**4) - b**3*x/(a**2*b**2 - b**4) + b**3*tanh(x)/ (a**2*b**2 - b**4), True))
Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.11 \[ \int \frac {\tanh ^3(x)}{a+b \tanh (x)} \, dx=\frac {a^{3} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{2} b^{2} - b^{4}} + \frac {x}{a + b} - \frac {a \log \left (e^{\left (-2 \, x\right )} + 1\right )}{b^{2}} - \frac {2}{b e^{\left (-2 \, x\right )} + b} \]
a^3*log(-(a - b)*e^(-2*x) - a - b)/(a^2*b^2 - b^4) + x/(a + b) - a*log(e^( -2*x) + 1)/b^2 - 2/(b*e^(-2*x) + b)
Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.17 \[ \int \frac {\tanh ^3(x)}{a+b \tanh (x)} \, dx=\frac {a^{3} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{2} b^{2} - b^{4}} - \frac {x}{a - b} - \frac {a \log \left (e^{\left (2 \, x\right )} + 1\right )}{b^{2}} + \frac {2}{b {\left (e^{\left (2 \, x\right )} + 1\right )}} \]
a^3*log(abs(a*e^(2*x) + b*e^(2*x) + a - b))/(a^2*b^2 - b^4) - x/(a - b) - a*log(e^(2*x) + 1)/b^2 + 2/(b*(e^(2*x) + 1))
Time = 0.13 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.92 \[ \int \frac {\tanh ^3(x)}{a+b \tanh (x)} \, dx=\frac {x}{a+b}-\frac {\mathrm {tanh}\left (x\right )}{b}-\frac {a\,\ln \left (\mathrm {tanh}\left (x\right )+1\right )}{a^2-b^2}+\frac {a^3\,\ln \left (a+b\,\mathrm {tanh}\left (x\right )\right )}{b^2\,\left (a^2-b^2\right )} \]