Integrand size = 12, antiderivative size = 100 \[ \int (a+a \tanh (c+d x))^5 \, dx=16 a^5 x+\frac {16 a^5 \log (\cosh (c+d x))}{d}-\frac {8 a^5 \tanh (c+d x)}{d}-\frac {2 a^2 (a+a \tanh (c+d x))^3}{3 d}-\frac {a (a+a \tanh (c+d x))^4}{4 d}-\frac {2 a \left (a^2+a^2 \tanh (c+d x)\right )^2}{d} \]
16*a^5*x+16*a^5*ln(cosh(d*x+c))/d-8*a^5*tanh(d*x+c)/d-2/3*a^2*(a+a*tanh(d* x+c))^3/d-1/4*a*(a+a*tanh(d*x+c))^4/d-2*a*(a^2+a^2*tanh(d*x+c))^2/d
Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.63 \[ \int (a+a \tanh (c+d x))^5 \, dx=-\frac {a^5 \left (35+192 \log (1-\tanh (c+d x))+180 \tanh (c+d x)+66 \tanh ^2(c+d x)+20 \tanh ^3(c+d x)+3 \tanh ^4(c+d x)\right )}{12 d} \]
-1/12*(a^5*(35 + 192*Log[1 - Tanh[c + d*x]] + 180*Tanh[c + d*x] + 66*Tanh[ c + d*x]^2 + 20*Tanh[c + d*x]^3 + 3*Tanh[c + d*x]^4))/d
Time = 0.56 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.08, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 3959, 3042, 3959, 3042, 3959, 3042, 3958, 26, 3042, 26, 3956}
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.
\(\displaystyle \int (a \tanh (c+d x)+a)^5 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a-i a \tan (i c+i d x))^5dx\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle 2 a \int (\tanh (c+d x) a+a)^4dx-\frac {a (a \tanh (c+d x)+a)^4}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a (a \tanh (c+d x)+a)^4}{4 d}+2 a \int (a-i a \tan (i c+i d x))^4dx\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle 2 a \left (2 a \int (\tanh (c+d x) a+a)^3dx-\frac {a (a \tanh (c+d x)+a)^3}{3 d}\right )-\frac {a (a \tanh (c+d x)+a)^4}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a (a \tanh (c+d x)+a)^4}{4 d}+2 a \left (-\frac {a (a \tanh (c+d x)+a)^3}{3 d}+2 a \int (a-i a \tan (i c+i d x))^3dx\right )\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle 2 a \left (2 a \left (2 a \int (\tanh (c+d x) a+a)^2dx-\frac {a (a \tanh (c+d x)+a)^2}{2 d}\right )-\frac {a (a \tanh (c+d x)+a)^3}{3 d}\right )-\frac {a (a \tanh (c+d x)+a)^4}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a (a \tanh (c+d x)+a)^4}{4 d}+2 a \left (-\frac {a (a \tanh (c+d x)+a)^3}{3 d}+2 a \left (-\frac {a (a \tanh (c+d x)+a)^2}{2 d}+2 a \int (a-i a \tan (i c+i d x))^2dx\right )\right )\) |
\(\Big \downarrow \) 3958 |
\(\displaystyle -\frac {a (a \tanh (c+d x)+a)^4}{4 d}+2 a \left (-\frac {a (a \tanh (c+d x)+a)^3}{3 d}+2 a \left (-\frac {a (a \tanh (c+d x)+a)^2}{2 d}+2 a \left (-2 i a^2 \int i \tanh (c+d x)dx-\frac {a^2 \tanh (c+d x)}{d}+2 a^2 x\right )\right )\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle 2 a \left (2 a \left (2 a \left (2 a^2 \int \tanh (c+d x)dx-\frac {a^2 \tanh (c+d x)}{d}+2 a^2 x\right )-\frac {a (a \tanh (c+d x)+a)^2}{2 d}\right )-\frac {a (a \tanh (c+d x)+a)^3}{3 d}\right )-\frac {a (a \tanh (c+d x)+a)^4}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a (a \tanh (c+d x)+a)^4}{4 d}+2 a \left (-\frac {a (a \tanh (c+d x)+a)^3}{3 d}+2 a \left (-\frac {a (a \tanh (c+d x)+a)^2}{2 d}+2 a \left (2 a^2 \int -i \tan (i c+i d x)dx-\frac {a^2 \tanh (c+d x)}{d}+2 a^2 x\right )\right )\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -\frac {a (a \tanh (c+d x)+a)^4}{4 d}+2 a \left (-\frac {a (a \tanh (c+d x)+a)^3}{3 d}+2 a \left (-\frac {a (a \tanh (c+d x)+a)^2}{2 d}+2 a \left (-2 i a^2 \int \tan (i c+i d x)dx-\frac {a^2 \tanh (c+d x)}{d}+2 a^2 x\right )\right )\right )\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle 2 a \left (2 a \left (2 a \left (-\frac {a^2 \tanh (c+d x)}{d}+\frac {2 a^2 \log (\cosh (c+d x))}{d}+2 a^2 x\right )-\frac {a (a \tanh (c+d x)+a)^2}{2 d}\right )-\frac {a (a \tanh (c+d x)+a)^3}{3 d}\right )-\frac {a (a \tanh (c+d x)+a)^4}{4 d}\) |
-1/4*(a*(a + a*Tanh[c + d*x])^4)/d + 2*a*(-1/3*(a*(a + a*Tanh[c + d*x])^3) /d + 2*a*(-1/2*(a*(a + a*Tanh[c + d*x])^2)/d + 2*a*(2*a^2*x + (2*a^2*Log[C osh[c + d*x]])/d - (a^2*Tanh[c + d*x])/d)))
3.1.41.3.1 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[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2) *x, x] + (Simp[b^2*(Tan[c + d*x]/d), x] + Simp[2*a*b Int[Tan[c + d*x], x] , x]) /; FreeQ[{a, b, c, d}, x]
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[2*a Int[(a + b*Tan[c + d* x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && GtQ[n , 1]
Time = 0.10 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.58
method | result | size |
derivativedivides | \(\frac {a^{5} \left (-\frac {\tanh \left (d x +c \right )^{4}}{4}-\frac {5 \tanh \left (d x +c \right )^{3}}{3}-\frac {11 \tanh \left (d x +c \right )^{2}}{2}-15 \tanh \left (d x +c \right )-16 \ln \left (\tanh \left (d x +c \right )-1\right )\right )}{d}\) | \(58\) |
default | \(\frac {a^{5} \left (-\frac {\tanh \left (d x +c \right )^{4}}{4}-\frac {5 \tanh \left (d x +c \right )^{3}}{3}-\frac {11 \tanh \left (d x +c \right )^{2}}{2}-15 \tanh \left (d x +c \right )-16 \ln \left (\tanh \left (d x +c \right )-1\right )\right )}{d}\) | \(58\) |
parallelrisch | \(-\frac {3 \tanh \left (d x +c \right )^{4} a^{5}+20 \tanh \left (d x +c \right )^{3} a^{5}+66 \tanh \left (d x +c \right )^{2} a^{5}+192 \ln \left (1-\tanh \left (d x +c \right )\right ) a^{5}+180 a^{5} \tanh \left (d x +c \right )}{12 d}\) | \(73\) |
risch | \(-\frac {32 a^{5} c}{d}+\frac {4 a^{5} \left (48 \,{\mathrm e}^{6 d x +6 c}+108 \,{\mathrm e}^{4 d x +4 c}+88 \,{\mathrm e}^{2 d x +2 c}+25\right )}{3 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{4}}+\frac {16 a^{5} \ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{d}\) | \(87\) |
parts | \(a^{5} x +\frac {a^{5} \left (-\frac {\tanh \left (d x +c \right )^{4}}{4}-\frac {\tanh \left (d x +c \right )^{2}}{2}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}+\frac {5 a^{5} \ln \left (\cosh \left (d x +c \right )\right )}{d}+\frac {10 a^{5} \left (-\tanh \left (d x +c \right )-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}+\frac {10 a^{5} \left (-\frac {\tanh \left (d x +c \right )^{2}}{2}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}+\frac {5 a^{5} \left (-\frac {\tanh \left (d x +c \right )^{3}}{3}-\tanh \left (d x +c \right )-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}\) | \(201\) |
1/d*a^5*(-1/4*tanh(d*x+c)^4-5/3*tanh(d*x+c)^3-11/2*tanh(d*x+c)^2-15*tanh(d *x+c)-16*ln(tanh(d*x+c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 907 vs. \(2 (96) = 192\).
Time = 0.26 (sec) , antiderivative size = 907, normalized size of antiderivative = 9.07 \[ \int (a+a \tanh (c+d x))^5 \, dx=\text {Too large to display} \]
4/3*(48*a^5*cosh(d*x + c)^6 + 288*a^5*cosh(d*x + c)*sinh(d*x + c)^5 + 48*a ^5*sinh(d*x + c)^6 + 108*a^5*cosh(d*x + c)^4 + 88*a^5*cosh(d*x + c)^2 + 25 *a^5 + 36*(20*a^5*cosh(d*x + c)^2 + 3*a^5)*sinh(d*x + c)^4 + 48*(20*a^5*co sh(d*x + c)^3 + 9*a^5*cosh(d*x + c))*sinh(d*x + c)^3 + 8*(90*a^5*cosh(d*x + c)^4 + 81*a^5*cosh(d*x + c)^2 + 11*a^5)*sinh(d*x + c)^2 + 12*(a^5*cosh(d *x + c)^8 + 8*a^5*cosh(d*x + c)*sinh(d*x + c)^7 + a^5*sinh(d*x + c)^8 + 4* a^5*cosh(d*x + c)^6 + 6*a^5*cosh(d*x + c)^4 + 4*a^5*cosh(d*x + c)^2 + 4*(7 *a^5*cosh(d*x + c)^2 + a^5)*sinh(d*x + c)^6 + 8*(7*a^5*cosh(d*x + c)^3 + 3 *a^5*cosh(d*x + c))*sinh(d*x + c)^5 + a^5 + 2*(35*a^5*cosh(d*x + c)^4 + 30 *a^5*cosh(d*x + c)^2 + 3*a^5)*sinh(d*x + c)^4 + 8*(7*a^5*cosh(d*x + c)^5 + 10*a^5*cosh(d*x + c)^3 + 3*a^5*cosh(d*x + c))*sinh(d*x + c)^3 + 4*(7*a^5* cosh(d*x + c)^6 + 15*a^5*cosh(d*x + c)^4 + 9*a^5*cosh(d*x + c)^2 + a^5)*si nh(d*x + c)^2 + 8*(a^5*cosh(d*x + c)^7 + 3*a^5*cosh(d*x + c)^5 + 3*a^5*cos h(d*x + c)^3 + a^5*cosh(d*x + c))*sinh(d*x + c))*log(2*cosh(d*x + c)/(cosh (d*x + c) - sinh(d*x + c))) + 16*(18*a^5*cosh(d*x + c)^5 + 27*a^5*cosh(d*x + c)^3 + 11*a^5*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^8 + 8*d*co sh(d*x + c)*sinh(d*x + c)^7 + d*sinh(d*x + c)^8 + 4*d*cosh(d*x + c)^6 + 4* (7*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^6 + 8*(7*d*cosh(d*x + c)^3 + 3*d*c osh(d*x + c))*sinh(d*x + c)^5 + 6*d*cosh(d*x + c)^4 + 2*(35*d*cosh(d*x + c )^4 + 30*d*cosh(d*x + c)^2 + 3*d)*sinh(d*x + c)^4 + 8*(7*d*cosh(d*x + c...
Time = 0.13 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.95 \[ \int (a+a \tanh (c+d x))^5 \, dx=\begin {cases} 32 a^{5} x - \frac {16 a^{5} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {a^{5} \tanh ^{4}{\left (c + d x \right )}}{4 d} - \frac {5 a^{5} \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac {11 a^{5} \tanh ^{2}{\left (c + d x \right )}}{2 d} - \frac {15 a^{5} \tanh {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \tanh {\left (c \right )} + a\right )^{5} & \text {otherwise} \end {cases} \]
Piecewise((32*a**5*x - 16*a**5*log(tanh(c + d*x) + 1)/d - a**5*tanh(c + d* x)**4/(4*d) - 5*a**5*tanh(c + d*x)**3/(3*d) - 11*a**5*tanh(c + d*x)**2/(2* d) - 15*a**5*tanh(c + d*x)/d, Ne(d, 0)), (x*(a*tanh(c) + a)**5, True))
Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (96) = 192\).
Time = 0.28 (sec) , antiderivative size = 302, normalized size of antiderivative = 3.02 \[ \int (a+a \tanh (c+d x))^5 \, dx=\frac {5}{3} \, a^{5} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + a^{5} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {4 \, {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}}\right )} + 10 \, a^{5} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + 10 \, a^{5} {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + a^{5} x + \frac {5 \, a^{5} \log \left (\cosh \left (d x + c\right )\right )}{d} \]
5/3*a^5*(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))) + a^5*( x + c/d + log(e^(-2*d*x - 2*c) + 1)/d + 4*(e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))/(d*(4*e^(-2*d*x - 2*c) + 6*e^(-4*d*x - 4*c) + 4*e ^(-6*d*x - 6*c) + e^(-8*d*x - 8*c) + 1))) + 10*a^5*(x + c/d + log(e^(-2*d* x - 2*c) + 1)/d + 2*e^(-2*d*x - 2*c)/(d*(2*e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) + 1))) + 10*a^5*(x + c/d - 2/(d*(e^(-2*d*x - 2*c) + 1))) + a^5*x + 5* a^5*log(cosh(d*x + c))/d
Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.85 \[ \int (a+a \tanh (c+d x))^5 \, dx=\frac {4 \, {\left (12 \, a^{5} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) + \frac {48 \, a^{5} e^{\left (6 \, d x + 6 \, c\right )} + 108 \, a^{5} e^{\left (4 \, d x + 4 \, c\right )} + 88 \, a^{5} e^{\left (2 \, d x + 2 \, c\right )} + 25 \, a^{5}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{4}}\right )}}{3 \, d} \]
4/3*(12*a^5*log(e^(2*d*x + 2*c) + 1) + (48*a^5*e^(6*d*x + 6*c) + 108*a^5*e ^(4*d*x + 4*c) + 88*a^5*e^(2*d*x + 2*c) + 25*a^5)/(e^(2*d*x + 2*c) + 1)^4) /d
Time = 0.15 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.65 \[ \int (a+a \tanh (c+d x))^5 \, dx=32\,a^5\,x-\frac {a^5\,\left (192\,\ln \left (\mathrm {tanh}\left (c+d\,x\right )+1\right )+180\,\mathrm {tanh}\left (c+d\,x\right )+66\,{\mathrm {tanh}\left (c+d\,x\right )}^2+20\,{\mathrm {tanh}\left (c+d\,x\right )}^3+3\,{\mathrm {tanh}\left (c+d\,x\right )}^4\right )}{12\,d} \]