Integrand size = 14, antiderivative size = 88 \[ \int \left (-\tanh ^2(c+d x)\right )^{5/2} \, dx=\frac {\coth (c+d x) \log (\cosh (c+d x)) \sqrt {-\tanh ^2(c+d x)}}{d}-\frac {\tanh (c+d x) \sqrt {-\tanh ^2(c+d x)}}{2 d}-\frac {\tanh ^3(c+d x) \sqrt {-\tanh ^2(c+d x)}}{4 d} \] Output:
coth(d*x+c)*ln(cosh(d*x+c))*(-tanh(d*x+c)^2)^(1/2)/d-1/2*tanh(d*x+c)*(-tan h(d*x+c)^2)^(1/2)/d-1/4*tanh(d*x+c)^3*(-tanh(d*x+c)^2)^(1/2)/d
Time = 0.06 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.67 \[ \int \left (-\tanh ^2(c+d x)\right )^{5/2} \, dx=\frac {\coth ^5(c+d x) \left (4 \log (\cosh (c+d x))+4 \text {sech}^2(c+d x)-\text {sech}^4(c+d x)\right ) \left (-\tanh ^2(c+d x)\right )^{5/2}}{4 d} \] Input:
Integrate[(-Tanh[c + d*x]^2)^(5/2),x]
Output:
(Coth[c + d*x]^5*(4*Log[Cosh[c + d*x]] + 4*Sech[c + d*x]^2 - Sech[c + d*x] ^4)*(-Tanh[c + d*x]^2)^(5/2))/(4*d)
Result contains complex when optimal does not.
Time = 0.48 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.83, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.929, Rules used = {3042, 4141, 3042, 26, 3954, 26, 3042, 26, 3954, 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.
\(\displaystyle \int \left (-\tanh ^2(c+d x)\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (\tan (i c+i d x)^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle \sqrt {-\tanh ^2(c+d x)} \coth (c+d x) \int \tanh ^5(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {-\tanh ^2(c+d x)} \coth (c+d x) \int -i \tan (i c+i d x)^5dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \sqrt {-\tanh ^2(c+d x)} \coth (c+d x) \int \tan (i c+i d x)^5dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -i \sqrt {-\tanh ^2(c+d x)} \coth (c+d x) \left (-\int -i \tanh ^3(c+d x)dx-\frac {i \tanh ^4(c+d x)}{4 d}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \sqrt {-\tanh ^2(c+d x)} \coth (c+d x) \left (i \int \tanh ^3(c+d x)dx-\frac {i \tanh ^4(c+d x)}{4 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i \sqrt {-\tanh ^2(c+d x)} \coth (c+d x) \left (i \int i \tan (i c+i d x)^3dx-\frac {i \tanh ^4(c+d x)}{4 d}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \sqrt {-\tanh ^2(c+d x)} \coth (c+d x) \left (-\int \tan (i c+i d x)^3dx-\frac {i \tanh ^4(c+d x)}{4 d}\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -i \sqrt {-\tanh ^2(c+d x)} \coth (c+d x) \left (\int i \tanh (c+d x)dx-\frac {i \tanh ^4(c+d x)}{4 d}-\frac {i \tanh ^2(c+d x)}{2 d}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \sqrt {-\tanh ^2(c+d x)} \coth (c+d x) \left (i \int \tanh (c+d x)dx-\frac {i \tanh ^4(c+d x)}{4 d}-\frac {i \tanh ^2(c+d x)}{2 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i \sqrt {-\tanh ^2(c+d x)} \coth (c+d x) \left (i \int -i \tan (i c+i d x)dx-\frac {i \tanh ^4(c+d x)}{4 d}-\frac {i \tanh ^2(c+d x)}{2 d}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \sqrt {-\tanh ^2(c+d x)} \coth (c+d x) \left (\int \tan (i c+i d x)dx-\frac {i \tanh ^4(c+d x)}{4 d}-\frac {i \tanh ^2(c+d x)}{2 d}\right )\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -i \sqrt {-\tanh ^2(c+d x)} \coth (c+d x) \left (-\frac {i \tanh ^4(c+d x)}{4 d}-\frac {i \tanh ^2(c+d x)}{2 d}+\frac {i \log (\cosh (c+d x))}{d}\right )\) |
Input:
Int[(-Tanh[c + d*x]^2)^(5/2),x]
Output:
(-I)*Coth[c + d*x]*Sqrt[-Tanh[c + d*x]^2]*((I*Log[Cosh[c + d*x]])/d - ((I/ 2)*Tanh[c + d*x]^2)/d - ((I/4)*Tanh[c + d*x]^4)/d)
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[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d *x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2 Int[(b*Tan[c + d*x])^(n - 2), x] , x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^ n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Ta n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / ; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
Time = 0.40 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(-\frac {\left (-\tanh \left (d x +c \right )^{2}\right )^{\frac {5}{2}} \left (\tanh \left (d x +c \right )^{4}+2 \tanh \left (d x +c \right )^{2}+2 \ln \left (\tanh \left (d x +c \right )-1\right )+2 \ln \left (\tanh \left (d x +c \right )+1\right )\right )}{4 d \tanh \left (d x +c \right )^{5}}\) | \(67\) |
default | \(-\frac {\left (-\tanh \left (d x +c \right )^{2}\right )^{\frac {5}{2}} \left (\tanh \left (d x +c \right )^{4}+2 \tanh \left (d x +c \right )^{2}+2 \ln \left (\tanh \left (d x +c \right )-1\right )+2 \ln \left (\tanh \left (d x +c \right )+1\right )\right )}{4 d \tanh \left (d x +c \right )^{5}}\) | \(67\) |
risch | \(\frac {\left ({\mathrm e}^{2 d x +2 c}+1\right ) \sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}}\, x}{{\mathrm e}^{2 d x +2 c}-1}-\frac {2 \left ({\mathrm e}^{2 d x +2 c}+1\right ) \sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}}\, \left (d x +c \right )}{\left ({\mathrm e}^{2 d x +2 c}-1\right ) d}+\frac {4 \sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}}\, {\mathrm e}^{2 d x +2 c} \left ({\mathrm e}^{4 d x +4 c}+{\mathrm e}^{2 d x +2 c}+1\right )}{\left ({\mathrm e}^{2 d x +2 c}-1\right ) \left ({\mathrm e}^{2 d x +2 c}+1\right )^{3} d}+\frac {\left ({\mathrm e}^{2 d x +2 c}+1\right ) \sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}}\, \ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{\left ({\mathrm e}^{2 d x +2 c}-1\right ) d}\) | \(282\) |
Input:
int((-tanh(d*x+c)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/4/d*(-tanh(d*x+c)^2)^(5/2)*(tanh(d*x+c)^4+2*tanh(d*x+c)^2+2*ln(tanh(d*x +c)-1)+2*ln(tanh(d*x+c)+1))/tanh(d*x+c)^5
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.05 \[ \int \left (-\tanh ^2(c+d x)\right )^{5/2} \, dx=\frac {-i \, d x e^{\left (8 \, d x + 8 \, c\right )} - i \, d x - 4 \, {\left (i \, d x - i\right )} e^{\left (6 \, d x + 6 \, c\right )} - 2 \, {\left (3 i \, d x - 2 i\right )} e^{\left (4 \, d x + 4 \, c\right )} - 4 \, {\left (i \, d x - i\right )} e^{\left (2 \, d x + 2 \, c\right )} + {\left (i \, e^{\left (8 \, d x + 8 \, c\right )} + 4 i \, e^{\left (6 \, d x + 6 \, c\right )} + 6 i \, e^{\left (4 \, d x + 4 \, c\right )} + 4 i \, e^{\left (2 \, d x + 2 \, c\right )} + i\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{d e^{\left (8 \, d x + 8 \, c\right )} + 4 \, d e^{\left (6 \, d x + 6 \, c\right )} + 6 \, d e^{\left (4 \, d x + 4 \, c\right )} + 4 \, d e^{\left (2 \, d x + 2 \, c\right )} + d} \] Input:
integrate((-tanh(d*x+c)^2)^(5/2),x, algorithm="fricas")
Output:
(-I*d*x*e^(8*d*x + 8*c) - I*d*x - 4*(I*d*x - I)*e^(6*d*x + 6*c) - 2*(3*I*d *x - 2*I)*e^(4*d*x + 4*c) - 4*(I*d*x - I)*e^(2*d*x + 2*c) + (I*e^(8*d*x + 8*c) + 4*I*e^(6*d*x + 6*c) + 6*I*e^(4*d*x + 4*c) + 4*I*e^(2*d*x + 2*c) + I )*log(e^(2*d*x + 2*c) + 1))/(d*e^(8*d*x + 8*c) + 4*d*e^(6*d*x + 6*c) + 6*d *e^(4*d*x + 4*c) + 4*d*e^(2*d*x + 2*c) + d)
\[ \int \left (-\tanh ^2(c+d x)\right )^{5/2} \, dx=\int \left (- \tanh ^{2}{\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \] Input:
integrate((-tanh(d*x+c)**2)**(5/2),x)
Output:
Integral((-tanh(c + d*x)**2)**(5/2), x)
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.28 \[ \int \left (-\tanh ^2(c+d x)\right )^{5/2} \, dx=-\frac {i \, {\left (d x + c\right )}}{d} - \frac {i \, \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {4 \, {\left (-i \, e^{\left (-2 \, d x - 2 \, c\right )} - i \, e^{\left (-4 \, d x - 4 \, c\right )} - i \, e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}} \] Input:
integrate((-tanh(d*x+c)^2)^(5/2),x, algorithm="maxima")
Output:
-I*(d*x + c)/d - I*log(e^(-2*d*x - 2*c) + 1)/d + 4*(-I*e^(-2*d*x - 2*c) - I*e^(-4*d*x - 4*c) - I*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))
Result contains complex when optimal does not.
Time = 0.16 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.61 \[ \int \left (-\tanh ^2(c+d x)\right )^{5/2} \, dx=\frac {i \, {\left (d x + c\right )} \mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right ) - i \, \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) \mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right ) - \frac {4 i \, {\left (e^{\left (6 \, d x + 6 \, c\right )} \mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right ) + e^{\left (4 \, d x + 4 \, c\right )} \mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right ) + e^{\left (2 \, d x + 2 \, c\right )} \mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right )\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{4}}}{d} \] Input:
integrate((-tanh(d*x+c)^2)^(5/2),x, algorithm="giac")
Output:
(I*(d*x + c)*sgn(-e^(4*d*x + 4*c) + 1) - I*log(e^(2*d*x + 2*c) + 1)*sgn(-e ^(4*d*x + 4*c) + 1) - 4*I*(e^(6*d*x + 6*c)*sgn(-e^(4*d*x + 4*c) + 1) + e^( 4*d*x + 4*c)*sgn(-e^(4*d*x + 4*c) + 1) + e^(2*d*x + 2*c)*sgn(-e^(4*d*x + 4 *c) + 1))/(e^(2*d*x + 2*c) + 1)^4)/d
Timed out. \[ \int \left (-\tanh ^2(c+d x)\right )^{5/2} \, dx=\int {\left (-{\mathrm {tanh}\left (c+d\,x\right )}^2\right )}^{5/2} \,d x \] Input:
int((-tanh(c + d*x)^2)^(5/2),x)
Output:
int((-tanh(c + d*x)^2)^(5/2), x)
Time = 0.25 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.88 \[ \int \left (-\tanh ^2(c+d x)\right )^{5/2} \, dx=\frac {i \left (e^{8 d x +8 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right )-e^{8 d x +8 c} d x -e^{8 d x +8 c}+4 e^{6 d x +6 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right )-4 e^{6 d x +6 c} d x +6 e^{4 d x +4 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right )-6 e^{4 d x +4 c} d x -2 e^{4 d x +4 c}+4 e^{2 d x +2 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right )-4 e^{2 d x +2 c} d x +\mathrm {log}\left (e^{2 d x +2 c}+1\right )-d x -1\right )}{d \left (e^{8 d x +8 c}+4 e^{6 d x +6 c}+6 e^{4 d x +4 c}+4 e^{2 d x +2 c}+1\right )} \] Input:
int((-tanh(d*x+c)^2)^(5/2),x)
Output:
(i*(e**(8*c + 8*d*x)*log(e**(2*c + 2*d*x) + 1) - e**(8*c + 8*d*x)*d*x - e* *(8*c + 8*d*x) + 4*e**(6*c + 6*d*x)*log(e**(2*c + 2*d*x) + 1) - 4*e**(6*c + 6*d*x)*d*x + 6*e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x) + 1) - 6*e**(4*c + 4*d*x)*d*x - 2*e**(4*c + 4*d*x) + 4*e**(2*c + 2*d*x)*log(e**(2*c + 2*d*x) + 1) - 4*e**(2*c + 2*d*x)*d*x + log(e**(2*c + 2*d*x) + 1) - d*x - 1))/(d*( e**(8*c + 8*d*x) + 4*e**(6*c + 6*d*x) + 6*e**(4*c + 4*d*x) + 4*e**(2*c + 2 *d*x) + 1))