3.1.32 \(\int \frac {1}{(-\tanh ^2(c+d x))^{5/2}} \, dx\) [32]

Optimal. Leaf size=88 \[ -\frac {\coth (c+d x)}{2 d \sqrt {-\tanh ^2(c+d x)}}-\frac {\coth ^3(c+d x)}{4 d \sqrt {-\tanh ^2(c+d x)}}+\frac {\log (\sinh (c+d x)) \tanh (c+d x)}{d \sqrt {-\tanh ^2(c+d x)}} \]

[Out]

-1/2*coth(d*x+c)/d/(-tanh(d*x+c)^2)^(1/2)-1/4*coth(d*x+c)^3/d/(-tanh(d*x+c)^2)^(1/2)+ln(sinh(d*x+c))*tanh(d*x+
c)/d/(-tanh(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3739, 3554, 3556} \begin {gather*} -\frac {\coth ^3(c+d x)}{4 d \sqrt {-\tanh ^2(c+d x)}}-\frac {\coth (c+d x)}{2 d \sqrt {-\tanh ^2(c+d x)}}+\frac {\tanh (c+d x) \log (\sinh (c+d x))}{d \sqrt {-\tanh ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-Tanh[c + d*x]^2)^(-5/2),x]

[Out]

-1/2*Coth[c + d*x]/(d*Sqrt[-Tanh[c + d*x]^2]) - Coth[c + d*x]^3/(4*d*Sqrt[-Tanh[c + d*x]^2]) + (Log[Sinh[c + d
*x]]*Tanh[c + d*x])/(d*Sqrt[-Tanh[c + d*x]^2])

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3739

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Tan[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
]])

Rubi steps

\begin {align*} \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{5/2}} \, dx &=\frac {\tanh (c+d x) \int \coth ^5(c+d x) \, dx}{\sqrt {-\tanh ^2(c+d x)}}\\ &=-\frac {\coth ^3(c+d x)}{4 d \sqrt {-\tanh ^2(c+d x)}}+\frac {\tanh (c+d x) \int \coth ^3(c+d x) \, dx}{\sqrt {-\tanh ^2(c+d x)}}\\ &=-\frac {\coth (c+d x)}{2 d \sqrt {-\tanh ^2(c+d x)}}-\frac {\coth ^3(c+d x)}{4 d \sqrt {-\tanh ^2(c+d x)}}+\frac {\tanh (c+d x) \int \coth (c+d x) \, dx}{\sqrt {-\tanh ^2(c+d x)}}\\ &=-\frac {\coth (c+d x)}{2 d \sqrt {-\tanh ^2(c+d x)}}-\frac {\coth ^3(c+d x)}{4 d \sqrt {-\tanh ^2(c+d x)}}+\frac {\log (\sinh (c+d x)) \tanh (c+d x)}{d \sqrt {-\tanh ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 63, normalized size = 0.72 \begin {gather*} \frac {-2 \coth (c+d x)-\coth ^3(c+d x)+4 (\log (\cosh (c+d x))+\log (\tanh (c+d x))) \tanh (c+d x)}{4 d \sqrt {-\tanh ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-Tanh[c + d*x]^2)^(-5/2),x]

[Out]

(-2*Coth[c + d*x] - Coth[c + d*x]^3 + 4*(Log[Cosh[c + d*x]] + Log[Tanh[c + d*x]])*Tanh[c + d*x])/(4*d*Sqrt[-Ta
nh[c + d*x]^2])

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Maple [A]
time = 1.74, size = 91, normalized size = 1.03

method	result	size
derivativedivides	\(\frac {\tanh \left (d x +c \right ) \left (4 \ln \left (\tanh \left (d x +c \right )\right ) \left (\tanh ^{4}\left (d x +c \right )\right )-2 \ln \left (\tanh \left (d x +c \right )+1\right ) \left (\tanh ^{4}\left (d x +c \right )\right )-2 \ln \left (\tanh \left (d x +c \right )-1\right ) \left (\tanh ^{4}\left (d x +c \right )\right )-2 \left (\tanh ^{2}\left (d x +c \right )\right )-1\right )}{4 d \left (-\left (\tanh ^{2}\left (d x +c \right )\right )\right )^{\frac {5}{2}}}\)	\(91\)
default	\(\frac {\tanh \left (d x +c \right ) \left (4 \ln \left (\tanh \left (d x +c \right )\right ) \left (\tanh ^{4}\left (d x +c \right )\right )-2 \ln \left (\tanh \left (d x +c \right )+1\right ) \left (\tanh ^{4}\left (d x +c \right )\right )-2 \ln \left (\tanh \left (d x +c \right )-1\right ) \left (\tanh ^{4}\left (d x +c \right )\right )-2 \left (\tanh ^{2}\left (d x +c \right )\right )-1\right )}{4 d \left (-\left (\tanh ^{2}\left (d x +c \right )\right )\right )^{\frac {5}{2}}}\)	\(91\)
risch	\(\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right ) x}{\sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left (1+{\mathrm e}^{2 d x +2 c}\right )^{2}}}\, \left (1+{\mathrm e}^{2 d x +2 c}\right )}-\frac {2 \left ({\mathrm e}^{2 d x +2 c}-1\right ) \left (d x +c \right )}{\sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left (1+{\mathrm e}^{2 d x +2 c}\right )^{2}}}\, \left (1+{\mathrm e}^{2 d x +2 c}\right ) d}-\frac {4 \,{\mathrm e}^{2 d x +2 c} \left ({\mathrm e}^{4 d x +4 c}-{\mathrm e}^{2 d x +2 c}+1\right )}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{3} \left (1+{\mathrm e}^{2 d x +2 c}\right ) \sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left (1+{\mathrm e}^{2 d x +2 c}\right )^{2}}}\, d}+\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right ) \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{\sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left (1+{\mathrm e}^{2 d x +2 c}\right )^{2}}}\, \left (1+{\mathrm e}^{2 d x +2 c}\right ) d}\)	\(284\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-tanh(d*x+c)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/4/d*tanh(d*x+c)*(4*ln(tanh(d*x+c))*tanh(d*x+c)^4-2*ln(tanh(d*x+c)+1)*tanh(d*x+c)^4-2*ln(tanh(d*x+c)-1)*tanh(
d*x+c)^4-2*tanh(d*x+c)^2-1)/(-tanh(d*x+c)^2)^(5/2)

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Maxima [C] Result contains complex when optimal does not.
time = 0.48, size = 132, normalized size = 1.50 \begin {gather*} \frac {i \, {\left (d x + c\right )}}{d} + \frac {i \, \log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {i \, \log \left (e^{\left (-d x - 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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-tanh(d*x+c)^2)^(5/2),x, algorithm="maxima")

[Out]

I*(d*x + c)/d + I*log(e^(-d*x - c) + 1)/d + I*log(e^(-d*x - 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))

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Fricas [C] Result contains complex when optimal does not.
time = 0.42, size = 180, normalized size = 2.05 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-tanh(d*x+c)^2)^(5/2),x, algorithm="fricas")

[Out]

(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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (- \tanh ^{2}{\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-tanh(d*x+c)**2)**(5/2),x)

[Out]

Integral((-tanh(c + d*x)**2)**(-5/2), x)

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Giac [C] Result contains complex when optimal does not.
time = 0.42, size = 126, normalized size = 1.43 \begin {gather*} -\frac {\frac {i \, d x + i \, c}{\mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right )} - \frac {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 \, {\left (i \, e^{\left (6 \, d x + 6 \, c\right )} - i \, e^{\left (4 \, d x + 4 \, c\right )} + i \, e^{\left (2 \, d x + 2 \, c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{4} \mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-tanh(d*x+c)^2)^(5/2),x, algorithm="giac")

[Out]

-((I*d*x + I*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) - I*e^(4*d*x + 4*c) + I*e^(2*d*x + 2*c))/((e^(2*d*x + 2*c) - 1)^4*sgn(-e^(4*d*x + 4*c) + 1)))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (-{\mathrm {tanh}\left (c+d\,x\right )}^2\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-tanh(c + d*x)^2)^(5/2),x)

[Out]

int(1/(-tanh(c + d*x)^2)^(5/2), x)

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