3.31 \(\int \frac {1}{(-\tanh ^2(c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=60 \[ \frac {\coth (c+d x)}{2 d \sqrt {-\tanh ^2(c+d x)}}-\frac {\tanh (c+d x) \log (\sinh (c+d x))}{d \sqrt {-\tanh ^2(c+d x)}} \]

[Out]

1/2*coth(d*x+c)/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 = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3658, 3473, 3475} \[ \frac {\coth (c+d x)}{2 d \sqrt {-\tanh ^2(c+d x)}}-\frac {\tanh (c+d x) \log (\sinh (c+d x))}{d \sqrt {-\tanh ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(-Tanh[c + d*x]^2)^(-3/2),x]

[Out]

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

Rule 3473

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 3475

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

Rule 3658

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 )^{3/2}} \, dx &=-\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 {\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 {\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 = 0.15, size = 51, normalized size = 0.85 \[ \frac {\coth (c+d x)-2 \tanh (c+d x) (\log (\tanh (c+d x))+\log (\cosh (c+d x)))}{2 d \sqrt {-\tanh ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-Tanh[c + d*x]^2)^(-3/2),x]

[Out]

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

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fricas [C]  time = 0.56, size = 99, normalized size = 1.65 \[ \frac {-i \, d x e^{\left (4 \, d x + 4 \, c\right )} - i \, d x + {\left (2 i \, d x - 2 i\right )} e^{\left (2 \, d x + 2 \, c\right )} + {\left (i \, e^{\left (4 \, d x + 4 \, c\right )} - 2 i \, e^{\left (2 \, d x + 2 \, c\right )} + i\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}{d e^{\left (4 \, d x + 4 \, c\right )} - 2 \, d e^{\left (2 \, d x + 2 \, c\right )} + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-tanh(d*x+c)^2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [C]  time = 0.80, size = 60, normalized size = 1.00 \[ -\frac {\frac {4 i \, e^{\left (2 \, d x + 2 \, c\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2}} + i \, \log \left (-i \, e^{\left (2 \, d x + 2 \, c\right )}\right ) - 2 i \, \log \left (i \, e^{\left (2 \, d x + 2 \, c\right )} - i\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-tanh(d*x+c)^2)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.12, size = 79, normalized size = 1.32 \[ -\frac {\tanh \left (d x +c \right ) \left (\ln \left (\tanh \left (d x +c \right )-1\right ) \left (\tanh ^{2}\left (d x +c \right )\right )+\ln \left (1+\tanh \left (d x +c \right )\right ) \left (\tanh ^{2}\left (d x +c \right )\right )-2 \ln \left (\tanh \left (d x +c \right )\right ) \left (\tanh ^{2}\left (d x +c \right )\right )+1\right )}{2 d \left (-\left (\tanh ^{2}\left (d x +c \right )\right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-tanh(d*x+c)^2)^(3/2),x)

[Out]

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

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maxima [C]  time = 0.44, size = 85, normalized size = 1.42 \[ -\frac {i \, {\left (d x + c\right )}}{d} - \frac {i \, \log \left (e^{\left (-d x - c\right )} + 1\right )}{d} - \frac {i \, \log \left (e^{\left (-d x - c\right )} - 1\right )}{d} - \frac {2 i \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-tanh(d*x+c)^2)^(3/2),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 - 2*I*e^(-2*d*x - 2*c)/(d*(2*e^(-2*d*x
- 2*c) - e^(-4*d*x - 4*c) - 1))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (-{\mathrm {tanh}\left (c+d\,x\right )}^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-tanh(c + d*x)^2)^(3/2),x)

[Out]

int(1/(-tanh(c + d*x)^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (- \tanh ^{2}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-tanh(d*x+c)**2)**(3/2),x)

[Out]

Integral((-tanh(c + d*x)**2)**(-3/2), x)

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