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

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

[Out]

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

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Rubi [A]  time = 0.0322464, antiderivative size = 60, normalized size of antiderivative = 1., 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} \[ \frac{\tanh (c+d x) \sqrt{-\tanh ^2(c+d x)}}{2 d}-\frac{\sqrt{-\tanh ^2(c+d x)} \coth (c+d x) \log (\cosh (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(-Tanh[c + d*x]^2)^(3/2),x]

[Out]

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

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
]])

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]

Rubi steps

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

Mathematica [A]  time = 0.0941762, size = 46, normalized size = 0.77 \[ \frac{\left (-\tanh ^2(c+d x)\right )^{3/2} \coth (c+d x) \left (2 \coth ^2(c+d x) \log (\cosh (c+d x))-1\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(-Tanh[c + d*x]^2)^(3/2),x]

[Out]

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

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Maple [A]  time = 0.015, size = 53, normalized size = 0.9 \begin{align*} -{\frac{ \left ( \tanh \left ( dx+c \right ) \right ) ^{2}+\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) +\ln \left ( \tanh \left ( dx+c \right ) +1 \right ) }{2\,d \left ( \tanh \left ( dx+c \right ) \right ) ^{3}} \left ( - \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.60142, size = 89, normalized size = 1.48 \begin{align*} \frac{i \,{\left (d x + c\right )}}{d} + \frac{i \, \log \left (e^{\left (-2 \, d x - 2 \, 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*(d*x + c)/d + I*log(e^(-2*d*x - 2*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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Fricas [C]  time = 2.26999, size = 252, normalized size = 4.2 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (- \tanh ^{2}{\left (c + d x \right )}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.29772, size = 124, normalized size = 2.07 \begin{align*} \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{2 i \, e^{\left (2 \, d x + 2 \, c\right )} \mathrm{sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right )}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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