3.27 \(\int (-\tanh ^2(c+d x))^{5/2} \, dx\)

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

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

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Rubi [A]  time = 0.0485525, antiderivative size = 88, normalized size of antiderivative = 1., 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 = {3658, 3473, 3475} \[ -\frac{\sqrt{-\tanh ^2(c+d x)} \tanh ^3(c+d x)}{4 d}-\frac{\sqrt{-\tanh ^2(c+d x)} \tanh (c+d x)}{2 d}+\frac{\sqrt{-\tanh ^2(c+d x)} \coth (c+d x) \log (\cosh (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(-Tanh[c + d*x]^2)^(5/2),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) - (
Tanh[c + d*x]^3*Sqrt[-Tanh[c + d*x]^2])/(4*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 )^{5/2} \, dx &=\left (\coth (c+d x) \sqrt{-\tanh ^2(c+d x)}\right ) \int \tanh ^5(c+d x) \, dx\\ &=-\frac{\tanh ^3(c+d x) \sqrt{-\tanh ^2(c+d x)}}{4 d}+\left (\coth (c+d x) \sqrt{-\tanh ^2(c+d x)}\right ) \int \tanh ^3(c+d x) \, dx\\ &=-\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}+\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}-\frac{\tanh ^3(c+d x) \sqrt{-\tanh ^2(c+d x)}}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.29524, size = 56, normalized size = 0.64 \[ \frac{\left (-\tanh ^2(c+d x)\right )^{5/2} \coth (c+d x) \left (-2 \coth ^2(c+d x)+4 \coth ^4(c+d x) \log (\cosh (c+d x))-1\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(-Tanh[c + d*x]^2)^(5/2),x]

[Out]

(Coth[c + d*x]*(-1 - 2*Coth[c + d*x]^2 + 4*Coth[c + d*x]^4*Log[Cosh[c + d*x]])*(-Tanh[c + d*x]^2)^(5/2))/(4*d)

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Maple [A]  time = 0.026, size = 67, normalized size = 0.8 \begin{align*} -{\frac{ \left ( \tanh \left ( dx+c \right ) \right ) ^{4}+2\, \left ( \tanh \left ( dx+c \right ) \right ) ^{2}+2\,\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) +2\,\ln \left ( \tanh \left ( dx+c \right ) +1 \right ) }{4\,d \left ( \tanh \left ( dx+c \right ) \right ) ^{5}} \left ( - \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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

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Maxima [C]  time = 1.69352, size = 153, normalized size = 1.74 \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{4 i \, e^{\left (-2 \, d x - 2 \, c\right )} + 4 i \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 i \, e^{\left (-6 \, d x - 6 \, c\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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*(d*x + c)/d - I*log(e^(-2*d*x - 2*c) + 1)/d - (4*I*e^(-2*d*x - 2*c) + 4*I*e^(-4*d*x - 4*c) + 4*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]  time = 2.33754, size = 467, normalized size = 5.31 \begin{align*} \frac{-i \, d x e^{\left (8 \, d x + 8 \, c\right )} - i \, d x +{\left (-4 i \, d x + 4 i\right )} e^{\left (6 \, d x + 6 \, c\right )} +{\left (-6 i \, d x + 4 i\right )} e^{\left (4 \, d x + 4 \, c\right )} +{\left (-4 i \, d x + 4 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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-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 + 4*I)*e^(6*d*x + 6*c) + (-6*I*d*x + 4*I)*e^(4*d*x + 4*c) + (-4*I*
d*x + 4*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., size = 0, normalized size = 0. \begin{align*} \int \left (- \tanh ^{2}{\left (c + d x \right )}\right )^{\frac{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.3601, size = 192, normalized size = 2.18 \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{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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-tanh(d*x+c)^2)^(5/2),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) - 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