3.452 \(\int \frac {\tanh ^2(e+f x)}{(a+a \sinh ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=106 \[ \frac {\tanh (e+f x)}{8 a f \sqrt {a \cosh ^2(e+f x)}}+\frac {\cosh (e+f x) \tan ^{-1}(\sinh (e+f x))}{8 a f \sqrt {a \cosh ^2(e+f x)}}-\frac {\tanh (e+f x) \text {sech}^2(e+f x)}{4 a f \sqrt {a \cosh ^2(e+f x)}} \]

[Out]

1/8*arctan(sinh(f*x+e))*cosh(f*x+e)/a/f/(a*cosh(f*x+e)^2)^(1/2)+1/8*tanh(f*x+e)/a/f/(a*cosh(f*x+e)^2)^(1/2)-1/
4*sech(f*x+e)^2*tanh(f*x+e)/a/f/(a*cosh(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3176, 3207, 2611, 3768, 3770} \[ \frac {\tanh (e+f x)}{8 a f \sqrt {a \cosh ^2(e+f x)}}+\frac {\cosh (e+f x) \tan ^{-1}(\sinh (e+f x))}{8 a f \sqrt {a \cosh ^2(e+f x)}}-\frac {\tanh (e+f x) \text {sech}^2(e+f x)}{4 a f \sqrt {a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[e + f*x]^2/(a + a*Sinh[e + f*x]^2)^(3/2),x]

[Out]

(ArcTan[Sinh[e + f*x]]*Cosh[e + f*x])/(8*a*f*Sqrt[a*Cosh[e + f*x]^2]) + Tanh[e + f*x]/(8*a*f*Sqrt[a*Cosh[e + f
*x]^2]) - (Sech[e + f*x]^2*Tanh[e + f*x])/(4*a*f*Sqrt[a*Cosh[e + f*x]^2])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3176

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rubi steps

\begin {align*} \int \frac {\tanh ^2(e+f x)}{\left (a+a \sinh ^2(e+f x)\right )^{3/2}} \, dx &=\int \frac {\tanh ^2(e+f x)}{\left (a \cosh ^2(e+f x)\right )^{3/2}} \, dx\\ &=\frac {\cosh (e+f x) \int \text {sech}^3(e+f x) \tanh ^2(e+f x) \, dx}{a \sqrt {a \cosh ^2(e+f x)}}\\ &=-\frac {\text {sech}^2(e+f x) \tanh (e+f x)}{4 a f \sqrt {a \cosh ^2(e+f x)}}+\frac {\cosh (e+f x) \int \text {sech}^3(e+f x) \, dx}{4 a \sqrt {a \cosh ^2(e+f x)}}\\ &=\frac {\tanh (e+f x)}{8 a f \sqrt {a \cosh ^2(e+f x)}}-\frac {\text {sech}^2(e+f x) \tanh (e+f x)}{4 a f \sqrt {a \cosh ^2(e+f x)}}+\frac {\cosh (e+f x) \int \text {sech}(e+f x) \, dx}{8 a \sqrt {a \cosh ^2(e+f x)}}\\ &=\frac {\tan ^{-1}(\sinh (e+f x)) \cosh (e+f x)}{8 a f \sqrt {a \cosh ^2(e+f x)}}+\frac {\tanh (e+f x)}{8 a f \sqrt {a \cosh ^2(e+f x)}}-\frac {\text {sech}^2(e+f x) \tanh (e+f x)}{4 a f \sqrt {a \cosh ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 58, normalized size = 0.55 \[ \frac {\tanh (e+f x) \left (1-2 \text {sech}^2(e+f x)\right )+\cosh (e+f x) \tan ^{-1}(\sinh (e+f x))}{8 a f \sqrt {a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[e + f*x]^2/(a + a*Sinh[e + f*x]^2)^(3/2),x]

[Out]

(ArcTan[Sinh[e + f*x]]*Cosh[e + f*x] + (1 - 2*Sech[e + f*x]^2)*Tanh[e + f*x])/(8*a*f*Sqrt[a*Cosh[e + f*x]^2])

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fricas [B]  time = 0.53, size = 1423, normalized size = 13.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)^2/(a+a*sinh(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

1/4*(7*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^6 + e^(f*x + e)*sinh(f*x + e)^7 + 7*(3*cosh(f*x + e)^2 - 1)*e^(
f*x + e)*sinh(f*x + e)^5 + 35*(cosh(f*x + e)^3 - cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^4 + 7*(5*cosh(f*x +
e)^4 - 10*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e)^3 + 7*(3*cosh(f*x + e)^5 - 10*cosh(f*x + e)^3 + 3*cos
h(f*x + e))*e^(f*x + e)*sinh(f*x + e)^2 + (7*cosh(f*x + e)^6 - 35*cosh(f*x + e)^4 + 21*cosh(f*x + e)^2 - 1)*e^
(f*x + e)*sinh(f*x + e) + (8*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^7 + e^(f*x + e)*sinh(f*x + e)^8 + 4*(7*co
sh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e)^6 + 8*(7*cosh(f*x + e)^3 + 3*cosh(f*x + e))*e^(f*x + e)*sinh(f*x
+ e)^5 + 2*(35*cosh(f*x + e)^4 + 30*cosh(f*x + e)^2 + 3)*e^(f*x + e)*sinh(f*x + e)^4 + 8*(7*cosh(f*x + e)^5 +
10*cosh(f*x + e)^3 + 3*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^3 + 4*(7*cosh(f*x + e)^6 + 15*cosh(f*x + e)^4
+ 9*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e)^2 + 8*(cosh(f*x + e)^7 + 3*cosh(f*x + e)^5 + 3*cosh(f*x + e
)^3 + cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e) + (cosh(f*x + e)^8 + 4*cosh(f*x + e)^6 + 6*cosh(f*x + e)^4 + 4*
cosh(f*x + e)^2 + 1)*e^(f*x + e))*arctan(cosh(f*x + e) + sinh(f*x + e)) + (cosh(f*x + e)^7 - 7*cosh(f*x + e)^5
+ 7*cosh(f*x + e)^3 - cosh(f*x + e))*e^(f*x + e))*sqrt(a*e^(4*f*x + 4*e) + 2*a*e^(2*f*x + 2*e) + a)*e^(-f*x -
e)/(a^2*f*cosh(f*x + e)^8 + 4*a^2*f*cosh(f*x + e)^6 + (a^2*f*e^(2*f*x + 2*e) + a^2*f)*sinh(f*x + e)^8 + 8*(a^
2*f*cosh(f*x + e)*e^(2*f*x + 2*e) + a^2*f*cosh(f*x + e))*sinh(f*x + e)^7 + 6*a^2*f*cosh(f*x + e)^4 + 4*(7*a^2*
f*cosh(f*x + e)^2 + a^2*f + (7*a^2*f*cosh(f*x + e)^2 + a^2*f)*e^(2*f*x + 2*e))*sinh(f*x + e)^6 + 8*(7*a^2*f*co
sh(f*x + e)^3 + 3*a^2*f*cosh(f*x + e) + (7*a^2*f*cosh(f*x + e)^3 + 3*a^2*f*cosh(f*x + e))*e^(2*f*x + 2*e))*sin
h(f*x + e)^5 + 4*a^2*f*cosh(f*x + e)^2 + 2*(35*a^2*f*cosh(f*x + e)^4 + 30*a^2*f*cosh(f*x + e)^2 + 3*a^2*f + (3
5*a^2*f*cosh(f*x + e)^4 + 30*a^2*f*cosh(f*x + e)^2 + 3*a^2*f)*e^(2*f*x + 2*e))*sinh(f*x + e)^4 + 8*(7*a^2*f*co
sh(f*x + e)^5 + 10*a^2*f*cosh(f*x + e)^3 + 3*a^2*f*cosh(f*x + e) + (7*a^2*f*cosh(f*x + e)^5 + 10*a^2*f*cosh(f*
x + e)^3 + 3*a^2*f*cosh(f*x + e))*e^(2*f*x + 2*e))*sinh(f*x + e)^3 + a^2*f + 4*(7*a^2*f*cosh(f*x + e)^6 + 15*a
^2*f*cosh(f*x + e)^4 + 9*a^2*f*cosh(f*x + e)^2 + a^2*f + (7*a^2*f*cosh(f*x + e)^6 + 15*a^2*f*cosh(f*x + e)^4 +
9*a^2*f*cosh(f*x + e)^2 + a^2*f)*e^(2*f*x + 2*e))*sinh(f*x + e)^2 + (a^2*f*cosh(f*x + e)^8 + 4*a^2*f*cosh(f*x
+ e)^6 + 6*a^2*f*cosh(f*x + e)^4 + 4*a^2*f*cosh(f*x + e)^2 + a^2*f)*e^(2*f*x + 2*e) + 8*(a^2*f*cosh(f*x + e)^
7 + 3*a^2*f*cosh(f*x + e)^5 + 3*a^2*f*cosh(f*x + e)^3 + a^2*f*cosh(f*x + e) + (a^2*f*cosh(f*x + e)^7 + 3*a^2*f
*cosh(f*x + e)^5 + 3*a^2*f*cosh(f*x + e)^3 + a^2*f*cosh(f*x + e))*e^(2*f*x + 2*e))*sinh(f*x + e))

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giac [A]  time = 0.37, size = 93, normalized size = 0.88 \[ \frac {\frac {\arctan \left (e^{\left (f x + e\right )}\right )}{a^{\frac {3}{2}}} + \frac {\sqrt {a} e^{\left (7 \, f x + 7 \, e\right )} - 7 \, \sqrt {a} e^{\left (5 \, f x + 5 \, e\right )} + 7 \, \sqrt {a} e^{\left (3 \, f x + 3 \, e\right )} - \sqrt {a} e^{\left (f x + e\right )}}{a^{2} {\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}^{4}}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)^2/(a+a*sinh(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

1/4*(arctan(e^(f*x + e))/a^(3/2) + (sqrt(a)*e^(7*f*x + 7*e) - 7*sqrt(a)*e^(5*f*x + 5*e) + 7*sqrt(a)*e^(3*f*x +
3*e) - sqrt(a)*e^(f*x + e))/(a^2*(e^(2*f*x + 2*e) + 1)^4))/f

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maple [A]  time = 0.22, size = 69, normalized size = 0.65 \[ \frac {\arctan \left (\sinh \left (f x +e \right )\right ) \left (\cosh ^{4}\left (f x +e \right )\right )+\left (\cosh ^{2}\left (f x +e \right )\right ) \sinh \left (f x +e \right )-2 \sinh \left (f x +e \right )}{8 a \cosh \left (f x +e \right )^{3} \sqrt {a \left (\cosh ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(f*x+e)^2/(a+a*sinh(f*x+e)^2)^(3/2),x)

[Out]

1/8/a*(arctan(sinh(f*x+e))*cosh(f*x+e)^4+cosh(f*x+e)^2*sinh(f*x+e)-2*sinh(f*x+e))/cosh(f*x+e)^3/(a*cosh(f*x+e)
^2)^(1/2)/f

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maxima [B]  time = 0.57, size = 369, normalized size = 3.48 \[ -\frac {\frac {3 \, e^{\left (-f x - e\right )} + 11 \, e^{\left (-3 \, f x - 3 \, e\right )} - 11 \, e^{\left (-5 \, f x - 5 \, e\right )} - 3 \, e^{\left (-7 \, f x - 7 \, e\right )}}{4 \, a^{\frac {3}{2}} e^{\left (-2 \, f x - 2 \, e\right )} + 6 \, a^{\frac {3}{2}} e^{\left (-4 \, f x - 4 \, e\right )} + 4 \, a^{\frac {3}{2}} e^{\left (-6 \, f x - 6 \, e\right )} + a^{\frac {3}{2}} e^{\left (-8 \, f x - 8 \, e\right )} + a^{\frac {3}{2}}} - \frac {3 \, \arctan \left (e^{\left (-f x - e\right )}\right )}{a^{\frac {3}{2}}}}{8 \, f} + \frac {15 \, e^{\left (-f x - e\right )} + 55 \, e^{\left (-3 \, f x - 3 \, e\right )} + 73 \, e^{\left (-5 \, f x - 5 \, e\right )} - 15 \, e^{\left (-7 \, f x - 7 \, e\right )}}{48 \, {\left (4 \, a^{\frac {3}{2}} e^{\left (-2 \, f x - 2 \, e\right )} + 6 \, a^{\frac {3}{2}} e^{\left (-4 \, f x - 4 \, e\right )} + 4 \, a^{\frac {3}{2}} e^{\left (-6 \, f x - 6 \, e\right )} + a^{\frac {3}{2}} e^{\left (-8 \, f x - 8 \, e\right )} + a^{\frac {3}{2}}\right )} f} + \frac {15 \, e^{\left (-f x - e\right )} - 73 \, e^{\left (-3 \, f x - 3 \, e\right )} - 55 \, e^{\left (-5 \, f x - 5 \, e\right )} - 15 \, e^{\left (-7 \, f x - 7 \, e\right )}}{48 \, {\left (4 \, a^{\frac {3}{2}} e^{\left (-2 \, f x - 2 \, e\right )} + 6 \, a^{\frac {3}{2}} e^{\left (-4 \, f x - 4 \, e\right )} + 4 \, a^{\frac {3}{2}} e^{\left (-6 \, f x - 6 \, e\right )} + a^{\frac {3}{2}} e^{\left (-8 \, f x - 8 \, e\right )} + a^{\frac {3}{2}}\right )} f} - \frac {5 \, \arctan \left (e^{\left (-f x - e\right )}\right )}{8 \, a^{\frac {3}{2}} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)^2/(a+a*sinh(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

-1/8*((3*e^(-f*x - e) + 11*e^(-3*f*x - 3*e) - 11*e^(-5*f*x - 5*e) - 3*e^(-7*f*x - 7*e))/(4*a^(3/2)*e^(-2*f*x -
2*e) + 6*a^(3/2)*e^(-4*f*x - 4*e) + 4*a^(3/2)*e^(-6*f*x - 6*e) + a^(3/2)*e^(-8*f*x - 8*e) + a^(3/2)) - 3*arct
an(e^(-f*x - e))/a^(3/2))/f + 1/48*(15*e^(-f*x - e) + 55*e^(-3*f*x - 3*e) + 73*e^(-5*f*x - 5*e) - 15*e^(-7*f*x
- 7*e))/((4*a^(3/2)*e^(-2*f*x - 2*e) + 6*a^(3/2)*e^(-4*f*x - 4*e) + 4*a^(3/2)*e^(-6*f*x - 6*e) + a^(3/2)*e^(-
8*f*x - 8*e) + a^(3/2))*f) + 1/48*(15*e^(-f*x - e) - 73*e^(-3*f*x - 3*e) - 55*e^(-5*f*x - 5*e) - 15*e^(-7*f*x
- 7*e))/((4*a^(3/2)*e^(-2*f*x - 2*e) + 6*a^(3/2)*e^(-4*f*x - 4*e) + 4*a^(3/2)*e^(-6*f*x - 6*e) + a^(3/2)*e^(-8
*f*x - 8*e) + a^(3/2))*f) - 5/8*arctan(e^(-f*x - e))/(a^(3/2)*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tanh}\left (e+f\,x\right )}^2}{{\left (a\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(e + f*x)^2/(a + a*sinh(e + f*x)^2)^(3/2),x)

[Out]

int(tanh(e + f*x)^2/(a + a*sinh(e + f*x)^2)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)**2/(a+a*sinh(f*x+e)**2)**(3/2),x)

[Out]

Integral(tanh(e + f*x)**2/(a*(sinh(e + f*x)**2 + 1))**(3/2), x)

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