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

Optimal. Leaf size=42 \[ \frac{a}{3 f \left (a \cosh ^2(e+f x)\right )^{3/2}}-\frac{1}{f \sqrt{a \cosh ^2(e+f x)}} \]

[Out]

a/(3*f*(a*Cosh[e + f*x]^2)^(3/2)) - 1/(f*Sqrt[a*Cosh[e + f*x]^2])

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Rubi [A]  time = 0.117641, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3176, 3205, 16, 43} \[ \frac{a}{3 f \left (a \cosh ^2(e+f x)\right )^{3/2}}-\frac{1}{f \sqrt{a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a/(3*f*(a*Cosh[e + f*x]^2)^(3/2)) - 1/(f*Sqrt[a*Cosh[e + f*x]^2])

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 3205

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFact
ors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(b*ff^(n/2)*x^(n/2))^p)/(1 - ff*x
)^((m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{b, e, f, p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2
]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\tanh ^3(e+f x)}{\sqrt{a+a \sinh ^2(e+f x)}} \, dx &=\int \frac{\tanh ^3(e+f x)}{\sqrt{a \cosh ^2(e+f x)}} \, dx\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1-x}{x^2 \sqrt{a x}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=-\frac{a^2 \operatorname{Subst}\left (\int \frac{1-x}{(a x)^{5/2}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=-\frac{a^2 \operatorname{Subst}\left (\int \left (\frac{1}{(a x)^{5/2}}-\frac{1}{a (a x)^{3/2}}\right ) \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=\frac{a}{3 f \left (a \cosh ^2(e+f x)\right )^{3/2}}-\frac{1}{f \sqrt{a \cosh ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.0701354, size = 31, normalized size = 0.74 \[ \frac{\text{sech}^2(e+f x)-3}{3 f \sqrt{a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3 + Sech[e + f*x]^2)/(3*f*Sqrt[a*Cosh[e + f*x]^2])

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Maple [C]  time = 0.098, size = 41, normalized size = 1. \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ({\frac{ \left ( \sinh \left ( fx+e \right ) \right ) ^{3}}{ \left ( \cosh \left ( fx+e \right ) \right ) ^{4}}{\frac{1}{\sqrt{a \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}}}},\sinh \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

`int/indef0`(sinh(f*x+e)^3/cosh(f*x+e)^4/(a*cosh(f*x+e)^2)^(1/2),sinh(f*x+e))/f

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Maxima [B]  time = 1.89918, size = 248, normalized size = 5.9 \begin{align*} -\frac{2 \, e^{\left (-f x - e\right )}}{{\left (3 \, \sqrt{a} e^{\left (-2 \, f x - 2 \, e\right )} + 3 \, \sqrt{a} e^{\left (-4 \, f x - 4 \, e\right )} + \sqrt{a} e^{\left (-6 \, f x - 6 \, e\right )} + \sqrt{a}\right )} f} - \frac{4 \, e^{\left (-3 \, f x - 3 \, e\right )}}{3 \,{\left (3 \, \sqrt{a} e^{\left (-2 \, f x - 2 \, e\right )} + 3 \, \sqrt{a} e^{\left (-4 \, f x - 4 \, e\right )} + \sqrt{a} e^{\left (-6 \, f x - 6 \, e\right )} + \sqrt{a}\right )} f} - \frac{2 \, e^{\left (-5 \, f x - 5 \, e\right )}}{{\left (3 \, \sqrt{a} e^{\left (-2 \, f x - 2 \, e\right )} + 3 \, \sqrt{a} e^{\left (-4 \, f x - 4 \, e\right )} + \sqrt{a} e^{\left (-6 \, f x - 6 \, e\right )} + \sqrt{a}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)^3/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

-2*e^(-f*x - e)/((3*sqrt(a)*e^(-2*f*x - 2*e) + 3*sqrt(a)*e^(-4*f*x - 4*e) + sqrt(a)*e^(-6*f*x - 6*e) + sqrt(a)
)*f) - 4/3*e^(-3*f*x - 3*e)/((3*sqrt(a)*e^(-2*f*x - 2*e) + 3*sqrt(a)*e^(-4*f*x - 4*e) + sqrt(a)*e^(-6*f*x - 6*
e) + sqrt(a))*f) - 2*e^(-5*f*x - 5*e)/((3*sqrt(a)*e^(-2*f*x - 2*e) + 3*sqrt(a)*e^(-4*f*x - 4*e) + sqrt(a)*e^(-
6*f*x - 6*e) + sqrt(a))*f)

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Fricas [B]  time = 1.76068, size = 1692, normalized size = 40.29 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)^3/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

-2/3*(15*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^4 + 3*e^(f*x + e)*sinh(f*x + e)^5 + 2*(15*cosh(f*x + e)^2 + 1
)*e^(f*x + e)*sinh(f*x + e)^3 + 6*(5*cosh(f*x + e)^3 + cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^2 + 3*(5*cosh(
f*x + e)^4 + 2*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e) + (3*cosh(f*x + e)^5 + 2*cosh(f*x + e)^3 + 3*cos
h(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*f*cosh(f*x + e)^6 +
(a*f*e^(2*f*x + 2*e) + a*f)*sinh(f*x + e)^6 + 3*a*f*cosh(f*x + e)^4 + 6*(a*f*cosh(f*x + e)*e^(2*f*x + 2*e) +
a*f*cosh(f*x + e))*sinh(f*x + e)^5 + 3*(5*a*f*cosh(f*x + e)^2 + a*f + (5*a*f*cosh(f*x + e)^2 + a*f)*e^(2*f*x +
2*e))*sinh(f*x + e)^4 + 3*a*f*cosh(f*x + e)^2 + 4*(5*a*f*cosh(f*x + e)^3 + 3*a*f*cosh(f*x + e) + (5*a*f*cosh(
f*x + e)^3 + 3*a*f*cosh(f*x + e))*e^(2*f*x + 2*e))*sinh(f*x + e)^3 + 3*(5*a*f*cosh(f*x + e)^4 + 6*a*f*cosh(f*x
+ e)^2 + a*f + (5*a*f*cosh(f*x + e)^4 + 6*a*f*cosh(f*x + e)^2 + a*f)*e^(2*f*x + 2*e))*sinh(f*x + e)^2 + a*f +
(a*f*cosh(f*x + e)^6 + 3*a*f*cosh(f*x + e)^4 + 3*a*f*cosh(f*x + e)^2 + a*f)*e^(2*f*x + 2*e) + 6*(a*f*cosh(f*x
+ e)^5 + 2*a*f*cosh(f*x + e)^3 + a*f*cosh(f*x + e) + (a*f*cosh(f*x + e)^5 + 2*a*f*cosh(f*x + e)^3 + a*f*cosh(
f*x + e))*e^(2*f*x + 2*e))*sinh(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{3}{\left (e + f x \right )}}{\sqrt{a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.34448, size = 88, normalized size = 2.1 \begin{align*} -\frac{2 \,{\left (3 \, \sqrt{a} e^{\left (5 \, f x + 5 \, e\right )} + 2 \, \sqrt{a} e^{\left (3 \, f x + 3 \, e\right )} + 3 \, \sqrt{a} e^{\left (f x + e\right )}\right )}}{3 \, a f{\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)^3/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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