3.5.38 \(\int \frac {\tanh ^3(e+f x)}{\sqrt {a+a \sinh ^2(e+f x)}} \, dx\) [438]

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]

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

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Rubi [A]
time = 0.08, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3255, 3284, 16, 45} \begin {gather*} \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 {gather*}

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 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 45

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])

Rule 3255

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 3284

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
]

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 {\text {Subst}\left (\int \frac {1-x}{x^2 \sqrt {a x}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=-\frac {a^2 \text {Subst}\left (\int \frac {1-x}{(a x)^{5/2}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=-\frac {a^2 \text {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*}

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Mathematica [A]
time = 0.09, size = 31, normalized size = 0.74 \begin {gather*} \frac {-3+\text {sech}^2(e+f x)}{3 f \sqrt {a \cosh ^2(e+f x)}} \end {gather*}

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] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.12, size = 41, normalized size = 0.98

method	result	size
default	\(\frac {\mathit {`\,int/indef0`\,}\left (\frac {\sinh ^{3}\left (f x +e \right )}{\cosh \left (f x +e \right )^{4} \sqrt {a \left (\cosh ^{2}\left (f x +e \right )\right )}}, \sinh \left (f x +e \right )\right )}{f}\)	\(41\)
risch	\(-\frac {2 \left (3 \,{\mathrm e}^{4 f x +4 e}+2 \,{\mathrm e}^{2 f x +2 e}+3\right )}{3 \sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, \left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} f}\)	\(69\)

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,method=_RETURNVERBOSE)

[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] Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (38) = 76\).
time = 0.52, size = 196, normalized size = 4.67 \begin {gather*} -\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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 641 vs. \(2 (36) = 72\).
time = 0.47, size = 641, normalized size = 15.26 \begin {gather*} -\frac {2 \, {\left (15 \, \cosh \left (f x + e\right ) e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{4} + 3 \, e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{5} + 2 \, {\left (15 \, \cosh \left (f x + e\right )^{2} + 1\right )} e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{3} + 6 \, {\left (5 \, \cosh \left (f x + e\right )^{3} + \cosh \left (f x + e\right )\right )} e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{2} + 3 \, {\left (5 \, \cosh \left (f x + e\right )^{4} + 2 \, \cosh \left (f x + e\right )^{2} + 1\right )} e^{\left (f x + e\right )} \sinh \left (f x + e\right ) + {\left (3 \, \cosh \left (f x + e\right )^{5} + 2 \, \cosh \left (f x + e\right )^{3} + 3 \, \cosh \left (f x + e\right )\right )} e^{\left (f x + e\right )}\right )} \sqrt {a e^{\left (4 \, f x + 4 \, e\right )} + 2 \, a e^{\left (2 \, f x + 2 \, e\right )} + a} e^{\left (-f x - e\right )}}{3 \, {\left (a f \cosh \left (f x + e\right )^{6} + {\left (a f e^{\left (2 \, f x + 2 \, e\right )} + a f\right )} \sinh \left (f x + e\right )^{6} + 3 \, a f \cosh \left (f x + e\right )^{4} + 6 \, {\left (a f \cosh \left (f x + e\right ) e^{\left (2 \, f x + 2 \, e\right )} + a f \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )^{5} + 3 \, {\left (5 \, a f \cosh \left (f x + e\right )^{2} + a f + {\left (5 \, a f \cosh \left (f x + e\right )^{2} + a f\right )} e^{\left (2 \, f x + 2 \, e\right )}\right )} \sinh \left (f x + e\right )^{4} + 3 \, a f \cosh \left (f x + e\right )^{2} + 4 \, {\left (5 \, a f \cosh \left (f x + e\right )^{3} + 3 \, a f \cosh \left (f x + e\right ) + {\left (5 \, a f \cosh \left (f x + e\right )^{3} + 3 \, a f \cosh \left (f x + e\right )\right )} e^{\left (2 \, f x + 2 \, e\right )}\right )} \sinh \left (f x + e\right )^{3} + 3 \, {\left (5 \, a f \cosh \left (f x + e\right )^{4} + 6 \, a f \cosh \left (f x + e\right )^{2} + a f + {\left (5 \, a f \cosh \left (f x + e\right )^{4} + 6 \, a f \cosh \left (f x + e\right )^{2} + a f\right )} e^{\left (2 \, f x + 2 \, e\right )}\right )} \sinh \left (f x + e\right )^{2} + a f + {\left (a f \cosh \left (f x + e\right )^{6} + 3 \, a f \cosh \left (f x + e\right )^{4} + 3 \, a f \cosh \left (f x + e\right )^{2} + a f\right )} e^{\left (2 \, f x + 2 \, e\right )} + 6 \, {\left (a f \cosh \left (f x + e\right )^{5} + 2 \, a f \cosh \left (f x + e\right )^{3} + a f \cosh \left (f x + e\right ) + {\left (a f \cosh \left (f x + e\right )^{5} + 2 \, a f \cosh \left (f x + e\right )^{3} + a f \cosh \left (f x + e\right )\right )} e^{\left (2 \, f x + 2 \, e\right )}\right )} \sinh \left (f x + e\right )\right )}} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tanh ^{3}{\left (e + f x \right )}}{\sqrt {a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )}}\, dx \end {gather*}

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 [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [B]
time = 0.12, size = 82, normalized size = 1.95 \begin {gather*} -\frac {4\,{\mathrm {e}}^{2\,e+2\,f\,x}\,\sqrt {a+a\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}\,\left (2\,{\mathrm {e}}^{2\,e+2\,f\,x}+3\,{\mathrm {e}}^{4\,e+4\,f\,x}+3\right )}{3\,a\,f\,{\left ({\mathrm {e}}^{2\,e+2\,f\,x}+1\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*exp(2*e + 2*f*x)*(a + a*(exp(e + f*x)/2 - exp(- e - f*x)/2)^2)^(1/2)*(2*exp(2*e + 2*f*x) + 3*exp(4*e + 4*f
*x) + 3))/(3*a*f*(exp(2*e + 2*f*x) + 1)^4)

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