3.5.48 \(\int \frac {\tanh ^3(e+f x)}{(a+a \sinh ^2(e+f x))^{3/2}} \, dx\) [448]

Optimal. Leaf size=44 \[ \frac {a}{5 f \left (a \cosh ^2(e+f x)\right )^{5/2}}-\frac {1}{3 f \left (a \cosh ^2(e+f x)\right )^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 44, 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}{5 f \left (a \cosh ^2(e+f x)\right )^{5/2}}-\frac {1}{3 f \left (a \cosh ^2(e+f x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a/(5*f*(a*Cosh[e + f*x]^2)^(5/2)) - 1/(3*f*(a*Cosh[e + f*x]^2)^(3/2))

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)}{\left (a+a \sinh ^2(e+f x)\right )^{3/2}} \, dx &=\int \frac {\tanh ^3(e+f x)}{\left (a \cosh ^2(e+f x)\right )^{3/2}} \, dx\\ &=-\frac {\text {Subst}\left (\int \frac {1-x}{x^2 (a x)^{3/2}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=-\frac {a^2 \text {Subst}\left (\int \frac {1-x}{(a x)^{7/2}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=-\frac {a^2 \text {Subst}\left (\int \left (\frac {1}{(a x)^{7/2}}-\frac {1}{a (a x)^{5/2}}\right ) \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=\frac {a}{5 f \left (a \cosh ^2(e+f x)\right )^{5/2}}-\frac {1}{3 f \left (a \cosh ^2(e+f x)\right )^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 34, normalized size = 0.77 \begin {gather*} \frac {a \left (3-5 \cosh ^2(e+f x)\right )}{15 f \left (a \cosh ^2(e+f x)\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(3 - 5*Cosh[e + f*x]^2))/(15*f*(a*Cosh[e + f*x]^2)^(5/2))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.18, size = 44, normalized size = 1.00

method	result	size
default	\(\frac {\mathit {`\,int/indef0`\,}\left (\frac {\sinh ^{3}\left (f x +e \right )}{\cosh \left (f x +e \right )^{6} a \sqrt {a \left (\cosh ^{2}\left (f x +e \right )\right )}}, \sinh \left (f x +e \right )\right )}{f}\)	\(44\)
risch	\(-\frac {8 \left (5 \,{\mathrm e}^{4 f x +4 e}-2 \,{\mathrm e}^{2 f x +2 e}+5\right ) {\mathrm e}^{2 f x +2 e}}{15 f \sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, \left ({\mathrm e}^{2 f x +2 e}+1\right )^{4} a}\)	\(81\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

`int/indef0`(sinh(f*x+e)^3/cosh(f*x+e)^6/a/(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. 286 vs. \(2 (38) = 76\).
time = 0.52, size = 286, normalized size = 6.50 \begin {gather*} -\frac {8 \, e^{\left (-3 \, f x - 3 \, e\right )}}{3 \, {\left (5 \, a^{\frac {3}{2}} e^{\left (-2 \, f x - 2 \, e\right )} + 10 \, a^{\frac {3}{2}} e^{\left (-4 \, f x - 4 \, e\right )} + 10 \, a^{\frac {3}{2}} e^{\left (-6 \, f x - 6 \, e\right )} + 5 \, a^{\frac {3}{2}} e^{\left (-8 \, f x - 8 \, e\right )} + a^{\frac {3}{2}} e^{\left (-10 \, f x - 10 \, e\right )} + a^{\frac {3}{2}}\right )} f} + \frac {16 \, e^{\left (-5 \, f x - 5 \, e\right )}}{15 \, {\left (5 \, a^{\frac {3}{2}} e^{\left (-2 \, f x - 2 \, e\right )} + 10 \, a^{\frac {3}{2}} e^{\left (-4 \, f x - 4 \, e\right )} + 10 \, a^{\frac {3}{2}} e^{\left (-6 \, f x - 6 \, e\right )} + 5 \, a^{\frac {3}{2}} e^{\left (-8 \, f x - 8 \, e\right )} + a^{\frac {3}{2}} e^{\left (-10 \, f x - 10 \, e\right )} + a^{\frac {3}{2}}\right )} f} - \frac {8 \, e^{\left (-7 \, f x - 7 \, e\right )}}{3 \, {\left (5 \, a^{\frac {3}{2}} e^{\left (-2 \, f x - 2 \, e\right )} + 10 \, a^{\frac {3}{2}} e^{\left (-4 \, f x - 4 \, e\right )} + 10 \, a^{\frac {3}{2}} e^{\left (-6 \, f x - 6 \, e\right )} + 5 \, a^{\frac {3}{2}} e^{\left (-8 \, f x - 8 \, e\right )} + a^{\frac {3}{2}} e^{\left (-10 \, f x - 10 \, e\right )} + a^{\frac {3}{2}}\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)^(3/2),x, algorithm="maxima")

[Out]

-8/3*e^(-3*f*x - 3*e)/((5*a^(3/2)*e^(-2*f*x - 2*e) + 10*a^(3/2)*e^(-4*f*x - 4*e) + 10*a^(3/2)*e^(-6*f*x - 6*e)
+ 5*a^(3/2)*e^(-8*f*x - 8*e) + a^(3/2)*e^(-10*f*x - 10*e) + a^(3/2))*f) + 16/15*e^(-5*f*x - 5*e)/((5*a^(3/2)*
e^(-2*f*x - 2*e) + 10*a^(3/2)*e^(-4*f*x - 4*e) + 10*a^(3/2)*e^(-6*f*x - 6*e) + 5*a^(3/2)*e^(-8*f*x - 8*e) + a^
(3/2)*e^(-10*f*x - 10*e) + a^(3/2))*f) - 8/3*e^(-7*f*x - 7*e)/((5*a^(3/2)*e^(-2*f*x - 2*e) + 10*a^(3/2)*e^(-4*
f*x - 4*e) + 10*a^(3/2)*e^(-6*f*x - 6*e) + 5*a^(3/2)*e^(-8*f*x - 8*e) + a^(3/2)*e^(-10*f*x - 10*e) + a^(3/2))*
f)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1400 vs. \(2 (36) = 72\).
time = 0.43, size = 1400, normalized size = 31.82 \begin {gather*} \text {Too large to display} \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)^(3/2),x, algorithm="fricas")

[Out]

-8/15*(35*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^6 + 5*e^(f*x + e)*sinh(f*x + e)^7 + (105*cosh(f*x + e)^2 - 2
)*e^(f*x + e)*sinh(f*x + e)^5 + 5*(35*cosh(f*x + e)^3 - 2*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^4 + 5*(35*c
osh(f*x + e)^4 - 4*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e)^3 + 5*(21*cosh(f*x + e)^5 - 4*cosh(f*x + e)^
3 + 3*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^2 + 5*(7*cosh(f*x + e)^6 - 2*cosh(f*x + e)^4 + 3*cosh(f*x + e)^
2)*e^(f*x + e)*sinh(f*x + e) + (5*cosh(f*x + e)^7 - 2*cosh(f*x + e)^5 + 5*cosh(f*x + e)^3)*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)^10 + 5*a^2*f*cosh(f*x + e)^8 + (
a^2*f*e^(2*f*x + 2*e) + a^2*f)*sinh(f*x + e)^10 + 10*(a^2*f*cosh(f*x + e)*e^(2*f*x + 2*e) + a^2*f*cosh(f*x + e
))*sinh(f*x + e)^9 + 10*a^2*f*cosh(f*x + e)^6 + 5*(9*a^2*f*cosh(f*x + e)^2 + a^2*f + (9*a^2*f*cosh(f*x + e)^2
+ a^2*f)*e^(2*f*x + 2*e))*sinh(f*x + e)^8 + 40*(3*a^2*f*cosh(f*x + e)^3 + a^2*f*cosh(f*x + e) + (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)^7 + 10*a^2*f*cosh(f*x + e)^4 + 10*(21*a^2*f*c
osh(f*x + e)^4 + 14*a^2*f*cosh(f*x + e)^2 + a^2*f + (21*a^2*f*cosh(f*x + e)^4 + 14*a^2*f*cosh(f*x + e)^2 + a^2
*f)*e^(2*f*x + 2*e))*sinh(f*x + e)^6 + 4*(63*a^2*f*cosh(f*x + e)^5 + 70*a^2*f*cosh(f*x + e)^3 + 15*a^2*f*cosh(
f*x + e) + (63*a^2*f*cosh(f*x + e)^5 + 70*a^2*f*cosh(f*x + e)^3 + 15*a^2*f*cosh(f*x + e))*e^(2*f*x + 2*e))*sin
h(f*x + e)^5 + 5*a^2*f*cosh(f*x + e)^2 + 10*(21*a^2*f*cosh(f*x + e)^6 + 35*a^2*f*cosh(f*x + e)^4 + 15*a^2*f*co
sh(f*x + e)^2 + a^2*f + (21*a^2*f*cosh(f*x + e)^6 + 35*a^2*f*cosh(f*x + e)^4 + 15*a^2*f*cosh(f*x + e)^2 + a^2*
f)*e^(2*f*x + 2*e))*sinh(f*x + e)^4 + 40*(3*a^2*f*cosh(f*x + e)^7 + 7*a^2*f*cosh(f*x + e)^5 + 5*a^2*f*cosh(f*x
+ e)^3 + a^2*f*cosh(f*x + e) + (3*a^2*f*cosh(f*x + e)^7 + 7*a^2*f*cosh(f*x + e)^5 + 5*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)^3 + a^2*f + 5*(9*a^2*f*cosh(f*x + e)^8 + 28*a^2*f*cosh(f*
x + e)^6 + 30*a^2*f*cosh(f*x + e)^4 + 12*a^2*f*cosh(f*x + e)^2 + a^2*f + (9*a^2*f*cosh(f*x + e)^8 + 28*a^2*f*c
osh(f*x + e)^6 + 30*a^2*f*cosh(f*x + e)^4 + 12*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)^10 + 5*a^2*f*cosh(f*x + e)^8 + 10*a^2*f*cosh(f*x + e)^6 + 10*a^2*f*cosh(f*x + e)^4 + 5
*a^2*f*cosh(f*x + e)^2 + a^2*f)*e^(2*f*x + 2*e) + 10*(a^2*f*cosh(f*x + e)^9 + 4*a^2*f*cosh(f*x + e)^7 + 6*a^2*
f*cosh(f*x + e)^5 + 4*a^2*f*cosh(f*x + e)^3 + a^2*f*cosh(f*x + e) + (a^2*f*cosh(f*x + e)^9 + 4*a^2*f*cosh(f*x
+ e)^7 + 6*a^2*f*cosh(f*x + e)^5 + 4*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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tanh ^{3}{\left (e + f x \right )}}{\left (a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, 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)**(3/2),x)

[Out]

Integral(tanh(e + f*x)**3/(a*(sinh(e + f*x)**2 + 1))**(3/2), 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)^(3/2),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 = 0.93, size = 305, normalized size = 6.93 \begin {gather*} \frac {272\,{\mathrm {e}}^{3\,e+3\,f\,x}\,\sqrt {a+a\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}}{15\,a^2\,f\,{\left ({\mathrm {e}}^{2\,e+2\,f\,x}+1\right )}^3\,\left ({\mathrm {e}}^{e+f\,x}+{\mathrm {e}}^{3\,e+3\,f\,x}\right )}-\frac {16\,{\mathrm {e}}^{3\,e+3\,f\,x}\,\sqrt {a+a\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}}{3\,a^2\,f\,{\left ({\mathrm {e}}^{2\,e+2\,f\,x}+1\right )}^2\,\left ({\mathrm {e}}^{e+f\,x}+{\mathrm {e}}^{3\,e+3\,f\,x}\right )}-\frac {128\,{\mathrm {e}}^{3\,e+3\,f\,x}\,\sqrt {a+a\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}}{5\,a^2\,f\,{\left ({\mathrm {e}}^{2\,e+2\,f\,x}+1\right )}^4\,\left ({\mathrm {e}}^{e+f\,x}+{\mathrm {e}}^{3\,e+3\,f\,x}\right )}+\frac {64\,{\mathrm {e}}^{3\,e+3\,f\,x}\,\sqrt {a+a\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}}{5\,a^2\,f\,{\left ({\mathrm {e}}^{2\,e+2\,f\,x}+1\right )}^5\,\left ({\mathrm {e}}^{e+f\,x}+{\mathrm {e}}^{3\,e+3\,f\,x}\right )} \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)^(3/2),x)

[Out]

(272*exp(3*e + 3*f*x)*(a + a*(exp(e + f*x)/2 - exp(- e - f*x)/2)^2)^(1/2))/(15*a^2*f*(exp(2*e + 2*f*x) + 1)^3*
(exp(e + f*x) + exp(3*e + 3*f*x))) - (16*exp(3*e + 3*f*x)*(a + a*(exp(e + f*x)/2 - exp(- e - f*x)/2)^2)^(1/2))
/(3*a^2*f*(exp(2*e + 2*f*x) + 1)^2*(exp(e + f*x) + exp(3*e + 3*f*x))) - (128*exp(3*e + 3*f*x)*(a + a*(exp(e +
f*x)/2 - exp(- e - f*x)/2)^2)^(1/2))/(5*a^2*f*(exp(2*e + 2*f*x) + 1)^4*(exp(e + f*x) + exp(3*e + 3*f*x))) + (6
4*exp(3*e + 3*f*x)*(a + a*(exp(e + f*x)/2 - exp(- e - f*x)/2)^2)^(1/2))/(5*a^2*f*(exp(2*e + 2*f*x) + 1)^5*(exp
(e + f*x) + exp(3*e + 3*f*x)))

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