3.5.37 \(\int \frac {\tanh ^5(e+f x)}{\sqrt {a+a \sinh ^2(e+f x)}} \, dx\) [437]

Optimal. Leaf size=66 \[ -\frac {a^2}{5 f \left (a \cosh ^2(e+f x)\right )^{5/2}}+\frac {2 a}{3 f \left (a \cosh ^2(e+f x)\right )^{3/2}}-\frac {1}{f \sqrt {a \cosh ^2(e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 66, 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^2}{5 f \left (a \cosh ^2(e+f x)\right )^{5/2}}+\frac {2 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]^5/Sqrt[a + a*Sinh[e + f*x]^2],x]

[Out]

-1/5*a^2/(f*(a*Cosh[e + f*x]^2)^(5/2)) + (2*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 ^5(e+f x)}{\sqrt {a+a \sinh ^2(e+f x)}} \, dx &=\int \frac {\tanh ^5(e+f x)}{\sqrt {a \cosh ^2(e+f x)}} \, dx\\ &=\frac {\text {Subst}\left (\int \frac {(1-x)^2}{x^3 \sqrt {a x}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=\frac {a^3 \text {Subst}\left (\int \frac {(1-x)^2}{(a x)^{7/2}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=\frac {a^3 \text {Subst}\left (\int \left (\frac {1}{(a x)^{7/2}}-\frac {2}{a (a x)^{5/2}}+\frac {1}{a^2 (a x)^{3/2}}\right ) \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=-\frac {a^2}{5 f \left (a \cosh ^2(e+f x)\right )^{5/2}}+\frac {2 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 = 0.11, size = 43, normalized size = 0.65 \begin {gather*} \frac {-15+10 \text {sech}^2(e+f x)-3 \text {sech}^4(e+f x)}{15 f \sqrt {a \cosh ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tanh[e + f*x]^5/Sqrt[a + a*Sinh[e + f*x]^2],x]

[Out]

(-15 + 10*Sech[e + f*x]^2 - 3*Sech[e + f*x]^4)/(15*f*Sqrt[a*Cosh[e + f*x]^2])

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.26, size = 41, normalized size = 0.62

method result size
default \(\frac {\mathit {`\,int/indef0`\,}\left (\frac {\sinh ^{5}\left (f x +e \right )}{\cosh \left (f x +e \right )^{6} \sqrt {a \left (\cosh ^{2}\left (f x +e \right )\right )}}, \sinh \left (f x +e \right )\right )}{f}\) \(41\)
risch \(-\frac {2 \left (15 \,{\mathrm e}^{8 f x +8 e}+20 \,{\mathrm e}^{6 f x +6 e}+58 \,{\mathrm e}^{4 f x +4 e}+20 \,{\mathrm e}^{2 f x +2 e}+15\right )}{15 \sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, \left ({\mathrm e}^{2 f x +2 e}+1\right )^{4} f}\) \(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(f*x+e)^5/(a+a*sinh(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

`int/indef0`(sinh(f*x+e)^5/cosh(f*x+e)^6/(a*cosh(f*x+e)^2)^(1/2),sinh(f*x+e))/f

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 476 vs. \(2 (59) = 118\).
time = 0.54, size = 476, normalized size = 7.21 \begin {gather*} -\frac {2 \, e^{\left (-f x - e\right )}}{{\left (5 \, \sqrt {a} e^{\left (-2 \, f x - 2 \, e\right )} + 10 \, \sqrt {a} e^{\left (-4 \, f x - 4 \, e\right )} + 10 \, \sqrt {a} e^{\left (-6 \, f x - 6 \, e\right )} + 5 \, \sqrt {a} e^{\left (-8 \, f x - 8 \, e\right )} + \sqrt {a} e^{\left (-10 \, f x - 10 \, e\right )} + \sqrt {a}\right )} f} - \frac {8 \, e^{\left (-3 \, f x - 3 \, e\right )}}{3 \, {\left (5 \, \sqrt {a} e^{\left (-2 \, f x - 2 \, e\right )} + 10 \, \sqrt {a} e^{\left (-4 \, f x - 4 \, e\right )} + 10 \, \sqrt {a} e^{\left (-6 \, f x - 6 \, e\right )} + 5 \, \sqrt {a} e^{\left (-8 \, f x - 8 \, e\right )} + \sqrt {a} e^{\left (-10 \, f x - 10 \, e\right )} + \sqrt {a}\right )} f} - \frac {116 \, e^{\left (-5 \, f x - 5 \, e\right )}}{15 \, {\left (5 \, \sqrt {a} e^{\left (-2 \, f x - 2 \, e\right )} + 10 \, \sqrt {a} e^{\left (-4 \, f x - 4 \, e\right )} + 10 \, \sqrt {a} e^{\left (-6 \, f x - 6 \, e\right )} + 5 \, \sqrt {a} e^{\left (-8 \, f x - 8 \, e\right )} + \sqrt {a} e^{\left (-10 \, f x - 10 \, e\right )} + \sqrt {a}\right )} f} - \frac {8 \, e^{\left (-7 \, f x - 7 \, e\right )}}{3 \, {\left (5 \, \sqrt {a} e^{\left (-2 \, f x - 2 \, e\right )} + 10 \, \sqrt {a} e^{\left (-4 \, f x - 4 \, e\right )} + 10 \, \sqrt {a} e^{\left (-6 \, f x - 6 \, e\right )} + 5 \, \sqrt {a} e^{\left (-8 \, f x - 8 \, e\right )} + \sqrt {a} e^{\left (-10 \, f x - 10 \, e\right )} + \sqrt {a}\right )} f} - \frac {2 \, e^{\left (-9 \, f x - 9 \, e\right )}}{{\left (5 \, \sqrt {a} e^{\left (-2 \, f x - 2 \, e\right )} + 10 \, \sqrt {a} e^{\left (-4 \, f x - 4 \, e\right )} + 10 \, \sqrt {a} e^{\left (-6 \, f x - 6 \, e\right )} + 5 \, \sqrt {a} e^{\left (-8 \, f x - 8 \, e\right )} + \sqrt {a} e^{\left (-10 \, f x - 10 \, e\right )} + \sqrt {a}\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)^5/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

-2*e^(-f*x - e)/((5*sqrt(a)*e^(-2*f*x - 2*e) + 10*sqrt(a)*e^(-4*f*x - 4*e) + 10*sqrt(a)*e^(-6*f*x - 6*e) + 5*s
qrt(a)*e^(-8*f*x - 8*e) + sqrt(a)*e^(-10*f*x - 10*e) + sqrt(a))*f) - 8/3*e^(-3*f*x - 3*e)/((5*sqrt(a)*e^(-2*f*
x - 2*e) + 10*sqrt(a)*e^(-4*f*x - 4*e) + 10*sqrt(a)*e^(-6*f*x - 6*e) + 5*sqrt(a)*e^(-8*f*x - 8*e) + sqrt(a)*e^
(-10*f*x - 10*e) + sqrt(a))*f) - 116/15*e^(-5*f*x - 5*e)/((5*sqrt(a)*e^(-2*f*x - 2*e) + 10*sqrt(a)*e^(-4*f*x -
 4*e) + 10*sqrt(a)*e^(-6*f*x - 6*e) + 5*sqrt(a)*e^(-8*f*x - 8*e) + sqrt(a)*e^(-10*f*x - 10*e) + sqrt(a))*f) -
8/3*e^(-7*f*x - 7*e)/((5*sqrt(a)*e^(-2*f*x - 2*e) + 10*sqrt(a)*e^(-4*f*x - 4*e) + 10*sqrt(a)*e^(-6*f*x - 6*e)
+ 5*sqrt(a)*e^(-8*f*x - 8*e) + sqrt(a)*e^(-10*f*x - 10*e) + sqrt(a))*f) - 2*e^(-9*f*x - 9*e)/((5*sqrt(a)*e^(-2
*f*x - 2*e) + 10*sqrt(a)*e^(-4*f*x - 4*e) + 10*sqrt(a)*e^(-6*f*x - 6*e) + 5*sqrt(a)*e^(-8*f*x - 8*e) + sqrt(a)
*e^(-10*f*x - 10*e) + sqrt(a))*f)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1387 vs. \(2 (56) = 112\).
time = 0.59, size = 1387, normalized size = 21.02 \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)^5/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

-2/15*(135*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^8 + 15*e^(f*x + e)*sinh(f*x + e)^9 + 20*(27*cosh(f*x + e)^2
 + 1)*e^(f*x + e)*sinh(f*x + e)^7 + 140*(9*cosh(f*x + e)^3 + cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^6 + 2*(9
45*cosh(f*x + e)^4 + 210*cosh(f*x + e)^2 + 29)*e^(f*x + e)*sinh(f*x + e)^5 + 10*(189*cosh(f*x + e)^5 + 70*cosh
(f*x + e)^3 + 29*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^4 + 20*(63*cosh(f*x + e)^6 + 35*cosh(f*x + e)^4 + 29
*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e)^3 + 20*(27*cosh(f*x + e)^7 + 21*cosh(f*x + e)^5 + 29*cosh(f*x
+ e)^3 + 3*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^2 + 5*(27*cosh(f*x + e)^8 + 28*cosh(f*x + e)^6 + 58*cosh(f
*x + e)^4 + 12*cosh(f*x + e)^2 + 3)*e^(f*x + e)*sinh(f*x + e) + (15*cosh(f*x + e)^9 + 20*cosh(f*x + e)^7 + 58*
cosh(f*x + e)^5 + 20*cosh(f*x + e)^3 + 15*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*f*cosh(f*x + e)^10 + (a*f*e^(2*f*x + 2*e) + a*f)*sinh(f*x + e)^10 + 5*a*f*cosh(f*x +
 e)^8 + 10*(a*f*cosh(f*x + e)*e^(2*f*x + 2*e) + a*f*cosh(f*x + e))*sinh(f*x + e)^9 + 5*(9*a*f*cosh(f*x + e)^2
+ a*f + (9*a*f*cosh(f*x + e)^2 + a*f)*e^(2*f*x + 2*e))*sinh(f*x + e)^8 + 10*a*f*cosh(f*x + e)^6 + 40*(3*a*f*co
sh(f*x + e)^3 + a*f*cosh(f*x + e) + (3*a*f*cosh(f*x + e)^3 + a*f*cosh(f*x + e))*e^(2*f*x + 2*e))*sinh(f*x + e)
^7 + 10*(21*a*f*cosh(f*x + e)^4 + 14*a*f*cosh(f*x + e)^2 + a*f + (21*a*f*cosh(f*x + e)^4 + 14*a*f*cosh(f*x + e
)^2 + a*f)*e^(2*f*x + 2*e))*sinh(f*x + e)^6 + 10*a*f*cosh(f*x + e)^4 + 4*(63*a*f*cosh(f*x + e)^5 + 70*a*f*cosh
(f*x + e)^3 + 15*a*f*cosh(f*x + e) + (63*a*f*cosh(f*x + e)^5 + 70*a*f*cosh(f*x + e)^3 + 15*a*f*cosh(f*x + e))*
e^(2*f*x + 2*e))*sinh(f*x + e)^5 + 10*(21*a*f*cosh(f*x + e)^6 + 35*a*f*cosh(f*x + e)^4 + 15*a*f*cosh(f*x + e)^
2 + a*f + (21*a*f*cosh(f*x + e)^6 + 35*a*f*cosh(f*x + e)^4 + 15*a*f*cosh(f*x + e)^2 + a*f)*e^(2*f*x + 2*e))*si
nh(f*x + e)^4 + 5*a*f*cosh(f*x + e)^2 + 40*(3*a*f*cosh(f*x + e)^7 + 7*a*f*cosh(f*x + e)^5 + 5*a*f*cosh(f*x + e
)^3 + a*f*cosh(f*x + e) + (3*a*f*cosh(f*x + e)^7 + 7*a*f*cosh(f*x + e)^5 + 5*a*f*cosh(f*x + e)^3 + a*f*cosh(f*
x + e))*e^(2*f*x + 2*e))*sinh(f*x + e)^3 + 5*(9*a*f*cosh(f*x + e)^8 + 28*a*f*cosh(f*x + e)^6 + 30*a*f*cosh(f*x
 + e)^4 + 12*a*f*cosh(f*x + e)^2 + a*f + (9*a*f*cosh(f*x + e)^8 + 28*a*f*cosh(f*x + e)^6 + 30*a*f*cosh(f*x + e
)^4 + 12*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)^10 + 5*a*f*cos
h(f*x + e)^8 + 10*a*f*cosh(f*x + e)^6 + 10*a*f*cosh(f*x + e)^4 + 5*a*f*cosh(f*x + e)^2 + a*f)*e^(2*f*x + 2*e)
+ 10*(a*f*cosh(f*x + e)^9 + 4*a*f*cosh(f*x + e)^7 + 6*a*f*cosh(f*x + e)^5 + 4*a*f*cosh(f*x + e)^3 + a*f*cosh(f
*x + e) + (a*f*cosh(f*x + e)^9 + 4*a*f*cosh(f*x + e)^7 + 6*a*f*cosh(f*x + e)^5 + 4*a*f*cosh(f*x + e)^3 + a*f*c
osh(f*x + e))*e^(2*f*x + 2*e))*sinh(f*x + e))

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tanh ^{5}{\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)**5/(a+a*sinh(f*x+e)**2)**(1/2),x)

[Out]

Integral(tanh(e + f*x)**5/sqrt(a*(sinh(e + f*x)**2 + 1)), x)

________________________________________________________________________________________

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)^5/(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

________________________________________________________________________________________

Mupad [B]
time = 0.89, size = 381, normalized size = 5.77 \begin {gather*} \frac {32\,{\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\,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 {4\,{\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}}{a\,f\,\left ({\mathrm {e}}^{2\,e+2\,f\,x}+1\right )\,\left ({\mathrm {e}}^{e+f\,x}+{\mathrm {e}}^{3\,e+3\,f\,x}\right )}-\frac {352\,{\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\,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 {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\,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\,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)^5/(a + a*sinh(e + f*x)^2)^(1/2),x)

[Out]

(32*exp(3*e + 3*f*x)*(a + a*(exp(e + f*x)/2 - exp(- e - f*x)/2)^2)^(1/2))/(3*a*f*(exp(2*e + 2*f*x) + 1)^2*(exp
(e + f*x) + exp(3*e + 3*f*x))) - (4*exp(3*e + 3*f*x)*(a + a*(exp(e + f*x)/2 - exp(- e - f*x)/2)^2)^(1/2))/(a*f
*(exp(2*e + 2*f*x) + 1)*(exp(e + f*x) + exp(3*e + 3*f*x))) - (352*exp(3*e + 3*f*x)*(a + a*(exp(e + f*x)/2 - ex
p(- e - f*x)/2)^2)^(1/2))/(15*a*f*(exp(2*e + 2*f*x) + 1)^3*(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*f*(exp(2*e + 2*f*x) + 1)^4*(exp(e + f*x) + e
xp(3*e + 3*f*x))) - (64*exp(3*e + 3*f*x)*(a + a*(exp(e + f*x)/2 - exp(- e - f*x)/2)^2)^(1/2))/(5*a*f*(exp(2*e
+ 2*f*x) + 1)^5*(exp(e + f*x) + exp(3*e + 3*f*x)))

________________________________________________________________________________________