3.437 \(\int \frac{\tanh ^5(e+f x)}{\sqrt{a+a \sinh ^2(e+f x)}} \, dx\)
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]
-a^2/(5*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])
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Rubi [A] time = 0.129055, antiderivative size = 66, 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} \[ -\frac{a^2}{5 f \left (a \cosh ^2(e+f x)\right )^{5/2}}+\frac{2 a}{3 f \left (a \cosh ^2(e+f x)\right )^{3/2}}-\frac{1}{f \sqrt{a \cosh ^2(e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[Tanh[e + f*x]^5/Sqrt[a + a*Sinh[e + f*x]^2],x]
[Out]
-a^2/(5*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 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 ^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{\operatorname{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 \operatorname{Subst}\left (\int \frac{(1-x)^2}{(a x)^{7/2}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=\frac{a^3 \operatorname{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 = 0.0896333, size = 43, normalized size = 0.65 \[ \frac{-3 \text{sech}^4(e+f x)+10 \text{sech}^2(e+f x)-15}{15 f \sqrt{a \cosh ^2(e+f x)}} \]
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])
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Maple [C] time = 0.102, size = 41, normalized size = 0.6 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ({\frac{ \left ( \sinh \left ( fx+e \right ) \right ) ^{5}}{ \left ( \cosh \left ( fx+e \right ) \right ) ^{6}}{\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)^5/(a+a*sinh(f*x+e)^2)^(1/2),x)
[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
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Maxima [B] time = 2.08391, size = 602, normalized size = 9.12 \begin{align*} -\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{align*}
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)
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Fricas [B] time = 1.89517, size = 3776, normalized size = 57.21 \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)^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))
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{5}{\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)**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)
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Giac [A] time = 1.39911, size = 128, normalized size = 1.94 \begin{align*} -\frac{2 \,{\left (15 \, \sqrt{a} e^{\left (9 \, f x + 9 \, e\right )} + 20 \, \sqrt{a} e^{\left (7 \, f x + 7 \, e\right )} + 58 \, \sqrt{a} e^{\left (5 \, f x + 5 \, e\right )} + 20 \, \sqrt{a} e^{\left (3 \, f x + 3 \, e\right )} + 15 \, \sqrt{a} e^{\left (f x + e\right )}\right )}}{15 \, a f{\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}^{5}} \end{align*}
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]
-2/15*(15*sqrt(a)*e^(9*f*x + 9*e) + 20*sqrt(a)*e^(7*f*x + 7*e) + 58*sqrt(a)*e^(5*f*x + 5*e) + 20*sqrt(a)*e^(3*
f*x + 3*e) + 15*sqrt(a)*e^(f*x + e))/(a*f*(e^(2*f*x + 2*e) + 1)^5)