3.448 \(\int \frac{\tanh ^3(e+f x)}{(a+a \sinh ^2(e+f x))^{3/2}} \, dx\)
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]
a/(5*f*(a*Cosh[e + f*x]^2)^(5/2)) - 1/(3*f*(a*Cosh[e + f*x]^2)^(3/2))
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Rubi [A] time = 0.126134, antiderivative size = 44, 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} \[ \frac{a}{5 f \left (a \cosh ^2(e+f x)\right )^{5/2}}-\frac{1}{3 f \left (a \cosh ^2(e+f x)\right )^{3/2}} \]
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 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)}{\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{\operatorname{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 \operatorname{Subst}\left (\int \frac{1-x}{(a x)^{7/2}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=-\frac{a^2 \operatorname{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*}
Mathematica [A] time = 0.122513, size = 34, normalized size = 0.77 \[ \frac{a \left (3-5 \cosh ^2(e+f x)\right )}{15 f \left (a \cosh ^2(e+f x)\right )^{5/2}} \]
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] time = 0.098, size = 44, 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 ) ^{6}a}{\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)^(3/2),x)
[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] time = 1.87989, size = 362, normalized size = 8.23 \begin{align*} -\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{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)^(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] time = 1.89151, size = 3571, normalized size = 81.16 \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)^(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., size = 0, normalized size = 0. \begin{align*} \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{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)**(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 [A] time = 1.53619, size = 92, normalized size = 2.09 \begin{align*} -\frac{8 \,{\left (5 \, \sqrt{a} e^{\left (7 \, f x + 7 \, e\right )} - 2 \, \sqrt{a} e^{\left (5 \, f x + 5 \, e\right )} + 5 \, \sqrt{a} e^{\left (3 \, f x + 3 \, e\right )}\right )}}{15 \, a^{2} 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)^3/(a+a*sinh(f*x+e)^2)^(3/2),x, algorithm="giac")
[Out]
-8/15*(5*sqrt(a)*e^(7*f*x + 7*e) - 2*sqrt(a)*e^(5*f*x + 5*e) + 5*sqrt(a)*e^(3*f*x + 3*e))/(a^2*f*(e^(2*f*x + 2
*e) + 1)^5)