3.426 \(\int \sqrt {a+a \sinh ^2(e+f x)} \tanh ^5(e+f x) \, dx\)

Optimal. Leaf size=63 \[ -\frac {a^2}{3 f \left (a \cosh ^2(e+f x)\right )^{3/2}}+\frac {2 a}{f \sqrt {a \cosh ^2(e+f x)}}+\frac {\sqrt {a \cosh ^2(e+f x)}}{f} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 63, 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 = {3176, 3205, 16, 43} \[ -\frac {a^2}{3 f \left (a \cosh ^2(e+f x)\right )^{3/2}}+\frac {2 a}{f \sqrt {a \cosh ^2(e+f x)}}+\frac {\sqrt {a \cosh ^2(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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
]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 51, normalized size = 0.81 \[ \frac {\left (3 \cosh ^4(e+f x)+6 \cosh ^2(e+f x)-1\right ) \text {sech}^4(e+f x) \sqrt {a \cosh ^2(e+f x)}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*Cosh[e + f*x]^2]*(-1 + 6*Cosh[e + f*x]^2 + 3*Cosh[e + f*x]^4)*Sech[e + f*x]^4)/(3*f)

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fricas [B]  time = 2.95, size = 875, normalized size = 13.89 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(24*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^7 + 3*e^(f*x + e)*sinh(f*x + e)^8 + 12*(7*cosh(f*x + e)^2 + 3)
*e^(f*x + e)*sinh(f*x + e)^6 + 24*(7*cosh(f*x + e)^3 + 9*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^5 + 10*(21*c
osh(f*x + e)^4 + 54*cosh(f*x + e)^2 + 5)*e^(f*x + e)*sinh(f*x + e)^4 + 8*(21*cosh(f*x + e)^5 + 90*cosh(f*x + e
)^3 + 25*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^3 + 12*(7*cosh(f*x + e)^6 + 45*cosh(f*x + e)^4 + 25*cosh(f*x
+ e)^2 + 3)*e^(f*x + e)*sinh(f*x + e)^2 + 8*(3*cosh(f*x + e)^7 + 27*cosh(f*x + e)^5 + 25*cosh(f*x + e)^3 + 9*
cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e) + (3*cosh(f*x + e)^8 + 36*cosh(f*x + e)^6 + 50*cosh(f*x + e)^4 + 36*c
osh(f*x + e)^2 + 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)/(f*cosh(f*x +
e)^7 + (f*e^(2*f*x + 2*e) + f)*sinh(f*x + e)^7 + 7*(f*cosh(f*x + e)*e^(2*f*x + 2*e) + f*cosh(f*x + e))*sinh(f*
x + e)^6 + 3*f*cosh(f*x + e)^5 + 3*(7*f*cosh(f*x + e)^2 + (7*f*cosh(f*x + e)^2 + f)*e^(2*f*x + 2*e) + f)*sinh(
f*x + e)^5 + 5*(7*f*cosh(f*x + e)^3 + 3*f*cosh(f*x + e) + (7*f*cosh(f*x + e)^3 + 3*f*cosh(f*x + e))*e^(2*f*x +
2*e))*sinh(f*x + e)^4 + 3*f*cosh(f*x + e)^3 + (35*f*cosh(f*x + e)^4 + 30*f*cosh(f*x + e)^2 + (35*f*cosh(f*x +
e)^4 + 30*f*cosh(f*x + e)^2 + 3*f)*e^(2*f*x + 2*e) + 3*f)*sinh(f*x + e)^3 + 3*(7*f*cosh(f*x + e)^5 + 10*f*cos
h(f*x + e)^3 + 3*f*cosh(f*x + e) + (7*f*cosh(f*x + e)^5 + 10*f*cosh(f*x + e)^3 + 3*f*cosh(f*x + e))*e^(2*f*x +
2*e))*sinh(f*x + e)^2 + f*cosh(f*x + e) + (f*cosh(f*x + e)^7 + 3*f*cosh(f*x + e)^5 + 3*f*cosh(f*x + e)^3 + f*
cosh(f*x + e))*e^(2*f*x + 2*e) + (7*f*cosh(f*x + e)^6 + 15*f*cosh(f*x + e)^4 + 9*f*cosh(f*x + e)^2 + (7*f*cosh
(f*x + e)^6 + 15*f*cosh(f*x + e)^4 + 9*f*cosh(f*x + e)^2 + f)*e^(2*f*x + 2*e) + f)*sinh(f*x + e))

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giac [A]  time = 0.21, size = 80, normalized size = 1.27 \[ \frac {\sqrt {a} {\left (\frac {8 \, {\left (3 \, e^{\left (5 \, f x + 5 \, e\right )} + 4 \, e^{\left (3 \, f x + 3 \, e\right )} + 3 \, e^{\left (f x + e\right )}\right )}}{{\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}^{3}} + 3 \, e^{\left (f x + e\right )} + 3 \, e^{\left (-f x - e\right )}\right )}}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(a)*(8*(3*e^(5*f*x + 5*e) + 4*e^(3*f*x + 3*e) + 3*e^(f*x + e))/(e^(2*f*x + 2*e) + 1)^3 + 3*e^(f*x + e)
+ 3*e^(-f*x - e))/f

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maple [C]  time = 0.24, size = 42, normalized size = 0.67 \[ \frac {\mathit {`\,int/indef0`\,}\left (\frac {\left (\sinh ^{5}\left (f x +e \right )\right ) a}{\cosh \left (f x +e \right )^{4} \sqrt {a \left (\cosh ^{2}\left (f x +e \right )\right )}}, \sinh \left (f x +e \right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

`int/indef0`(sinh(f*x+e)^5*a/cosh(f*x+e)^4/(a*cosh(f*x+e)^2)^(1/2),sinh(f*x+e))/f

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maxima [B]  time = 0.88, size = 292, normalized size = 4.63 \[ \frac {6 \, \sqrt {a} e^{\left (-2 \, f x - 2 \, e\right )}}{f {\left (e^{\left (-f x - e\right )} + 3 \, e^{\left (-3 \, f x - 3 \, e\right )} + 3 \, e^{\left (-5 \, f x - 5 \, e\right )} + e^{\left (-7 \, f x - 7 \, e\right )}\right )}} + \frac {25 \, \sqrt {a} e^{\left (-4 \, f x - 4 \, e\right )}}{3 \, f {\left (e^{\left (-f x - e\right )} + 3 \, e^{\left (-3 \, f x - 3 \, e\right )} + 3 \, e^{\left (-5 \, f x - 5 \, e\right )} + e^{\left (-7 \, f x - 7 \, e\right )}\right )}} + \frac {6 \, \sqrt {a} e^{\left (-6 \, f x - 6 \, e\right )}}{f {\left (e^{\left (-f x - e\right )} + 3 \, e^{\left (-3 \, f x - 3 \, e\right )} + 3 \, e^{\left (-5 \, f x - 5 \, e\right )} + e^{\left (-7 \, f x - 7 \, e\right )}\right )}} + \frac {\sqrt {a} e^{\left (-8 \, f x - 8 \, e\right )}}{2 \, f {\left (e^{\left (-f x - e\right )} + 3 \, e^{\left (-3 \, f x - 3 \, e\right )} + 3 \, e^{\left (-5 \, f x - 5 \, e\right )} + e^{\left (-7 \, f x - 7 \, e\right )}\right )}} + \frac {\sqrt {a}}{2 \, f {\left (e^{\left (-f x - e\right )} + 3 \, e^{\left (-3 \, f x - 3 \, e\right )} + 3 \, e^{\left (-5 \, f x - 5 \, e\right )} + e^{\left (-7 \, f x - 7 \, e\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*sqrt(a)*e^(-2*f*x - 2*e)/(f*(e^(-f*x - e) + 3*e^(-3*f*x - 3*e) + 3*e^(-5*f*x - 5*e) + e^(-7*f*x - 7*e))) + 2
5/3*sqrt(a)*e^(-4*f*x - 4*e)/(f*(e^(-f*x - e) + 3*e^(-3*f*x - 3*e) + 3*e^(-5*f*x - 5*e) + e^(-7*f*x - 7*e))) +
6*sqrt(a)*e^(-6*f*x - 6*e)/(f*(e^(-f*x - e) + 3*e^(-3*f*x - 3*e) + 3*e^(-5*f*x - 5*e) + e^(-7*f*x - 7*e))) +
1/2*sqrt(a)*e^(-8*f*x - 8*e)/(f*(e^(-f*x - e) + 3*e^(-3*f*x - 3*e) + 3*e^(-5*f*x - 5*e) + e^(-7*f*x - 7*e))) +
1/2*sqrt(a)/(f*(e^(-f*x - e) + 3*e^(-3*f*x - 3*e) + 3*e^(-5*f*x - 5*e) + e^(-7*f*x - 7*e)))

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mupad [B]  time = 0.94, size = 252, normalized size = 4.00 \[ \frac {\sqrt {a+a\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}}{f}+\frac {8\,{\mathrm {e}}^{3\,e+3\,f\,x}\,\sqrt {a+a\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}}{f\,\left ({\mathrm {e}}^{2\,e+2\,f\,x}+1\right )\,\left ({\mathrm {e}}^{e+f\,x}+{\mathrm {e}}^{3\,e+3\,f\,x}\right )}-\frac {16\,{\mathrm {e}}^{3\,e+3\,f\,x}\,\sqrt {a+a\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}}{3\,f\,{\left ({\mathrm {e}}^{2\,e+2\,f\,x}+1\right )}^2\,\left ({\mathrm {e}}^{e+f\,x}+{\mathrm {e}}^{3\,e+3\,f\,x}\right )}+\frac {16\,{\mathrm {e}}^{3\,e+3\,f\,x}\,\sqrt {a+a\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}}{3\,f\,{\left ({\mathrm {e}}^{2\,e+2\,f\,x}+1\right )}^3\,\left ({\mathrm {e}}^{e+f\,x}+{\mathrm {e}}^{3\,e+3\,f\,x}\right )} \]

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]

(a + a*(exp(e + f*x)/2 - exp(- e - f*x)/2)^2)^(1/2)/f + (8*exp(3*e + 3*f*x)*(a + a*(exp(e + f*x)/2 - exp(- e -
f*x)/2)^2)^(1/2))/(f*(exp(2*e + 2*f*x) + 1)*(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*f*(exp(2*e + 2*f*x) + 1)^2*(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*f*(exp(2*e + 2*f*x) + 1)^3*(ex
p(e + f*x) + exp(3*e + 3*f*x)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )} \tanh ^{5}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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