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]

-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

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Rubi [A]  time = 0.128025, antiderivative size = 63, 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} \[ -\frac{a^2}{3 f \left (a \cosh ^2(e+f x)\right )^{3/2}}+\frac{2 a}{f \sqrt{a \cosh ^2(e+f x)}}+\frac{\sqrt{a \cosh ^2(e+f x)}}{f} \]

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 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 \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*}

Mathematica [A]  time = 0.100224, 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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Maple [C]  time = 0.145, size = 42, normalized size = 0.7 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ({\frac{ \left ( \sinh \left ( fx+e \right ) \right ) ^{5}a}{ \left ( \cosh \left ( fx+e \right ) \right ) ^{4}}{\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((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 = 1.79859, size = 394, normalized size = 6.25 \begin{align*} \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 )}} \end{align*}

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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Fricas [B]  time = 1.8592, size = 2361, normalized size = 37.48 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [A]  time = 1.30062, size = 108, normalized size = 1.71 \begin{align*} \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} \end{align*}

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