3.432 \(\int \sqrt{a+a \sinh ^2(e+f x)} \tanh ^4(e+f x) \, dx\)

Optimal. Leaf size=91 \[ -\frac{\tanh ^3(e+f x) \sqrt{a \cosh ^2(e+f x)}}{2 f}+\frac{3 \tanh (e+f x) \sqrt{a \cosh ^2(e+f x)}}{2 f}-\frac{3 \text{sech}(e+f x) \sqrt{a \cosh ^2(e+f x)} \tan ^{-1}(\sinh (e+f x))}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.124577, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3176, 3207, 2592, 288, 321, 203} \[ -\frac{\tanh ^3(e+f x) \sqrt{a \cosh ^2(e+f x)}}{2 f}+\frac{3 \tanh (e+f x) \sqrt{a \cosh ^2(e+f x)}}{2 f}-\frac{3 \text{sech}(e+f x) \sqrt{a \cosh ^2(e+f x)} \tan ^{-1}(\sinh (e+f x))}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sinh[e + f*x]^2]*Tanh[e + f*x]^4,x]

[Out]

(-3*ArcTan[Sinh[e + f*x]]*Sqrt[a*Cosh[e + f*x]^2]*Sech[e + f*x])/(2*f) + (3*Sqrt[a*Cosh[e + f*x]^2]*Tanh[e + f
*x])/(2*f) - (Sqrt[a*Cosh[e + f*x]^2]*Tanh[e + f*x]^3)/(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 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \sqrt{a+a \sinh ^2(e+f x)} \tanh ^4(e+f x) \, dx &=\int \sqrt{a \cosh ^2(e+f x)} \tanh ^4(e+f x) \, dx\\ &=\left (\sqrt{a \cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \int \sinh (e+f x) \tanh ^3(e+f x) \, dx\\ &=\frac{\left (\sqrt{a \cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^2} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=-\frac{\sqrt{a \cosh ^2(e+f x)} \tanh ^3(e+f x)}{2 f}+\frac{\left (3 \sqrt{a \cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^2}{1+x^2} \, dx,x,\sinh (e+f x)\right )}{2 f}\\ &=\frac{3 \sqrt{a \cosh ^2(e+f x)} \tanh (e+f x)}{2 f}-\frac{\sqrt{a \cosh ^2(e+f x)} \tanh ^3(e+f x)}{2 f}-\frac{\left (3 \sqrt{a \cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (e+f x)\right )}{2 f}\\ &=-\frac{3 \tan ^{-1}(\sinh (e+f x)) \sqrt{a \cosh ^2(e+f x)} \text{sech}(e+f x)}{2 f}+\frac{3 \sqrt{a \cosh ^2(e+f x)} \tanh (e+f x)}{2 f}-\frac{\sqrt{a \cosh ^2(e+f x)} \tanh ^3(e+f x)}{2 f}\\ \end{align*}

Mathematica [A]  time = 0.194464, size = 55, normalized size = 0.6 \[ \frac{a \left ((\cosh (2 (e+f x))+2) \tanh (e+f x)-3 \cosh (e+f x) \tan ^{-1}(\sinh (e+f x))\right )}{2 f \sqrt{a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Sinh[e + f*x]^2]*Tanh[e + f*x]^4,x]

[Out]

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

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Maple [A]  time = 0.098, size = 69, normalized size = 0.8 \begin{align*} -{\frac{a \left ( 3\,\arctan \left ( \sinh \left ( fx+e \right ) \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2}-2\, \left ( \cosh \left ( fx+e \right ) \right ) ^{2}\sinh \left ( fx+e \right ) -\sinh \left ( fx+e \right ) \right ) }{2\,f\cosh \left ( fx+e \right ) }{\frac{1}{\sqrt{a \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}}}} \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)^4,x)

[Out]

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

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Maxima [B]  time = 1.67472, size = 522, normalized size = 5.74 \begin{align*} \frac{15 \, \sqrt{a} \arctan \left (e^{\left (-f x - e\right )}\right )}{8 \, f} + \frac{3 \, \sqrt{a} \arctan \left (e^{\left (-f x - e\right )}\right ) + \frac{5 \, \sqrt{a} e^{\left (-f x - e\right )} + 3 \, \sqrt{a} e^{\left (-3 \, f x - 3 \, e\right )}}{2 \, e^{\left (-2 \, f x - 2 \, e\right )} + e^{\left (-4 \, f x - 4 \, e\right )} + 1}}{4 \, f} + \frac{3 \, \sqrt{a} \arctan \left (e^{\left (-f x - e\right )}\right ) - \frac{3 \, \sqrt{a} e^{\left (-f x - e\right )} + 5 \, \sqrt{a} e^{\left (-3 \, f x - 3 \, e\right )}}{2 \, e^{\left (-2 \, f x - 2 \, e\right )} + e^{\left (-4 \, f x - 4 \, e\right )} + 1}}{4 \, f} - \frac{3 \,{\left (\sqrt{a} \arctan \left (e^{\left (-f x - e\right )}\right ) - \frac{\sqrt{a} e^{\left (-f x - e\right )} - \sqrt{a} e^{\left (-3 \, f x - 3 \, e\right )}}{2 \, e^{\left (-2 \, f x - 2 \, e\right )} + e^{\left (-4 \, f x - 4 \, e\right )} + 1}\right )}}{8 \, f} + \frac{25 \, \sqrt{a} e^{\left (-2 \, f x - 2 \, e\right )} + 15 \, \sqrt{a} e^{\left (-4 \, f x - 4 \, e\right )} + 8 \, \sqrt{a}}{16 \, f{\left (e^{\left (-f x - e\right )} + 2 \, e^{\left (-3 \, f x - 3 \, e\right )} + e^{\left (-5 \, f x - 5 \, e\right )}\right )}} - \frac{15 \, \sqrt{a} e^{\left (-f x - e\right )} + 25 \, \sqrt{a} e^{\left (-3 \, f x - 3 \, e\right )} + 8 \, \sqrt{a} e^{\left (-5 \, f x - 5 \, e\right )}}{16 \, f{\left (2 \, e^{\left (-2 \, f x - 2 \, e\right )} + e^{\left (-4 \, f x - 4 \, e\right )} + 1\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)^4,x, algorithm="maxima")

[Out]

15/8*sqrt(a)*arctan(e^(-f*x - e))/f + 1/4*(3*sqrt(a)*arctan(e^(-f*x - e)) + (5*sqrt(a)*e^(-f*x - e) + 3*sqrt(a
)*e^(-3*f*x - 3*e))/(2*e^(-2*f*x - 2*e) + e^(-4*f*x - 4*e) + 1))/f + 1/4*(3*sqrt(a)*arctan(e^(-f*x - e)) - (3*
sqrt(a)*e^(-f*x - e) + 5*sqrt(a)*e^(-3*f*x - 3*e))/(2*e^(-2*f*x - 2*e) + e^(-4*f*x - 4*e) + 1))/f - 3/8*(sqrt(
a)*arctan(e^(-f*x - e)) - (sqrt(a)*e^(-f*x - e) - sqrt(a)*e^(-3*f*x - 3*e))/(2*e^(-2*f*x - 2*e) + e^(-4*f*x -
4*e) + 1))/f + 1/16*(25*sqrt(a)*e^(-2*f*x - 2*e) + 15*sqrt(a)*e^(-4*f*x - 4*e) + 8*sqrt(a))/(f*(e^(-f*x - e) +
2*e^(-3*f*x - 3*e) + e^(-5*f*x - 5*e))) - 1/16*(15*sqrt(a)*e^(-f*x - e) + 25*sqrt(a)*e^(-3*f*x - 3*e) + 8*sqr
t(a)*e^(-5*f*x - 5*e))/(f*(2*e^(-2*f*x - 2*e) + e^(-4*f*x - 4*e) + 1))

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Fricas [B]  time = 1.96191, size = 2003, normalized size = 22.01 \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)^4,x, algorithm="fricas")

[Out]

1/2*(6*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^5 + e^(f*x + e)*sinh(f*x + e)^6 + 3*(5*cosh(f*x + e)^2 + 1)*e^(
f*x + e)*sinh(f*x + e)^4 + 4*(5*cosh(f*x + e)^3 + 3*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^3 + 3*(5*cosh(f*x
+ e)^4 + 6*cosh(f*x + e)^2 - 1)*e^(f*x + e)*sinh(f*x + e)^2 + 6*(cosh(f*x + e)^5 + 2*cosh(f*x + e)^3 - cosh(f
*x + e))*e^(f*x + e)*sinh(f*x + e) - 6*(5*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^4 + e^(f*x + e)*sinh(f*x + e
)^5 + 2*(5*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e)^3 + 2*(5*cosh(f*x + e)^3 + 3*cosh(f*x + e))*e^(f*x +
e)*sinh(f*x + e)^2 + (5*cosh(f*x + e)^4 + 6*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e) + (cosh(f*x + e)^5
+ 2*cosh(f*x + e)^3 + cosh(f*x + e))*e^(f*x + e))*arctan(cosh(f*x + e) + sinh(f*x + e)) + (cosh(f*x + e)^6 +
3*cosh(f*x + e)^4 - 3*cosh(f*x + e)^2 - 1)*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)^5 + (f*e^(2*f*x + 2*e) + f)*sinh(f*x + e)^5 + 5*(f*cosh(f*x + e)*e^(2*f*x + 2*e) + f
*cosh(f*x + e))*sinh(f*x + e)^4 + 2*f*cosh(f*x + e)^3 + 2*(5*f*cosh(f*x + e)^2 + (5*f*cosh(f*x + e)^2 + f)*e^(
2*f*x + 2*e) + f)*sinh(f*x + e)^3 + 2*(5*f*cosh(f*x + e)^3 + 3*f*cosh(f*x + e) + (5*f*cosh(f*x + e)^3 + 3*f*co
sh(f*x + e))*e^(2*f*x + 2*e))*sinh(f*x + e)^2 + f*cosh(f*x + e) + (f*cosh(f*x + e)^5 + 2*f*cosh(f*x + e)^3 + f
*cosh(f*x + e))*e^(2*f*x + 2*e) + (5*f*cosh(f*x + e)^4 + 6*f*cosh(f*x + e)^2 + (5*f*cosh(f*x + e)^4 + 6*f*cosh
(f*x + e)^2 + f)*e^(2*f*x + 2*e) + f)*sinh(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )} \tanh ^{4}{\left (e + f x \right )}\, dx \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)**4,x)

[Out]

Integral(sqrt(a*(sinh(e + f*x)**2 + 1))*tanh(e + f*x)**4, x)

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Giac [A]  time = 1.28528, size = 100, normalized size = 1.1 \begin{align*} \frac{\sqrt{a}{\left (\frac{2 \,{\left (e^{\left (3 \, f x + 3 \, e\right )} - e^{\left (f x + e\right )}\right )}}{{\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}^{2}} - 6 \, \arctan \left (e^{\left (f x + e\right )}\right ) + e^{\left (f x + e\right )} - e^{\left (-f x - e\right )}\right )}}{2 \, 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)^4,x, algorithm="giac")

[Out]

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