3.430 \(\int \coth ^3(e+f x) \sqrt{a+a \sinh ^2(e+f x)} \, dx\)

Optimal. Leaf size=87 \[ \frac{3 \sqrt{a \cosh ^2(e+f x)}}{2 f}-\frac{3 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \cosh ^2(e+f x)}}{\sqrt{a}}\right )}{2 f}-\frac{\text{csch}^2(e+f x) \left (a \cosh ^2(e+f x)\right )^{3/2}}{2 a f} \]

[Out]

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

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Rubi [A]  time = 0.142148, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3176, 3205, 16, 47, 50, 63, 206} \[ \frac{3 \sqrt{a \cosh ^2(e+f x)}}{2 f}-\frac{3 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \cosh ^2(e+f x)}}{\sqrt{a}}\right )}{2 f}-\frac{\text{csch}^2(e+f x) \left (a \cosh ^2(e+f x)\right )^{3/2}}{2 a f} \]

Antiderivative was successfully verified.

[In]

Int[Coth[e + f*x]^3*Sqrt[a + a*Sinh[e + f*x]^2],x]

[Out]

(-3*Sqrt[a]*ArcTanh[Sqrt[a*Cosh[e + f*x]^2]/Sqrt[a]])/(2*f) + (3*Sqrt[a*Cosh[e + f*x]^2])/(2*f) - ((a*Cosh[e +
f*x]^2)^(3/2)*Csch[e + f*x]^2)/(2*a*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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.260867, size = 77, normalized size = 0.89 \[ \frac{\text{sech}(e+f x) \sqrt{a \cosh ^2(e+f x)} \left (8 \cosh (e+f x)-\text{csch}^2\left (\frac{1}{2} (e+f x)\right )-\text{sech}^2\left (\frac{1}{2} (e+f x)\right )+12 \log \left (\tanh \left (\frac{1}{2} (e+f x)\right )\right )\right )}{8 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[e + f*x]^3*Sqrt[a + a*Sinh[e + f*x]^2],x]

[Out]

(Sqrt[a*Cosh[e + f*x]^2]*(8*Cosh[e + f*x] - Csch[(e + f*x)/2]^2 + 12*Log[Tanh[(e + f*x)/2]] - Sech[(e + f*x)/2
]^2)*Sech[e + f*x])/(8*f)

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Maple [C]  time = 0.092, size = 54, normalized size = 0.6 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ({\frac{a \left ( \cosh \left ( fx+e \right ) \right ) ^{4}}{\sinh \left ( fx+e \right ) \left ( \left ( \cosh \left ( fx+e \right ) \right ) ^{2}-1 \right ) }{\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(coth(f*x+e)^3*(a+a*sinh(f*x+e)^2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.94735, size = 170, normalized size = 1.95 \begin{align*} -\frac{3 \, \sqrt{a} \log \left (e^{\left (-f x - e\right )} + 1\right )}{2 \, f} + \frac{3 \, \sqrt{a} \log \left (e^{\left (-f x - e\right )} - 1\right )}{2 \, f} - \frac{3 \, \sqrt{a} e^{\left (-2 \, f x - 2 \, e\right )} + 3 \, \sqrt{a} e^{\left (-4 \, f x - 4 \, e\right )} - \sqrt{a} e^{\left (-6 \, f x - 6 \, e\right )} - \sqrt{a}}{2 \, 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^3*(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.97631, size = 2056, normalized size = 23.63 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^3*(a+a*sinh(f*x+e)^2)^(1/2),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) + (cosh(f*x + e)^6 - 3*cosh(f*x + e)^4 - 3*cosh(f*x + e)^2 + 1)*e^(f*x + e)
+ 3*(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))*log((cosh(f*x + e) + sinh(f*x + e) - 1)/(cosh(f*x + e) + sinh(f*x + e) + 1)))*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*cosh(f*x + e))*e^(2*f*x + 2*e))*sinh(f*x + e)^2 + f*cosh(f*x + e) + (f*co
sh(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(-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(coth(f*x+e)**3*(a+a*sinh(f*x+e)**2)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.25896, size = 120, normalized size = 1.38 \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}} - e^{\left (f x + e\right )} - e^{\left (-f x - e\right )} + 3 \, \log \left (e^{\left (f x + e\right )} + 1\right ) - 3 \, \log \left ({\left | e^{\left (f x + e\right )} - 1 \right |}\right )\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^3*(a+a*sinh(f*x+e)^2)^(1/2),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 - e^(f*x + e) - e^(-f*x - e) + 3*log(e
^(f*x + e) + 1) - 3*log(abs(e^(f*x + e) - 1)))/f