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

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

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 91, 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, 3207, 2590, 270} \[ \frac {\tanh (e+f x) \sqrt {a \cosh ^2(e+f x)}}{f}-\frac {\text {csch}^3(e+f x) \text {sech}(e+f x) \sqrt {a \cosh ^2(e+f x)}}{3 f}-\frac {2 \text {csch}(e+f x) \text {sech}(e+f x) \sqrt {a \cosh ^2(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 270

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

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/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 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
]])

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 47, normalized size = 0.52 \[ -\frac {\tanh (e+f x) \left (\text {csch}^4(e+f x)+6 \text {csch}^2(e+f x)-3\right ) \sqrt {a \cosh ^2(e+f x)}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.68, size = 885, normalized size = 9.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^4*(a+a*sinh(f*x+e)^2)^(1/2),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 = 0.17, size = 80, normalized size = 0.88 \[ -\frac {\sqrt {a} {\left (\frac {8 \, {\left (3 \, e^{\left (5 \, f x + 5 \, e\right )} - 4 \, e^{\left (3 \, f x + 3 \, e\right )} + 3 \, e^{\left (f x + e\right )}\right )}}{{\left (e^{\left (2 \, f x + 2 \, e\right )} - 1\right )}^{3}} - 3 \, e^{\left (f x + e\right )} + 3 \, e^{\left (-f x - e\right )}\right )}}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^4*(a+a*sinh(f*x+e)^2)^(1/2),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 [A]  time = 0.20, size = 55, normalized size = 0.60 \[ \frac {\cosh \left (f x +e \right ) a \left (3 \left (\sinh ^{4}\left (f x +e \right )\right )-6 \left (\sinh ^{2}\left (f x +e \right )\right )-1\right )}{3 \sinh \left (f x +e \right )^{3} \sqrt {a \left (\cosh ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(f*x+e)^4*(a+a*sinh(f*x+e)^2)^(1/2),x)

[Out]

1/3*cosh(f*x+e)*a*(3*sinh(f*x+e)^4-6*sinh(f*x+e)^2-1)/sinh(f*x+e)^3/(a*cosh(f*x+e)^2)^(1/2)/f

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maxima [B]  time = 1.11, size = 487, normalized size = 5.35 \[ -\frac {3 \, \sqrt {a} \log \left (e^{\left (-f x - e\right )} + 1\right ) - 3 \, \sqrt {a} \log \left (e^{\left (-f x - e\right )} - 1\right ) - \frac {2 \, {\left (9 \, \sqrt {a} e^{\left (-f x - e\right )} - 8 \, \sqrt {a} e^{\left (-3 \, f x - 3 \, e\right )} + 3 \, \sqrt {a} e^{\left (-5 \, f x - 5 \, e\right )}\right )}}{3 \, e^{\left (-2 \, f x - 2 \, e\right )} - 3 \, e^{\left (-4 \, f x - 4 \, e\right )} + e^{\left (-6 \, f x - 6 \, e\right )} - 1}}{12 \, f} + \frac {3 \, \sqrt {a} \log \left (e^{\left (-f x - e\right )} + 1\right ) - 3 \, \sqrt {a} \log \left (e^{\left (-f x - e\right )} - 1\right ) + \frac {2 \, {\left (3 \, \sqrt {a} e^{\left (-f x - e\right )} - 8 \, \sqrt {a} e^{\left (-3 \, f x - 3 \, e\right )} + 9 \, \sqrt {a} e^{\left (-5 \, f x - 5 \, e\right )}\right )}}{3 \, e^{\left (-2 \, f x - 2 \, e\right )} - 3 \, e^{\left (-4 \, f x - 4 \, e\right )} + e^{\left (-6 \, f x - 6 \, e\right )} - 1}}{12 \, f} + \frac {\sqrt {a} e^{\left (-3 \, f x - 3 \, e\right )}}{f {\left (3 \, e^{\left (-2 \, f x - 2 \, e\right )} - 3 \, e^{\left (-4 \, f x - 4 \, e\right )} + e^{\left (-6 \, f x - 6 \, e\right )} - 1\right )}} - \frac {33 \, \sqrt {a} e^{\left (-2 \, f x - 2 \, e\right )} - 40 \, \sqrt {a} e^{\left (-4 \, f x - 4 \, e\right )} + 15 \, \sqrt {a} e^{\left (-6 \, f x - 6 \, e\right )} - 6 \, \sqrt {a}}{12 \, f {\left (e^{\left (-f x - e\right )} - 3 \, e^{\left (-3 \, f x - 3 \, e\right )} + 3 \, e^{\left (-5 \, f x - 5 \, e\right )} - e^{\left (-7 \, f x - 7 \, e\right )}\right )}} + \frac {15 \, \sqrt {a} e^{\left (-f x - e\right )} - 40 \, \sqrt {a} e^{\left (-3 \, f x - 3 \, e\right )} + 33 \, \sqrt {a} e^{\left (-5 \, f x - 5 \, e\right )} - 6 \, \sqrt {a} e^{\left (-7 \, f x - 7 \, e\right )}}{12 \, f {\left (3 \, e^{\left (-2 \, f x - 2 \, e\right )} - 3 \, e^{\left (-4 \, f x - 4 \, e\right )} + e^{\left (-6 \, f x - 6 \, e\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^4*(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/12*(3*sqrt(a)*log(e^(-f*x - e) + 1) - 3*sqrt(a)*log(e^(-f*x - e) - 1) - 2*(9*sqrt(a)*e^(-f*x - e) - 8*sqrt(
a)*e^(-3*f*x - 3*e) + 3*sqrt(a)*e^(-5*f*x - 5*e))/(3*e^(-2*f*x - 2*e) - 3*e^(-4*f*x - 4*e) + e^(-6*f*x - 6*e)
- 1))/f + 1/12*(3*sqrt(a)*log(e^(-f*x - e) + 1) - 3*sqrt(a)*log(e^(-f*x - e) - 1) + 2*(3*sqrt(a)*e^(-f*x - e)
- 8*sqrt(a)*e^(-3*f*x - 3*e) + 9*sqrt(a)*e^(-5*f*x - 5*e))/(3*e^(-2*f*x - 2*e) - 3*e^(-4*f*x - 4*e) + e^(-6*f*
x - 6*e) - 1))/f + sqrt(a)*e^(-3*f*x - 3*e)/(f*(3*e^(-2*f*x - 2*e) - 3*e^(-4*f*x - 4*e) + e^(-6*f*x - 6*e) - 1
)) - 1/12*(33*sqrt(a)*e^(-2*f*x - 2*e) - 40*sqrt(a)*e^(-4*f*x - 4*e) + 15*sqrt(a)*e^(-6*f*x - 6*e) - 6*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))) + 1/12*(15*sqrt(a)*e^(-f*x -
e) - 40*sqrt(a)*e^(-3*f*x - 3*e) + 33*sqrt(a)*e^(-5*f*x - 5*e) - 6*sqrt(a)*e^(-7*f*x - 7*e))/(f*(3*e^(-2*f*x
- 2*e) - 3*e^(-4*f*x - 4*e) + e^(-6*f*x - 6*e) - 1))

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mupad [B]  time = 0.94, size = 281, normalized size = 3.09 \[ -\frac {\left (\frac {1}{f}-\frac {{\mathrm {e}}^{2\,e+2\,f\,x}}{f}\right )\,\sqrt {a+a\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}}{{\mathrm {e}}^{2\,e+2\,f\,x}+1}-\frac {8\,{\mathrm {e}}^{3\,e+3\,f\,x}\,\sqrt {a+a\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}}{f\,\left ({\mathrm {e}}^{2\,e+2\,f\,x}-1\right )\,\left ({\mathrm {e}}^{e+f\,x}+{\mathrm {e}}^{3\,e+3\,f\,x}\right )}-\frac {16\,{\mathrm {e}}^{3\,e+3\,f\,x}\,\sqrt {a+a\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}}{3\,f\,{\left ({\mathrm {e}}^{2\,e+2\,f\,x}-1\right )}^2\,\left ({\mathrm {e}}^{e+f\,x}+{\mathrm {e}}^{3\,e+3\,f\,x}\right )}-\frac {16\,{\mathrm {e}}^{3\,e+3\,f\,x}\,\sqrt {a+a\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}}{3\,f\,{\left ({\mathrm {e}}^{2\,e+2\,f\,x}-1\right )}^3\,\left ({\mathrm {e}}^{e+f\,x}+{\mathrm {e}}^{3\,e+3\,f\,x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(e + f*x)^4*(a + a*sinh(e + f*x)^2)^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )} \coth ^{4}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)**4*(a+a*sinh(f*x+e)**2)**(1/2),x)

[Out]

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

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