3.5.83 \(\int \frac {\coth ^3(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx\) [483]

Optimal. Leaf size=77 \[ -\frac {(2 a-b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{3/2} f}-\frac {\text {csch}^2(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{2 a f} \]

[Out]

-1/2*(2*a-b)*arctanh((a+b*sinh(f*x+e)^2)^(1/2)/a^(1/2))/a^(3/2)/f-1/2*csch(f*x+e)^2*(a+b*sinh(f*x+e)^2)^(1/2)/
a/f

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Rubi [A]
time = 0.07, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3273, 79, 65, 214} \begin {gather*} -\frac {(2 a-b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{3/2} f}-\frac {\text {csch}^2(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{2 a f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((2*a - b)*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a]])/(a^(3/2)*f) - (Csch[e + f*x]^2*Sqrt[a + b*Sinh[e
+ f*x]^2])/(2*a*f)

Rule 65

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 214

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

Rule 3273

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^p/(1 - ff*x)^((m
+ 1)/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [A]
time = 0.14, size = 72, normalized size = 0.94 \begin {gather*} -\frac {\frac {(2 a-b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2}}+\frac {\text {csch}^2(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{a}}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(((2*a - b)*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a]])/a^(3/2) + (Csch[e + f*x]^2*Sqrt[a + b*Sinh[e +
f*x]^2])/a)/f

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.43, size = 44, normalized size = 0.57

method	result	size
default	\(\frac {\mathit {`\,int/indef0`\,}\left (\frac {\frac {1}{\sinh \left (f x +e \right )}+\frac {1}{\sinh \left (f x +e \right )^{3}}}{\sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}}, \sinh \left (f x +e \right )\right )}{f}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

`int/indef0`((1/sinh(f*x+e)+1/sinh(f*x+e)^3)/(a+b*sinh(f*x+e)^2)^(1/2),sinh(f*x+e))/f

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(coth(f*x + e)^3/sqrt(b*sinh(f*x + e)^2 + a), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 471 vs. \(2 (65) = 130\).
time = 0.51, size = 1144, normalized size = 14.86 \begin {gather*} \left [-\frac {{\left ({\left (2 \, a - b\right )} \cosh \left (f x + e\right )^{4} + 4 \, {\left (2 \, a - b\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{3} + {\left (2 \, a - b\right )} \sinh \left (f x + e\right )^{4} - 2 \, {\left (2 \, a - b\right )} \cosh \left (f x + e\right )^{2} + 2 \, {\left (3 \, {\left (2 \, a - b\right )} \cosh \left (f x + e\right )^{2} - 2 \, a + b\right )} \sinh \left (f x + e\right )^{2} + 4 \, {\left ({\left (2 \, a - b\right )} \cosh \left (f x + e\right )^{3} - {\left (2 \, a - b\right )} \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right ) + 2 \, a - b\right )} \sqrt {a} \log \left (\frac {b \cosh \left (f x + e\right )^{4} + 4 \, b \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{3} + b \sinh \left (f x + e\right )^{4} + 2 \, {\left (4 \, a - b\right )} \cosh \left (f x + e\right )^{2} + 2 \, {\left (3 \, b \cosh \left (f x + e\right )^{2} + 4 \, a - b\right )} \sinh \left (f x + e\right )^{2} + 4 \, \sqrt {2} \sqrt {a} \sqrt {\frac {b \cosh \left (f x + e\right )^{2} + b \sinh \left (f x + e\right )^{2} + 2 \, a - b}{\cosh \left (f x + e\right )^{2} - 2 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + \sinh \left (f x + e\right )^{2}}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )} + 4 \, {\left (b \cosh \left (f x + e\right )^{3} + {\left (4 \, a - b\right )} \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right ) + b}{\cosh \left (f x + e\right )^{4} + 4 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{3} + \sinh \left (f x + e\right )^{4} + 2 \, {\left (3 \, \cosh \left (f x + e\right )^{2} - 1\right )} \sinh \left (f x + e\right )^{2} - 2 \, \cosh \left (f x + e\right )^{2} + 4 \, {\left (\cosh \left (f x + e\right )^{3} - \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right ) + 1}\right ) + 4 \, \sqrt {2} {\left (a \cosh \left (f x + e\right ) + a \sinh \left (f x + e\right )\right )} \sqrt {\frac {b \cosh \left (f x + e\right )^{2} + b \sinh \left (f x + e\right )^{2} + 2 \, a - b}{\cosh \left (f x + e\right )^{2} - 2 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + \sinh \left (f x + e\right )^{2}}}}{4 \, {\left (a^{2} f \cosh \left (f x + e\right )^{4} + 4 \, a^{2} f \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{3} + a^{2} f \sinh \left (f x + e\right )^{4} - 2 \, a^{2} f \cosh \left (f x + e\right )^{2} + a^{2} f + 2 \, {\left (3 \, a^{2} f \cosh \left (f x + e\right )^{2} - a^{2} f\right )} \sinh \left (f x + e\right )^{2} + 4 \, {\left (a^{2} f \cosh \left (f x + e\right )^{3} - a^{2} f \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )\right )}}, \frac {{\left ({\left (2 \, a - b\right )} \cosh \left (f x + e\right )^{4} + 4 \, {\left (2 \, a - b\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{3} + {\left (2 \, a - b\right )} \sinh \left (f x + e\right )^{4} - 2 \, {\left (2 \, a - b\right )} \cosh \left (f x + e\right )^{2} + 2 \, {\left (3 \, {\left (2 \, a - b\right )} \cosh \left (f x + e\right )^{2} - 2 \, a + b\right )} \sinh \left (f x + e\right )^{2} + 4 \, {\left ({\left (2 \, a - b\right )} \cosh \left (f x + e\right )^{3} - {\left (2 \, a - b\right )} \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right ) + 2 \, a - b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {b \cosh \left (f x + e\right )^{2} + b \sinh \left (f x + e\right )^{2} + 2 \, a - b}{\cosh \left (f x + e\right )^{2} - 2 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + \sinh \left (f x + e\right )^{2}}}}{2 \, {\left (a \cosh \left (f x + e\right ) + a \sinh \left (f x + e\right )\right )}}\right ) - 2 \, \sqrt {2} {\left (a \cosh \left (f x + e\right ) + a \sinh \left (f x + e\right )\right )} \sqrt {\frac {b \cosh \left (f x + e\right )^{2} + b \sinh \left (f x + e\right )^{2} + 2 \, a - b}{\cosh \left (f x + e\right )^{2} - 2 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + \sinh \left (f x + e\right )^{2}}}}{2 \, {\left (a^{2} f \cosh \left (f x + e\right )^{4} + 4 \, a^{2} f \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{3} + a^{2} f \sinh \left (f x + e\right )^{4} - 2 \, a^{2} f \cosh \left (f x + e\right )^{2} + a^{2} f + 2 \, {\left (3 \, a^{2} f \cosh \left (f x + e\right )^{2} - a^{2} f\right )} \sinh \left (f x + e\right )^{2} + 4 \, {\left (a^{2} f \cosh \left (f x + e\right )^{3} - a^{2} f \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(((2*a - b)*cosh(f*x + e)^4 + 4*(2*a - b)*cosh(f*x + e)*sinh(f*x + e)^3 + (2*a - b)*sinh(f*x + e)^4 - 2*
(2*a - b)*cosh(f*x + e)^2 + 2*(3*(2*a - b)*cosh(f*x + e)^2 - 2*a + b)*sinh(f*x + e)^2 + 4*((2*a - b)*cosh(f*x
+ e)^3 - (2*a - b)*cosh(f*x + e))*sinh(f*x + e) + 2*a - b)*sqrt(a)*log((b*cosh(f*x + e)^4 + 4*b*cosh(f*x + e)*
sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(4*a - b)*cosh(f*x + e)^2 + 2*(3*b*cosh(f*x + e)^2 + 4*a - b)*sinh(f*x
+ e)^2 + 4*sqrt(2)*sqrt(a)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f
*x + e)*sinh(f*x + e) + sinh(f*x + e)^2))*(cosh(f*x + e) + sinh(f*x + e)) + 4*(b*cosh(f*x + e)^3 + (4*a - b)*c
osh(f*x + e))*sinh(f*x + e) + b)/(cosh(f*x + e)^4 + 4*cosh(f*x + e)*sinh(f*x + e)^3 + sinh(f*x + e)^4 + 2*(3*c
osh(f*x + e)^2 - 1)*sinh(f*x + e)^2 - 2*cosh(f*x + e)^2 + 4*(cosh(f*x + e)^3 - cosh(f*x + e))*sinh(f*x + e) +
1)) + 4*sqrt(2)*(a*cosh(f*x + e) + a*sinh(f*x + e))*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(co
sh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2)))/(a^2*f*cosh(f*x + e)^4 + 4*a^2*f*cosh(f*x +
e)*sinh(f*x + e)^3 + a^2*f*sinh(f*x + e)^4 - 2*a^2*f*cosh(f*x + e)^2 + a^2*f + 2*(3*a^2*f*cosh(f*x + e)^2 - a
^2*f)*sinh(f*x + e)^2 + 4*(a^2*f*cosh(f*x + e)^3 - a^2*f*cosh(f*x + e))*sinh(f*x + e)), 1/2*(((2*a - b)*cosh(f
*x + e)^4 + 4*(2*a - b)*cosh(f*x + e)*sinh(f*x + e)^3 + (2*a - b)*sinh(f*x + e)^4 - 2*(2*a - b)*cosh(f*x + e)^
2 + 2*(3*(2*a - b)*cosh(f*x + e)^2 - 2*a + b)*sinh(f*x + e)^2 + 4*((2*a - b)*cosh(f*x + e)^3 - (2*a - b)*cosh(
f*x + e))*sinh(f*x + e) + 2*a - b)*sqrt(-a)*arctan(1/2*sqrt(2)*sqrt(-a)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x +
e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2))/(a*cosh(f*x + e) + a*sin
h(f*x + e))) - 2*sqrt(2)*(a*cosh(f*x + e) + a*sinh(f*x + e))*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a
- b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2)))/(a^2*f*cosh(f*x + e)^4 + 4*a^2*f*c
osh(f*x + e)*sinh(f*x + e)^3 + a^2*f*sinh(f*x + e)^4 - 2*a^2*f*cosh(f*x + e)^2 + a^2*f + 2*(3*a^2*f*cosh(f*x +
e)^2 - a^2*f)*sinh(f*x + e)^2 + 4*(a^2*f*cosh(f*x + e)^3 - a^2*f*cosh(f*x + e))*sinh(f*x + e))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\coth ^{3}{\left (e + f x \right )}}{\sqrt {a + b \sinh ^{2}{\left (e + f x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)**3/(a+b*sinh(f*x+e)**2)**(1/2),x)

[Out]

Integral(coth(e + f*x)**3/sqrt(a + b*sinh(e + f*x)**2), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{256,[6,8,6]%%%}+%%%{%%%{-768,[1]%%%},[6,8,5]%%%}+%%%{%%%
{768,[2]%%%

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {coth}\left (e+f\,x\right )}^3}{\sqrt {b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(e + f*x)^3/(a + b*sinh(e + f*x)^2)^(1/2),x)

[Out]

int(coth(e + f*x)^3/(a + b*sinh(e + f*x)^2)^(1/2), x)

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