3.471 \(\int \coth (e+f x) (a+b \sinh ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=78 \[ -\frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {a \sqrt {a+b \sinh ^2(e+f x)}}{f}+\frac {\left (a+b \sinh ^2(e+f x)\right )^{3/2}}{3 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3194, 50, 63, 208} \[ -\frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {a \sqrt {a+b \sinh ^2(e+f x)}}{f}+\frac {\left (a+b \sinh ^2(e+f x)\right )^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 208

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

Rule 3194

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

________________________________________________________________________________________

Mathematica [A]  time = 0.14, size = 69, normalized size = 0.88 \[ \frac {\sqrt {a+b \sinh ^2(e+f x)} \left (4 a+b \sinh ^2(e+f x)\right )-3 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a}}\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^(3/2)*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a]] + Sqrt[a + b*Sinh[e + f*x]^2]*(4*a + b*Sinh[e + f*x]^
2))/(3*f)

________________________________________________________________________________________

fricas [B]  time = 0.75, size = 1000, normalized size = 12.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(12*(a*cosh(f*x + e)^3 + 3*a*cosh(f*x + e)^2*sinh(f*x + e) + 3*a*cosh(f*x + e)*sinh(f*x + e)^2 + a*sinh(
f*x + e)^3)*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)*cosh(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*cosh(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)) + sqrt(2)*(b*cosh(f*x + e)^4 + 4*b*cosh(f*x
+ e)*sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(8*a - b)*cosh(f*x + e)^2 + 2*(3*b*cosh(f*x + e)^2 + 8*a - b)*si
nh(f*x + e)^2 + 4*(b*cosh(f*x + e)^3 + (8*a - b)*cosh(f*x + e))*sinh(f*x + e) + b)*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)))/(f*cosh(f*x
+ e)^3 + 3*f*cosh(f*x + e)^2*sinh(f*x + e) + 3*f*cosh(f*x + e)*sinh(f*x + e)^2 + f*sinh(f*x + e)^3), 1/24*(24*
(a*cosh(f*x + e)^3 + 3*a*cosh(f*x + e)^2*sinh(f*x + e) + 3*a*cosh(f*x + e)*sinh(f*x + e)^2 + a*sinh(f*x + e)^3
)*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*sinh(f*x + e))) + sqrt(2)*(b*cosh(f*
x + e)^4 + 4*b*cosh(f*x + e)*sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(8*a - b)*cosh(f*x + e)^2 + 2*(3*b*cosh(f
*x + e)^2 + 8*a - b)*sinh(f*x + e)^2 + 4*(b*cosh(f*x + e)^3 + (8*a - b)*cosh(f*x + e))*sinh(f*x + e) + b)*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)))/(f*cosh(f*x + e)^3 + 3*f*cosh(f*x + e)^2*sinh(f*x + e) + 3*f*cosh(f*x + e)*sinh(f*x + e)^2 + f*sinh
(f*x + e)^3)]

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regular val
ue [0] was discarded and replaced randomly by 0=[45]Warning, need to choose a branch for the root of a polynom
ial with parameters. This might be wrong.Non regular value [0] was discarded and replaced randomly by 0=[-8]Wa
rning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regular v
alue [0] was discarded and replaced randomly by 0=[87]Warning, need to choose a branch for the root of a polyn
omial with parameters. This might be wrong.Non regular value [0] was discarded and replaced randomly by 0=[51]
Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regular
value [0] was discarded and replaced randomly by 0=[-90]Warning, need to choose a branch for the root of a po
lynomial with parameters. This might be wrong.Non regular value [0] was discarded and replaced randomly by 0=[
-27]Evaluation time: 3.63index.cc index_m operator + Error: Bad Argument Value

________________________________________________________________________________________

maple [C]  time = 0.17, size = 62, normalized size = 0.79 \[ \frac {\mathit {`\,int/indef0`\,}\left (\frac {b^{2} \left (\sinh ^{3}\left (f x +e \right )\right )+2 a b \sinh \left (f x +e \right )+\frac {a^{2}}{\sinh \left (f x +e \right )}}{\sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}}, \sinh \left (f x +e \right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \coth \left (f x + e\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {coth}\left (e+f\,x\right )\,{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________