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[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)
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Rubi [A] time = 0.0924509, antiderivative size = 78, normalized size of antiderivative = 1.,
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 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]
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]
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 = 0.122469, 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)
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Maple [C] time = 0.078, size = 62, normalized size = 0.8 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ({ \left ({b}^{2} \left ( \sinh \left ( fx+e \right ) \right ) ^{3}+2\,ab\sinh \left ( fx+e \right ) +{\frac{{a}^{2}}{\sinh \left ( fx+e \right ) }} \right ){\frac{1}{\sqrt{a+b \left ( \sinh \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)*(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
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \coth \left (f x + e\right )\,{d x} \end{align*}
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)
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Fricas [B] time = 3.79683, size = 2696, normalized size = 34.56 \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)*(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)]
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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)*(a+b*sinh(f*x+e)**2)**(3/2),x)
[Out]
Timed out
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
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