3.38 \(\int \frac{1}{(b \coth ^3(c+d x))^{4/3}} \, dx\)
Optimal. Leaf size=80 \[ \frac{x \coth (c+d x)}{b \sqrt [3]{b \coth ^3(c+d x)}}-\frac{1}{b d \sqrt [3]{b \coth ^3(c+d x)}}-\frac{\tanh ^2(c+d x)}{3 b d \sqrt [3]{b \coth ^3(c+d x)}} \]
[Out]
-(1/(b*d*(b*Coth[c + d*x]^3)^(1/3))) + (x*Coth[c + d*x])/(b*(b*Coth[c + d*x]^3)^(1/3)) - Tanh[c + d*x]^2/(3*b*
d*(b*Coth[c + d*x]^3)^(1/3))
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Rubi [A] time = 0.0353556, antiderivative size = 80, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used =
{3658, 3473, 8} \[ \frac{x \coth (c+d x)}{b \sqrt [3]{b \coth ^3(c+d x)}}-\frac{1}{b d \sqrt [3]{b \coth ^3(c+d x)}}-\frac{\tanh ^2(c+d x)}{3 b d \sqrt [3]{b \coth ^3(c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[(b*Coth[c + d*x]^3)^(-4/3),x]
[Out]
-(1/(b*d*(b*Coth[c + d*x]^3)^(1/3))) + (x*Coth[c + d*x])/(b*(b*Coth[c + d*x]^3)^(1/3)) - Tanh[c + d*x]^2/(3*b*
d*(b*Coth[c + d*x]^3)^(1/3))
Rule 3658
Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[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
]])
Rule 3473
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rubi steps
\begin{align*} \int \frac{1}{\left (b \coth ^3(c+d x)\right )^{4/3}} \, dx &=\frac{\coth (c+d x) \int \tanh ^4(c+d x) \, dx}{b \sqrt [3]{b \coth ^3(c+d x)}}\\ &=-\frac{\tanh ^2(c+d x)}{3 b d \sqrt [3]{b \coth ^3(c+d x)}}+\frac{\coth (c+d x) \int \tanh ^2(c+d x) \, dx}{b \sqrt [3]{b \coth ^3(c+d x)}}\\ &=-\frac{1}{b d \sqrt [3]{b \coth ^3(c+d x)}}-\frac{\tanh ^2(c+d x)}{3 b d \sqrt [3]{b \coth ^3(c+d x)}}+\frac{\coth (c+d x) \int 1 \, dx}{b \sqrt [3]{b \coth ^3(c+d x)}}\\ &=-\frac{1}{b d \sqrt [3]{b \coth ^3(c+d x)}}+\frac{x \coth (c+d x)}{b \sqrt [3]{b \coth ^3(c+d x)}}-\frac{\tanh ^2(c+d x)}{3 b d \sqrt [3]{b \coth ^3(c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.0878472, size = 51, normalized size = 0.64 \[ \frac{-\tanh ^2(c+d x)+3 \tanh ^{-1}(\tanh (c+d x)) \coth (c+d x)-3}{3 b d \sqrt [3]{b \coth ^3(c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(b*Coth[c + d*x]^3)^(-4/3),x]
[Out]
(-3 + 3*ArcTanh[Tanh[c + d*x]]*Coth[c + d*x] - Tanh[c + d*x]^2)/(3*b*d*(b*Coth[c + d*x]^3)^(1/3))
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Maple [B] time = 0.099, size = 149, normalized size = 1.9 \begin{align*}{\frac{ \left ( 1+{{\rm e}^{2\,dx+2\,c}} \right ) x}{b \left ({{\rm e}^{2\,dx+2\,c}}-1 \right ) }{\frac{1}{\sqrt [3]{{\frac{b \left ( 1+{{\rm e}^{2\,dx+2\,c}} \right ) ^{3}}{ \left ({{\rm e}^{2\,dx+2\,c}}-1 \right ) ^{3}}}}}}}+{\frac{12\,{{\rm e}^{4\,dx+4\,c}}+12\,{{\rm e}^{2\,dx+2\,c}}+8}{3\,b \left ( 1+{{\rm e}^{2\,dx+2\,c}} \right ) ^{2} \left ({{\rm e}^{2\,dx+2\,c}}-1 \right ) d}{\frac{1}{\sqrt [3]{{\frac{b \left ( 1+{{\rm e}^{2\,dx+2\,c}} \right ) ^{3}}{ \left ({{\rm e}^{2\,dx+2\,c}}-1 \right ) ^{3}}}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(b*coth(d*x+c)^3)^(4/3),x)
[Out]
1/b*(1+exp(2*d*x+2*c))/(exp(2*d*x+2*c)-1)/(b*(1+exp(2*d*x+2*c))^3/(exp(2*d*x+2*c)-1)^3)^(1/3)*x+4/3/b/(1+exp(2
*d*x+2*c))^2/(exp(2*d*x+2*c)-1)/(b*(1+exp(2*d*x+2*c))^3/(exp(2*d*x+2*c)-1)^3)^(1/3)*(3*exp(4*d*x+4*c)+3*exp(2*
d*x+2*c)+2)/d
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Maxima [A] time = 1.77013, size = 120, normalized size = 1.5 \begin{align*} -\frac{4 \,{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{3 \,{\left (3 \, b^{\frac{4}{3}} e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, b^{\frac{4}{3}} e^{\left (-4 \, d x - 4 \, c\right )} + b^{\frac{4}{3}} e^{\left (-6 \, d x - 6 \, c\right )} + b^{\frac{4}{3}}\right )} d} + \frac{d x + c}{b^{\frac{4}{3}} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*coth(d*x+c)^3)^(4/3),x, algorithm="maxima")
[Out]
-4/3*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + 2)/((3*b^(4/3)*e^(-2*d*x - 2*c) + 3*b^(4/3)*e^(-4*d*x - 4*c) +
b^(4/3)*e^(-6*d*x - 6*c) + b^(4/3))*d) + (d*x + c)/(b^(4/3)*d)
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Fricas [B] time = 2.37196, size = 4035, normalized size = 50.44 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*coth(d*x+c)^3)^(4/3),x, algorithm="fricas")
[Out]
1/3*(3*d*x*cosh(d*x + c)^6 + 3*(d*x*e^(4*d*x + 4*c) - 2*d*x*e^(2*d*x + 2*c) + d*x)*sinh(d*x + c)^6 + 18*(d*x*c
osh(d*x + c)*e^(4*d*x + 4*c) - 2*d*x*cosh(d*x + c)*e^(2*d*x + 2*c) + d*x*cosh(d*x + c))*sinh(d*x + c)^5 + 3*(3
*d*x + 4)*cosh(d*x + c)^4 + 3*(15*d*x*cosh(d*x + c)^2 + 3*d*x + (15*d*x*cosh(d*x + c)^2 + 3*d*x + 4)*e^(4*d*x
+ 4*c) - 2*(15*d*x*cosh(d*x + c)^2 + 3*d*x + 4)*e^(2*d*x + 2*c) + 4)*sinh(d*x + c)^4 + 12*(5*d*x*cosh(d*x + c)
^3 + (3*d*x + 4)*cosh(d*x + c) + (5*d*x*cosh(d*x + c)^3 + (3*d*x + 4)*cosh(d*x + c))*e^(4*d*x + 4*c) - 2*(5*d*
x*cosh(d*x + c)^3 + (3*d*x + 4)*cosh(d*x + c))*e^(2*d*x + 2*c))*sinh(d*x + c)^3 + 3*(3*d*x + 4)*cosh(d*x + c)^
2 + 3*(15*d*x*cosh(d*x + c)^4 + 6*(3*d*x + 4)*cosh(d*x + c)^2 + 3*d*x + (15*d*x*cosh(d*x + c)^4 + 6*(3*d*x + 4
)*cosh(d*x + c)^2 + 3*d*x + 4)*e^(4*d*x + 4*c) - 2*(15*d*x*cosh(d*x + c)^4 + 6*(3*d*x + 4)*cosh(d*x + c)^2 + 3
*d*x + 4)*e^(2*d*x + 2*c) + 4)*sinh(d*x + c)^2 + 3*d*x + (3*d*x*cosh(d*x + c)^6 + 3*(3*d*x + 4)*cosh(d*x + c)^
4 + 3*(3*d*x + 4)*cosh(d*x + c)^2 + 3*d*x + 8)*e^(4*d*x + 4*c) - 2*(3*d*x*cosh(d*x + c)^6 + 3*(3*d*x + 4)*cosh
(d*x + c)^4 + 3*(3*d*x + 4)*cosh(d*x + c)^2 + 3*d*x + 8)*e^(2*d*x + 2*c) + 6*(3*d*x*cosh(d*x + c)^5 + 2*(3*d*x
+ 4)*cosh(d*x + c)^3 + (3*d*x + 4)*cosh(d*x + c) + (3*d*x*cosh(d*x + c)^5 + 2*(3*d*x + 4)*cosh(d*x + c)^3 + (
3*d*x + 4)*cosh(d*x + c))*e^(4*d*x + 4*c) - 2*(3*d*x*cosh(d*x + c)^5 + 2*(3*d*x + 4)*cosh(d*x + c)^3 + (3*d*x
+ 4)*cosh(d*x + c))*e^(2*d*x + 2*c))*sinh(d*x + c) + 8)*((b*e^(6*d*x + 6*c) + 3*b*e^(4*d*x + 4*c) + 3*b*e^(2*d
*x + 2*c) + b)/(e^(6*d*x + 6*c) - 3*e^(4*d*x + 4*c) + 3*e^(2*d*x + 2*c) - 1))^(2/3)/(b^2*d*cosh(d*x + c)^6 + 3
*b^2*d*cosh(d*x + c)^4 + (b^2*d*e^(4*d*x + 4*c) + 2*b^2*d*e^(2*d*x + 2*c) + b^2*d)*sinh(d*x + c)^6 + 6*(b^2*d*
cosh(d*x + c)*e^(4*d*x + 4*c) + 2*b^2*d*cosh(d*x + c)*e^(2*d*x + 2*c) + b^2*d*cosh(d*x + c))*sinh(d*x + c)^5 +
3*b^2*d*cosh(d*x + c)^2 + 3*(5*b^2*d*cosh(d*x + c)^2 + b^2*d + (5*b^2*d*cosh(d*x + c)^2 + b^2*d)*e^(4*d*x + 4
*c) + 2*(5*b^2*d*cosh(d*x + c)^2 + b^2*d)*e^(2*d*x + 2*c))*sinh(d*x + c)^4 + 4*(5*b^2*d*cosh(d*x + c)^3 + 3*b^
2*d*cosh(d*x + c) + (5*b^2*d*cosh(d*x + c)^3 + 3*b^2*d*cosh(d*x + c))*e^(4*d*x + 4*c) + 2*(5*b^2*d*cosh(d*x +
c)^3 + 3*b^2*d*cosh(d*x + c))*e^(2*d*x + 2*c))*sinh(d*x + c)^3 + b^2*d + 3*(5*b^2*d*cosh(d*x + c)^4 + 6*b^2*d*
cosh(d*x + c)^2 + b^2*d + (5*b^2*d*cosh(d*x + c)^4 + 6*b^2*d*cosh(d*x + c)^2 + b^2*d)*e^(4*d*x + 4*c) + 2*(5*b
^2*d*cosh(d*x + c)^4 + 6*b^2*d*cosh(d*x + c)^2 + b^2*d)*e^(2*d*x + 2*c))*sinh(d*x + c)^2 + (b^2*d*cosh(d*x + c
)^6 + 3*b^2*d*cosh(d*x + c)^4 + 3*b^2*d*cosh(d*x + c)^2 + b^2*d)*e^(4*d*x + 4*c) + 2*(b^2*d*cosh(d*x + c)^6 +
3*b^2*d*cosh(d*x + c)^4 + 3*b^2*d*cosh(d*x + c)^2 + b^2*d)*e^(2*d*x + 2*c) + 6*(b^2*d*cosh(d*x + c)^5 + 2*b^2*
d*cosh(d*x + c)^3 + b^2*d*cosh(d*x + c) + (b^2*d*cosh(d*x + c)^5 + 2*b^2*d*cosh(d*x + c)^3 + b^2*d*cosh(d*x +
c))*e^(4*d*x + 4*c) + 2*(b^2*d*cosh(d*x + c)^5 + 2*b^2*d*cosh(d*x + c)^3 + b^2*d*cosh(d*x + c))*e^(2*d*x + 2*c
))*sinh(d*x + c))
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \coth ^{3}{\left (c + d x \right )}\right )^{\frac{4}{3}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*coth(d*x+c)**3)**(4/3),x)
[Out]
Integral((b*coth(c + d*x)**3)**(-4/3), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \coth \left (d x + c\right )^{3}\right )^{\frac{4}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*coth(d*x+c)^3)^(4/3),x, algorithm="giac")
[Out]
integrate((b*coth(d*x + c)^3)^(-4/3), x)