3.3.19 \(\int \frac {\coth ^2(x)}{(a+b \text {sech}^2(x))^{5/2}} \, dx\) [219]
Optimal. Leaf size=133 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{a^{5/2}}-\frac {b \coth (x)}{3 a (a+b) \left (a+b-b \tanh ^2(x)\right )^{3/2}}-\frac {b (7 a+3 b) \coth (x)}{3 a^2 (a+b)^2 \sqrt {a+b-b \tanh ^2(x)}}-\frac {(a-3 b) (3 a+b) \coth (x) \sqrt {a+b-b \tanh ^2(x)}}{3 a^2 (a+b)^3} \]
[Out]
arctanh(a^(1/2)*tanh(x)/(a+b-b*tanh(x)^2)^(1/2))/a^(5/2)-1/3*b*(7*a+3*b)*coth(x)/a^2/(a+b)^2/(a+b-b*tanh(x)^2)
^(1/2)-1/3*(a-3*b)*(3*a+b)*coth(x)*(a+b-b*tanh(x)^2)^(1/2)/a^2/(a+b)^3-1/3*b*coth(x)/a/(a+b)/(a+b-b*tanh(x)^2)
^(3/2)
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Rubi [A]
time = 0.25, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 8, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {4226, 2000,
483, 593, 597, 12, 385, 212} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{a^{5/2}}-\frac {(a-3 b) (3 a+b) \coth (x) \sqrt {a-b \tanh ^2(x)+b}}{3 a^2 (a+b)^3}-\frac {b (7 a+3 b) \coth (x)}{3 a^2 (a+b)^2 \sqrt {a-b \tanh ^2(x)+b}}-\frac {b \coth (x)}{3 a (a+b) \left (a-b \tanh ^2(x)+b\right )^{3/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Coth[x]^2/(a + b*Sech[x]^2)^(5/2),x]
[Out]
ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]]/a^(5/2) - (b*Coth[x])/(3*a*(a + b)*(a + b - b*Tanh[x]^2)^
(3/2)) - (b*(7*a + 3*b)*Coth[x])/(3*a^2*(a + b)^2*Sqrt[a + b - b*Tanh[x]^2]) - ((a - 3*b)*(3*a + b)*Coth[x]*Sq
rt[a + b - b*Tanh[x]^2])/(3*a^2*(a + b)^3)
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 212
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 385
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
Rule 483
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*(e*
x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*e*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a*d)
*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n
*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ
[p, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Rule 593
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*g*n*(b*c - a*d)*(p +
1))), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)
*(m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]
Rule 597
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*g*(m + 1))), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]
Rule 2000
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&& !BinomialMatchQ[{u, v}, x]
Rule 4226
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2
*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])
Rubi steps
\begin {align*} \int \frac {\coth ^2(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx &=\text {Subst}\left (\int \frac {1}{x^2 \left (1-x^2\right ) \left (a+b \left (1-x^2\right )\right )^{5/2}} \, dx,x,\tanh (x)\right )\\ &=\text {Subst}\left (\int \frac {1}{x^2 \left (1-x^2\right ) \left (a+b-b x^2\right )^{5/2}} \, dx,x,\tanh (x)\right )\\ &=-\frac {b \coth (x)}{3 a (a+b) \left (a+b-b \tanh ^2(x)\right )^{3/2}}-\frac {\text {Subst}\left (\int \frac {-3 a+b-4 b x^2}{x^2 \left (1-x^2\right ) \left (a+b-b x^2\right )^{3/2}} \, dx,x,\tanh (x)\right )}{3 a (a+b)}\\ &=-\frac {b \coth (x)}{3 a (a+b) \left (a+b-b \tanh ^2(x)\right )^{3/2}}-\frac {b (7 a+3 b) \coth (x)}{3 a^2 (a+b)^2 \sqrt {a+b-b \tanh ^2(x)}}+\frac {\text {Subst}\left (\int \frac {(a-3 b) (3 a+b)+2 b (7 a+3 b) x^2}{x^2 \left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\tanh (x)\right )}{3 a^2 (a+b)^2}\\ &=-\frac {b \coth (x)}{3 a (a+b) \left (a+b-b \tanh ^2(x)\right )^{3/2}}-\frac {b (7 a+3 b) \coth (x)}{3 a^2 (a+b)^2 \sqrt {a+b-b \tanh ^2(x)}}-\frac {(a-3 b) (3 a+b) \coth (x) \sqrt {a+b-b \tanh ^2(x)}}{3 a^2 (a+b)^3}-\frac {\text {Subst}\left (\int -\frac {3 (a+b)^3}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\tanh (x)\right )}{3 a^2 (a+b)^3}\\ &=-\frac {b \coth (x)}{3 a (a+b) \left (a+b-b \tanh ^2(x)\right )^{3/2}}-\frac {b (7 a+3 b) \coth (x)}{3 a^2 (a+b)^2 \sqrt {a+b-b \tanh ^2(x)}}-\frac {(a-3 b) (3 a+b) \coth (x) \sqrt {a+b-b \tanh ^2(x)}}{3 a^2 (a+b)^3}+\frac {\text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\tanh (x)\right )}{a^2}\\ &=-\frac {b \coth (x)}{3 a (a+b) \left (a+b-b \tanh ^2(x)\right )^{3/2}}-\frac {b (7 a+3 b) \coth (x)}{3 a^2 (a+b)^2 \sqrt {a+b-b \tanh ^2(x)}}-\frac {(a-3 b) (3 a+b) \coth (x) \sqrt {a+b-b \tanh ^2(x)}}{3 a^2 (a+b)^3}+\frac {\text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{a^2}\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{a^{5/2}}-\frac {b \coth (x)}{3 a (a+b) \left (a+b-b \tanh ^2(x)\right )^{3/2}}-\frac {b (7 a+3 b) \coth (x)}{3 a^2 (a+b)^2 \sqrt {a+b-b \tanh ^2(x)}}-\frac {(a-3 b) (3 a+b) \coth (x) \sqrt {a+b-b \tanh ^2(x)}}{3 a^2 (a+b)^3}\\ \end {align*}
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Mathematica [A]
time = 0.54, size = 155, normalized size = 1.17 \begin {gather*} \frac {\text {sech}^5(x) \left (\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sinh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right ) (a+2 b+a \cosh (2 x))^{5/2}}{a^{5/2}}-\frac {(a+2 b+a \cosh (2 x)) \left (3 a^2 (a+2 b+a \cosh (2 x))^2 \text {csch}(x)-4 b^3 (a+b) \sinh (x)+2 b^2 (9 a+4 b) (a+2 b+a \cosh (2 x)) \sinh (x)\right )}{3 a^2 (a+b)^3}\right )}{8 \left (a+b \text {sech}^2(x)\right )^{5/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Coth[x]^2/(a + b*Sech[x]^2)^(5/2),x]
[Out]
(Sech[x]^5*((Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sinh[x])/Sqrt[a + 2*b + a*Cosh[2*x]]]*(a + 2*b + a*Cosh[2*x])^(5
/2))/a^(5/2) - ((a + 2*b + a*Cosh[2*x])*(3*a^2*(a + 2*b + a*Cosh[2*x])^2*Csch[x] - 4*b^3*(a + b)*Sinh[x] + 2*b
^2*(9*a + 4*b)*(a + 2*b + a*Cosh[2*x])*Sinh[x]))/(3*a^2*(a + b)^3)))/(8*(a + b*Sech[x]^2)^(5/2))
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Maple [F]
time = 1.68, size = 0, normalized size = 0.00 \[\int \frac {\coth ^{2}\left (x \right )}{\left (a +b \mathrm {sech}\left (x \right )^{2}\right )^{\frac {5}{2}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(coth(x)^2/(a+b*sech(x)^2)^(5/2),x)
[Out]
int(coth(x)^2/(a+b*sech(x)^2)^(5/2),x)
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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(x)^2/(a+b*sech(x)^2)^(5/2),x, algorithm="maxima")
[Out]
integrate(coth(x)^2/(b*sech(x)^2 + a)^(5/2), x)
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 5323 vs.
\(2 (115) = 230\).
time = 0.99, size = 11205, normalized size = 84.25 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)^2/(a+b*sech(x)^2)^(5/2),x, algorithm="fricas")
[Out]
[1/12*(3*((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cosh(x)^10 + 10*(a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cosh(x)*
sinh(x)^9 + (a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*sinh(x)^10 + (3*a^5 + 17*a^4*b + 33*a^3*b^2 + 27*a^2*b^3 + 8
*a*b^4)*cosh(x)^8 + (3*a^5 + 17*a^4*b + 33*a^3*b^2 + 27*a^2*b^3 + 8*a*b^4 + 45*(a^5 + 3*a^4*b + 3*a^3*b^2 + a^
2*b^3)*cosh(x)^2)*sinh(x)^8 + 8*(15*(a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cosh(x)^3 + (3*a^5 + 17*a^4*b + 33*a
^3*b^2 + 27*a^2*b^3 + 8*a*b^4)*cosh(x))*sinh(x)^7 + 2*(a^5 + 7*a^4*b + 23*a^3*b^2 + 37*a^2*b^3 + 28*a*b^4 + 8*
b^5)*cosh(x)^6 + 2*(a^5 + 7*a^4*b + 23*a^3*b^2 + 37*a^2*b^3 + 28*a*b^4 + 8*b^5 + 105*(a^5 + 3*a^4*b + 3*a^3*b^
2 + a^2*b^3)*cosh(x)^4 + 14*(3*a^5 + 17*a^4*b + 33*a^3*b^2 + 27*a^2*b^3 + 8*a*b^4)*cosh(x)^2)*sinh(x)^6 + 4*(6
3*(a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cosh(x)^5 + 14*(3*a^5 + 17*a^4*b + 33*a^3*b^2 + 27*a^2*b^3 + 8*a*b^4)*
cosh(x)^3 + 3*(a^5 + 7*a^4*b + 23*a^3*b^2 + 37*a^2*b^3 + 28*a*b^4 + 8*b^5)*cosh(x))*sinh(x)^5 - a^5 - 3*a^4*b
- 3*a^3*b^2 - a^2*b^3 - 2*(a^5 + 7*a^4*b + 23*a^3*b^2 + 37*a^2*b^3 + 28*a*b^4 + 8*b^5)*cosh(x)^4 + 2*(105*(a^5
+ 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cosh(x)^6 - a^5 - 7*a^4*b - 23*a^3*b^2 - 37*a^2*b^3 - 28*a*b^4 - 8*b^5 + 35*
(3*a^5 + 17*a^4*b + 33*a^3*b^2 + 27*a^2*b^3 + 8*a*b^4)*cosh(x)^4 + 15*(a^5 + 7*a^4*b + 23*a^3*b^2 + 37*a^2*b^3
+ 28*a*b^4 + 8*b^5)*cosh(x)^2)*sinh(x)^4 + 8*(15*(a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cosh(x)^7 + 7*(3*a^5 +
17*a^4*b + 33*a^3*b^2 + 27*a^2*b^3 + 8*a*b^4)*cosh(x)^5 + 5*(a^5 + 7*a^4*b + 23*a^3*b^2 + 37*a^2*b^3 + 28*a*b
^4 + 8*b^5)*cosh(x)^3 - (a^5 + 7*a^4*b + 23*a^3*b^2 + 37*a^2*b^3 + 28*a*b^4 + 8*b^5)*cosh(x))*sinh(x)^3 - (3*a
^5 + 17*a^4*b + 33*a^3*b^2 + 27*a^2*b^3 + 8*a*b^4)*cosh(x)^2 + (45*(a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cosh(
x)^8 + 28*(3*a^5 + 17*a^4*b + 33*a^3*b^2 + 27*a^2*b^3 + 8*a*b^4)*cosh(x)^6 - 3*a^5 - 17*a^4*b - 33*a^3*b^2 - 2
7*a^2*b^3 - 8*a*b^4 + 30*(a^5 + 7*a^4*b + 23*a^3*b^2 + 37*a^2*b^3 + 28*a*b^4 + 8*b^5)*cosh(x)^4 - 12*(a^5 + 7*
a^4*b + 23*a^3*b^2 + 37*a^2*b^3 + 28*a*b^4 + 8*b^5)*cosh(x)^2)*sinh(x)^2 + 2*(5*(a^5 + 3*a^4*b + 3*a^3*b^2 + a
^2*b^3)*cosh(x)^9 + 4*(3*a^5 + 17*a^4*b + 33*a^3*b^2 + 27*a^2*b^3 + 8*a*b^4)*cosh(x)^7 + 6*(a^5 + 7*a^4*b + 23
*a^3*b^2 + 37*a^2*b^3 + 28*a*b^4 + 8*b^5)*cosh(x)^5 - 4*(a^5 + 7*a^4*b + 23*a^3*b^2 + 37*a^2*b^3 + 28*a*b^4 +
8*b^5)*cosh(x)^3 - (3*a^5 + 17*a^4*b + 33*a^3*b^2 + 27*a^2*b^3 + 8*a*b^4)*cosh(x))*sinh(x))*sqrt(a)*log((a*b^2
*cosh(x)^8 + 8*a*b^2*cosh(x)*sinh(x)^7 + a*b^2*sinh(x)^8 - 2*(a*b^2 - b^3)*cosh(x)^6 + 2*(14*a*b^2*cosh(x)^2 -
a*b^2 + b^3)*sinh(x)^6 + 4*(14*a*b^2*cosh(x)^3 - 3*(a*b^2 - b^3)*cosh(x))*sinh(x)^5 + (a^3 + 4*a^2*b + 9*a*b^
2)*cosh(x)^4 + (70*a*b^2*cosh(x)^4 + a^3 + 4*a^2*b + 9*a*b^2 - 30*(a*b^2 - b^3)*cosh(x)^2)*sinh(x)^4 + 4*(14*a
*b^2*cosh(x)^5 - 10*(a*b^2 - b^3)*cosh(x)^3 + (a^3 + 4*a^2*b + 9*a*b^2)*cosh(x))*sinh(x)^3 + a^3 + 2*(a^3 + 3*
a^2*b)*cosh(x)^2 + 2*(14*a*b^2*cosh(x)^6 - 15*(a*b^2 - b^3)*cosh(x)^4 + a^3 + 3*a^2*b + 3*(a^3 + 4*a^2*b + 9*a
*b^2)*cosh(x)^2)*sinh(x)^2 + sqrt(2)*(b^2*cosh(x)^6 + 6*b^2*cosh(x)*sinh(x)^5 + b^2*sinh(x)^6 - 3*b^2*cosh(x)^
4 + 3*(5*b^2*cosh(x)^2 - b^2)*sinh(x)^4 + 4*(5*b^2*cosh(x)^3 - 3*b^2*cosh(x))*sinh(x)^3 - (a^2 + 4*a*b)*cosh(x
)^2 + (15*b^2*cosh(x)^4 - 18*b^2*cosh(x)^2 - a^2 - 4*a*b)*sinh(x)^2 - a^2 + 2*(3*b^2*cosh(x)^5 - 6*b^2*cosh(x)
^3 - (a^2 + 4*a*b)*cosh(x))*sinh(x))*sqrt(a)*sqrt((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)
*sinh(x) + sinh(x)^2)) + 4*(2*a*b^2*cosh(x)^7 - 3*(a*b^2 - b^3)*cosh(x)^5 + (a^3 + 4*a^2*b + 9*a*b^2)*cosh(x)^
3 + (a^3 + 3*a^2*b)*cosh(x))*sinh(x))/(cosh(x)^6 + 6*cosh(x)^5*sinh(x) + 15*cosh(x)^4*sinh(x)^2 + 20*cosh(x)^3
*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6)) + 3*((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*
b^3)*cosh(x)^10 + 10*(a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cosh(x)*sinh(x)^9 + (a^5 + 3*a^4*b + 3*a^3*b^2 + a^
2*b^3)*sinh(x)^10 + (3*a^5 + 17*a^4*b + 33*a^3*b^2 + 27*a^2*b^3 + 8*a*b^4)*cosh(x)^8 + (3*a^5 + 17*a^4*b + 33*
a^3*b^2 + 27*a^2*b^3 + 8*a*b^4 + 45*(a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cosh(x)^2)*sinh(x)^8 + 8*(15*(a^5 +
3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cosh(x)^3 + (3*a^5 + 17*a^4*b + 33*a^3*b^2 + 27*a^2*b^3 + 8*a*b^4)*cosh(x))*sin
h(x)^7 + 2*(a^5 + 7*a^4*b + 23*a^3*b^2 + 37*a^2*b^3 + 28*a*b^4 + 8*b^5)*cosh(x)^6 + 2*(a^5 + 7*a^4*b + 23*a^3*
b^2 + 37*a^2*b^3 + 28*a*b^4 + 8*b^5 + 105*(a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cosh(x)^4 + 14*(3*a^5 + 17*a^4
*b + 33*a^3*b^2 + 27*a^2*b^3 + 8*a*b^4)*cosh(x)^2)*sinh(x)^6 + 4*(63*(a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cos
h(x)^5 + 14*(3*a^5 + 17*a^4*b + 33*a^3*b^2 + 27*a^2*b^3 + 8*a*b^4)*cosh(x)^3 + 3*(a^5 + 7*a^4*b + 23*a^3*b^2 +
37*a^2*b^3 + 28*a*b^4 + 8*b^5)*cosh(x))*sinh(x)^5 - a^5 - 3*a^4*b - 3*a^3*b^2 - a^2*b^3 - 2*(a^5 + 7*a^4*b +
23*a^3*b^2 + 37*a^2*b^3 + 28*a*b^4 + 8*b^5)*cosh(x)^4 + 2*(105*(a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cosh(x)^6
- a^5 - 7*a^4*b - 23*a^3*b^2 - 37*a^2*b^3 - 28*a*b^4 - 8*b^5 + 35*(3*a^5 + 17*a^4*b + 33*a^3*b^2 + 27*a^2*b^3
+ 8*a*b^4)*cosh(x)^4 + 15*(a^5 + 7*a^4*b + 23*...
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)**2/(a+b*sech(x)**2)**(5/2),x)
[Out]
Timed out
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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(x)^2/(a+b*sech(x)^2)^(5/2),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {coth}\left (x\right )}^2}{{\left (a+\frac {b}{{\mathrm {cosh}\left (x\right )}^2}\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(coth(x)^2/(a + b/cosh(x)^2)^(5/2),x)
[Out]
int(coth(x)^2/(a + b/cosh(x)^2)^(5/2), x)
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