3.3.11 \(\int \frac {\tanh ^6(x)}{(a+b \text {sech}^2(x))^{5/2}} \, dx\) [211]
Optimal. Leaf size=118 \[ -\frac {\text {ArcTan}\left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{b^{5/2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{a^{5/2}}-\frac {(a+b) \tanh ^3(x)}{3 a b \left (a+b-b \tanh ^2(x)\right )^{3/2}}-\frac {\left (\frac {1}{a^2}-\frac {1}{b^2}\right ) \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}} \]
[Out]
-arctan(b^(1/2)*tanh(x)/(a+b-b*tanh(x)^2)^(1/2))/b^(5/2)+arctanh(a^(1/2)*tanh(x)/(a+b-b*tanh(x)^2)^(1/2))/a^(5
/2)-(1/a^2-1/b^2)*tanh(x)/(a+b-b*tanh(x)^2)^(1/2)-1/3*(a+b)*tanh(x)^3/a/b/(a+b-b*tanh(x)^2)^(3/2)
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Rubi [A]
time = 0.23, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 9, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {4226, 2000,
481, 592, 537, 223, 209, 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 {\left (\frac {1}{a^2}-\frac {1}{b^2}\right ) \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}-\frac {\text {ArcTan}\left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{b^{5/2}}-\frac {(a+b) \tanh ^3(x)}{3 a b \left (a-b \tanh ^2(x)+b\right )^{3/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Tanh[x]^6/(a + b*Sech[x]^2)^(5/2),x]
[Out]
-(ArcTan[(Sqrt[b]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]]/b^(5/2)) + ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b - b*Tanh
[x]^2]]/a^(5/2) - ((a + b)*Tanh[x]^3)/(3*a*b*(a + b - b*Tanh[x]^2)^(3/2)) - ((a^(-2) - b^(-2))*Tanh[x])/Sqrt[a
+ b - b*Tanh[x]^2]
Rule 209
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
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 223
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] && !GtQ[a, 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 481
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-a)*e^(
2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c - a*d)*(p + 1))), x] + Dist[e^
(2*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1)
+ (a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Rule 537
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]
Rule 592
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c -
a*d)*(p + 1))), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]
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 {\tanh ^6(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx &=\text {Subst}\left (\int \frac {x^6}{\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 {x^6}{\left (1-x^2\right ) \left (a+b-b x^2\right )^{5/2}} \, dx,x,\tanh (x)\right )\\ &=-\frac {(a+b) \tanh ^3(x)}{3 a b \left (a+b-b \tanh ^2(x)\right )^{3/2}}+\frac {\text {Subst}\left (\int \frac {x^2 \left (3 (a+b)-3 a x^2\right )}{\left (1-x^2\right ) \left (a+b-b x^2\right )^{3/2}} \, dx,x,\tanh (x)\right )}{3 a b}\\ &=-\frac {(a+b) \tanh ^3(x)}{3 a b \left (a+b-b \tanh ^2(x)\right )^{3/2}}-\frac {\left (\frac {1}{a^2}-\frac {1}{b^2}\right ) \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}-\frac {\text {Subst}\left (\int \frac {3 \left (a^2-b^2\right )-3 a^2 x^2}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\tanh (x)\right )}{3 a^2 b^2}\\ &=-\frac {(a+b) \tanh ^3(x)}{3 a b \left (a+b-b \tanh ^2(x)\right )^{3/2}}-\frac {\left (\frac {1}{a^2}-\frac {1}{b^2}\right ) \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}+\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 {\text {Subst}\left (\int \frac {1}{\sqrt {a+b-b x^2}} \, dx,x,\tanh (x)\right )}{b^2}\\ &=-\frac {(a+b) \tanh ^3(x)}{3 a b \left (a+b-b \tanh ^2(x)\right )^{3/2}}-\frac {\left (\frac {1}{a^2}-\frac {1}{b^2}\right ) \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}+\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 {\text {Subst}\left (\int \frac {1}{1+b x^2} \, dx,x,\frac {\tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{b^2}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{b^{5/2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{a^{5/2}}-\frac {(a+b) \tanh ^3(x)}{3 a b \left (a+b-b \tanh ^2(x)\right )^{3/2}}-\frac {\left (\frac {1}{a^2}-\frac {1}{b^2}\right ) \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\\ \end {align*}
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Mathematica [A]
time = 0.49, size = 178, normalized size = 1.51 \begin {gather*} \frac {\text {sech}^5(x) \left (\frac {\sqrt {2} \left (-a^{5/2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {b} \sinh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right )+b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sinh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right )\right ) (a+2 b+a \cosh (2 x))^{5/2}}{a^{5/2} b^{5/2}}+\frac {2 (a+b) (a+2 b+a \cosh (2 x)) \left (3 a^2+4 a b-6 b^2+a (3 a-4 b) \cosh (2 x)\right ) \sinh (x)}{3 a^2 b^2}\right )}{8 \left (a+b \text {sech}^2(x)\right )^{5/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Tanh[x]^6/(a + b*Sech[x]^2)^(5/2),x]
[Out]
(Sech[x]^5*((Sqrt[2]*(-(a^(5/2)*ArcTan[(Sqrt[2]*Sqrt[b]*Sinh[x])/Sqrt[a + 2*b + a*Cosh[2*x]]]) + b^(5/2)*ArcTa
nh[(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)*b^(5/2)) +
(2*(a + b)*(a + 2*b + a*Cosh[2*x])*(3*a^2 + 4*a*b - 6*b^2 + a*(3*a - 4*b)*Cosh[2*x])*Sinh[x])/(3*a^2*b^2)))/(8
*(a + b*Sech[x]^2)^(5/2))
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Maple [F]
time = 1.57, size = 0, normalized size = 0.00 \[\int \frac {\tanh ^{6}\left (x \right )}{\left (a +b \mathrm {sech}\left (x \right )^{2}\right )^{\frac {5}{2}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tanh(x)^6/(a+b*sech(x)^2)^(5/2),x)
[Out]
int(tanh(x)^6/(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(tanh(x)^6/(a+b*sech(x)^2)^(5/2),x, algorithm="maxima")
[Out]
integrate(tanh(x)^6/(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. 2666 vs.
\(2 (100) = 200\).
time = 0.87, size = 11939, normalized size = 101.18 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(x)^6/(a+b*sech(x)^2)^(5/2),x, algorithm="fricas")
[Out]
[1/12*(3*(a^2*b^3*cosh(x)^8 + 8*a^2*b^3*cosh(x)*sinh(x)^7 + a^2*b^3*sinh(x)^8 + 4*(a^2*b^3 + 2*a*b^4)*cosh(x)^
6 + 4*(7*a^2*b^3*cosh(x)^2 + a^2*b^3 + 2*a*b^4)*sinh(x)^6 + 8*(7*a^2*b^3*cosh(x)^3 + 3*(a^2*b^3 + 2*a*b^4)*cos
h(x))*sinh(x)^5 + a^2*b^3 + 2*(3*a^2*b^3 + 8*a*b^4 + 8*b^5)*cosh(x)^4 + 2*(35*a^2*b^3*cosh(x)^4 + 3*a^2*b^3 +
8*a*b^4 + 8*b^5 + 30*(a^2*b^3 + 2*a*b^4)*cosh(x)^2)*sinh(x)^4 + 8*(7*a^2*b^3*cosh(x)^5 + 10*(a^2*b^3 + 2*a*b^4
)*cosh(x)^3 + (3*a^2*b^3 + 8*a*b^4 + 8*b^5)*cosh(x))*sinh(x)^3 + 4*(a^2*b^3 + 2*a*b^4)*cosh(x)^2 + 4*(7*a^2*b^
3*cosh(x)^6 + a^2*b^3 + 2*a*b^4 + 15*(a^2*b^3 + 2*a*b^4)*cosh(x)^4 + 3*(3*a^2*b^3 + 8*a*b^4 + 8*b^5)*cosh(x)^2
)*sinh(x)^2 + 8*(a^2*b^3*cosh(x)^7 + 3*(a^2*b^3 + 2*a*b^4)*cosh(x)^5 + (3*a^2*b^3 + 8*a*b^4 + 8*b^5)*cosh(x)^3
+ (a^2*b^3 + 2*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)*sin
h(x)^5 + sinh(x)^6)) - 6*(a^5*cosh(x)^8 + 8*a^5*cosh(x)*sinh(x)^7 + a^5*sinh(x)^8 + 4*(a^5 + 2*a^4*b)*cosh(x)^
6 + 4*(7*a^5*cosh(x)^2 + a^5 + 2*a^4*b)*sinh(x)^6 + 8*(7*a^5*cosh(x)^3 + 3*(a^5 + 2*a^4*b)*cosh(x))*sinh(x)^5
+ a^5 + 2*(3*a^5 + 8*a^4*b + 8*a^3*b^2)*cosh(x)^4 + 2*(35*a^5*cosh(x)^4 + 3*a^5 + 8*a^4*b + 8*a^3*b^2 + 30*(a^
5 + 2*a^4*b)*cosh(x)^2)*sinh(x)^4 + 8*(7*a^5*cosh(x)^5 + 10*(a^5 + 2*a^4*b)*cosh(x)^3 + (3*a^5 + 8*a^4*b + 8*a
^3*b^2)*cosh(x))*sinh(x)^3 + 4*(a^5 + 2*a^4*b)*cosh(x)^2 + 4*(7*a^5*cosh(x)^6 + a^5 + 2*a^4*b + 15*(a^5 + 2*a^
4*b)*cosh(x)^4 + 3*(3*a^5 + 8*a^4*b + 8*a^3*b^2)*cosh(x)^2)*sinh(x)^2 + 8*(a^5*cosh(x)^7 + 3*(a^5 + 2*a^4*b)*c
osh(x)^5 + (3*a^5 + 8*a^4*b + 8*a^3*b^2)*cosh(x)^3 + (a^5 + 2*a^4*b)*cosh(x))*sinh(x))*sqrt(-b)*log(-((a - b)*
cosh(x)^4 + 4*(a - b)*cosh(x)*sinh(x)^3 + (a - b)*sinh(x)^4 + 2*(a + 3*b)*cosh(x)^2 + 2*(3*(a - b)*cosh(x)^2 +
a + 3*b)*sinh(x)^2 - 2*sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*sqrt(-b)*sqrt((a*cosh(x)^2 + a
*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*((a - b)*cosh(x)^3 + (a + 3*b)*cosh(x))
*sinh(x) + a - b)/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2 + 1)*sinh(x)^2 + 2*cosh(x)^2 +
4*(cosh(x)^3 + cosh(x))*sinh(x) + 1)) + 3*(a^2*b^3*cosh(x)^8 + 8*a^2*b^3*cosh(x)*sinh(x)^7 + a^2*b^3*sinh(x)^
8 + 4*(a^2*b^3 + 2*a*b^4)*cosh(x)^6 + 4*(7*a^2*b^3*cosh(x)^2 + a^2*b^3 + 2*a*b^4)*sinh(x)^6 + 8*(7*a^2*b^3*cos
h(x)^3 + 3*(a^2*b^3 + 2*a*b^4)*cosh(x))*sinh(x)^5 + a^2*b^3 + 2*(3*a^2*b^3 + 8*a*b^4 + 8*b^5)*cosh(x)^4 + 2*(3
5*a^2*b^3*cosh(x)^4 + 3*a^2*b^3 + 8*a*b^4 + 8*b^5 + 30*(a^2*b^3 + 2*a*b^4)*cosh(x)^2)*sinh(x)^4 + 8*(7*a^2*b^3
*cosh(x)^5 + 10*(a^2*b^3 + 2*a*b^4)*cosh(x)^3 + (3*a^2*b^3 + 8*a*b^4 + 8*b^5)*cosh(x))*sinh(x)^3 + 4*(a^2*b^3
+ 2*a*b^4)*cosh(x)^2 + 4*(7*a^2*b^3*cosh(x)^6 + a^2*b^3 + 2*a*b^4 + 15*(a^2*b^3 + 2*a*b^4)*cosh(x)^4 + 3*(3*a^
2*b^3 + 8*a*b^4 + 8*b^5)*cosh(x)^2)*sinh(x)^2 + 8*(a^2*b^3*cosh(x)^7 + 3*(a^2*b^3 + 2*a*b^4)*cosh(x)^5 + (3*a^
2*b^3 + 8*a*b^4 + 8*b^5)*cosh(x)^3 + (a^2*b^3 + 2*a*b^4)*cosh(x))*sinh(x))*sqrt(a)*log(-(a*cosh(x)^4 + 4*a*cos
h(x)*sinh(x)^3 + a*sinh(x)^4 + 2*(a + b)*cosh(x)^2 + 2*(3*a*cosh(x)^2 + a + b)*sinh(x)^2 + sqrt(2)*(cosh(x)^2
+ 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*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*(a*cosh(x)^3 + (a + b)*cosh(x))*sinh(x) + a)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(
x)^2)) + 4*sqrt(2)*((3*a^4*b - a^3*b^2 - 4*a^2*b^3)*cosh(x)^6 + 6*(3*a^4*b - a^3*b^2 - 4*a^2*b^3)*cosh(x)*sinh
(x)^5 + (3*a^4*b - a^3*b^2 - 4*a^2*b^3)*sinh(x)^6 - 3*a^4*b + a^3*b^2 + 4*a^2*b^3 + 3*(a^4*b + 5*a^3*b^2 - 4*a
*b^4)*cosh(x)^4 + 3*(a^4*b + 5*a^3*b^2 - 4*a*b^4 + 5*(3*a^4*b - a^3*b^2 - 4*a^2*b^3)*cosh(x)^2)*sinh(x)^4 + 4*
(5*(3*a^4*b - a^3*b^2 - 4*a^2*b^3)*cosh(x)^3 + 3*(a^4*b + 5*a^3*b^2 - 4*a*b^4)*cosh(x))*sinh(x)^3 - 3*(a^4*b +
5*a^3*b^2 - 4*a*b^4)*cosh(x)^2 - 3*(a^4*b + 5*...
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tanh ^{6}{\left (x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(x)**6/(a+b*sech(x)**2)**(5/2),x)
[Out]
Integral(tanh(x)**6/(a + b*sech(x)**2)**(5/2), 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(tanh(x)^6/(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 {tanh}\left (x\right )}^6}{{\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(tanh(x)^6/(a + b/cosh(x)^2)^(5/2),x)
[Out]
int(tanh(x)^6/(a + b/cosh(x)^2)^(5/2), x)
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