3.2.95 \(\int \frac {\tanh ^4(x)}{\sqrt {a+b \text {sech}^2(x)}} \, dx\) [195]
Optimal. Leaf size=90 \[ -\frac {(a+3 b) \text {ArcTan}\left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{2 b^{3/2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{\sqrt {a}}+\frac {\tanh (x) \sqrt {a+b-b \tanh ^2(x)}}{2 b} \]
[Out]
-1/2*(a+3*b)*arctan(b^(1/2)*tanh(x)/(a+b-b*tanh(x)^2)^(1/2))/b^(3/2)+arctanh(a^(1/2)*tanh(x)/(a+b-b*tanh(x)^2)
^(1/2))/a^(1/2)+1/2*(a+b-b*tanh(x)^2)^(1/2)*tanh(x)/b
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Rubi [A]
time = 0.16, antiderivative size = 90, 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,
490, 537, 223, 209, 385, 212} \begin {gather*} -\frac {(a+3 b) \text {ArcTan}\left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{2 b^{3/2}}+\frac {\tanh (x) \sqrt {a-b \tanh ^2(x)+b}}{2 b}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{\sqrt {a}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Tanh[x]^4/Sqrt[a + b*Sech[x]^2],x]
[Out]
-1/2*((a + 3*b)*ArcTan[(Sqrt[b]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]])/b^(3/2) + ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[
a + b - b*Tanh[x]^2]]/Sqrt[a] + (Tanh[x]*Sqrt[a + b - b*Tanh[x]^2])/(2*b)
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 490
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(2*n -
1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*d*(m + n*(p + q) + 1))), x] - Dist[e^(2*n)
/(b*d*(m + n*(p + q) + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) + (a*d*(m +
n*(q - 1) + 1) + b*c*(m + n*(p - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d
, 0] && IGtQ[n, 0] && 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 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 ^4(x)}{\sqrt {a+b \text {sech}^2(x)}} \, dx &=\text {Subst}\left (\int \frac {x^4}{\left (1-x^2\right ) \sqrt {a+b \left (1-x^2\right )}} \, dx,x,\tanh (x)\right )\\ &=\text {Subst}\left (\int \frac {x^4}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\tanh (x)\right )\\ &=\frac {\tanh (x) \sqrt {a+b-b \tanh ^2(x)}}{2 b}-\frac {\text {Subst}\left (\int \frac {a+b+(-a-3 b) x^2}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\tanh (x)\right )}{2 b}\\ &=\frac {\tanh (x) \sqrt {a+b-b \tanh ^2(x)}}{2 b}-\frac {(a+3 b) \text {Subst}\left (\int \frac {1}{\sqrt {a+b-b x^2}} \, dx,x,\tanh (x)\right )}{2 b}+\text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\tanh (x)\right )\\ &=\frac {\tanh (x) \sqrt {a+b-b \tanh ^2(x)}}{2 b}-\frac {(a+3 b) \text {Subst}\left (\int \frac {1}{1+b x^2} \, dx,x,\frac {\tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{2 b}+\text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )\\ &=-\frac {(a+3 b) \tan ^{-1}\left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{2 b^{3/2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{\sqrt {a}}+\frac {\tanh (x) \sqrt {a+b-b \tanh ^2(x)}}{2 b}\\ \end {align*}
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Mathematica [A]
time = 0.36, size = 169, normalized size = 1.88 \begin {gather*} \frac {\text {sech}(x) \left (2 \sqrt {2} b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sinh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right ) \sqrt {a+2 b+a \cosh (2 x)}+\sqrt {a} \left (-\sqrt {2} (a+3 b) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {b} \sinh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right ) \sqrt {a+2 b+a \cosh (2 x)}+\sqrt {b} (a+2 b+a \cosh (2 x)) \text {sech}(x) \tanh (x)\right )\right )}{4 \sqrt {a} b^{3/2} \sqrt {a+b \text {sech}^2(x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Tanh[x]^4/Sqrt[a + b*Sech[x]^2],x]
[Out]
(Sech[x]*(2*Sqrt[2]*b^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sinh[x])/Sqrt[a + 2*b + a*Cosh[2*x]]]*Sqrt[a + 2*b + a*Co
sh[2*x]] + Sqrt[a]*(-(Sqrt[2]*(a + 3*b)*ArcTan[(Sqrt[2]*Sqrt[b]*Sinh[x])/Sqrt[a + 2*b + a*Cosh[2*x]]]*Sqrt[a +
2*b + a*Cosh[2*x]]) + Sqrt[b]*(a + 2*b + a*Cosh[2*x])*Sech[x]*Tanh[x])))/(4*Sqrt[a]*b^(3/2)*Sqrt[a + b*Sech[x
]^2])
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Maple [F]
time = 1.71, size = 0, normalized size = 0.00 \[\int \frac {\tanh ^{4}\left (x \right )}{\sqrt {a +b \mathrm {sech}\left (x \right )^{2}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tanh(x)^4/(a+b*sech(x)^2)^(1/2),x)
[Out]
int(tanh(x)^4/(a+b*sech(x)^2)^(1/2),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)^4/(a+b*sech(x)^2)^(1/2),x, algorithm="maxima")
[Out]
integrate(tanh(x)^4/sqrt(b*sech(x)^2 + a), x)
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 822 vs.
\(2 (72) = 144\).
time = 0.58, size = 4569, normalized size = 50.77 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(x)^4/(a+b*sech(x)^2)^(1/2),x, algorithm="fricas")
[Out]
[1/4*((b^2*cosh(x)^4 + 4*b^2*cosh(x)*sinh(x)^3 + b^2*sinh(x)^4 + 2*b^2*cosh(x)^2 + 2*(3*b^2*cosh(x)^2 + b^2)*s
inh(x)^2 + b^2 + 4*(b^2*cosh(x)^3 + b^2*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*cos
h(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*c
osh(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)) - ((a^2 + 3*a*b)*cosh(x)^4 + 4*(a^2 + 3*a*b)*cosh(x)*sinh(x)^3 + (a^2 +
3*a*b)*sinh(x)^4 + 2*(a^2 + 3*a*b)*cosh(x)^2 + 2*(3*(a^2 + 3*a*b)*cosh(x)^2 + a^2 + 3*a*b)*sinh(x)^2 + a^2 + 3
*a*b + 4*((a^2 + 3*a*b)*cosh(x)^3 + (a^2 + 3*a*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 + cos
h(x))*sinh(x) + 1)) + (b^2*cosh(x)^4 + 4*b^2*cosh(x)*sinh(x)^3 + b^2*sinh(x)^4 + 2*b^2*cosh(x)^2 + 2*(3*b^2*co
sh(x)^2 + b^2)*sinh(x)^2 + b^2 + 4*(b^2*cosh(x)^3 + b^2*cosh(x))*sinh(x))*sqrt(a)*log(-(a*cosh(x)^4 + 4*a*cosh
(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)) + 2*sqrt(2)*(a*b*cosh(x)^2 + 2*a*b*cosh(x)*sinh(x) + a*b*sinh(x)^2 - a*b)*sqrt((a*cosh(x)^2 + a*sinh(x)^
2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)))/(a*b^2*cosh(x)^4 + 4*a*b^2*cosh(x)*sinh(x)^3 + a*b^
2*sinh(x)^4 + 2*a*b^2*cosh(x)^2 + a*b^2 + 2*(3*a*b^2*cosh(x)^2 + a*b^2)*sinh(x)^2 + 4*(a*b^2*cosh(x)^3 + a*b^2
*cosh(x))*sinh(x)), -1/4*(2*((a^2 + 3*a*b)*cosh(x)^4 + 4*(a^2 + 3*a*b)*cosh(x)*sinh(x)^3 + (a^2 + 3*a*b)*sinh(
x)^4 + 2*(a^2 + 3*a*b)*cosh(x)^2 + 2*(3*(a^2 + 3*a*b)*cosh(x)^2 + a^2 + 3*a*b)*sinh(x)^2 + a^2 + 3*a*b + 4*((a
^2 + 3*a*b)*cosh(x)^3 + (a^2 + 3*a*b)*cosh(x))*sinh(x))*sqrt(b)*arctan(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
))/(a*cosh(x)^4 + 4*a*cosh(x)*sinh(x)^3 + a*sinh(x)^4 + 2*(a + 2*b)*cosh(x)^2 + 2*(3*a*cosh(x)^2 + a + 2*b)*si
nh(x)^2 + 4*(a*cosh(x)^3 + (a + 2*b)*cosh(x))*sinh(x) + a)) - (b^2*cosh(x)^4 + 4*b^2*cosh(x)*sinh(x)^3 + b^2*s
inh(x)^4 + 2*b^2*cosh(x)^2 + 2*(3*b^2*cosh(x)^2 + b^2)*sinh(x)^2 + b^2 + 4*(b^2*cosh(x)^3 + b^2*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*cos
h(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(...
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tanh ^{4}{\left (x \right )}}{\sqrt {a + b \operatorname {sech}^{2}{\left (x \right )}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(x)**4/(a+b*sech(x)**2)**(1/2),x)
[Out]
Integral(tanh(x)**4/sqrt(a + b*sech(x)**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)^4/(a+b*sech(x)^2)^(1/2),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 )}^4}{\sqrt {a+\frac {b}{{\mathrm {cosh}\left (x\right )}^2}}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tanh(x)^4/(a + b/cosh(x)^2)^(1/2),x)
[Out]
int(tanh(x)^4/(a + b/cosh(x)^2)^(1/2), x)
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