3.116 \(\int \frac {\tanh ^4(x)}{a+b \text {sech}(x)} \, dx\)

Optimal. Leaf size=94 \[ \frac {\left (2 a^2-3 b^2\right ) \tan ^{-1}(\sinh (x))}{2 b^3}-\frac {2 (a-b)^{3/2} (a+b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a b^3}-\frac {a \tanh (x)}{b^2}+\frac {x}{a}+\frac {\tanh (x) \text {sech}(x)}{2 b} \]

[Out]

x/a+1/2*(2*a^2-3*b^2)*arctan(sinh(x))/b^3-2*(a-b)^(3/2)*(a+b)^(3/2)*arctan((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2)
)/a/b^3-a*tanh(x)/b^2+1/2*sech(x)*tanh(x)/b

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Rubi [A]  time = 0.32, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3898, 2893, 3057, 2659, 205, 3770} \[ \frac {\left (2 a^2-3 b^2\right ) \tan ^{-1}(\sinh (x))}{2 b^3}-\frac {a \tanh (x)}{b^2}-\frac {2 (a-b)^{3/2} (a+b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a b^3}+\frac {x}{a}+\frac {\tanh (x) \text {sech}(x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^4/(a + b*Sech[x]),x]

[Out]

x/a + ((2*a^2 - 3*b^2)*ArcTan[Sinh[x]])/(2*b^3) - (2*(a - b)^(3/2)*(a + b)^(3/2)*ArcTan[(Sqrt[a - b]*Tanh[x/2]
)/Sqrt[a + b]])/(a*b^3) - (a*Tanh[x])/b^2 + (Sech[x]*Tanh[x])/(2*b)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 2893

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 1))/(a*d*f*(n + 1)), x] +
(-Dist[1/(a^2*d^2*(n + 1)*(n + 2)), Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 2)*Simp[a^2*n*(n + 2) -
b^2*(m + n + 2)*(m + n + 3) + a*b*m*Sin[e + f*x] - (a^2*(n + 1)*(n + 2) - b^2*(m + n + 2)*(m + n + 4))*Sin[e +
f*x]^2, x], x], x] - Simp[(b*(m + n + 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 2))/
(a^2*d^2*f*(n + 1)*(n + 2)), x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || Intege
rsQ[2*m, 2*n]) &&  !m < -1 && LtQ[n, -1] && (LtQ[n, -2] || EqQ[m + n + 4, 0])

Rule 3057

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(C*x)/(b*d), x] + (Dist[(A*b^2 - a*b*B + a
^2*C)/(b*(b*c - a*d)), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[(c^2*C - B*c*d + A*d^2)/(d*(b*c - a*d)), Int[
1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2
, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3898

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[(Cos[c + d*x]^
m*(b + a*Sin[c + d*x])^n)/Sin[c + d*x]^(m + n), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[
n] && IntegerQ[m] && (IntegerQ[m/2] || LeQ[m, 1])

Rubi steps

\begin {align*} \int \frac {\tanh ^4(x)}{a+b \text {sech}(x)} \, dx &=\int \frac {\sinh (x) \tanh ^3(x)}{b+a \cosh (x)} \, dx\\ &=-\frac {a \tanh (x)}{b^2}+\frac {\text {sech}(x) \tanh (x)}{2 b}-\frac {\int \frac {\left (-2 a^2+3 b^2-a b \cosh (x)-2 b^2 \cosh ^2(x)\right ) \text {sech}(x)}{b+a \cosh (x)} \, dx}{2 b^2}\\ &=\frac {x}{a}-\frac {a \tanh (x)}{b^2}+\frac {\text {sech}(x) \tanh (x)}{2 b}-\frac {\left (a^2-b^2\right )^2 \int \frac {1}{b+a \cosh (x)} \, dx}{a b^3}-\frac {\left (-2 a^2+3 b^2\right ) \int \text {sech}(x) \, dx}{2 b^3}\\ &=\frac {x}{a}+\frac {\left (2 a^2-3 b^2\right ) \tan ^{-1}(\sinh (x))}{2 b^3}-\frac {a \tanh (x)}{b^2}+\frac {\text {sech}(x) \tanh (x)}{2 b}-\frac {\left (2 \left (a^2-b^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+b-(-a+b) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a b^3}\\ &=\frac {x}{a}+\frac {\left (2 a^2-3 b^2\right ) \tan ^{-1}(\sinh (x))}{2 b^3}-\frac {2 (a-b)^{3/2} (a+b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a b^3}-\frac {a \tanh (x)}{b^2}+\frac {\text {sech}(x) \tanh (x)}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.42, size = 113, normalized size = 1.20 \[ \frac {\text {sech}^2(x) (a \cosh (x)+b) \left (2 \cosh (x) \left (a \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )+2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )+b^3 x\right )+a b (b \tanh (x)-2 a \sinh (x))\right )}{2 a b^3 (a+b \text {sech}(x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^4/(a + b*Sech[x]),x]

[Out]

((b + a*Cosh[x])*Sech[x]^2*(2*(b^3*x + a*(2*a^2 - 3*b^2)*ArcTan[Tanh[x/2]] + 2*(a^2 - b^2)^(3/2)*ArcTan[((-a +
b)*Tanh[x/2])/Sqrt[a^2 - b^2]])*Cosh[x] + a*b*(-2*a*Sinh[x] + b*Tanh[x])))/(2*a*b^3*(a + b*Sech[x]))

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fricas [B]  time = 0.51, size = 1254, normalized size = 13.34 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^4/(a+b*sech(x)),x, algorithm="fricas")

[Out]

[(b^3*x*cosh(x)^4 + b^3*x*sinh(x)^4 + a*b^2*cosh(x)^3 + b^3*x - a*b^2*cosh(x) + (4*b^3*x*cosh(x) + a*b^2)*sinh
(x)^3 + 2*a^2*b + 2*(b^3*x + a^2*b)*cosh(x)^2 + (6*b^3*x*cosh(x)^2 + 2*b^3*x + 3*a*b^2*cosh(x) + 2*a^2*b)*sinh
(x)^2 - ((a^2 - b^2)*cosh(x)^4 + 4*(a^2 - b^2)*cosh(x)*sinh(x)^3 + (a^2 - b^2)*sinh(x)^4 + 2*(a^2 - b^2)*cosh(
x)^2 + 2*(3*(a^2 - b^2)*cosh(x)^2 + a^2 - b^2)*sinh(x)^2 + a^2 - b^2 + 4*((a^2 - b^2)*cosh(x)^3 + (a^2 - b^2)*
cosh(x))*sinh(x))*sqrt(-a^2 + b^2)*log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2*a*b*cosh(x) - a^2 + 2*b^2 + 2*(a^2*c
osh(x) + a*b)*sinh(x) + 2*sqrt(-a^2 + b^2)*(a*cosh(x) + a*sinh(x) + b))/(a*cosh(x)^2 + a*sinh(x)^2 + 2*b*cosh(
x) + 2*(a*cosh(x) + b)*sinh(x) + a)) + ((2*a^3 - 3*a*b^2)*cosh(x)^4 + 4*(2*a^3 - 3*a*b^2)*cosh(x)*sinh(x)^3 +
(2*a^3 - 3*a*b^2)*sinh(x)^4 + 2*a^3 - 3*a*b^2 + 2*(2*a^3 - 3*a*b^2)*cosh(x)^2 + 2*(2*a^3 - 3*a*b^2 + 3*(2*a^3
- 3*a*b^2)*cosh(x)^2)*sinh(x)^2 + 4*((2*a^3 - 3*a*b^2)*cosh(x)^3 + (2*a^3 - 3*a*b^2)*cosh(x))*sinh(x))*arctan(
cosh(x) + sinh(x)) + (4*b^3*x*cosh(x)^3 + 3*a*b^2*cosh(x)^2 - a*b^2 + 4*(b^3*x + a^2*b)*cosh(x))*sinh(x))/(a*b
^3*cosh(x)^4 + 4*a*b^3*cosh(x)*sinh(x)^3 + a*b^3*sinh(x)^4 + 2*a*b^3*cosh(x)^2 + a*b^3 + 2*(3*a*b^3*cosh(x)^2
+ a*b^3)*sinh(x)^2 + 4*(a*b^3*cosh(x)^3 + a*b^3*cosh(x))*sinh(x)), (b^3*x*cosh(x)^4 + b^3*x*sinh(x)^4 + a*b^2*
cosh(x)^3 + b^3*x - a*b^2*cosh(x) + (4*b^3*x*cosh(x) + a*b^2)*sinh(x)^3 + 2*a^2*b + 2*(b^3*x + a^2*b)*cosh(x)^
2 + (6*b^3*x*cosh(x)^2 + 2*b^3*x + 3*a*b^2*cosh(x) + 2*a^2*b)*sinh(x)^2 + 2*((a^2 - b^2)*cosh(x)^4 + 4*(a^2 -
b^2)*cosh(x)*sinh(x)^3 + (a^2 - b^2)*sinh(x)^4 + 2*(a^2 - b^2)*cosh(x)^2 + 2*(3*(a^2 - b^2)*cosh(x)^2 + a^2 -
b^2)*sinh(x)^2 + a^2 - b^2 + 4*((a^2 - b^2)*cosh(x)^3 + (a^2 - b^2)*cosh(x))*sinh(x))*sqrt(a^2 - b^2)*arctan(-
(a*cosh(x) + a*sinh(x) + b)/sqrt(a^2 - b^2)) + ((2*a^3 - 3*a*b^2)*cosh(x)^4 + 4*(2*a^3 - 3*a*b^2)*cosh(x)*sinh
(x)^3 + (2*a^3 - 3*a*b^2)*sinh(x)^4 + 2*a^3 - 3*a*b^2 + 2*(2*a^3 - 3*a*b^2)*cosh(x)^2 + 2*(2*a^3 - 3*a*b^2 + 3
*(2*a^3 - 3*a*b^2)*cosh(x)^2)*sinh(x)^2 + 4*((2*a^3 - 3*a*b^2)*cosh(x)^3 + (2*a^3 - 3*a*b^2)*cosh(x))*sinh(x))
*arctan(cosh(x) + sinh(x)) + (4*b^3*x*cosh(x)^3 + 3*a*b^2*cosh(x)^2 - a*b^2 + 4*(b^3*x + a^2*b)*cosh(x))*sinh(
x))/(a*b^3*cosh(x)^4 + 4*a*b^3*cosh(x)*sinh(x)^3 + a*b^3*sinh(x)^4 + 2*a*b^3*cosh(x)^2 + a*b^3 + 2*(3*a*b^3*co
sh(x)^2 + a*b^3)*sinh(x)^2 + 4*(a*b^3*cosh(x)^3 + a*b^3*cosh(x))*sinh(x))]

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giac [A]  time = 0.14, size = 111, normalized size = 1.18 \[ \frac {x}{a} + \frac {{\left (2 \, a^{2} - 3 \, b^{2}\right )} \arctan \left (e^{x}\right )}{b^{3}} - \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \arctan \left (\frac {a e^{x} + b}{\sqrt {a^{2} - b^{2}}}\right )}{\sqrt {a^{2} - b^{2}} a b^{3}} + \frac {b e^{\left (3 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - b e^{x} + 2 \, a}{b^{2} {\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^4/(a+b*sech(x)),x, algorithm="giac")

[Out]

x/a + (2*a^2 - 3*b^2)*arctan(e^x)/b^3 - 2*(a^4 - 2*a^2*b^2 + b^4)*arctan((a*e^x + b)/sqrt(a^2 - b^2))/(sqrt(a^
2 - b^2)*a*b^3) + (b*e^(3*x) + 2*a*e^(2*x) - b*e^x + 2*a)/(b^2*(e^(2*x) + 1)^2)

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maple [B]  time = 0.14, size = 248, normalized size = 2.64 \[ -\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}-\frac {2 a^{3} \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {4 a \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 b \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a}-\frac {2 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) a}{b^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {\tanh ^{3}\left (\frac {x}{2}\right )}{b \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {2 \tanh \left (\frac {x}{2}\right ) a}{b^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\tanh \left (\frac {x}{2}\right )}{b \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right ) a^{2}}{b^{3}}-\frac {3 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^4/(a+b*sech(x)),x)

[Out]

-1/a*ln(tanh(1/2*x)-1)-2/b^3*a^3/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))+4*a/b/((a+b
)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))-2*b/a/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*x
)/((a+b)*(a-b))^(1/2))+1/a*ln(tanh(1/2*x)+1)-2/b^2/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^3*a-1/b/(tanh(1/2*x)^2+1)^2
*tanh(1/2*x)^3-2/b^2/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)*a+1/b/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)+2/b^3*arctan(tanh(1
/2*x))*a^2-3/b*arctan(tanh(1/2*x))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^4/(a+b*sech(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 7.26, size = 700, normalized size = 7.45 \[ \frac {\frac {2\,a}{b^2}+\frac {{\mathrm {e}}^x}{b}}{{\mathrm {e}}^{2\,x}+1}+\frac {x}{a}-\frac {\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,\left (a^2\,2{}\mathrm {i}-b^2\,3{}\mathrm {i}\right )}{2\,b^3}+\frac {\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,\left (a^2\,2{}\mathrm {i}-b^2\,3{}\mathrm {i}\right )}{2\,b^3}-\frac {2\,{\mathrm {e}}^x}{b\,\left (2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1\right )}+\frac {\ln \left (\frac {\left (\frac {64\,a^8+96\,{\mathrm {e}}^x\,a^7\,b-288\,a^6\,b^2-416\,{\mathrm {e}}^x\,a^5\,b^3+456\,a^4\,b^4+600\,{\mathrm {e}}^x\,a^3\,b^5-272\,a^2\,b^6-288\,{\mathrm {e}}^x\,a\,b^7+32\,b^8}{a^6\,b^4}-\frac {\left (\frac {16\,\left (a^2-b^2\right )\,\left (-4\,a^3-7\,{\mathrm {e}}^x\,a^2\,b+4\,a\,b^2+8\,{\mathrm {e}}^x\,b^3\right )}{a^6}+\frac {32\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}\,\left (-2\,a^3-3\,{\mathrm {e}}^x\,a^2\,b+3\,a\,b^2+4\,{\mathrm {e}}^x\,b^3\right )}{a^6\,b}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{a\,b^3}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{a\,b^3}-\frac {8\,{\left (a^2-b^2\right )}^2\,\left (2\,a^2-3\,b^2\right )\,\left (-4\,a^3-7\,{\mathrm {e}}^x\,a^2\,b+6\,a\,b^2+10\,{\mathrm {e}}^x\,b^3\right )}{a^6\,b^6}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{a\,b^3}-\frac {\ln \left (-\frac {\left (\frac {64\,a^8+96\,{\mathrm {e}}^x\,a^7\,b-288\,a^6\,b^2-416\,{\mathrm {e}}^x\,a^5\,b^3+456\,a^4\,b^4+600\,{\mathrm {e}}^x\,a^3\,b^5-272\,a^2\,b^6-288\,{\mathrm {e}}^x\,a\,b^7+32\,b^8}{a^6\,b^4}+\frac {\left (\frac {16\,\left (a^2-b^2\right )\,\left (-4\,a^3-7\,{\mathrm {e}}^x\,a^2\,b+4\,a\,b^2+8\,{\mathrm {e}}^x\,b^3\right )}{a^6}-\frac {32\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}\,\left (-2\,a^3-3\,{\mathrm {e}}^x\,a^2\,b+3\,a\,b^2+4\,{\mathrm {e}}^x\,b^3\right )}{a^6\,b}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{a\,b^3}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{a\,b^3}-\frac {8\,{\left (a^2-b^2\right )}^2\,\left (2\,a^2-3\,b^2\right )\,\left (-4\,a^3-7\,{\mathrm {e}}^x\,a^2\,b+6\,a\,b^2+10\,{\mathrm {e}}^x\,b^3\right )}{a^6\,b^6}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{a\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^4/(a + b/cosh(x)),x)

[Out]

((2*a)/b^2 + exp(x)/b)/(exp(2*x) + 1) + x/a - (log(exp(x) - 1i)*(a^2*2i - b^2*3i))/(2*b^3) + (log(exp(x) + 1i)
*(a^2*2i - b^2*3i))/(2*b^3) - (2*exp(x))/(b*(2*exp(2*x) + exp(4*x) + 1)) + (log((((64*a^8 + 32*b^8 - 272*a^2*b
^6 + 456*a^4*b^4 - 288*a^6*b^2 - 288*a*b^7*exp(x) + 96*a^7*b*exp(x) + 600*a^3*b^5*exp(x) - 416*a^5*b^3*exp(x))
/(a^6*b^4) - (((16*(a^2 - b^2)*(4*a*b^2 - 4*a^3 + 8*b^3*exp(x) - 7*a^2*b*exp(x)))/a^6 + (32*(-(a + b)^3*(a - b
)^3)^(1/2)*(3*a*b^2 - 2*a^3 + 4*b^3*exp(x) - 3*a^2*b*exp(x)))/(a^6*b))*(-(a + b)^3*(a - b)^3)^(1/2))/(a*b^3))*
(-(a + b)^3*(a - b)^3)^(1/2))/(a*b^3) - (8*(a^2 - b^2)^2*(2*a^2 - 3*b^2)*(6*a*b^2 - 4*a^3 + 10*b^3*exp(x) - 7*
a^2*b*exp(x)))/(a^6*b^6))*(-(a + b)^3*(a - b)^3)^(1/2))/(a*b^3) - (log(- (((64*a^8 + 32*b^8 - 272*a^2*b^6 + 45
6*a^4*b^4 - 288*a^6*b^2 - 288*a*b^7*exp(x) + 96*a^7*b*exp(x) + 600*a^3*b^5*exp(x) - 416*a^5*b^3*exp(x))/(a^6*b
^4) + (((16*(a^2 - b^2)*(4*a*b^2 - 4*a^3 + 8*b^3*exp(x) - 7*a^2*b*exp(x)))/a^6 - (32*(-(a + b)^3*(a - b)^3)^(1
/2)*(3*a*b^2 - 2*a^3 + 4*b^3*exp(x) - 3*a^2*b*exp(x)))/(a^6*b))*(-(a + b)^3*(a - b)^3)^(1/2))/(a*b^3))*(-(a +
b)^3*(a - b)^3)^(1/2))/(a*b^3) - (8*(a^2 - b^2)^2*(2*a^2 - 3*b^2)*(6*a*b^2 - 4*a^3 + 10*b^3*exp(x) - 7*a^2*b*e
xp(x)))/(a^6*b^6))*(-(a + b)^3*(a - b)^3)^(1/2))/(a*b^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{4}{\relax (x )}}{a + b \operatorname {sech}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**4/(a+b*sech(x)),x)

[Out]

Integral(tanh(x)**4/(a + b*sech(x)), x)

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