3.82 \(\int \frac {\text {sech}^6(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx\)

Optimal. Leaf size=77 \[ \frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{b^{5/2} d \sqrt {a+b}}-\frac {(a-b) \tanh (c+d x)}{b^2 d}-\frac {\tanh ^3(c+d x)}{3 b d} \]

[Out]

a^2*arctanh(b^(1/2)*tanh(d*x+c)/(a+b)^(1/2))/b^(5/2)/d/(a+b)^(1/2)-(a-b)*tanh(d*x+c)/b^2/d-1/3*tanh(d*x+c)^3/b
/d

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Rubi [A]  time = 0.09, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4146, 390, 208} \[ \frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{b^{5/2} d \sqrt {a+b}}-\frac {(a-b) \tanh (c+d x)}{b^2 d}-\frac {\tanh ^3(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sech[c + d*x]^6/(a + b*Sech[c + d*x]^2),x]

[Out]

(a^2*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(b^(5/2)*Sqrt[a + b]*d) - ((a - b)*Tanh[c + d*x])/(b^2*d) -
Tanh[c + d*x]^3/(3*b*d)

Rule 208

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

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 4146

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/
2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]

Rubi steps

\begin {align*} \int \frac {\text {sech}^6(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2}{a+b-b x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-\frac {a-b}{b^2}-\frac {x^2}{b}+\frac {a^2}{b^2 \left (a+b-b x^2\right )}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac {(a-b) \tanh (c+d x)}{b^2 d}-\frac {\tanh ^3(c+d x)}{3 b d}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{a+b-b x^2} \, dx,x,\tanh (c+d x)\right )}{b^2 d}\\ &=\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{b^{5/2} \sqrt {a+b} d}-\frac {(a-b) \tanh (c+d x)}{b^2 d}-\frac {\tanh ^3(c+d x)}{3 b d}\\ \end {align*}

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Mathematica [B]  time = 2.18, size = 214, normalized size = 2.78 \[ \frac {\text {sech}^2(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (3 a^2 (\cosh (2 c)-\sinh (2 c)) \tanh ^{-1}\left (\frac {(\cosh (2 c)-\sinh (2 c)) \text {sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right )+\sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4} \text {sech}(c+d x) \left (\text {sech}(c) \sinh (d x) \left (-3 a+b \text {sech}^2(c+d x)+2 b\right )+b \tanh (c) \text {sech}(c+d x)\right )\right )}{6 b^2 d \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4} \left (a+b \text {sech}^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[c + d*x]^6/(a + b*Sech[c + d*x]^2),x]

[Out]

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(3*a^2*ArcTanh[(Sech[d*x]*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*
Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4])]*(Cosh[2*c] - Sinh[2*c]) + Sqrt[
a + b]*Sech[c + d*x]*Sqrt[b*(Cosh[c] - Sinh[c])^4]*(Sech[c]*(-3*a + 2*b + b*Sech[c + d*x]^2)*Sinh[d*x] + b*Sec
h[c + d*x]*Tanh[c])))/(6*b^2*Sqrt[a + b]*d*(a + b*Sech[c + d*x]^2)*Sqrt[b*(Cosh[c] - Sinh[c])^4])

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fricas [B]  time = 0.49, size = 1905, normalized size = 24.74 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^6/(a+b*sech(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/6*(12*(a^2*b + a*b^2)*cosh(d*x + c)^4 + 48*(a^2*b + a*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + 12*(a^2*b + a*b^
2)*sinh(d*x + c)^4 + 12*a^2*b + 4*a*b^2 - 8*b^3 + 24*(a^2*b - b^3)*cosh(d*x + c)^2 + 24*(a^2*b - b^3 + 3*(a^2*
b + a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 3*(a^2*cosh(d*x + c)^6 + 6*a^2*cosh(d*x + c)*sinh(d*x + c)^5 + a
^2*sinh(d*x + c)^6 + 3*a^2*cosh(d*x + c)^4 + 3*(5*a^2*cosh(d*x + c)^2 + a^2)*sinh(d*x + c)^4 + 3*a^2*cosh(d*x
+ c)^2 + 4*(5*a^2*cosh(d*x + c)^3 + 3*a^2*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*a^2*cosh(d*x + c)^4 + 6*a^2*co
sh(d*x + c)^2 + a^2)*sinh(d*x + c)^2 + a^2 + 6*(a^2*cosh(d*x + c)^5 + 2*a^2*cosh(d*x + c)^3 + a^2*cosh(d*x + c
))*sinh(d*x + c))*sqrt(a*b + b^2)*log((a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*
x + c)^4 + 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 + 8
*a*b + 8*b^2 + 4*(a^2*cosh(d*x + c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c) - 4*(a*cosh(d*x + c)^2 + 2*
a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(a*b + b^2))/(a*cosh(d*x + c)^4 + 4*a*cosh(d*
x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 + a + 2*b)*s
inh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a)) + 48*((a^2*b + a*b^2)*cos
h(d*x + c)^3 + (a^2*b - b^3)*cosh(d*x + c))*sinh(d*x + c))/((a*b^3 + b^4)*d*cosh(d*x + c)^6 + 6*(a*b^3 + b^4)*
d*cosh(d*x + c)*sinh(d*x + c)^5 + (a*b^3 + b^4)*d*sinh(d*x + c)^6 + 3*(a*b^3 + b^4)*d*cosh(d*x + c)^4 + 3*(5*(
a*b^3 + b^4)*d*cosh(d*x + c)^2 + (a*b^3 + b^4)*d)*sinh(d*x + c)^4 + 3*(a*b^3 + b^4)*d*cosh(d*x + c)^2 + 4*(5*(
a*b^3 + b^4)*d*cosh(d*x + c)^3 + 3*(a*b^3 + b^4)*d*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*(a*b^3 + b^4)*d*cosh(
d*x + c)^4 + 6*(a*b^3 + b^4)*d*cosh(d*x + c)^2 + (a*b^3 + b^4)*d)*sinh(d*x + c)^2 + (a*b^3 + b^4)*d + 6*((a*b^
3 + b^4)*d*cosh(d*x + c)^5 + 2*(a*b^3 + b^4)*d*cosh(d*x + c)^3 + (a*b^3 + b^4)*d*cosh(d*x + c))*sinh(d*x + c))
, 1/3*(6*(a^2*b + a*b^2)*cosh(d*x + c)^4 + 24*(a^2*b + a*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + 6*(a^2*b + a*b^2
)*sinh(d*x + c)^4 + 6*a^2*b + 2*a*b^2 - 4*b^3 + 12*(a^2*b - b^3)*cosh(d*x + c)^2 + 12*(a^2*b - b^3 + 3*(a^2*b
+ a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 3*(a^2*cosh(d*x + c)^6 + 6*a^2*cosh(d*x + c)*sinh(d*x + c)^5 + a^2
*sinh(d*x + c)^6 + 3*a^2*cosh(d*x + c)^4 + 3*(5*a^2*cosh(d*x + c)^2 + a^2)*sinh(d*x + c)^4 + 3*a^2*cosh(d*x +
c)^2 + 4*(5*a^2*cosh(d*x + c)^3 + 3*a^2*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*a^2*cosh(d*x + c)^4 + 6*a^2*cosh
(d*x + c)^2 + a^2)*sinh(d*x + c)^2 + a^2 + 6*(a^2*cosh(d*x + c)^5 + 2*a^2*cosh(d*x + c)^3 + a^2*cosh(d*x + c))
*sinh(d*x + c))*sqrt(-a*b - b^2)*arctan(1/2*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x
+ c)^2 + a + 2*b)*sqrt(-a*b - b^2)/(a*b + b^2)) + 24*((a^2*b + a*b^2)*cosh(d*x + c)^3 + (a^2*b - b^3)*cosh(d*x
+ c))*sinh(d*x + c))/((a*b^3 + b^4)*d*cosh(d*x + c)^6 + 6*(a*b^3 + b^4)*d*cosh(d*x + c)*sinh(d*x + c)^5 + (a*
b^3 + b^4)*d*sinh(d*x + c)^6 + 3*(a*b^3 + b^4)*d*cosh(d*x + c)^4 + 3*(5*(a*b^3 + b^4)*d*cosh(d*x + c)^2 + (a*b
^3 + b^4)*d)*sinh(d*x + c)^4 + 3*(a*b^3 + b^4)*d*cosh(d*x + c)^2 + 4*(5*(a*b^3 + b^4)*d*cosh(d*x + c)^3 + 3*(a
*b^3 + b^4)*d*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*(a*b^3 + b^4)*d*cosh(d*x + c)^4 + 6*(a*b^3 + b^4)*d*cosh(d
*x + c)^2 + (a*b^3 + b^4)*d)*sinh(d*x + c)^2 + (a*b^3 + b^4)*d + 6*((a*b^3 + b^4)*d*cosh(d*x + c)^5 + 2*(a*b^3
+ b^4)*d*cosh(d*x + c)^3 + (a*b^3 + b^4)*d*cosh(d*x + c))*sinh(d*x + c))]

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giac [A]  time = 0.64, size = 118, normalized size = 1.53 \[ \frac {\frac {3 \, a^{2} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} b^{2}} + \frac {2 \, {\left (3 \, a e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a e^{\left (2 \, d x + 2 \, c\right )} - 6 \, b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a - 2 \, b\right )}}{b^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^6/(a+b*sech(d*x+c)^2),x, algorithm="giac")

[Out]

1/3*(3*a^2*arctan(1/2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2))/(sqrt(-a*b - b^2)*b^2) + 2*(3*a*e^(4*d*x
+ 4*c) + 6*a*e^(2*d*x + 2*c) - 6*b*e^(2*d*x + 2*c) + 3*a - 2*b)/(b^2*(e^(2*d*x + 2*c) + 1)^3))/d

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maple [B]  time = 0.28, size = 316, normalized size = 4.10 \[ -\frac {a^{2} \ln \left (-\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {a +b}\right )}{2 d \,b^{\frac {5}{2}} \sqrt {a +b}}+\frac {a^{2} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d \,b^{\frac {5}{2}} \sqrt {a +b}}-\frac {2 \left (\tanh ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{2} \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {2 \left (\tanh ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {4 \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{2} \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {4 \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d b \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{d \,b^{2} \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d b \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)^6/(a+b*sech(d*x+c)^2),x)

[Out]

-1/2/d*a^2/b^(5/2)/(a+b)^(1/2)*ln(-(a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*b^(1/2)*tanh(1/2*d*x+1/2*c)-(a+b)^(1/2)
)+1/2/d*a^2/b^(5/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*b^(1/2)*tanh(1/2*d*x+1/2*c)+(a+b)^(1/2)
)-2/d/b^2/(tanh(1/2*d*x+1/2*c)^2+1)^3*tanh(1/2*d*x+1/2*c)^5*a+2/d/b/(tanh(1/2*d*x+1/2*c)^2+1)^3*tanh(1/2*d*x+1
/2*c)^5-4/d/b^2/(tanh(1/2*d*x+1/2*c)^2+1)^3*tanh(1/2*d*x+1/2*c)^3*a+4/3/d/b/(tanh(1/2*d*x+1/2*c)^2+1)^3*tanh(1
/2*d*x+1/2*c)^3-2/d/b^2/(tanh(1/2*d*x+1/2*c)^2+1)^3*tanh(1/2*d*x+1/2*c)*a+2/d/b/(tanh(1/2*d*x+1/2*c)^2+1)^3*ta
nh(1/2*d*x+1/2*c)

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maxima [B]  time = 0.46, size = 160, normalized size = 2.08 \[ -\frac {a^{2} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{2 \, \sqrt {{\left (a + b\right )} b} b^{2} d} - \frac {2 \, {\left (6 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, a e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, a - 2 \, b\right )}}{3 \, {\left (3 \, b^{2} e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, b^{2} e^{\left (-4 \, d x - 4 \, c\right )} + b^{2} e^{\left (-6 \, d x - 6 \, c\right )} + b^{2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^6/(a+b*sech(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/2*a^2*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(-2*d*x - 2*c) + a + 2*b + 2*sqrt((a + b)
*b)))/(sqrt((a + b)*b)*b^2*d) - 2/3*(6*(a - b)*e^(-2*d*x - 2*c) + 3*a*e^(-4*d*x - 4*c) + 3*a - 2*b)/((3*b^2*e^
(-2*d*x - 2*c) + 3*b^2*e^(-4*d*x - 4*c) + b^2*e^(-6*d*x - 6*c) + b^2)*d)

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mupad [B]  time = 1.99, size = 334, normalized size = 4.34 \[ \frac {8}{3\,b\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}-\frac {4}{b\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}+\frac {2\,a}{b^2\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {a^2\,\ln \left (\frac {4\,a^2\,\left (2\,a\,b+a^2+a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{b^5\,\left (a+b\right )}-\frac {8\,a^2\,\left (a+2\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}+4\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{b^{9/2}\,\sqrt {a+b}}\right )}{2\,b^{5/2}\,d\,\sqrt {a+b}}+\frac {a^2\,\ln \left (\frac {8\,a^2\,\left (a+2\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}+4\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{b^{9/2}\,\sqrt {a+b}}+\frac {4\,a^2\,\left (2\,a\,b+a^2+a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{b^5\,\left (a+b\right )}\right )}{2\,b^{5/2}\,d\,\sqrt {a+b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(c + d*x)^6*(a + b/cosh(c + d*x)^2)),x)

[Out]

8/(3*b*d*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)) - 4/(b*d*(2*exp(2*c + 2*d*x) + exp(
4*c + 4*d*x) + 1)) + (2*a)/(b^2*d*(exp(2*c + 2*d*x) + 1)) - (a^2*log((4*a^2*(2*a*b + a^2 + a^2*exp(2*c + 2*d*x
) + 8*b^2*exp(2*c + 2*d*x) + 8*a*b*exp(2*c + 2*d*x)))/(b^5*(a + b)) - (8*a^2*(a + 2*a*exp(2*c + 2*d*x) + 4*b*e
xp(2*c + 2*d*x)))/(b^(9/2)*(a + b)^(1/2))))/(2*b^(5/2)*d*(a + b)^(1/2)) + (a^2*log((8*a^2*(a + 2*a*exp(2*c + 2
*d*x) + 4*b*exp(2*c + 2*d*x)))/(b^(9/2)*(a + b)^(1/2)) + (4*a^2*(2*a*b + a^2 + a^2*exp(2*c + 2*d*x) + 8*b^2*ex
p(2*c + 2*d*x) + 8*a*b*exp(2*c + 2*d*x)))/(b^5*(a + b))))/(2*b^(5/2)*d*(a + b)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{6}{\left (c + d x \right )}}{a + b \operatorname {sech}^{2}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)**6/(a+b*sech(d*x+c)**2),x)

[Out]

Integral(sech(c + d*x)**6/(a + b*sech(c + d*x)**2), x)

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