3.89 \(\int \frac {\text {sech}^4(c+d x)}{(a+b \text {sech}^2(c+d x))^2} \, dx\)

Optimal. Leaf size=83 \[ \frac {(a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 b^{3/2} d (a+b)^{3/2}}-\frac {a \tanh (c+d x)}{2 b d (a+b) \left (a-b \tanh ^2(c+d x)+b\right )} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4146, 385, 208} \[ \frac {(a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 b^{3/2} d (a+b)^{3/2}}-\frac {a \tanh (c+d x)}{2 b d (a+b) \left (a-b \tanh ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sech[c + d*x]^4/(a + b*Sech[c + d*x]^2)^2,x]

[Out]

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

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

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

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Mathematica [A]  time = 0.22, size = 88, normalized size = 1.06 \[ 4 \left (\frac {(a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{8 b^{3/2} d (a+b)^{3/2}}-\frac {a \sinh (2 (c+d x))}{8 b d (a+b) (a \cosh (2 (c+d x))+a+2 b)}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[c + d*x]^4/(a + b*Sech[c + d*x]^2)^2,x]

[Out]

4*(((a + 2*b)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(8*b^(3/2)*(a + b)^(3/2)*d) - (a*Sinh[2*(c + d*x)]
)/(8*b*(a + b)*d*(a + 2*b + a*Cosh[2*(c + d*x)])))

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fricas [B]  time = 0.45, size = 1569, normalized size = 18.90 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^4/(a+b*sech(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[1/4*(4*a^2*b + 4*a*b^2 + 4*(a^2*b + 3*a*b^2 + 2*b^3)*cosh(d*x + c)^2 + 8*(a^2*b + 3*a*b^2 + 2*b^3)*cosh(d*x +
c)*sinh(d*x + c) + 4*(a^2*b + 3*a*b^2 + 2*b^3)*sinh(d*x + c)^2 + ((a^2 + 2*a*b)*cosh(d*x + c)^4 + 4*(a^2 + 2*
a*b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2*a*b)*sinh(d*x + c)^4 + 2*(a^2 + 4*a*b + 4*b^2)*cosh(d*x + c)^2 +
2*(3*(a^2 + 2*a*b)*cosh(d*x + c)^2 + a^2 + 4*a*b + 4*b^2)*sinh(d*x + c)^2 + a^2 + 2*a*b + 4*((a^2 + 2*a*b)*co
sh(d*x + c)^3 + (a^2 + 4*a*b + 4*b^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)*cos
h(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 + a + 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x +
c))*sinh(d*x + c) + a)))/((a^3*b^2 + 2*a^2*b^3 + a*b^4)*d*cosh(d*x + c)^4 + 4*(a^3*b^2 + 2*a^2*b^3 + a*b^4)*d
*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3*b^2 + 2*a^2*b^3 + a*b^4)*d*sinh(d*x + c)^4 + 2*(a^3*b^2 + 4*a^2*b^3 + 5*
a*b^4 + 2*b^5)*d*cosh(d*x + c)^2 + 2*(3*(a^3*b^2 + 2*a^2*b^3 + a*b^4)*d*cosh(d*x + c)^2 + (a^3*b^2 + 4*a^2*b^3
+ 5*a*b^4 + 2*b^5)*d)*sinh(d*x + c)^2 + (a^3*b^2 + 2*a^2*b^3 + a*b^4)*d + 4*((a^3*b^2 + 2*a^2*b^3 + a*b^4)*d*
cosh(d*x + c)^3 + (a^3*b^2 + 4*a^2*b^3 + 5*a*b^4 + 2*b^5)*d*cosh(d*x + c))*sinh(d*x + c)), 1/2*(2*a^2*b + 2*a*
b^2 + 2*(a^2*b + 3*a*b^2 + 2*b^3)*cosh(d*x + c)^2 + 4*(a^2*b + 3*a*b^2 + 2*b^3)*cosh(d*x + c)*sinh(d*x + c) +
2*(a^2*b + 3*a*b^2 + 2*b^3)*sinh(d*x + c)^2 + ((a^2 + 2*a*b)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b)*cosh(d*x + c)*s
inh(d*x + c)^3 + (a^2 + 2*a*b)*sinh(d*x + c)^4 + 2*(a^2 + 4*a*b + 4*b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b)*
cosh(d*x + c)^2 + a^2 + 4*a*b + 4*b^2)*sinh(d*x + c)^2 + a^2 + 2*a*b + 4*((a^2 + 2*a*b)*cosh(d*x + c)^3 + (a^2
+ 4*a*b + 4*b^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)))/((a^3*b^2 + 2*a^2*b^3 + a*b^4
)*d*cosh(d*x + c)^4 + 4*(a^3*b^2 + 2*a^2*b^3 + a*b^4)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3*b^2 + 2*a^2*b^3 +
a*b^4)*d*sinh(d*x + c)^4 + 2*(a^3*b^2 + 4*a^2*b^3 + 5*a*b^4 + 2*b^5)*d*cosh(d*x + c)^2 + 2*(3*(a^3*b^2 + 2*a^
2*b^3 + a*b^4)*d*cosh(d*x + c)^2 + (a^3*b^2 + 4*a^2*b^3 + 5*a*b^4 + 2*b^5)*d)*sinh(d*x + c)^2 + (a^3*b^2 + 2*a
^2*b^3 + a*b^4)*d + 4*((a^3*b^2 + 2*a^2*b^3 + a*b^4)*d*cosh(d*x + c)^3 + (a^3*b^2 + 4*a^2*b^3 + 5*a*b^4 + 2*b^
5)*d*cosh(d*x + c))*sinh(d*x + c))]

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giac [A]  time = 0.77, size = 139, normalized size = 1.67 \[ \frac {\frac {{\left (a + 2 \, b\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{{\left (a b + b^{2}\right )} \sqrt {-a b - b^{2}}} + \frac {2 \, {\left (a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}}{{\left (a b + b^{2}\right )} {\left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^4/(a+b*sech(d*x+c)^2)^2,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.29, size = 374, normalized size = 4.51 \[ -\frac {a \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right ) b \left (a +b \right )}-\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right ) b \left (a +b \right )}-\frac {a \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{4 d \left (a +b \right )^{\frac {3}{2}} b^{\frac {3}{2}}}+\frac {a \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{4 d \left (a +b \right )^{\frac {3}{2}} b^{\frac {3}{2}}}-\frac {\ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d \left (a +b \right )^{\frac {3}{2}} \sqrt {b}}+\frac {\ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d \left (a +b \right )^{\frac {3}{2}} \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)^4/(a+b*sech(d*x+c)^2)^2,x)

[Out]

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

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maxima [B]  time = 0.48, size = 165, normalized size = 1.99 \[ -\frac {{\left (a + 2 \, b\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{4 \, \sqrt {{\left (a + b\right )} b} {\left (a b + b^{2}\right )} d} - \frac {{\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a}{{\left (a^{2} b + a b^{2} + 2 \, {\left (a^{2} b + 3 \, a b^{2} + 2 \, b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{2} b + a b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^4/(a+b*sech(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^4\,{\left (a+\frac {b}{{\mathrm {cosh}\left (c+d\,x\right )}^2}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(c + d*x)^4*(a + b/cosh(c + d*x)^2)^2),x)

[Out]

int(1/(cosh(c + d*x)^4*(a + b/cosh(c + d*x)^2)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{4}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)**4/(a+b*sech(d*x+c)**2)**2,x)

[Out]

Integral(sech(c + d*x)**4/(a + b*sech(c + d*x)**2)**2, x)

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