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[(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)

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Rubi [A]  time = 0.0891387, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, 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 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]

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 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]

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*}

Mathematica [B]  time = 2.23284, 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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Maple [B]  time = 0.051, size = 313, normalized size = 4.1 \begin{align*}{\frac{{a}^{2}}{2\,d}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}+\sqrt{a+b} \right ){b}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{a+b}}}}-{\frac{{a}^{2}}{2\,d}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}-2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}+\sqrt{a+b} \right ){b}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{a+b}}}}-2\,{\frac{ \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}a}{d{b}^{2} \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+2\,{\frac{ \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{bd \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-4\,{\frac{ \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}a}{d{b}^{2} \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{4}{3\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-3}}-2\,{\frac{\tanh \left ( 1/2\,dx+c/2 \right ) a}{d{b}^{2} \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+2\,{\frac{\tanh \left ( 1/2\,dx+c/2 \right ) }{bd \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{3}}} \end{align*}

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*tanh(1/2*d*x+1/2*c)*b^(1/2)+(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*tanh(1/2*d*x+1/2*c)*b^(1/2)+(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*tanh
(1/2*d*x+1/2*c)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 2.38192, size = 4709, normalized size = 61.16 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{6}{\left (c + d x \right )}}{a + b \operatorname{sech}^{2}{\left (c + d x \right )}}\, dx \end{align*}

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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Giac [A]  time = 1.25143, size = 159, normalized size = 2.06 \begin{align*} \frac{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} d} + \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 )}}{3 \, b^{2} d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}} \end{align*}

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]

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*d) + 2/3*(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*d*(e^(2*d*x + 2*c) + 1)^3)