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

Optimal. Leaf size=101 \[ \frac{a^2 \tanh (c+d x)}{2 b^2 d (a+b) \left (a-b \tanh ^2(c+d x)+b\right )}-\frac{a (3 a+4 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{2 b^{5/2} d (a+b)^{3/2}}+\frac{\tanh (c+d x)}{b^2 d} \]

[Out]

-(a*(3*a + 4*b)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(2*b^(5/2)*(a + b)^(3/2)*d) + Tanh[c + d*x]/(b^2
*d) + (a^2*Tanh[c + d*x])/(2*b^2*(a + b)*d*(a + b - b*Tanh[c + d*x]^2))

________________________________________________________________________________________

Rubi [A]  time = 0.141507, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {4146, 390, 385, 208} \[ \frac{a^2 \tanh (c+d x)}{2 b^2 d (a+b) \left (a-b \tanh ^2(c+d x)+b\right )}-\frac{a (3 a+4 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{2 b^{5/2} d (a+b)^{3/2}}+\frac{\tanh (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(3*a + 4*b)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(2*b^(5/2)*(a + b)^(3/2)*d) + Tanh[c + d*x]/(b^2
*d) + (a^2*Tanh[c + d*x])/(2*b^2*(a + b)*d*(a + b - b*Tanh[c + d*x]^2))

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 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 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)}{\left (a+b \text{sech}^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{\left (a+b-b x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{b^2}-\frac{a (a+2 b)-2 a b x^2}{b^2 \left (a+b-b x^2\right )^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\tanh (c+d x)}{b^2 d}-\frac{\operatorname{Subst}\left (\int \frac{a (a+2 b)-2 a b x^2}{\left (a+b-b x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{b^2 d}\\ &=\frac{\tanh (c+d x)}{b^2 d}+\frac{a^2 \tanh (c+d x)}{2 b^2 (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )}-\frac{(a (3 a+4 b)) \operatorname{Subst}\left (\int \frac{1}{a+b-b x^2} \, dx,x,\tanh (c+d x)\right )}{2 b^2 (a+b) d}\\ &=-\frac{a (3 a+4 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{2 b^{5/2} (a+b)^{3/2} d}+\frac{\tanh (c+d x)}{b^2 d}+\frac{a^2 \tanh (c+d x)}{2 b^2 (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 3.78992, size = 229, normalized size = 2.27 \[ \frac{\text{sech}^4(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (2 \text{sech}(c) \sinh (d x) \text{sech}(c+d x) (a \cosh (2 (c+d x))+a+2 b)+\frac{a (a \text{sech}(2 c) \sinh (2 d x)-(a+2 b) \tanh (2 c))}{a+b}-\frac{a (3 a+4 b) (\cosh (2 c)-\sinh (2 c)) (a \cosh (2 (c+d x))+a+2 b) \tanh ^{-1}\left (\frac{(\cosh (2 c)-\sinh (2 c)) \text{sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt{a+b} \sqrt{b (\cosh (c)-\sinh (c))^4}}\right )}{(a+b)^{3/2} \sqrt{b (\cosh (c)-\sinh (c))^4}}\right )}{8 b^2 d \left (a+b \text{sech}^2(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.063, size = 413, normalized size = 4.1 \begin{align*}{\frac{{a}^{2}}{d{b}^{2} \left ( a+b \right ) } \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}a+b \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) ^{-1}}+{\frac{{a}^{2}}{d{b}^{2} \left ( a+b \right ) }\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}a+b \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) ^{-1}}-{\frac{3\,{a}^{2}}{4\,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}}} \left ( a+b \right ) ^{-{\frac{3}{2}}}}+{\frac{3\,{a}^{2}}{4\,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}}} \left ( a+b \right ) ^{-{\frac{3}{2}}}}-{\frac{a}{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 ) \left ( a+b \right ) ^{-{\frac{3}{2}}}{b}^{-{\frac{3}{2}}}}+{\frac{a}{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 ) \left ( a+b \right ) ^{-{\frac{3}{2}}}{b}^{-{\frac{3}{2}}}}+2\,{\frac{\tanh \left ( 1/2\,dx+c/2 \right ) }{d{b}^{2} \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) }} \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)^2,x)

[Out]

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

________________________________________________________________________________________

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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.49778, size = 6961, normalized size = 68.92 \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)^2,x, algorithm="fricas")

[Out]

[-1/4*(4*(3*a^3*b + 7*a^2*b^2 + 4*a*b^3)*cosh(d*x + c)^4 + 16*(3*a^3*b + 7*a^2*b^2 + 4*a*b^3)*cosh(d*x + c)*si
nh(d*x + c)^3 + 4*(3*a^3*b + 7*a^2*b^2 + 4*a*b^3)*sinh(d*x + c)^4 + 12*a^3*b + 20*a^2*b^2 + 8*a*b^3 + 8*(3*a^3
*b + 10*a^2*b^2 + 11*a*b^3 + 4*b^4)*cosh(d*x + c)^2 + 8*(3*a^3*b + 10*a^2*b^2 + 11*a*b^3 + 4*b^4 + 3*(3*a^3*b
+ 7*a^2*b^2 + 4*a*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - ((3*a^3 + 4*a^2*b)*cosh(d*x + c)^6 + 6*(3*a^3 + 4*a^
2*b)*cosh(d*x + c)*sinh(d*x + c)^5 + (3*a^3 + 4*a^2*b)*sinh(d*x + c)^6 + (9*a^3 + 24*a^2*b + 16*a*b^2)*cosh(d*
x + c)^4 + (9*a^3 + 24*a^2*b + 16*a*b^2 + 15*(3*a^3 + 4*a^2*b)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 4*(5*(3*a^3
+ 4*a^2*b)*cosh(d*x + c)^3 + (9*a^3 + 24*a^2*b + 16*a*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 3*a^3 + 4*a^2*b +
(9*a^3 + 24*a^2*b + 16*a*b^2)*cosh(d*x + c)^2 + (15*(3*a^3 + 4*a^2*b)*cosh(d*x + c)^4 + 9*a^3 + 24*a^2*b + 16*
a*b^2 + 6*(9*a^3 + 24*a^2*b + 16*a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 2*(3*(3*a^3 + 4*a^2*b)*cosh(d*x + c
)^5 + 2*(9*a^3 + 24*a^2*b + 16*a*b^2)*cosh(d*x + c)^3 + (9*a^3 + 24*a^2*b + 16*a*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)*cosh(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)) + 16*((3*a^3*b + 7*a^2*b^2 + 4*a*b^3)
*cosh(d*x + c)^3 + (3*a^3*b + 10*a^2*b^2 + 11*a*b^3 + 4*b^4)*cosh(d*x + c))*sinh(d*x + c))/((a^3*b^3 + 2*a^2*b
^4 + a*b^5)*d*cosh(d*x + c)^6 + 6*(a^3*b^3 + 2*a^2*b^4 + a*b^5)*d*cosh(d*x + c)*sinh(d*x + c)^5 + (a^3*b^3 + 2
*a^2*b^4 + a*b^5)*d*sinh(d*x + c)^6 + (3*a^3*b^3 + 10*a^2*b^4 + 11*a*b^5 + 4*b^6)*d*cosh(d*x + c)^4 + (15*(a^3
*b^3 + 2*a^2*b^4 + a*b^5)*d*cosh(d*x + c)^2 + (3*a^3*b^3 + 10*a^2*b^4 + 11*a*b^5 + 4*b^6)*d)*sinh(d*x + c)^4 +
(3*a^3*b^3 + 10*a^2*b^4 + 11*a*b^5 + 4*b^6)*d*cosh(d*x + c)^2 + 4*(5*(a^3*b^3 + 2*a^2*b^4 + a*b^5)*d*cosh(d*x
+ c)^3 + (3*a^3*b^3 + 10*a^2*b^4 + 11*a*b^5 + 4*b^6)*d*cosh(d*x + c))*sinh(d*x + c)^3 + (15*(a^3*b^3 + 2*a^2*
b^4 + a*b^5)*d*cosh(d*x + c)^4 + 6*(3*a^3*b^3 + 10*a^2*b^4 + 11*a*b^5 + 4*b^6)*d*cosh(d*x + c)^2 + (3*a^3*b^3
+ 10*a^2*b^4 + 11*a*b^5 + 4*b^6)*d)*sinh(d*x + c)^2 + (a^3*b^3 + 2*a^2*b^4 + a*b^5)*d + 2*(3*(a^3*b^3 + 2*a^2*
b^4 + a*b^5)*d*cosh(d*x + c)^5 + 2*(3*a^3*b^3 + 10*a^2*b^4 + 11*a*b^5 + 4*b^6)*d*cosh(d*x + c)^3 + (3*a^3*b^3
+ 10*a^2*b^4 + 11*a*b^5 + 4*b^6)*d*cosh(d*x + c))*sinh(d*x + c)), -1/2*(2*(3*a^3*b + 7*a^2*b^2 + 4*a*b^3)*cosh
(d*x + c)^4 + 8*(3*a^3*b + 7*a^2*b^2 + 4*a*b^3)*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(3*a^3*b + 7*a^2*b^2 + 4*a*b
^3)*sinh(d*x + c)^4 + 6*a^3*b + 10*a^2*b^2 + 4*a*b^3 + 4*(3*a^3*b + 10*a^2*b^2 + 11*a*b^3 + 4*b^4)*cosh(d*x +
c)^2 + 4*(3*a^3*b + 10*a^2*b^2 + 11*a*b^3 + 4*b^4 + 3*(3*a^3*b + 7*a^2*b^2 + 4*a*b^3)*cosh(d*x + c)^2)*sinh(d*
x + c)^2 + ((3*a^3 + 4*a^2*b)*cosh(d*x + c)^6 + 6*(3*a^3 + 4*a^2*b)*cosh(d*x + c)*sinh(d*x + c)^5 + (3*a^3 + 4
*a^2*b)*sinh(d*x + c)^6 + (9*a^3 + 24*a^2*b + 16*a*b^2)*cosh(d*x + c)^4 + (9*a^3 + 24*a^2*b + 16*a*b^2 + 15*(3
*a^3 + 4*a^2*b)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 4*(5*(3*a^3 + 4*a^2*b)*cosh(d*x + c)^3 + (9*a^3 + 24*a^2*b
+ 16*a*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 3*a^3 + 4*a^2*b + (9*a^3 + 24*a^2*b + 16*a*b^2)*cosh(d*x + c)^2 +
(15*(3*a^3 + 4*a^2*b)*cosh(d*x + c)^4 + 9*a^3 + 24*a^2*b + 16*a*b^2 + 6*(9*a^3 + 24*a^2*b + 16*a*b^2)*cosh(d*
x + c)^2)*sinh(d*x + c)^2 + 2*(3*(3*a^3 + 4*a^2*b)*cosh(d*x + c)^5 + 2*(9*a^3 + 24*a^2*b + 16*a*b^2)*cosh(d*x
+ c)^3 + (9*a^3 + 24*a^2*b + 16*a*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)) + 8*((3*a
^3*b + 7*a^2*b^2 + 4*a*b^3)*cosh(d*x + c)^3 + (3*a^3*b + 10*a^2*b^2 + 11*a*b^3 + 4*b^4)*cosh(d*x + c))*sinh(d*
x + c))/((a^3*b^3 + 2*a^2*b^4 + a*b^5)*d*cosh(d*x + c)^6 + 6*(a^3*b^3 + 2*a^2*b^4 + a*b^5)*d*cosh(d*x + c)*sin
h(d*x + c)^5 + (a^3*b^3 + 2*a^2*b^4 + a*b^5)*d*sinh(d*x + c)^6 + (3*a^3*b^3 + 10*a^2*b^4 + 11*a*b^5 + 4*b^6)*d
*cosh(d*x + c)^4 + (15*(a^3*b^3 + 2*a^2*b^4 + a*b^5)*d*cosh(d*x + c)^2 + (3*a^3*b^3 + 10*a^2*b^4 + 11*a*b^5 +
4*b^6)*d)*sinh(d*x + c)^4 + (3*a^3*b^3 + 10*a^2*b^4 + 11*a*b^5 + 4*b^6)*d*cosh(d*x + c)^2 + 4*(5*(a^3*b^3 + 2*
a^2*b^4 + a*b^5)*d*cosh(d*x + c)^3 + (3*a^3*b^3 + 10*a^2*b^4 + 11*a*b^5 + 4*b^6)*d*cosh(d*x + c))*sinh(d*x + c
)^3 + (15*(a^3*b^3 + 2*a^2*b^4 + a*b^5)*d*cosh(d*x + c)^4 + 6*(3*a^3*b^3 + 10*a^2*b^4 + 11*a*b^5 + 4*b^6)*d*co
sh(d*x + c)^2 + (3*a^3*b^3 + 10*a^2*b^4 + 11*a*b^5 + 4*b^6)*d)*sinh(d*x + c)^2 + (a^3*b^3 + 2*a^2*b^4 + a*b^5)
*d + 2*(3*(a^3*b^3 + 2*a^2*b^4 + a*b^5)*d*cosh(d*x + c)^5 + 2*(3*a^3*b^3 + 10*a^2*b^4 + 11*a*b^5 + 4*b^6)*d*co
sh(d*x + c)^3 + (3*a^3*b^3 + 10*a^2*b^4 + 11*a*b^5 + 4*b^6)*d*cosh(d*x + c))*sinh(d*x + c))]

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.2742, size = 306, normalized size = 3.03 \begin{align*} -\frac{{\left (3 \, a^{2} + 4 \, a b\right )} \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt{-a b - b^{2}}}\right )}{2 \,{\left (a b^{2} d + b^{3} d\right )} \sqrt{-a b - b^{2}}} - \frac{3 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 14 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 8 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a^{2} + 2 \, a b}{{\left (a b^{2} d + b^{3} d\right )}{\left (a e^{\left (6 \, d x + 6 \, c\right )} + 3 \, a e^{\left (4 \, d x + 4 \, c\right )} + 4 \, b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}} \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)^2,x, algorithm="giac")

[Out]

-1/2*(3*a^2 + 4*a*b)*arctan(1/2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2))/((a*b^2*d + b^3*d)*sqrt(-a*b -
b^2)) - (3*a^2*e^(4*d*x + 4*c) + 4*a*b*e^(4*d*x + 4*c) + 6*a^2*e^(2*d*x + 2*c) + 14*a*b*e^(2*d*x + 2*c) + 8*b
^2*e^(2*d*x + 2*c) + 3*a^2 + 2*a*b)/((a*b^2*d + b^3*d)*(a*e^(6*d*x + 6*c) + 3*a*e^(4*d*x + 4*c) + 4*b*e^(4*d*x
+ 4*c) + 3*a*e^(2*d*x + 2*c) + 4*b*e^(2*d*x + 2*c) + a))