3.61 \(\int \text{sech}(c+d x) (a+b \text{sech}^2(c+d x))^2 \, dx\)
Optimal. Leaf size=90 \[ \frac{\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}+\frac{3 b (2 a+b) \tanh (c+d x) \text{sech}(c+d x)}{8 d}+\frac{b \tanh (c+d x) \text{sech}^3(c+d x) \left (a \sinh ^2(c+d x)+a+b\right )}{4 d} \]
[Out]
((8*a^2 + 8*a*b + 3*b^2)*ArcTan[Sinh[c + d*x]])/(8*d) + (3*b*(2*a + b)*Sech[c + d*x]*Tanh[c + d*x])/(8*d) + (b
*Sech[c + d*x]^3*(a + b + a*Sinh[c + d*x]^2)*Tanh[c + d*x])/(4*d)
________________________________________________________________________________________
Rubi [A] time = 0.0838808, antiderivative size = 90, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used =
{4147, 413, 385, 203} \[ \frac{\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}+\frac{3 b (2 a+b) \tanh (c+d x) \text{sech}(c+d x)}{8 d}+\frac{b \tanh (c+d x) \text{sech}^3(c+d x) \left (a \sinh ^2(c+d x)+a+b\right )}{4 d} \]
Antiderivative was successfully verified.
[In]
Int[Sech[c + d*x]*(a + b*Sech[c + d*x]^2)^2,x]
[Out]
((8*a^2 + 8*a*b + 3*b^2)*ArcTan[Sinh[c + d*x]])/(8*d) + (3*b*(2*a + b)*Sech[c + d*x]*Tanh[c + d*x])/(8*d) + (b
*Sech[c + d*x]^3*(a + b + a*Sinh[c + d*x]^2)*Tanh[c + d*x])/(4*d)
Rule 4147
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fr
eeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2*x^2)^
((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n
/2] && IntegerQ[p]
Rule 413
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]
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 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rubi steps
\begin{align*} \int \text{sech}(c+d x) \left (a+b \text{sech}^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b+a x^2\right )^2}{\left (1+x^2\right )^3} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{b \text{sech}^3(c+d x) \left (a+b+a \sinh ^2(c+d x)\right ) \tanh (c+d x)}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{(a+b) (4 a+3 b)+a (4 a+b) x^2}{\left (1+x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{4 d}\\ &=\frac{3 b (2 a+b) \text{sech}(c+d x) \tanh (c+d x)}{8 d}+\frac{b \text{sech}^3(c+d x) \left (a+b+a \sinh ^2(c+d x)\right ) \tanh (c+d x)}{4 d}+\frac{\left (8 a^2+8 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{8 d}\\ &=\frac{\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}+\frac{3 b (2 a+b) \text{sech}(c+d x) \tanh (c+d x)}{8 d}+\frac{b \text{sech}^3(c+d x) \left (a+b+a \sinh ^2(c+d x)\right ) \tanh (c+d x)}{4 d}\\ \end{align*}
Mathematica [A] time = 0.131312, size = 71, normalized size = 0.79 \[ \frac{\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}(\sinh (c+d x))+b (8 a+3 b) \tanh (c+d x) \text{sech}(c+d x)+2 b^2 \tanh (c+d x) \text{sech}^3(c+d x)}{8 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Sech[c + d*x]*(a + b*Sech[c + d*x]^2)^2,x]
[Out]
((8*a^2 + 8*a*b + 3*b^2)*ArcTan[Sinh[c + d*x]] + b*(8*a + 3*b)*Sech[c + d*x]*Tanh[c + d*x] + 2*b^2*Sech[c + d*
x]^3*Tanh[c + d*x])/(8*d)
________________________________________________________________________________________
Maple [A] time = 0.027, size = 106, normalized size = 1.2 \begin{align*} 2\,{\frac{{a}^{2}\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}+{\frac{ab{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{d}}+2\,{\frac{ab\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}+{\frac{{b}^{2}\tanh \left ( dx+c \right ) \left ({\rm sech} \left (dx+c\right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{b}^{2}{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{2}\arctan \left ({{\rm e}^{dx+c}} \right ) }{4\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sech(d*x+c)*(a+b*sech(d*x+c)^2)^2,x)
[Out]
2/d*a^2*arctan(exp(d*x+c))+1/d*a*b*sech(d*x+c)*tanh(d*x+c)+2/d*a*b*arctan(exp(d*x+c))+1/4/d*b^2*tanh(d*x+c)*se
ch(d*x+c)^3+3/8*b^2*sech(d*x+c)*tanh(d*x+c)/d+3/4/d*b^2*arctan(exp(d*x+c))
________________________________________________________________________________________
Maxima [B] time = 1.73507, size = 271, normalized size = 3.01 \begin{align*} -\frac{1}{4} \, b^{2}{\left (\frac{3 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac{3 \, e^{\left (-d x - c\right )} + 11 \, e^{\left (-3 \, d x - 3 \, c\right )} - 11 \, e^{\left (-5 \, d x - 5 \, c\right )} - 3 \, e^{\left (-7 \, d x - 7 \, c\right )}}{d{\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}}\right )} - 2 \, a b{\left (\frac{\arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac{e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d{\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + \frac{a^{2} \arctan \left (\sinh \left (d x + c\right )\right )}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(d*x+c)*(a+b*sech(d*x+c)^2)^2,x, algorithm="maxima")
[Out]
-1/4*b^2*(3*arctan(e^(-d*x - c))/d - (3*e^(-d*x - c) + 11*e^(-3*d*x - 3*c) - 11*e^(-5*d*x - 5*c) - 3*e^(-7*d*x
- 7*c))/(d*(4*e^(-2*d*x - 2*c) + 6*e^(-4*d*x - 4*c) + 4*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c) + 1))) - 2*a*b*(a
rctan(e^(-d*x - c))/d - (e^(-d*x - c) - e^(-3*d*x - 3*c))/(d*(2*e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) + 1))) + a
^2*arctan(sinh(d*x + c))/d
________________________________________________________________________________________
Fricas [B] time = 2.21768, size = 3445, normalized size = 38.28 \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)*(a+b*sech(d*x+c)^2)^2,x, algorithm="fricas")
[Out]
1/4*((8*a*b + 3*b^2)*cosh(d*x + c)^7 + 7*(8*a*b + 3*b^2)*cosh(d*x + c)*sinh(d*x + c)^6 + (8*a*b + 3*b^2)*sinh(
d*x + c)^7 + (8*a*b + 11*b^2)*cosh(d*x + c)^5 + (21*(8*a*b + 3*b^2)*cosh(d*x + c)^2 + 8*a*b + 11*b^2)*sinh(d*x
+ c)^5 + 5*(7*(8*a*b + 3*b^2)*cosh(d*x + c)^3 + (8*a*b + 11*b^2)*cosh(d*x + c))*sinh(d*x + c)^4 - (8*a*b + 11
*b^2)*cosh(d*x + c)^3 + (35*(8*a*b + 3*b^2)*cosh(d*x + c)^4 + 10*(8*a*b + 11*b^2)*cosh(d*x + c)^2 - 8*a*b - 11
*b^2)*sinh(d*x + c)^3 + (21*(8*a*b + 3*b^2)*cosh(d*x + c)^5 + 10*(8*a*b + 11*b^2)*cosh(d*x + c)^3 - 3*(8*a*b +
11*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 + ((8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^8 + 8*(8*a^2 + 8*a*b + 3*b^2)
*cosh(d*x + c)*sinh(d*x + c)^7 + (8*a^2 + 8*a*b + 3*b^2)*sinh(d*x + c)^8 + 4*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x
+ c)^6 + 4*(7*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^2 + 8*a^2 + 8*a*b + 3*b^2)*sinh(d*x + c)^6 + 8*(7*(8*a^2 +
8*a*b + 3*b^2)*cosh(d*x + c)^3 + 3*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c)^5 + 6*(8*a^2 + 8*a*b
+ 3*b^2)*cosh(d*x + c)^4 + 2*(35*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^4 + 30*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x
+ c)^2 + 24*a^2 + 24*a*b + 9*b^2)*sinh(d*x + c)^4 + 8*(7*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^5 + 10*(8*a^2
+ 8*a*b + 3*b^2)*cosh(d*x + c)^3 + 3*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 4*(8*a^2 + 8*a*b
+ 3*b^2)*cosh(d*x + c)^2 + 4*(7*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^6 + 15*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x
+ c)^4 + 9*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^2 + 8*a^2 + 8*a*b + 3*b^2)*sinh(d*x + c)^2 + 8*a^2 + 8*a*b +
3*b^2 + 8*((8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^7 + 3*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^5 + 3*(8*a^2 + 8
*a*b + 3*b^2)*cosh(d*x + c)^3 + (8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c))*arctan(cosh(d*x + c) + s
inh(d*x + c)) - (8*a*b + 3*b^2)*cosh(d*x + c) + (7*(8*a*b + 3*b^2)*cosh(d*x + c)^6 + 5*(8*a*b + 11*b^2)*cosh(d
*x + c)^4 - 3*(8*a*b + 11*b^2)*cosh(d*x + c)^2 - 8*a*b - 3*b^2)*sinh(d*x + c))/(d*cosh(d*x + c)^8 + 8*d*cosh(d
*x + c)*sinh(d*x + c)^7 + d*sinh(d*x + c)^8 + 4*d*cosh(d*x + c)^6 + 4*(7*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^
6 + 8*(7*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^5 + 6*d*cosh(d*x + c)^4 + 2*(35*d*cosh(d*x + c)^
4 + 30*d*cosh(d*x + c)^2 + 3*d)*sinh(d*x + c)^4 + 8*(7*d*cosh(d*x + c)^5 + 10*d*cosh(d*x + c)^3 + 3*d*cosh(d*x
+ c))*sinh(d*x + c)^3 + 4*d*cosh(d*x + c)^2 + 4*(7*d*cosh(d*x + c)^6 + 15*d*cosh(d*x + c)^4 + 9*d*cosh(d*x +
c)^2 + d)*sinh(d*x + c)^2 + 8*(d*cosh(d*x + c)^7 + 3*d*cosh(d*x + c)^5 + 3*d*cosh(d*x + c)^3 + d*cosh(d*x + c)
)*sinh(d*x + c) + d)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{sech}^{2}{\left (c + d x \right )}\right )^{2} \operatorname{sech}{\left (c + d x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(d*x+c)*(a+b*sech(d*x+c)**2)**2,x)
[Out]
Integral((a + b*sech(c + d*x)**2)**2*sech(c + d*x), x)
________________________________________________________________________________________
Giac [B] time = 1.15525, size = 232, normalized size = 2.58 \begin{align*} \frac{{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )}{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )}}{16 \, d} + \frac{8 \, a b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 3 \, b^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 32 \, a b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 20 \, b^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{4 \,{\left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}^{2} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(d*x+c)*(a+b*sech(d*x+c)^2)^2,x, algorithm="giac")
[Out]
1/16*(pi + 2*arctan(1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*(8*a^2 + 8*a*b + 3*b^2)/d + 1/4*(8*a*b*(e^(d*x +
c) - e^(-d*x - c))^3 + 3*b^2*(e^(d*x + c) - e^(-d*x - c))^3 + 32*a*b*(e^(d*x + c) - e^(-d*x - c)) + 20*b^2*(e^
(d*x + c) - e^(-d*x - c)))/(((e^(d*x + c) - e^(-d*x - c))^2 + 4)^2*d)