3.115 \(\int (a+b \text{sech}^2(c+d x))^2 \tanh (c+d x) \, dx\)
Optimal. Leaf size=48 \[ \frac{a^2 \log (\cosh (c+d x))}{d}-\frac{a b \text{sech}^2(c+d x)}{d}-\frac{b^2 \text{sech}^4(c+d x)}{4 d} \]
[Out]
(a^2*Log[Cosh[c + d*x]])/d - (a*b*Sech[c + d*x]^2)/d - (b^2*Sech[c + d*x]^4)/(4*d)
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Rubi [A] time = 0.0500898, antiderivative size = 48, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used =
{4138, 266, 43} \[ \frac{a^2 \log (\cosh (c+d x))}{d}-\frac{a b \text{sech}^2(c+d x)}{d}-\frac{b^2 \text{sech}^4(c+d x)}{4 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Sech[c + d*x]^2)^2*Tanh[c + d*x],x]
[Out]
(a^2*Log[Cosh[c + d*x]])/d - (a*b*Sech[c + d*x]^2)/d - (b^2*Sech[c + d*x]^4)/(4*d)
Rule 4138
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Module[{ff =
FreeFactors[Cos[e + f*x], x]}, -Dist[(f*ff^(m + n*p - 1))^(-1), Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*
(ff*x)^n)^p)/x^(m + n*p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n] && IntegerQ[p]
Rule 266
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Rule 43
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Rubi steps
\begin{align*} \int \left (a+b \text{sech}^2(c+d x)\right )^2 \tanh (c+d x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (b+a x^2\right )^2}{x^5} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(b+a x)^2}{x^3} \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{b^2}{x^3}+\frac{2 a b}{x^2}+\frac{a^2}{x}\right ) \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=\frac{a^2 \log (\cosh (c+d x))}{d}-\frac{a b \text{sech}^2(c+d x)}{d}-\frac{b^2 \text{sech}^4(c+d x)}{4 d}\\ \end{align*}
Mathematica [A] time = 0.131715, size = 81, normalized size = 1.69 \[ \frac{\text{sech}^4(c+d x) \left (a \cosh ^2(c+d x)+b\right )^2 \left (4 a^2 \cosh ^4(c+d x) \log (\cosh (c+d x))-4 a b \cosh ^2(c+d x)-b^2\right )}{d (a \cosh (2 (c+d x))+a+2 b)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Sech[c + d*x]^2)^2*Tanh[c + d*x],x]
[Out]
((b + a*Cosh[c + d*x]^2)^2*(-b^2 - 4*a*b*Cosh[c + d*x]^2 + 4*a^2*Cosh[c + d*x]^4*Log[Cosh[c + d*x]])*Sech[c +
d*x]^4)/(d*(a + 2*b + a*Cosh[2*(c + d*x)])^2)
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Maple [A] time = 0.011, size = 48, normalized size = 1. \begin{align*} -{\frac{{b}^{2} \left ({\rm sech} \left (dx+c\right ) \right ) ^{4}}{4\,d}}-{\frac{ab \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{d}}-{\frac{{a}^{2}\ln \left ({\rm sech} \left (dx+c\right ) \right ) }{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*sech(d*x+c)^2)^2*tanh(d*x+c),x)
[Out]
-1/4*b^2*sech(d*x+c)^4/d-a*b*sech(d*x+c)^2/d-1/d*a^2*ln(sech(d*x+c))
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Maxima [A] time = 1.11594, size = 74, normalized size = 1.54 \begin{align*} \frac{a b \tanh \left (d x + c\right )^{2}}{d} + \frac{a^{2} \log \left (\cosh \left (d x + c\right )\right )}{d} - \frac{4 \, b^{2}}{d{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sech(d*x+c)^2)^2*tanh(d*x+c),x, algorithm="maxima")
[Out]
a*b*tanh(d*x + c)^2/d + a^2*log(cosh(d*x + c))/d - 4*b^2/(d*(e^(d*x + c) + e^(-d*x - c))^4)
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Fricas [B] time = 2.18418, size = 3012, normalized size = 62.75 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sech(d*x+c)^2)^2*tanh(d*x+c),x, algorithm="fricas")
[Out]
-(a^2*d*x*cosh(d*x + c)^8 + 8*a^2*d*x*cosh(d*x + c)*sinh(d*x + c)^7 + a^2*d*x*sinh(d*x + c)^8 + 4*(a^2*d*x + a
*b)*cosh(d*x + c)^6 + 4*(7*a^2*d*x*cosh(d*x + c)^2 + a^2*d*x + a*b)*sinh(d*x + c)^6 + 8*(7*a^2*d*x*cosh(d*x +
c)^3 + 3*(a^2*d*x + a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(3*a^2*d*x + 4*a*b + 2*b^2)*cosh(d*x + c)^4 + 2*(3
5*a^2*d*x*cosh(d*x + c)^4 + 3*a^2*d*x + 30*(a^2*d*x + a*b)*cosh(d*x + c)^2 + 4*a*b + 2*b^2)*sinh(d*x + c)^4 +
a^2*d*x + 8*(7*a^2*d*x*cosh(d*x + c)^5 + 10*(a^2*d*x + a*b)*cosh(d*x + c)^3 + (3*a^2*d*x + 4*a*b + 2*b^2)*cosh
(d*x + c))*sinh(d*x + c)^3 + 4*(a^2*d*x + a*b)*cosh(d*x + c)^2 + 4*(7*a^2*d*x*cosh(d*x + c)^6 + 15*(a^2*d*x +
a*b)*cosh(d*x + c)^4 + a^2*d*x + 3*(3*a^2*d*x + 4*a*b + 2*b^2)*cosh(d*x + c)^2 + a*b)*sinh(d*x + c)^2 - (a^2*c
osh(d*x + c)^8 + 8*a^2*cosh(d*x + c)*sinh(d*x + c)^7 + a^2*sinh(d*x + c)^8 + 4*a^2*cosh(d*x + c)^6 + 4*(7*a^2*
cosh(d*x + c)^2 + a^2)*sinh(d*x + c)^6 + 6*a^2*cosh(d*x + c)^4 + 8*(7*a^2*cosh(d*x + c)^3 + 3*a^2*cosh(d*x + c
))*sinh(d*x + c)^5 + 2*(35*a^2*cosh(d*x + c)^4 + 30*a^2*cosh(d*x + c)^2 + 3*a^2)*sinh(d*x + c)^4 + 4*a^2*cosh(
d*x + c)^2 + 8*(7*a^2*cosh(d*x + c)^5 + 10*a^2*cosh(d*x + c)^3 + 3*a^2*cosh(d*x + c))*sinh(d*x + c)^3 + 4*(7*a
^2*cosh(d*x + c)^6 + 15*a^2*cosh(d*x + c)^4 + 9*a^2*cosh(d*x + c)^2 + a^2)*sinh(d*x + c)^2 + a^2 + 8*(a^2*cosh
(d*x + c)^7 + 3*a^2*cosh(d*x + c)^5 + 3*a^2*cosh(d*x + c)^3 + a^2*cosh(d*x + c))*sinh(d*x + c))*log(2*cosh(d*x
+ c)/(cosh(d*x + c) - sinh(d*x + c))) + 8*(a^2*d*x*cosh(d*x + c)^7 + 3*(a^2*d*x + a*b)*cosh(d*x + c)^5 + (3*a
^2*d*x + 4*a*b + 2*b^2)*cosh(d*x + c)^3 + (a^2*d*x + a*b)*cosh(d*x + c))*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*co
sh(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)
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Sympy [A] time = 2.91853, size = 63, normalized size = 1.31 \begin{align*} \begin{cases} a^{2} x - \frac{a^{2} \log{\left (\tanh{\left (c + d x \right )} + 1 \right )}}{d} - \frac{a b \operatorname{sech}^{2}{\left (c + d x \right )}}{d} - \frac{b^{2} \operatorname{sech}^{4}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\x \left (a + b \operatorname{sech}^{2}{\left (c \right )}\right )^{2} \tanh{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sech(d*x+c)**2)**2*tanh(d*x+c),x)
[Out]
Piecewise((a**2*x - a**2*log(tanh(c + d*x) + 1)/d - a*b*sech(c + d*x)**2/d - b**2*sech(c + d*x)**4/(4*d), Ne(d
, 0)), (x*(a + b*sech(c)**2)**2*tanh(c), True))
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Giac [B] time = 1.16215, size = 215, normalized size = 4.48 \begin{align*} -\frac{12 \, a^{2} d x - 12 \, a^{2} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) + \frac{25 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 100 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 48 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 150 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 96 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 48 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 100 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 48 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 25 \, a^{2}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{4}}}{12 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sech(d*x+c)^2)^2*tanh(d*x+c),x, algorithm="giac")
[Out]
-1/12*(12*a^2*d*x - 12*a^2*log(e^(2*d*x + 2*c) + 1) + (25*a^2*e^(8*d*x + 8*c) + 100*a^2*e^(6*d*x + 6*c) + 48*a
*b*e^(6*d*x + 6*c) + 150*a^2*e^(4*d*x + 4*c) + 96*a*b*e^(4*d*x + 4*c) + 48*b^2*e^(4*d*x + 4*c) + 100*a^2*e^(2*
d*x + 2*c) + 48*a*b*e^(2*d*x + 2*c) + 25*a^2)/(e^(2*d*x + 2*c) + 1)^4)/d