3.103 \(\int (a+b \text{sech}^2(c+d x)) \tanh ^3(c+d x) \, dx\)
Optimal. Leaf size=49 \[ \frac{(a-b) \text{sech}^2(c+d x)}{2 d}+\frac{a \log (\cosh (c+d x))}{d}+\frac{b \text{sech}^4(c+d x)}{4 d} \]
[Out]
(a*Log[Cosh[c + d*x]])/d + ((a - b)*Sech[c + d*x]^2)/(2*d) + (b*Sech[c + d*x]^4)/(4*d)
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Rubi [A] time = 0.0564093, antiderivative size = 49, 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, 446, 76} \[ \frac{(a-b) \text{sech}^2(c+d x)}{2 d}+\frac{a \log (\cosh (c+d x))}{d}+\frac{b \text{sech}^4(c+d x)}{4 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Sech[c + d*x]^2)*Tanh[c + d*x]^3,x]
[Out]
(a*Log[Cosh[c + d*x]])/d + ((a - b)*Sech[c + d*x]^2)/(2*d) + (b*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 446
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Rule 76
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
Rubi steps
\begin{align*} \int \left (a+b \text{sech}^2(c+d x)\right ) \tanh ^3(c+d x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right ) \left (b+a x^2\right )}{x^5} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{(1-x) (b+a x)}{x^3} \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{b}{x^3}+\frac{a-b}{x^2}-\frac{a}{x}\right ) \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=\frac{a \log (\cosh (c+d x))}{d}+\frac{(a-b) \text{sech}^2(c+d x)}{2 d}+\frac{b \text{sech}^4(c+d x)}{4 d}\\ \end{align*}
Mathematica [A] time = 0.0225026, size = 45, normalized size = 0.92 \[ -\frac{a \tanh ^2(c+d x)}{2 d}+\frac{a \log (\cosh (c+d x))}{d}+\frac{b \tanh ^4(c+d x)}{4 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Sech[c + d*x]^2)*Tanh[c + d*x]^3,x]
[Out]
(a*Log[Cosh[c + d*x]])/d - (a*Tanh[c + d*x]^2)/(2*d) + (b*Tanh[c + d*x]^4)/(4*d)
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Maple [A] time = 0.035, size = 72, normalized size = 1.5 \begin{align*}{\frac{a\ln \left ( \cosh \left ( dx+c \right ) \right ) }{d}}-{\frac{ \left ( \tanh \left ( dx+c \right ) \right ) ^{2}a}{2\,d}}-{\frac{b \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{4\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{4}}}+{\frac{b \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{4\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*sech(d*x+c)^2)*tanh(d*x+c)^3,x)
[Out]
a*ln(cosh(d*x+c))/d-1/2/d*a*tanh(d*x+c)^2-1/4/d*b*sinh(d*x+c)^2/cosh(d*x+c)^4+1/4/d*b*sinh(d*x+c)^2/cosh(d*x+c
)^2
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Maxima [A] time = 1.70759, size = 105, normalized size = 2.14 \begin{align*} \frac{b \tanh \left (d x + c\right )^{4}}{4 \, d} + a{\left (x + \frac{c}{d} + \frac{\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac{2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d{\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sech(d*x+c)^2)*tanh(d*x+c)^3,x, algorithm="maxima")
[Out]
1/4*b*tanh(d*x + c)^4/d + a*(x + c/d + log(e^(-2*d*x - 2*c) + 1)/d + 2*e^(-2*d*x - 2*c)/(d*(2*e^(-2*d*x - 2*c)
+ e^(-4*d*x - 4*c) + 1)))
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Fricas [B] time = 2.39513, size = 2871, normalized size = 58.59 \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)*tanh(d*x+c)^3,x, algorithm="fricas")
[Out]
-(a*d*x*cosh(d*x + c)^8 + 8*a*d*x*cosh(d*x + c)*sinh(d*x + c)^7 + a*d*x*sinh(d*x + c)^8 + 2*(2*a*d*x - a + b)*
cosh(d*x + c)^6 + 2*(14*a*d*x*cosh(d*x + c)^2 + 2*a*d*x - a + b)*sinh(d*x + c)^6 + 4*(14*a*d*x*cosh(d*x + c)^3
+ 3*(2*a*d*x - a + b)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(3*a*d*x - 2*a)*cosh(d*x + c)^4 + 2*(35*a*d*x*cosh(d
*x + c)^4 + 3*a*d*x + 15*(2*a*d*x - a + b)*cosh(d*x + c)^2 - 2*a)*sinh(d*x + c)^4 + 8*(7*a*d*x*cosh(d*x + c)^5
+ 5*(2*a*d*x - a + b)*cosh(d*x + c)^3 + (3*a*d*x - 2*a)*cosh(d*x + c))*sinh(d*x + c)^3 + a*d*x + 2*(2*a*d*x -
a + b)*cosh(d*x + c)^2 + 2*(14*a*d*x*cosh(d*x + c)^6 + 15*(2*a*d*x - a + b)*cosh(d*x + c)^4 + 2*a*d*x + 6*(3*
a*d*x - 2*a)*cosh(d*x + c)^2 - a + b)*sinh(d*x + c)^2 - (a*cosh(d*x + c)^8 + 8*a*cosh(d*x + c)*sinh(d*x + c)^7
+ a*sinh(d*x + c)^8 + 4*a*cosh(d*x + c)^6 + 4*(7*a*cosh(d*x + c)^2 + a)*sinh(d*x + c)^6 + 8*(7*a*cosh(d*x + c
)^3 + 3*a*cosh(d*x + c))*sinh(d*x + c)^5 + 6*a*cosh(d*x + c)^4 + 2*(35*a*cosh(d*x + c)^4 + 30*a*cosh(d*x + c)^
2 + 3*a)*sinh(d*x + c)^4 + 8*(7*a*cosh(d*x + c)^5 + 10*a*cosh(d*x + c)^3 + 3*a*cosh(d*x + c))*sinh(d*x + c)^3
+ 4*a*cosh(d*x + c)^2 + 4*(7*a*cosh(d*x + c)^6 + 15*a*cosh(d*x + c)^4 + 9*a*cosh(d*x + c)^2 + a)*sinh(d*x + c)
^2 + 8*(a*cosh(d*x + c)^7 + 3*a*cosh(d*x + c)^5 + 3*a*cosh(d*x + c)^3 + a*cosh(d*x + c))*sinh(d*x + c) + a)*lo
g(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 4*(2*a*d*x*cosh(d*x + c)^7 + 3*(2*a*d*x - a + b)*cosh(d*x
+ c)^5 + 2*(3*a*d*x - 2*a)*cosh(d*x + c)^3 + (2*a*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*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)
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Sympy [A] time = 2.73193, size = 80, normalized size = 1.63 \begin{align*} \begin{cases} a x - \frac{a \log{\left (\tanh{\left (c + d x \right )} + 1 \right )}}{d} - \frac{a \tanh ^{2}{\left (c + d x \right )}}{2 d} - \frac{b \tanh ^{2}{\left (c + d x \right )} \operatorname{sech}^{2}{\left (c + d x \right )}}{4 d} - \frac{b \operatorname{sech}^{2}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\x \left (a + b \operatorname{sech}^{2}{\left (c \right )}\right ) \tanh ^{3}{\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)*tanh(d*x+c)**3,x)
[Out]
Piecewise((a*x - a*log(tanh(c + d*x) + 1)/d - a*tanh(c + d*x)**2/(2*d) - b*tanh(c + d*x)**2*sech(c + d*x)**2/(
4*d) - b*sech(c + d*x)**2/(4*d), Ne(d, 0)), (x*(a + b*sech(c)**2)*tanh(c)**3, True))
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Giac [B] time = 1.25088, size = 157, normalized size = 3.2 \begin{align*} -\frac{12 \, a d x - 12 \, a \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) + \frac{25 \, a e^{\left (8 \, d x + 8 \, c\right )} + 76 \, a e^{\left (6 \, d x + 6 \, c\right )} + 24 \, b e^{\left (6 \, d x + 6 \, c\right )} + 102 \, a e^{\left (4 \, d x + 4 \, c\right )} + 76 \, a e^{\left (2 \, d x + 2 \, c\right )} + 24 \, b e^{\left (2 \, d x + 2 \, c\right )} + 25 \, a}{{\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)*tanh(d*x+c)^3,x, algorithm="giac")
[Out]
-1/12*(12*a*d*x - 12*a*log(e^(2*d*x + 2*c) + 1) + (25*a*e^(8*d*x + 8*c) + 76*a*e^(6*d*x + 6*c) + 24*b*e^(6*d*x
+ 6*c) + 102*a*e^(4*d*x + 4*c) + 76*a*e^(2*d*x + 2*c) + 24*b*e^(2*d*x + 2*c) + 25*a)/(e^(2*d*x + 2*c) + 1)^4)
/d