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*ln(cosh(d*x+c))/d+1/2*(a-b)*sech(d*x+c)^2/d+1/4*b*sech(d*x+c)^4/d

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Rubi [A]  time = 0.06, antiderivative size = 49, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [B]  time = 0.43, size = 1072, normalized size = 21.88 \[ \text {result too large to display} \]

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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giac [B]  time = 0.18, size = 116, normalized size = 2.37 \[ -\frac {12 \, a d x - 12 \, a \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) + \frac {25 \, a e^{\left (8 \, d x + 8 \, c\right )} + 76 \, a e^{\left (6 \, d x + 6 \, c\right )} + 24 \, b e^{\left (6 \, d x + 6 \, c\right )} + 102 \, a e^{\left (4 \, d x + 4 \, c\right )} + 76 \, a e^{\left (2 \, d x + 2 \, c\right )} + 24 \, b e^{\left (2 \, d x + 2 \, c\right )} + 25 \, a}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{4}}}{12 \, d} \]

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

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maple [A]  time = 0.20, size = 64, normalized size = 1.31 \[ \frac {a \ln \left (\cosh \left (d x +c \right )\right )}{d}-\frac {\left (\tanh ^{2}\left (d x +c \right )\right ) a}{2 d}-\frac {b \left (\sinh ^{2}\left (d x +c \right )\right )}{2 d \cosh \left (d x +c \right )^{4}}-\frac {b}{4 d \cosh \left (d x +c \right )^{4}} \]

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*tanh(d*x+c)^2*a-1/2/d*b*sinh(d*x+c)^2/cosh(d*x+c)^4-1/4/d*b/cosh(d*x+c)^4

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maxima [A]  time = 0.75, size = 78, normalized size = 1.59 \[ \frac {b \tanh \left (d x + c\right )^{4}}{4 \, d} + a {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} \]

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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mupad [B]  time = 0.12, size = 173, normalized size = 3.53 \[ \frac {2\,\left (a-b\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-a\,x-\frac {2\,\left (a-3\,b\right )}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}-\frac {8\,b}{d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}+\frac {4\,b}{d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}+\frac {a\,\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(c + d*x)^3*(a + b/cosh(c + d*x)^2),x)

[Out]

(2*(a - b))/(d*(exp(2*c + 2*d*x) + 1)) - a*x - (2*(a - 3*b))/(d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1)) -
(8*b)/(d*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)) + (4*b)/(d*(4*exp(2*c + 2*d*x) + 6
*exp(4*c + 4*d*x) + 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1)) + (a*log(exp(2*c)*exp(2*d*x) + 1))/d

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sympy [A]  time = 1.66, size = 80, normalized size = 1.63 \[ \begin {cases} a x - \frac {a \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {a \tanh ^{2}{\left (c + d x \right )}}{2 d} - \frac {b \tanh ^{2}{\left (c + d x \right )} \operatorname {sech}^{2}{\left (c + d x \right )}}{4 d} - \frac {b \operatorname {sech}^{2}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (a + b \operatorname {sech}^{2}{\relax (c )}\right ) \tanh ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

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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