3.55 \(\int \text {sech}^3(c+d x) (a+b \text {sech}^2(c+d x)) \, dx\)

Optimal. Leaf size=70 \[ \frac {(4 a+3 b) \tan ^{-1}(\sinh (c+d x))}{8 d}+\frac {(4 a+3 b) \tanh (c+d x) \text {sech}(c+d x)}{8 d}+\frac {b \tanh (c+d x) \text {sech}^3(c+d x)}{4 d} \]

[Out]

1/8*(4*a+3*b)*arctan(sinh(d*x+c))/d+1/8*(4*a+3*b)*sech(d*x+c)*tanh(d*x+c)/d+1/4*b*sech(d*x+c)^3*tanh(d*x+c)/d

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Rubi [A]  time = 0.05, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4046, 3768, 3770} \[ \frac {(4 a+3 b) \tan ^{-1}(\sinh (c+d x))}{8 d}+\frac {(4 a+3 b) \tanh (c+d x) \text {sech}(c+d x)}{8 d}+\frac {b \tanh (c+d x) \text {sech}^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sech[c + d*x]^3*(a + b*Sech[c + d*x]^2),x]

[Out]

((4*a + 3*b)*ArcTan[Sinh[c + d*x]])/(8*d) + ((4*a + 3*b)*Sech[c + d*x]*Tanh[c + d*x])/(8*d) + (b*Sech[c + d*x]
^3*Tanh[c + d*x])/(4*d)

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rubi steps

\begin {align*} \int \text {sech}^3(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx &=\frac {b \text {sech}^3(c+d x) \tanh (c+d x)}{4 d}+\frac {1}{4} (4 a+3 b) \int \text {sech}^3(c+d x) \, dx\\ &=\frac {(4 a+3 b) \text {sech}(c+d x) \tanh (c+d x)}{8 d}+\frac {b \text {sech}^3(c+d x) \tanh (c+d x)}{4 d}+\frac {1}{8} (4 a+3 b) \int \text {sech}(c+d x) \, dx\\ &=\frac {(4 a+3 b) \tan ^{-1}(\sinh (c+d x))}{8 d}+\frac {(4 a+3 b) \text {sech}(c+d x) \tanh (c+d x)}{8 d}+\frac {b \text {sech}^3(c+d x) \tanh (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 60, normalized size = 0.86 \[ \frac {(4 a+3 b) \tan ^{-1}(\sinh (c+d x))+(4 a+3 b) \tanh (c+d x) \text {sech}(c+d x)+2 b \tanh (c+d x) \text {sech}^3(c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[c + d*x]^3*(a + b*Sech[c + d*x]^2),x]

[Out]

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

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fricas [B]  time = 0.42, size = 1112, normalized size = 15.89 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^3*(a+b*sech(d*x+c)^2),x, algorithm="fricas")

[Out]

1/4*((4*a + 3*b)*cosh(d*x + c)^7 + 7*(4*a + 3*b)*cosh(d*x + c)*sinh(d*x + c)^6 + (4*a + 3*b)*sinh(d*x + c)^7 +
(4*a + 11*b)*cosh(d*x + c)^5 + (21*(4*a + 3*b)*cosh(d*x + c)^2 + 4*a + 11*b)*sinh(d*x + c)^5 + 5*(7*(4*a + 3*
b)*cosh(d*x + c)^3 + (4*a + 11*b)*cosh(d*x + c))*sinh(d*x + c)^4 - (4*a + 11*b)*cosh(d*x + c)^3 + (35*(4*a + 3
*b)*cosh(d*x + c)^4 + 10*(4*a + 11*b)*cosh(d*x + c)^2 - 4*a - 11*b)*sinh(d*x + c)^3 + (21*(4*a + 3*b)*cosh(d*x
+ c)^5 + 10*(4*a + 11*b)*cosh(d*x + c)^3 - 3*(4*a + 11*b)*cosh(d*x + c))*sinh(d*x + c)^2 + ((4*a + 3*b)*cosh(
d*x + c)^8 + 8*(4*a + 3*b)*cosh(d*x + c)*sinh(d*x + c)^7 + (4*a + 3*b)*sinh(d*x + c)^8 + 4*(4*a + 3*b)*cosh(d*
x + c)^6 + 4*(7*(4*a + 3*b)*cosh(d*x + c)^2 + 4*a + 3*b)*sinh(d*x + c)^6 + 8*(7*(4*a + 3*b)*cosh(d*x + c)^3 +
3*(4*a + 3*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 6*(4*a + 3*b)*cosh(d*x + c)^4 + 2*(35*(4*a + 3*b)*cosh(d*x + c)
^4 + 30*(4*a + 3*b)*cosh(d*x + c)^2 + 12*a + 9*b)*sinh(d*x + c)^4 + 8*(7*(4*a + 3*b)*cosh(d*x + c)^5 + 10*(4*a
+ 3*b)*cosh(d*x + c)^3 + 3*(4*a + 3*b)*cosh(d*x + c))*sinh(d*x + c)^3 + 4*(4*a + 3*b)*cosh(d*x + c)^2 + 4*(7*
(4*a + 3*b)*cosh(d*x + c)^6 + 15*(4*a + 3*b)*cosh(d*x + c)^4 + 9*(4*a + 3*b)*cosh(d*x + c)^2 + 4*a + 3*b)*sinh
(d*x + c)^2 + 8*((4*a + 3*b)*cosh(d*x + c)^7 + 3*(4*a + 3*b)*cosh(d*x + c)^5 + 3*(4*a + 3*b)*cosh(d*x + c)^3 +
(4*a + 3*b)*cosh(d*x + c))*sinh(d*x + c) + 4*a + 3*b)*arctan(cosh(d*x + c) + sinh(d*x + c)) - (4*a + 3*b)*cos
h(d*x + c) + (7*(4*a + 3*b)*cosh(d*x + c)^6 + 5*(4*a + 11*b)*cosh(d*x + c)^4 - 3*(4*a + 11*b)*cosh(d*x + c)^2
- 4*a - 3*b)*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*c
osh(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))*sin
h(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 = 0.16, size = 156, normalized size = 2.23 \[ \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} {\left (4 \, a + 3 \, b\right )} + \frac {4 \, {\left (4 \, a {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 3 \, b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 16 \, a {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 20 \, b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}\right )}}{{\left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^3*(a+b*sech(d*x+c)^2),x, algorithm="giac")

[Out]

1/16*((pi + 2*arctan(1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*(4*a + 3*b) + 4*(4*a*(e^(d*x + c) - e^(-d*x - c)
)^3 + 3*b*(e^(d*x + c) - e^(-d*x - c))^3 + 16*a*(e^(d*x + c) - e^(-d*x - c)) + 20*b*(e^(d*x + c) - e^(-d*x - c
)))/((e^(d*x + c) - e^(-d*x - c))^2 + 4)^2)/d

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maple [A]  time = 0.42, size = 83, normalized size = 1.19 \[ \frac {a \,\mathrm {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2 d}+\frac {a \arctan \left ({\mathrm e}^{d x +c}\right )}{d}+\frac {b \mathrm {sech}\left (d x +c \right )^{3} \tanh \left (d x +c \right )}{4 d}+\frac {3 b \,\mathrm {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{8 d}+\frac {3 b \arctan \left ({\mathrm e}^{d x +c}\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)^3*(a+b*sech(d*x+c)^2),x)

[Out]

1/2/d*a*sech(d*x+c)*tanh(d*x+c)+1/d*a*arctan(exp(d*x+c))+1/4*b*sech(d*x+c)^3*tanh(d*x+c)/d+3/8*b*sech(d*x+c)*t
anh(d*x+c)/d+3/4/d*b*arctan(exp(d*x+c))

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maxima [B]  time = 0.41, size = 184, normalized size = 2.63 \[ -\frac {1}{4} \, b {\left (\frac {3 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {3 \, e^{\left (-d x - c\right )} + 11 \, e^{\left (-3 \, d x - 3 \, c\right )} - 11 \, e^{\left (-5 \, d x - 5 \, c\right )} - 3 \, e^{\left (-7 \, d x - 7 \, c\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}}\right )} - a {\left (\frac {\arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^3*(a+b*sech(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/4*b*(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))) - a*(arctan(
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)))

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mupad [B]  time = 1.38, size = 283, normalized size = 4.04 \[ \frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (4\,a\,\sqrt {d^2}+3\,b\,\sqrt {d^2}\right )}{d\,\sqrt {16\,a^2+24\,a\,b+9\,b^2}}\right )\,\sqrt {16\,a^2+24\,a\,b+9\,b^2}}{4\,\sqrt {d^2}}-\frac {\frac {a\,{\mathrm {e}}^{5\,c+5\,d\,x}}{d}+\frac {2\,{\mathrm {e}}^{3\,c+3\,d\,x}\,\left (a+2\,b\right )}{d}+\frac {a\,{\mathrm {e}}^{c+d\,x}}{d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {{\mathrm {e}}^{c+d\,x}\,\left (2\,a-b\right )}{2\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}-\frac {2\,b\,{\mathrm {e}}^{c+d\,x}}{d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}+\frac {{\mathrm {e}}^{c+d\,x}\,\left (4\,a+3\,b\right )}{4\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((exp(d*x)*exp(c)*(4*a*(d^2)^(1/2) + 3*b*(d^2)^(1/2)))/(d*(24*a*b + 16*a^2 + 9*b^2)^(1/2)))*(24*a*b + 16*
a^2 + 9*b^2)^(1/2))/(4*(d^2)^(1/2)) - ((a*exp(5*c + 5*d*x))/d + (2*exp(3*c + 3*d*x)*(a + 2*b))/d + (a*exp(c +
d*x))/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) - (exp(c + d*x)
*(2*a - b))/(2*d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1)) - (2*b*exp(c + d*x))/(d*(3*exp(2*c + 2*d*x) + 3*
exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)) + (exp(c + d*x)*(4*a + 3*b))/(4*d*(exp(2*c + 2*d*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right ) \operatorname {sech}^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)**3*(a+b*sech(d*x+c)**2),x)

[Out]

Integral((a + b*sech(c + d*x)**2)*sech(c + d*x)**3, x)

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