\(\int (a+b \text {sech}(c+d x))^3 \, dx\) [88]

Optimal result

Integrand size = 12, antiderivative size = 73 \[ \int (a+b \text {sech}(c+d x))^3 \, dx=a^3 x+\frac {b \left (6 a^2+b^2\right ) \arctan (\sinh (c+d x))}{2 d}+\frac {5 a b^2 \tanh (c+d x)}{2 d}+\frac {b^2 (a+b \text {sech}(c+d x)) \tanh (c+d x)}{2 d} \]

[Out]

a^3*x+1/2*b*(6*a^2+b^2)*arctan(sinh(d*x+c))/d+5/2*a*b^2*tanh(d*x+c)/d+1/2*b^2*(a+b*sech(d*x+c))*tanh(d*x+c)/d

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3867, 3855, 3852, 8} \[ \int (a+b \text {sech}(c+d x))^3 \, dx=a^3 x+\frac {b \left (6 a^2+b^2\right ) \arctan (\sinh (c+d x))}{2 d}+\frac {5 a b^2 \tanh (c+d x)}{2 d}+\frac {b^2 \tanh (c+d x) (a+b \text {sech}(c+d x))}{2 d} \]

[In]

Int[(a + b*Sech[c + d*x])^3,x]

[Out]

a^3*x + (b*(6*a^2 + b^2)*ArcTan[Sinh[c + d*x]])/(2*d) + (5*a*b^2*Tanh[c + d*x])/(2*d) + (b^2*(a + b*Sech[c + d
*x])*Tanh[c + d*x])/(2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

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

Rule 3867

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-b^2)*Cot[c + d*x]*((a + b*Csc[c + d*x])^(
n - 2)/(d*(n - 1))), x] + Dist[1/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 3)*Simp[a^3*(n - 1) + (b*(b^2*(n - 2)
+ 3*a^2*(n - 1)))*Csc[c + d*x] + (a*b^2*(3*n - 4))*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ
[a^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = \frac {b^2 (a+b \text {sech}(c+d x)) \tanh (c+d x)}{2 d}+\frac {1}{2} \int \left (2 a^3+b \left (6 a^2+b^2\right ) \text {sech}(c+d x)+5 a b^2 \text {sech}^2(c+d x)\right ) \, dx \\ & = a^3 x+\frac {b^2 (a+b \text {sech}(c+d x)) \tanh (c+d x)}{2 d}+\frac {1}{2} \left (5 a b^2\right ) \int \text {sech}^2(c+d x) \, dx+\frac {1}{2} \left (b \left (6 a^2+b^2\right )\right ) \int \text {sech}(c+d x) \, dx \\ & = a^3 x+\frac {b \left (6 a^2+b^2\right ) \arctan (\sinh (c+d x))}{2 d}+\frac {b^2 (a+b \text {sech}(c+d x)) \tanh (c+d x)}{2 d}+\frac {\left (5 i a b^2\right ) \text {Subst}(\int 1 \, dx,x,-i \tanh (c+d x))}{2 d} \\ & = a^3 x+\frac {b \left (6 a^2+b^2\right ) \arctan (\sinh (c+d x))}{2 d}+\frac {5 a b^2 \tanh (c+d x)}{2 d}+\frac {b^2 (a+b \text {sech}(c+d x)) \tanh (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.75 \[ \int (a+b \text {sech}(c+d x))^3 \, dx=\frac {2 a^3 d x+b \left (6 a^2+b^2\right ) \arctan (\sinh (c+d x))+b^2 (6 a+b \text {sech}(c+d x)) \tanh (c+d x)}{2 d} \]

[In]

Integrate[(a + b*Sech[c + d*x])^3,x]

[Out]

(2*a^3*d*x + b*(6*a^2 + b^2)*ArcTan[Sinh[c + d*x]] + b^2*(6*a + b*Sech[c + d*x])*Tanh[c + d*x])/(2*d)

Maple [A] (verified)

Time = 0.91 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {a^{3} \left (d x +c \right )+6 a^{2} b \arctan \left ({\mathrm e}^{d x +c}\right )+3 a \,b^{2} \tanh \left (d x +c \right )+b^{3} \left (\frac {\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+\arctan \left ({\mathrm e}^{d x +c}\right )\right )}{d}\)	\(66\)
default	\(\frac {a^{3} \left (d x +c \right )+6 a^{2} b \arctan \left ({\mathrm e}^{d x +c}\right )+3 a \,b^{2} \tanh \left (d x +c \right )+b^{3} \left (\frac {\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+\arctan \left ({\mathrm e}^{d x +c}\right )\right )}{d}\)	\(66\)
parts	\(a^{3} x +\frac {b^{3} \left (\frac {\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+\arctan \left ({\mathrm e}^{d x +c}\right )\right )}{d}+\frac {3 a \,b^{2} \tanh \left (d x +c \right )}{d}+\frac {3 a^{2} b \arctan \left (\sinh \left (d x +c \right )\right )}{d}\)	\(67\)
parallelrisch	\(\frac {-3 i \left (1+\cosh \left (2 d x +2 c \right )\right ) b \left (a^{2}+\frac {b^{2}}{6}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )+3 i \left (1+\cosh \left (2 d x +2 c \right )\right ) b \left (a^{2}+\frac {b^{2}}{6}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )+a^{3} d x \cosh \left (2 d x +2 c \right )+a^{3} d x +3 a \,b^{2} \sinh \left (2 d x +2 c \right )+b^{3} \sinh \left (d x +c \right )}{d \left (1+\cosh \left (2 d x +2 c \right )\right )}\)	\(139\)
risch	\(a^{3} x -\frac {b^{2} \left (-{\mathrm e}^{3 d x +3 c} b +6 \,{\mathrm e}^{2 d x +2 c} a +{\mathrm e}^{d x +c} b +6 a \right )}{d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}+\frac {3 i b \ln \left ({\mathrm e}^{d x +c}+i\right ) a^{2}}{d}+\frac {i b^{3} \ln \left ({\mathrm e}^{d x +c}+i\right )}{2 d}-\frac {3 i b \ln \left ({\mathrm e}^{d x +c}-i\right ) a^{2}}{d}-\frac {i b^{3} \ln \left ({\mathrm e}^{d x +c}-i\right )}{2 d}\)	\(142\)

[In]

int((a+b*sech(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^3*(d*x+c)+6*a^2*b*arctan(exp(d*x+c))+3*a*b^2*tanh(d*x+c)+b^3*(1/2*sech(d*x+c)*tanh(d*x+c)+arctan(exp(d*
x+c))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 521 vs. \(2 (67) = 134\).

Time = 0.27 (sec) , antiderivative size = 521, normalized size of antiderivative = 7.14 \[ \int (a+b \text {sech}(c+d x))^3 \, dx=\frac {a^{3} d x \cosh \left (d x + c\right )^{4} + a^{3} d x \sinh \left (d x + c\right )^{4} + b^{3} \cosh \left (d x + c\right )^{3} + a^{3} d x - b^{3} \cosh \left (d x + c\right ) + {\left (4 \, a^{3} d x \cosh \left (d x + c\right ) + b^{3}\right )} \sinh \left (d x + c\right )^{3} - 6 \, a b^{2} + 2 \, {\left (a^{3} d x - 3 \, a b^{2}\right )} \cosh \left (d x + c\right )^{2} + {\left (6 \, a^{3} d x \cosh \left (d x + c\right )^{2} + 2 \, a^{3} d x + 3 \, b^{3} \cosh \left (d x + c\right ) - 6 \, a b^{2}\right )} \sinh \left (d x + c\right )^{2} + {\left ({\left (6 \, a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{4} + 4 \, {\left (6 \, a^{2} b + b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (6 \, a^{2} b + b^{3}\right )} \sinh \left (d x + c\right )^{4} + 6 \, a^{2} b + b^{3} + 2 \, {\left (6 \, a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (6 \, a^{2} b + b^{3} + 3 \, {\left (6 \, a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left ({\left (6 \, a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{3} + {\left (6 \, a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \arctan \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + {\left (4 \, a^{3} d x \cosh \left (d x + c\right )^{3} + 3 \, b^{3} \cosh \left (d x + c\right )^{2} - b^{3} + 4 \, {\left (a^{3} d x - 3 \, a b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{d \cosh \left (d x + c\right )^{4} + 4 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + d \sinh \left (d x + c\right )^{4} + 2 \, d \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, d \cosh \left (d x + c\right )^{2} + d\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (d \cosh \left (d x + c\right )^{3} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + d} \]

[In]

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

[Out]

(a^3*d*x*cosh(d*x + c)^4 + a^3*d*x*sinh(d*x + c)^4 + b^3*cosh(d*x + c)^3 + a^3*d*x - b^3*cosh(d*x + c) + (4*a^
3*d*x*cosh(d*x + c) + b^3)*sinh(d*x + c)^3 - 6*a*b^2 + 2*(a^3*d*x - 3*a*b^2)*cosh(d*x + c)^2 + (6*a^3*d*x*cosh
(d*x + c)^2 + 2*a^3*d*x + 3*b^3*cosh(d*x + c) - 6*a*b^2)*sinh(d*x + c)^2 + ((6*a^2*b + b^3)*cosh(d*x + c)^4 +
4*(6*a^2*b + b^3)*cosh(d*x + c)*sinh(d*x + c)^3 + (6*a^2*b + b^3)*sinh(d*x + c)^4 + 6*a^2*b + b^3 + 2*(6*a^2*b
+ b^3)*cosh(d*x + c)^2 + 2*(6*a^2*b + b^3 + 3*(6*a^2*b + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((6*a^2*b
+ b^3)*cosh(d*x + c)^3 + (6*a^2*b + b^3)*cosh(d*x + c))*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) +
(4*a^3*d*x*cosh(d*x + c)^3 + 3*b^3*cosh(d*x + c)^2 - b^3 + 4*(a^3*d*x - 3*a*b^2)*cosh(d*x + c))*sinh(d*x + c)
)/(d*cosh(d*x + c)^4 + 4*d*cosh(d*x + c)*sinh(d*x + c)^3 + d*sinh(d*x + c)^4 + 2*d*cosh(d*x + c)^2 + 2*(3*d*co
sh(d*x + c)^2 + d)*sinh(d*x + c)^2 + 4*(d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c) + d)

Sympy [F]

\[ \int (a+b \text {sech}(c+d x))^3 \, dx=\int \left (a + b \operatorname {sech}{\left (c + d x \right )}\right )^{3}\, dx \]

[In]

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

[Out]

Integral((a + b*sech(c + d*x))**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.56 \[ \int (a+b \text {sech}(c+d x))^3 \, dx=a^{3} x - b^{3} {\left (\frac {\arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + \frac {3 \, a^{2} b \arctan \left (\sinh \left (d x + c\right )\right )}{d} + \frac {6 \, a b^{2}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \]

[In]

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

[Out]

a^3*x - b^3*(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))) + 3*a^2*b*arctan(sinh(d*x + c))/d + 6*a*b^2/(d*(e^(-2*d*x - 2*c) + 1))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.26 \[ \int (a+b \text {sech}(c+d x))^3 \, dx=\frac {{\left (d x + c\right )} a^{3} + {\left (6 \, a^{2} b + b^{3}\right )} \arctan \left (e^{\left (d x + c\right )}\right ) + \frac {b^{3} e^{\left (3 \, d x + 3 \, c\right )} - 6 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - b^{3} e^{\left (d x + c\right )} - 6 \, a b^{2}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}}}{d} \]

[In]

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

[Out]

((d*x + c)*a^3 + (6*a^2*b + b^3)*arctan(e^(d*x + c)) + (b^3*e^(3*d*x + 3*c) - 6*a*b^2*e^(2*d*x + 2*c) - b^3*e^
(d*x + c) - 6*a*b^2)/(e^(2*d*x + 2*c) + 1)^2)/d

Mupad [B] (verification not implemented)

Time = 2.06 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.26 \[ \int (a+b \text {sech}(c+d x))^3 \, dx=a^3\,x-\frac {\frac {6\,a\,b^2}{d}-\frac {b^3\,{\mathrm {e}}^{c+d\,x}}{d}}{{\mathrm {e}}^{2\,c+2\,d\,x}+1}+\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (b^3\,\sqrt {d^2}+6\,a^2\,b\,\sqrt {d^2}\right )}{d\,\sqrt {36\,a^4\,b^2+12\,a^2\,b^4+b^6}}\right )\,\sqrt {36\,a^4\,b^2+12\,a^2\,b^4+b^6}}{\sqrt {d^2}}-\frac {2\,b^3\,{\mathrm {e}}^{c+d\,x}}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )} \]

[In]

int((a + b/cosh(c + d*x))^3,x)

[Out]

a^3*x - ((6*a*b^2)/d - (b^3*exp(c + d*x))/d)/(exp(2*c + 2*d*x) + 1) + (atan((exp(d*x)*exp(c)*(b^3*(d^2)^(1/2)
+ 6*a^2*b*(d^2)^(1/2)))/(d*(b^6 + 12*a^2*b^4 + 36*a^4*b^2)^(1/2)))*(b^6 + 12*a^2*b^4 + 36*a^4*b^2)^(1/2))/(d^2
)^(1/2) - (2*b^3*exp(c + d*x))/(d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1))