3.2.28 \(\int (a+b \text {sech}^2(c+d x))^3 \, dx\) [128]

Optimal. Leaf size=73 \[ a^3 x+\frac {b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)}{d}-\frac {b^2 (3 a+2 b) \tanh ^3(c+d x)}{3 d}+\frac {b^3 \tanh ^5(c+d x)}{5 d} \]

[Out]

a^3*x+b*(3*a^2+3*a*b+b^2)*tanh(d*x+c)/d-1/3*b^2*(3*a+2*b)*tanh(d*x+c)^3/d+1/5*b^3*tanh(d*x+c)^5/d

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Rubi [A]
time = 0.04, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4213, 398, 212} \begin {gather*} a^3 x+\frac {b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)}{d}-\frac {b^2 (3 a+2 b) \tanh ^3(c+d x)}{3 d}+\frac {b^3 \tanh ^5(c+d x)}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sech[c + d*x]^2)^3,x]

[Out]

a^3*x + (b*(3*a^2 + 3*a*b + b^2)*Tanh[c + d*x])/d - (b^2*(3*a + 2*b)*Tanh[c + d*x]^3)/(3*d) + (b^3*Tanh[c + d*
x]^5)/(5*d)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 398

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 4213

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rubi steps

\begin {align*} \int \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b-b x^2\right )^3}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (b \left (3 a^2+3 a b+b^2\right )-b^2 (3 a+2 b) x^2+b^3 x^4+\frac {a^3}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)}{d}-\frac {b^2 (3 a+2 b) \tanh ^3(c+d x)}{3 d}+\frac {b^3 \tanh ^5(c+d x)}{5 d}+\frac {a^3 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=a^3 x+\frac {b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)}{d}-\frac {b^2 (3 a+2 b) \tanh ^3(c+d x)}{3 d}+\frac {b^3 \tanh ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(268\) vs. \(2(73)=146\).
time = 0.62, size = 268, normalized size = 3.67 \begin {gather*} \frac {\text {sech}(c) \text {sech}^5(c+d x) \left (150 a^3 d x \cosh (d x)+150 a^3 d x \cosh (2 c+d x)+75 a^3 d x \cosh (2 c+3 d x)+75 a^3 d x \cosh (4 c+3 d x)+15 a^3 d x \cosh (4 c+5 d x)+15 a^3 d x \cosh (6 c+5 d x)+540 a^2 b \sinh (d x)+420 a b^2 \sinh (d x)+160 b^3 \sinh (d x)-360 a^2 b \sinh (2 c+d x)-180 a b^2 \sinh (2 c+d x)+360 a^2 b \sinh (2 c+3 d x)+300 a b^2 \sinh (2 c+3 d x)+80 b^3 \sinh (2 c+3 d x)-90 a^2 b \sinh (4 c+3 d x)+90 a^2 b \sinh (4 c+5 d x)+60 a b^2 \sinh (4 c+5 d x)+16 b^3 \sinh (4 c+5 d x)\right )}{480 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sech[c]*Sech[c + d*x]^5*(150*a^3*d*x*Cosh[d*x] + 150*a^3*d*x*Cosh[2*c + d*x] + 75*a^3*d*x*Cosh[2*c + 3*d*x] +
75*a^3*d*x*Cosh[4*c + 3*d*x] + 15*a^3*d*x*Cosh[4*c + 5*d*x] + 15*a^3*d*x*Cosh[6*c + 5*d*x] + 540*a^2*b*Sinh[d
*x] + 420*a*b^2*Sinh[d*x] + 160*b^3*Sinh[d*x] - 360*a^2*b*Sinh[2*c + d*x] - 180*a*b^2*Sinh[2*c + d*x] + 360*a^
2*b*Sinh[2*c + 3*d*x] + 300*a*b^2*Sinh[2*c + 3*d*x] + 80*b^3*Sinh[2*c + 3*d*x] - 90*a^2*b*Sinh[4*c + 3*d*x] +
90*a^2*b*Sinh[4*c + 5*d*x] + 60*a*b^2*Sinh[4*c + 5*d*x] + 16*b^3*Sinh[4*c + 5*d*x]))/(480*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(163\) vs. \(2(69)=138\).
time = 1.70, size = 164, normalized size = 2.25

method	result	size
risch	\(a^{3} x -\frac {2 b \left (45 a^{2} {\mathrm e}^{8 d x +8 c}+180 a^{2} {\mathrm e}^{6 d x +6 c}+90 a b \,{\mathrm e}^{6 d x +6 c}+270 a^{2} {\mathrm e}^{4 d x +4 c}+210 a b \,{\mathrm e}^{4 d x +4 c}+80 b^{2} {\mathrm e}^{4 d x +4 c}+180 a^{2} {\mathrm e}^{2 d x +2 c}+150 a b \,{\mathrm e}^{2 d x +2 c}+40 b^{2} {\mathrm e}^{2 d x +2 c}+45 a^{2}+30 a b +8 b^{2}\right )}{15 d \left (1+{\mathrm e}^{2 d x +2 c}\right )^{5}}\)	\(164\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x-2/15*b*(45*a^2*exp(8*d*x+8*c)+180*a^2*exp(6*d*x+6*c)+90*a*b*exp(6*d*x+6*c)+270*a^2*exp(4*d*x+4*c)+210*a*
b*exp(4*d*x+4*c)+80*b^2*exp(4*d*x+4*c)+180*a^2*exp(2*d*x+2*c)+150*a*b*exp(2*d*x+2*c)+40*b^2*exp(2*d*x+2*c)+45*
a^2+30*a*b+8*b^2)/d/(1+exp(2*d*x+2*c))^5

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 332 vs. \(2 (69) = 138\).
time = 0.28, size = 332, normalized size = 4.55 \begin {gather*} a^{3} x + \frac {16}{15} \, b^{3} {\left (\frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + 4 \, a b^{2} {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + \frac {6 \, a^{2} b}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x + 16/15*b^3*(5*e^(-2*d*x - 2*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e
^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 10*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c)
+ 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 1/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*
x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1))) + 4*a*b^2*(3*e^(-2*d*x - 2*c)/
(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1)) + 1/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x
- 4*c) + e^(-6*d*x - 6*c) + 1))) + 6*a^2*b/(d*(e^(-2*d*x - 2*c) + 1))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 470 vs. \(2 (69) = 138\).
time = 0.41, size = 470, normalized size = 6.44 \begin {gather*} \frac {{\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + {\left (45 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (27 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3} + 2 \, {\left (45 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right ) + 5 \, {\left ({\left (45 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 18 \, a^{2} b + 24 \, a b^{2} + 16 \, b^{3} + 3 \, {\left (27 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*((15*a^3*d*x - 45*a^2*b - 30*a*b^2 - 8*b^3)*cosh(d*x + c)^5 + 5*(15*a^3*d*x - 45*a^2*b - 30*a*b^2 - 8*b^3
)*cosh(d*x + c)*sinh(d*x + c)^4 + (45*a^2*b + 30*a*b^2 + 8*b^3)*sinh(d*x + c)^5 + 5*(15*a^3*d*x - 45*a^2*b - 3
0*a*b^2 - 8*b^3)*cosh(d*x + c)^3 + 5*(27*a^2*b + 30*a*b^2 + 8*b^3 + 2*(45*a^2*b + 30*a*b^2 + 8*b^3)*cosh(d*x +
c)^2)*sinh(d*x + c)^3 + 5*(2*(15*a^3*d*x - 45*a^2*b - 30*a*b^2 - 8*b^3)*cosh(d*x + c)^3 + 3*(15*a^3*d*x - 45*
a^2*b - 30*a*b^2 - 8*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(15*a^3*d*x - 45*a^2*b - 30*a*b^2 - 8*b^3)*cosh(
d*x + c) + 5*((45*a^2*b + 30*a*b^2 + 8*b^3)*cosh(d*x + c)^4 + 18*a^2*b + 24*a*b^2 + 16*b^3 + 3*(27*a^2*b + 30*
a*b^2 + 8*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + 5*d*co
sh(d*x + c)^3 + 5*(2*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^2 + 10*d*cosh(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (69) = 138\).
time = 0.41, size = 182, normalized size = 2.49 \begin {gather*} \frac {15 \, {\left (d x + c\right )} a^{3} - \frac {2 \, {\left (45 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 180 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 210 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 80 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 180 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 150 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 40 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*(d*x + c)*a^3 - 2*(45*a^2*b*e^(8*d*x + 8*c) + 180*a^2*b*e^(6*d*x + 6*c) + 90*a*b^2*e^(6*d*x + 6*c) +
270*a^2*b*e^(4*d*x + 4*c) + 210*a*b^2*e^(4*d*x + 4*c) + 80*b^3*e^(4*d*x + 4*c) + 180*a^2*b*e^(2*d*x + 2*c) + 1
50*a*b^2*e^(2*d*x + 2*c) + 40*b^3*e^(2*d*x + 2*c) + 45*a^2*b + 30*a*b^2 + 8*b^3)/(e^(2*d*x + 2*c) + 1)^5)/d

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Mupad [B]
time = 1.44, size = 502, normalized size = 6.88 \begin {gather*} a^3\,x-\frac {\frac {2\,\left (9\,a^2\,b+12\,a\,b^2+8\,b^3\right )}{15\,d}+\frac {12\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {6\,a^2\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {\frac {6\,a^2\,b}{5\,d}+\frac {24\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {24\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (9\,a^2\,b+12\,a\,b^2+8\,b^3\right )}{5\,d}+\frac {6\,a^2\,b\,{\mathrm {e}}^{8\,c+8\,d\,x}}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {\frac {6\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {18\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (9\,a^2\,b+12\,a\,b^2+8\,b^3\right )}{5\,d}+\frac {6\,a^2\,b\,{\mathrm {e}}^{6\,c+6\,d\,x}}{5\,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 {\frac {6\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {6\,a^2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {6\,a^2\,b}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cosh(c + d*x)^2)^3,x)

[Out]

a^3*x - ((2*(12*a*b^2 + 9*a^2*b + 8*b^3))/(15*d) + (12*exp(2*c + 2*d*x)*(a*b^2 + a^2*b))/(5*d) + (6*a^2*b*exp(
4*c + 4*d*x))/(5*d))/(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1) - ((6*a^2*b)/(5*d) + (24
*exp(2*c + 2*d*x)*(a*b^2 + a^2*b))/(5*d) + (24*exp(6*c + 6*d*x)*(a*b^2 + a^2*b))/(5*d) + (4*exp(4*c + 4*d*x)*(
12*a*b^2 + 9*a^2*b + 8*b^3))/(5*d) + (6*a^2*b*exp(8*c + 8*d*x))/(5*d))/(5*exp(2*c + 2*d*x) + 10*exp(4*c + 4*d*
x) + 10*exp(6*c + 6*d*x) + 5*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) + 1) - ((6*(a*b^2 + a^2*b))/(5*d) + (18*exp
(4*c + 4*d*x)*(a*b^2 + a^2*b))/(5*d) + (2*exp(2*c + 2*d*x)*(12*a*b^2 + 9*a^2*b + 8*b^3))/(5*d) + (6*a^2*b*exp(
6*c + 6*d*x))/(5*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) - (
(6*(a*b^2 + a^2*b))/(5*d) + (6*a^2*b*exp(2*c + 2*d*x))/(5*d))/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) - (6
*a^2*b)/(5*d*(exp(2*c + 2*d*x) + 1))

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