3.1.61 \(\int \text {sech}(c+d x) (a+b \text {sech}^2(c+d x))^2 \, dx\) [61]

Optimal. Leaf size=90 \[ \frac {\left (8 a^2+8 a b+3 b^2\right ) \text {ArcTan}(\sinh (c+d x))}{8 d}+\frac {3 b (2 a+b) \text {sech}(c+d x) \tanh (c+d x)}{8 d}+\frac {b \text {sech}^3(c+d x) \left (a+b+a \sinh ^2(c+d x)\right ) \tanh (c+d x)}{4 d} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {4232, 424, 393, 209} \begin {gather*} \frac {\left (8 a^2+8 a b+3 b^2\right ) \text {ArcTan}(\sinh (c+d x))}{8 d}+\frac {3 b (2 a+b) \tanh (c+d x) \text {sech}(c+d x)}{8 d}+\frac {b \tanh (c+d x) \text {sech}^3(c+d x) \left (a \sinh ^2(c+d x)+a+b\right )}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 209

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

Rule 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 424

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a*d - c*b)*x*(a + b*x^n)^(
p + 1)*((c + d*x^n)^(q - 1)/(a*b*n*(p + 1))), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

Rule 4232

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fr
eeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2*x^2)^
((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n
/2] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 71, normalized size = 0.79 \begin {gather*} \frac {\left (8 a^2+8 a b+3 b^2\right ) \text {ArcTan}(\sinh (c+d x))+b (8 a+3 b) \text {sech}(c+d x) \tanh (c+d x)+2 b^2 \text {sech}^3(c+d x) \tanh (c+d x)}{8 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 1.62, size = 218, normalized size = 2.42

method	result	size
risch	\(\frac {b \,{\mathrm e}^{d x +c} \left (8 a \,{\mathrm e}^{6 d x +6 c}+3 b \,{\mathrm e}^{6 d x +6 c}+8 a \,{\mathrm e}^{4 d x +4 c}+11 b \,{\mathrm e}^{4 d x +4 c}-8 a \,{\mathrm e}^{2 d x +2 c}-11 b \,{\mathrm e}^{2 d x +2 c}-8 a -3 b \right )}{4 d \left (1+{\mathrm e}^{2 d x +2 c}\right )^{4}}+\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right ) a^{2}}{d}+\frac {i b a \ln \left ({\mathrm e}^{d x +c}+i\right )}{d}+\frac {3 i b^{2} \ln \left ({\mathrm e}^{d x +c}+i\right )}{8 d}-\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right ) a^{2}}{d}-\frac {i b a \ln \left ({\mathrm e}^{d x +c}-i\right )}{d}-\frac {3 i b^{2} \ln \left ({\mathrm e}^{d x +c}-i\right )}{8 d}\)	\(218\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*exp(d*x+c)*(8*a*exp(6*d*x+6*c)+3*b*exp(6*d*x+6*c)+8*a*exp(4*d*x+4*c)+11*b*exp(4*d*x+4*c)-8*a*exp(2*d*x+2
*c)-11*b*exp(2*d*x+2*c)-8*a-3*b)/d/(1+exp(2*d*x+2*c))^4+I/d*ln(exp(d*x+c)+I)*a^2+I*b*a/d*ln(exp(d*x+c)+I)+3/8*
I*b^2/d*ln(exp(d*x+c)+I)-I/d*ln(exp(d*x+c)-I)*a^2-I*b*a/d*ln(exp(d*x+c)-I)-3/8*I/d*ln(exp(d*x+c)-I)*b^2

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (84) = 168\).
time = 0.51, size = 201, normalized size = 2.23 \begin {gather*} -\frac {1}{4} \, b^{2} {\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 )} - 2 \, a b {\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 {a^{2} \arctan \left (\sinh \left (d x + c\right )\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1372 vs. \(2 (84) = 168\).
time = 0.36, size = 1372, normalized size = 15.24 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((8*a*b + 3*b^2)*cosh(d*x + c)^7 + 7*(8*a*b + 3*b^2)*cosh(d*x + c)*sinh(d*x + c)^6 + (8*a*b + 3*b^2)*sinh(
d*x + c)^7 + (8*a*b + 11*b^2)*cosh(d*x + c)^5 + (21*(8*a*b + 3*b^2)*cosh(d*x + c)^2 + 8*a*b + 11*b^2)*sinh(d*x
+ c)^5 + 5*(7*(8*a*b + 3*b^2)*cosh(d*x + c)^3 + (8*a*b + 11*b^2)*cosh(d*x + c))*sinh(d*x + c)^4 - (8*a*b + 11
*b^2)*cosh(d*x + c)^3 + (35*(8*a*b + 3*b^2)*cosh(d*x + c)^4 + 10*(8*a*b + 11*b^2)*cosh(d*x + c)^2 - 8*a*b - 11
*b^2)*sinh(d*x + c)^3 + (21*(8*a*b + 3*b^2)*cosh(d*x + c)^5 + 10*(8*a*b + 11*b^2)*cosh(d*x + c)^3 - 3*(8*a*b +
11*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 + ((8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^8 + 8*(8*a^2 + 8*a*b + 3*b^2)
*cosh(d*x + c)*sinh(d*x + c)^7 + (8*a^2 + 8*a*b + 3*b^2)*sinh(d*x + c)^8 + 4*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x
+ c)^6 + 4*(7*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^2 + 8*a^2 + 8*a*b + 3*b^2)*sinh(d*x + c)^6 + 8*(7*(8*a^2 +
8*a*b + 3*b^2)*cosh(d*x + c)^3 + 3*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c)^5 + 6*(8*a^2 + 8*a*b
+ 3*b^2)*cosh(d*x + c)^4 + 2*(35*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^4 + 30*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x
+ c)^2 + 24*a^2 + 24*a*b + 9*b^2)*sinh(d*x + c)^4 + 8*(7*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^5 + 10*(8*a^2
+ 8*a*b + 3*b^2)*cosh(d*x + c)^3 + 3*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 4*(8*a^2 + 8*a*b
+ 3*b^2)*cosh(d*x + c)^2 + 4*(7*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^6 + 15*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x
+ c)^4 + 9*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^2 + 8*a^2 + 8*a*b + 3*b^2)*sinh(d*x + c)^2 + 8*a^2 + 8*a*b +
3*b^2 + 8*((8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^7 + 3*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^5 + 3*(8*a^2 + 8
*a*b + 3*b^2)*cosh(d*x + c)^3 + (8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c))*arctan(cosh(d*x + c) + s
inh(d*x + c)) - (8*a*b + 3*b^2)*cosh(d*x + c) + (7*(8*a*b + 3*b^2)*cosh(d*x + c)^6 + 5*(8*a*b + 11*b^2)*cosh(d
*x + c)^4 - 3*(8*a*b + 11*b^2)*cosh(d*x + c)^2 - 8*a*b - 3*b^2)*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 [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 )^{2} \operatorname {sech}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (84) = 168\).
time = 0.40, size = 170, normalized size = 1.89 \begin {gather*} \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 (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} + \frac {4 \, {\left (8 \, a b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 3 \, b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 32 \, a b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 20 \, b^{2} {\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((pi + 2*arctan(1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*(8*a^2 + 8*a*b + 3*b^2) + 4*(8*a*b*(e^(d*x + c)
- e^(-d*x - c))^3 + 3*b^2*(e^(d*x + c) - e^(-d*x - c))^3 + 32*a*b*(e^(d*x + c) - e^(-d*x - c)) + 20*b^2*(e^(d*
x + c) - e^(-d*x - c)))/((e^(d*x + c) - e^(-d*x - c))^2 + 4)^2)/d

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Mupad [B]
time = 1.52, size = 303, normalized size = 3.37 \begin {gather*} \frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (8\,a^2\,\sqrt {d^2}+3\,b^2\,\sqrt {d^2}+8\,a\,b\,\sqrt {d^2}\right )}{d\,\sqrt {64\,a^4+128\,a^3\,b+112\,a^2\,b^2+48\,a\,b^3+9\,b^4}}\right )\,\sqrt {64\,a^4+128\,a^3\,b+112\,a^2\,b^2+48\,a\,b^3+9\,b^4}}{4\,\sqrt {d^2}}-\frac {6\,b^2\,{\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 {4\,b^2\,{\mathrm {e}}^{c+d\,x}}{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 {{\mathrm {e}}^{c+d\,x}\,\left (3\,b^2+8\,a\,b\right )}{4\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{c+d\,x}\,\left (8\,a\,b-b^2\right )}{2\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,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)^2/cosh(c + d*x),x)

[Out]

(atan((exp(d*x)*exp(c)*(8*a^2*(d^2)^(1/2) + 3*b^2*(d^2)^(1/2) + 8*a*b*(d^2)^(1/2)))/(d*(48*a*b^3 + 128*a^3*b +
64*a^4 + 9*b^4 + 112*a^2*b^2)^(1/2)))*(48*a*b^3 + 128*a^3*b + 64*a^4 + 9*b^4 + 112*a^2*b^2)^(1/2))/(4*(d^2)^(
1/2)) - (6*b^2*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)) + (4*b^2*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)*(8*a*b + 3*b^2))/(4*d*(exp(2*c + 2*d*x) + 1)) - (exp(c + d*x)*(8*a*b - b^2))/(2*d*(2*exp(2*c + 2*d*x) +
exp(4*c + 4*d*x) + 1))

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