3.186 \(\int ((1-b^2 n^2) \text {sech}(a+b \log (c x^n))+2 b^2 n^2 \text {sech}^3(a+b \log (c x^n))) \, dx\)

Optimal. Leaf size=40 \[ x \text {sech}\left (a+b \log \left (c x^n\right )\right )+b n x \tanh \left (a+b \log \left (c x^n\right )\right ) \text {sech}\left (a+b \log \left (c x^n\right )\right ) \]

[Out]

x*sech(a+b*ln(c*x^n))+b*n*x*sech(a+b*ln(c*x^n))*tanh(a+b*ln(c*x^n))

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Rubi [C]  time = 0.14, antiderivative size = 139, normalized size of antiderivative = 3.48, number of steps used = 9, number of rules used = 4, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {5545, 5547, 263, 364} \[ \frac {16 e^{3 a} b^2 n^2 x \left (c x^n\right )^{3 b} \, _2F_1\left (3,\frac {3 b+\frac {1}{n}}{2 b};\frac {1}{2} \left (5+\frac {1}{b n}\right );-e^{2 a} \left (c x^n\right )^{2 b}\right )}{3 b n+1}+2 e^a x (1-b n) \left (c x^n\right )^b \, _2F_1\left (1,\frac {b+\frac {1}{n}}{2 b};\frac {1}{2} \left (3+\frac {1}{b n}\right );-e^{2 a} \left (c x^n\right )^{2 b}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Int[(1 - b^2*n^2)*Sech[a + b*Log[c*x^n]] + 2*b^2*n^2*Sech[a + b*Log[c*x^n]]^3,x]

[Out]

2*E^a*(1 - b*n)*x*(c*x^n)^b*Hypergeometric2F1[1, (b + n^(-1))/(2*b), (3 + 1/(b*n))/2, -(E^(2*a)*(c*x^n)^(2*b))
] + (16*b^2*E^(3*a)*n^2*x*(c*x^n)^(3*b)*Hypergeometric2F1[3, (3*b + n^(-1))/(2*b), (5 + 1/(b*n))/2, -(E^(2*a)*
(c*x^n)^(2*b))])/(1 + 3*b*n)

Rule 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 5545

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

Rule 5547

Int[((e_.)*(x_))^(m_.)*Sech[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[2^p/E^(a*d*p), Int[(e*x)^m
/(x^(b*d*p)*(1 + 1/(E^(2*a*d)*x^(2*b*d)))^p), x], x] /; FreeQ[{a, b, d, e, m}, x] && IntegerQ[p]

Rubi steps

\begin {align*} \int \left (\left (1-b^2 n^2\right ) \text {sech}\left (a+b \log \left (c x^n\right )\right )+2 b^2 n^2 \text {sech}^3\left (a+b \log \left (c x^n\right )\right )\right ) \, dx &=\left (2 b^2 n^2\right ) \int \text {sech}^3\left (a+b \log \left (c x^n\right )\right ) \, dx+\left (1-b^2 n^2\right ) \int \text {sech}\left (a+b \log \left (c x^n\right )\right ) \, dx\\ &=\left (2 b^2 n x \left (c x^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int x^{-1+\frac {1}{n}} \text {sech}^3(a+b \log (x)) \, dx,x,c x^n\right )+\frac {\left (\left (1-b^2 n^2\right ) x \left (c x^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int x^{-1+\frac {1}{n}} \text {sech}(a+b \log (x)) \, dx,x,c x^n\right )}{n}\\ &=\left (16 b^2 e^{-3 a} n x \left (c x^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int \frac {x^{-1-3 b+\frac {1}{n}}}{\left (1+e^{-2 a} x^{-2 b}\right )^3} \, dx,x,c x^n\right )+\frac {\left (2 e^{-a} \left (1-b^2 n^2\right ) x \left (c x^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int \frac {x^{-1-b+\frac {1}{n}}}{1+e^{-2 a} x^{-2 b}} \, dx,x,c x^n\right )}{n}\\ &=\left (16 b^2 e^{-3 a} n x \left (c x^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int \frac {x^{-1+3 b+\frac {1}{n}}}{\left (e^{-2 a}+x^{2 b}\right )^3} \, dx,x,c x^n\right )+\frac {\left (2 e^{-a} \left (1-b^2 n^2\right ) x \left (c x^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int \frac {x^{-1+b+\frac {1}{n}}}{e^{-2 a}+x^{2 b}} \, dx,x,c x^n\right )}{n}\\ &=2 e^a (1-b n) x \left (c x^n\right )^b \, _2F_1\left (1,\frac {b+\frac {1}{n}}{2 b};\frac {1}{2} \left (3+\frac {1}{b n}\right );-e^{2 a} \left (c x^n\right )^{2 b}\right )+\frac {16 b^2 e^{3 a} n^2 x \left (c x^n\right )^{3 b} \, _2F_1\left (3,\frac {3 b+\frac {1}{n}}{2 b};\frac {1}{2} \left (5+\frac {1}{b n}\right );-e^{2 a} \left (c x^n\right )^{2 b}\right )}{1+3 b n}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 29, normalized size = 0.72 \[ x \left (b n \tanh \left (a+b \log \left (c x^n\right )\right )+1\right ) \text {sech}\left (a+b \log \left (c x^n\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - b^2*n^2)*Sech[a + b*Log[c*x^n]] + 2*b^2*n^2*Sech[a + b*Log[c*x^n]]^3,x]

[Out]

x*Sech[a + b*Log[c*x^n]]*(1 + b*n*Tanh[a + b*Log[c*x^n]])

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fricas [B]  time = 0.43, size = 189, normalized size = 4.72 \[ \frac {2 \, {\left ({\left (b n + 1\right )} x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} + 2 \, {\left (b n + 1\right )} x \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) + {\left (b n + 1\right )} x \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} - {\left (b n - 1\right )} x\right )}}{\cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{3} + 3 \, \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} + \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{3} + {\left (3 \, \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} + 1\right )} \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) + 3 \, \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*n^2+1)*sech(a+b*log(c*x^n))+2*b^2*n^2*sech(a+b*log(c*x^n))^3,x, algorithm="fricas")

[Out]

2*((b*n + 1)*x*cosh(b*n*log(x) + b*log(c) + a)^2 + 2*(b*n + 1)*x*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(
x) + b*log(c) + a) + (b*n + 1)*x*sinh(b*n*log(x) + b*log(c) + a)^2 - (b*n - 1)*x)/(cosh(b*n*log(x) + b*log(c)
+ a)^3 + 3*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a)^2 + sinh(b*n*log(x) + b*log(c) + a)
^3 + (3*cosh(b*n*log(x) + b*log(c) + a)^2 + 1)*sinh(b*n*log(x) + b*log(c) + a) + 3*cosh(b*n*log(x) + b*log(c)
+ a))

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giac [B]  time = 1.22, size = 215, normalized size = 5.38 \[ \frac {2 \, b c^{3 \, b} n x x^{3 \, b n} e^{\left (3 \, a\right )}}{c^{4 \, b} x^{4 \, b n} e^{\left (4 \, a\right )} + 2 \, c^{2 \, b} x^{2 \, b n} e^{\left (2 \, a\right )} + 1} - \frac {2 \, b c^{b} n x x^{b n} e^{a}}{c^{4 \, b} x^{4 \, b n} e^{\left (4 \, a\right )} + 2 \, c^{2 \, b} x^{2 \, b n} e^{\left (2 \, a\right )} + 1} + \frac {2 \, c^{3 \, b} x x^{3 \, b n} e^{\left (3 \, a\right )}}{c^{4 \, b} x^{4 \, b n} e^{\left (4 \, a\right )} + 2 \, c^{2 \, b} x^{2 \, b n} e^{\left (2 \, a\right )} + 1} + \frac {2 \, c^{b} x x^{b n} e^{a}}{c^{4 \, b} x^{4 \, b n} e^{\left (4 \, a\right )} + 2 \, c^{2 \, b} x^{2 \, b n} e^{\left (2 \, a\right )} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*n^2+1)*sech(a+b*log(c*x^n))+2*b^2*n^2*sech(a+b*log(c*x^n))^3,x, algorithm="giac")

[Out]

2*b*c^(3*b)*n*x*x^(3*b*n)*e^(3*a)/(c^(4*b)*x^(4*b*n)*e^(4*a) + 2*c^(2*b)*x^(2*b*n)*e^(2*a) + 1) - 2*b*c^b*n*x*
x^(b*n)*e^a/(c^(4*b)*x^(4*b*n)*e^(4*a) + 2*c^(2*b)*x^(2*b*n)*e^(2*a) + 1) + 2*c^(3*b)*x*x^(3*b*n)*e^(3*a)/(c^(
4*b)*x^(4*b*n)*e^(4*a) + 2*c^(2*b)*x^(2*b*n)*e^(2*a) + 1) + 2*c^b*x*x^(b*n)*e^a/(c^(4*b)*x^(4*b*n)*e^(4*a) + 2
*c^(2*b)*x^(2*b*n)*e^(2*a) + 1)

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maple [C]  time = 1.06, size = 509, normalized size = 12.72 \[ \frac {2 c^{b} \left (x^{n}\right )^{b} x \left (n b \left (x^{n}\right )^{2 b} c^{2 b} {\mathrm e}^{3 a} {\mathrm e}^{-\frac {3 i b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \pi }{2}} {\mathrm e}^{\frac {3 i b \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i c \right ) \pi }{2}} {\mathrm e}^{\frac {3 i b \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i x^{n}\right ) \pi }{2}} {\mathrm e}^{-\frac {3 i b \,\mathrm {csgn}\left (i c \,x^{n}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \pi }{2}}-{\mathrm e}^{a} {\mathrm e}^{-\frac {i b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \pi }{2}} {\mathrm e}^{\frac {i b \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i c \right ) \pi }{2}} {\mathrm e}^{\frac {i b \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i x^{n}\right ) \pi }{2}} {\mathrm e}^{-\frac {i b \,\mathrm {csgn}\left (i c \,x^{n}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \pi }{2}} b n +\left (x^{n}\right )^{2 b} c^{2 b} {\mathrm e}^{3 a} {\mathrm e}^{-\frac {3 i b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \pi }{2}} {\mathrm e}^{\frac {3 i b \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i c \right ) \pi }{2}} {\mathrm e}^{\frac {3 i b \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i x^{n}\right ) \pi }{2}} {\mathrm e}^{-\frac {3 i b \,\mathrm {csgn}\left (i c \,x^{n}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \pi }{2}}+{\mathrm e}^{a} {\mathrm e}^{-\frac {i b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \pi }{2}} {\mathrm e}^{\frac {i b \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i c \right ) \pi }{2}} {\mathrm e}^{\frac {i b \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i x^{n}\right ) \pi }{2}} {\mathrm e}^{-\frac {i b \,\mathrm {csgn}\left (i c \,x^{n}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \pi }{2}}\right )}{\left (\left (x^{n}\right )^{2 b} c^{2 b} {\mathrm e}^{2 a} {\mathrm e}^{-i b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \pi } {\mathrm e}^{i b \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i c \right ) \pi } {\mathrm e}^{i b \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i x^{n}\right ) \pi } {\mathrm e}^{-i b \,\mathrm {csgn}\left (i c \,x^{n}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \pi }+1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b^2*n^2+1)*sech(a+b*ln(c*x^n))+2*b^2*n^2*sech(a+b*ln(c*x^n))^3,x)

[Out]

2*c^b*(x^n)^b*x/(((x^n)^b)^2*(c^b)^2*exp(2*a)*exp(-I*b*csgn(I*c*x^n)^3*Pi)*exp(I*b*csgn(I*c*x^n)^2*csgn(I*c)*P
i)*exp(I*b*csgn(I*c*x^n)^2*csgn(I*x^n)*Pi)*exp(-I*b*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)*Pi)+1)^2*(n*b*((x^n)^b
)^2*(c^b)^2*exp(3*a)*exp(-3/2*I*b*csgn(I*c*x^n)^3*Pi)*exp(3/2*I*b*csgn(I*c*x^n)^2*csgn(I*c)*Pi)*exp(3/2*I*b*cs
gn(I*c*x^n)^2*csgn(I*x^n)*Pi)*exp(-3/2*I*b*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)*Pi)-exp(a)*exp(-1/2*I*b*csgn(I*
c*x^n)^3*Pi)*exp(1/2*I*b*csgn(I*c*x^n)^2*csgn(I*c)*Pi)*exp(1/2*I*b*csgn(I*c*x^n)^2*csgn(I*x^n)*Pi)*exp(-1/2*I*
b*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)*Pi)*b*n+((x^n)^b)^2*(c^b)^2*exp(3*a)*exp(-3/2*I*b*csgn(I*c*x^n)^3*Pi)*ex
p(3/2*I*b*csgn(I*c*x^n)^2*csgn(I*c)*Pi)*exp(3/2*I*b*csgn(I*c*x^n)^2*csgn(I*x^n)*Pi)*exp(-3/2*I*b*csgn(I*c*x^n)
*csgn(I*c)*csgn(I*x^n)*Pi)+exp(a)*exp(-1/2*I*b*csgn(I*c*x^n)^3*Pi)*exp(1/2*I*b*csgn(I*c*x^n)^2*csgn(I*c)*Pi)*e
xp(1/2*I*b*csgn(I*c*x^n)^2*csgn(I*x^n)*Pi)*exp(-1/2*I*b*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)*Pi))

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maxima [B]  time = 0.61, size = 96, normalized size = 2.40 \[ \frac {2 \, {\left ({\left (b c^{3 \, b} n + c^{3 \, b}\right )} x e^{\left (3 \, b \log \left (x^{n}\right ) + 3 \, a\right )} - {\left (b c^{b} n - c^{b}\right )} x e^{\left (b \log \left (x^{n}\right ) + a\right )}\right )}}{c^{4 \, b} e^{\left (4 \, b \log \left (x^{n}\right ) + 4 \, a\right )} + 2 \, c^{2 \, b} e^{\left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*n^2+1)*sech(a+b*log(c*x^n))+2*b^2*n^2*sech(a+b*log(c*x^n))^3,x, algorithm="maxima")

[Out]

2*((b*c^(3*b)*n + c^(3*b))*x*e^(3*b*log(x^n) + 3*a) - (b*c^b*n - c^b)*x*e^(b*log(x^n) + a))/(c^(4*b)*e^(4*b*lo
g(x^n) + 4*a) + 2*c^(2*b)*e^(2*b*log(x^n) + 2*a) + 1)

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mupad [B]  time = 1.40, size = 66, normalized size = 1.65 \[ \frac {2\,x\,{\mathrm {e}}^a\,{\left (c\,x^n\right )}^b\,\left ({\mathrm {e}}^{2\,a}\,{\left (c\,x^n\right )}^{2\,b}-b\,n+b\,n\,{\mathrm {e}}^{2\,a}\,{\left (c\,x^n\right )}^{2\,b}+1\right )}{{\left ({\mathrm {e}}^{2\,a}\,{\left (c\,x^n\right )}^{2\,b}+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*b^2*n^2)/cosh(a + b*log(c*x^n))^3 - (b^2*n^2 - 1)/cosh(a + b*log(c*x^n)),x)

[Out]

(2*x*exp(a)*(c*x^n)^b*(exp(2*a)*(c*x^n)^(2*b) - b*n + b*n*exp(2*a)*(c*x^n)^(2*b) + 1))/(exp(2*a)*(c*x^n)^(2*b)
+ 1)^2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (2 b^{2} n^{2} \operatorname {sech}^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )} - b^{2} n^{2} + 1\right ) \operatorname {sech}{\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b**2*n**2+1)*sech(a+b*ln(c*x**n))+2*b**2*n**2*sech(a+b*ln(c*x**n))**3,x)

[Out]

Integral((2*b**2*n**2*sech(a + b*log(c*x**n))**2 - b**2*n**2 + 1)*sech(a + b*log(c*x**n)), x)

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