3.259 \(\int (-((1+b^2 n^2) \sec (a+b \log (c x^n)))+2 b^2 n^2 \sec ^3(a+b \log (c x^n))) \, dx\)

Optimal. Leaf size=41 \[ b n x \tan \left (a+b \log \left (c x^n\right )\right ) \sec \left (a+b \log \left (c x^n\right )\right )-x \sec \left (a+b \log \left (c x^n\right )\right ) \]

[Out]

-x*sec(a+b*ln(c*x^n))+b*n*x*sec(a+b*ln(c*x^n))*tan(a+b*ln(c*x^n))

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Rubi [C]  time = 0.13, antiderivative size = 175, normalized size of antiderivative = 4.27, number of steps used = 7, number of rules used = 3, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {4503, 4505, 364} \[ \frac {16 e^{3 i a} b^2 n^2 x \left (c x^n\right )^{3 i b} \, _2F_1\left (3,\frac {1}{2} \left (3-\frac {i}{b n}\right );\frac {1}{2} \left (5-\frac {i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+3 i b n}-2 e^{i a} x (1-i b n) \left (c x^n\right )^{i b} \, _2F_1\left (1,\frac {1}{2} \left (1-\frac {i}{b n}\right );\frac {1}{2} \left (3-\frac {i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-2*E^(I*a)*(1 - I*b*n)*x*(c*x^n)^(I*b)*Hypergeometric2F1[1, (1 - I/(b*n))/2, (3 - I/(b*n))/2, -(E^((2*I)*a)*(c
*x^n)^((2*I)*b))] + (16*b^2*E^((3*I)*a)*n^2*x*(c*x^n)^((3*I)*b)*Hypergeometric2F1[3, (3 - I/(b*n))/2, (5 - I/(
b*n))/2, -(E^((2*I)*a)*(c*x^n)^((2*I)*b))])/(1 + (3*I)*b*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 4503

Int[Sec[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[x
^(1/n - 1)*Sec[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 4505

Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[2^p*E^(I*a*d*p), Int[((e*x)
^m*x^(I*b*d*p))/(1 + E^(2*I*a*d)*x^(2*I*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 ) \sec \left (a+b \log \left (c x^n\right )\right )+2 b^2 n^2 \sec ^3\left (a+b \log \left (c x^n\right )\right )\right ) \, dx &=\left (2 b^2 n^2\right ) \int \sec ^3\left (a+b \log \left (c x^n\right )\right ) \, dx+\left (-1-b^2 n^2\right ) \int \sec \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}} \sec ^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}} \sec (a+b \log (x)) \, dx,x,c x^n\right )}{n}\\ &=\left (16 b^2 e^{3 i a} n x \left (c x^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int \frac {x^{-1+3 i b+\frac {1}{n}}}{\left (1+e^{2 i a} x^{2 i b}\right )^3} \, dx,x,c x^n\right )+\frac {\left (2 e^{i a} \left (-1-b^2 n^2\right ) x \left (c x^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int \frac {x^{-1+i b+\frac {1}{n}}}{1+e^{2 i a} x^{2 i b}} \, dx,x,c x^n\right )}{n}\\ &=-2 e^{i a} (1-i b n) x \left (c x^n\right )^{i b} \, _2F_1\left (1,\frac {1}{2} \left (1-\frac {i}{b n}\right );\frac {1}{2} \left (3-\frac {i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )+\frac {16 b^2 e^{3 i a} n x \left (c x^n\right )^{3 i b} \, _2F_1\left (3,\frac {1}{2} \left (3-\frac {i}{b n}\right );\frac {1}{2} \left (5-\frac {i}{b n}\right );-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{3 i b+\frac {1}{n}}\\ \end {align*}

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Mathematica [A]  time = 0.46, size = 29, normalized size = 0.71 \[ x \left (b n \tan \left (a+b \log \left (c x^n\right )\right )-1\right ) \sec \left (a+b \log \left (c x^n\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*Sec[a + b*Log[c*x^n]]*(-1 + b*n*Tan[a + b*Log[c*x^n]])

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fricas [A]  time = 3.98, size = 47, normalized size = 1.15 \[ \frac {b n x \sin \left (b n \log \relax (x) + b \log \relax (c) + a\right ) - x \cos \left (b n \log \relax (x) + b \log \relax (c) + a\right )}{\cos \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*n*x*sin(b*n*log(x) + b*log(c) + a) - x*cos(b*n*log(x) + b*log(c) + a))/cos(b*n*log(x) + b*log(c) + a)^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int 2 \, b^{2} n^{2} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{3} - {\left (b^{2} n^{2} + 1\right )} \sec \left (b \log \left (c x^{n}\right ) + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(2*b^2*n^2*sec(b*log(c*x^n) + a)^3 - (b^2*n^2 + 1)*sec(b*log(c*x^n) + a), x)

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maple [C]  time = 0.64, size = 525, normalized size = 12.80 \[ -\frac {2 i c^{i b} \left (x^{n}\right )^{i b} x \left (n b \,c^{2 i b} \left (x^{n}\right )^{2 i b} {\mathrm e}^{\frac {3 b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2}} {\mathrm e}^{-\frac {3 b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i c \right )}{2}} {\mathrm e}^{-\frac {3 b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i x^{n}\right )}{2}} {\mathrm e}^{\frac {3 b \pi \,\mathrm {csgn}\left (i c \,x^{n}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right )}{2}} {\mathrm e}^{3 i a}-b n \,{\mathrm e}^{\frac {b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2}} {\mathrm e}^{-\frac {b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i c \right )}{2}} {\mathrm e}^{-\frac {b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i x^{n}\right )}{2}} {\mathrm e}^{\frac {b \pi \,\mathrm {csgn}\left (i c \,x^{n}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right )}{2}} {\mathrm e}^{i a}-i c^{2 i b} \left (x^{n}\right )^{2 i b} {\mathrm e}^{\frac {3 b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2}} {\mathrm e}^{-\frac {3 b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i c \right )}{2}} {\mathrm e}^{-\frac {3 b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i x^{n}\right )}{2}} {\mathrm e}^{\frac {3 b \pi \,\mathrm {csgn}\left (i c \,x^{n}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right )}{2}} {\mathrm e}^{3 i a}-i {\mathrm e}^{\frac {b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2}} {\mathrm e}^{-\frac {b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i c \right )}{2}} {\mathrm e}^{-\frac {b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i x^{n}\right )}{2}} {\mathrm e}^{\frac {b \pi \,\mathrm {csgn}\left (i c \,x^{n}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right )}{2}} {\mathrm e}^{i a}\right )}{\left (\left (x^{n}\right )^{2 i b} c^{2 i b} {\mathrm e}^{b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3}} {\mathrm e}^{-b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i c \right )} {\mathrm e}^{-b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \mathrm {csgn}\left (i x^{n}\right )} {\mathrm e}^{b \pi \,\mathrm {csgn}\left (i c \,x^{n}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right )} {\mathrm e}^{2 i a}+1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*c^(I*b)*(x^n)^(I*b)*x/(((x^n)^(I*b))^2*(c^(I*b))^2*exp(b*Pi*csgn(I*c*x^n)^3)*exp(-b*Pi*csgn(I*c*x^n)^2*cs
gn(I*c))*exp(-b*Pi*csgn(I*c*x^n)^2*csgn(I*x^n))*exp(b*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n))*exp(2*I*a)+1)^2*
(n*b*(c^(I*b))^2*((x^n)^(I*b))^2*exp(3/2*b*Pi*csgn(I*c*x^n)^3)*exp(-3/2*b*Pi*csgn(I*c*x^n)^2*csgn(I*c))*exp(-3
/2*b*Pi*csgn(I*c*x^n)^2*csgn(I*x^n))*exp(3/2*b*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n))*exp(3*I*a)-b*n*exp(1/2*
b*Pi*csgn(I*c*x^n)^3)*exp(-1/2*b*Pi*csgn(I*c*x^n)^2*csgn(I*c))*exp(-1/2*b*Pi*csgn(I*c*x^n)^2*csgn(I*x^n))*exp(
1/2*b*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n))*exp(I*a)-I*(c^(I*b))^2*((x^n)^(I*b))^2*exp(3/2*b*Pi*csgn(I*c*x^n
)^3)*exp(-3/2*b*Pi*csgn(I*c*x^n)^2*csgn(I*c))*exp(-3/2*b*Pi*csgn(I*c*x^n)^2*csgn(I*x^n))*exp(3/2*b*Pi*csgn(I*c
*x^n)*csgn(I*c)*csgn(I*x^n))*exp(3*I*a)-I*exp(1/2*b*Pi*csgn(I*c*x^n)^3)*exp(-1/2*b*Pi*csgn(I*c*x^n)^2*csgn(I*c
))*exp(-1/2*b*Pi*csgn(I*c*x^n)^2*csgn(I*x^n))*exp(1/2*b*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n))*exp(I*a))

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maxima [B]  time = 2.33, size = 1696, normalized size = 41.37 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((b*n*sin(b*log(c)) + cos(b*log(c)))*x*cos(b*log(x^n) + a) + (b*n*cos(b*log(c)) - sin(b*log(c)))*x*sin(b*lo
g(x^n) + a) + (((b*cos(3*b*log(c))*sin(4*b*log(c)) - b*cos(4*b*log(c))*sin(3*b*log(c)))*n + cos(4*b*log(c))*co
s(3*b*log(c)) + sin(4*b*log(c))*sin(3*b*log(c)))*x*cos(3*b*log(x^n) + 3*a) - ((b*cos(b*log(c))*sin(4*b*log(c))
- b*cos(4*b*log(c))*sin(b*log(c)))*n - cos(4*b*log(c))*cos(b*log(c)) - sin(4*b*log(c))*sin(b*log(c)))*x*cos(b
*log(x^n) + a) - ((b*cos(4*b*log(c))*cos(3*b*log(c)) + b*sin(4*b*log(c))*sin(3*b*log(c)))*n - cos(3*b*log(c))*
sin(4*b*log(c)) + cos(4*b*log(c))*sin(3*b*log(c)))*x*sin(3*b*log(x^n) + 3*a) + ((b*cos(4*b*log(c))*cos(b*log(c
)) + b*sin(4*b*log(c))*sin(b*log(c)))*n + cos(b*log(c))*sin(4*b*log(c)) - cos(4*b*log(c))*sin(b*log(c)))*x*sin
(b*log(x^n) + a))*cos(4*b*log(x^n) + 4*a) - (2*((b*cos(2*b*log(c))*sin(3*b*log(c)) - b*cos(3*b*log(c))*sin(2*b
*log(c)))*n - cos(3*b*log(c))*cos(2*b*log(c)) - sin(3*b*log(c))*sin(2*b*log(c)))*x*cos(2*b*log(x^n) + 2*a) - 2
*((b*cos(3*b*log(c))*cos(2*b*log(c)) + b*sin(3*b*log(c))*sin(2*b*log(c)))*n + cos(2*b*log(c))*sin(3*b*log(c))
- cos(3*b*log(c))*sin(2*b*log(c)))*x*sin(2*b*log(x^n) + 2*a) + (b*n*sin(3*b*log(c)) - cos(3*b*log(c)))*x)*cos(
3*b*log(x^n) + 3*a) - 2*(((b*cos(b*log(c))*sin(2*b*log(c)) - b*cos(2*b*log(c))*sin(b*log(c)))*n - cos(2*b*log(
c))*cos(b*log(c)) - sin(2*b*log(c))*sin(b*log(c)))*x*cos(b*log(x^n) + a) - ((b*cos(2*b*log(c))*cos(b*log(c)) +
b*sin(2*b*log(c))*sin(b*log(c)))*n + cos(b*log(c))*sin(2*b*log(c)) - cos(2*b*log(c))*sin(b*log(c)))*x*sin(b*l
og(x^n) + a))*cos(2*b*log(x^n) + 2*a) + (((b*cos(4*b*log(c))*cos(3*b*log(c)) + b*sin(4*b*log(c))*sin(3*b*log(c
)))*n - cos(3*b*log(c))*sin(4*b*log(c)) + cos(4*b*log(c))*sin(3*b*log(c)))*x*cos(3*b*log(x^n) + 3*a) - ((b*cos
(4*b*log(c))*cos(b*log(c)) + b*sin(4*b*log(c))*sin(b*log(c)))*n + cos(b*log(c))*sin(4*b*log(c)) - cos(4*b*log(
c))*sin(b*log(c)))*x*cos(b*log(x^n) + a) + ((b*cos(3*b*log(c))*sin(4*b*log(c)) - b*cos(4*b*log(c))*sin(3*b*log
(c)))*n + cos(4*b*log(c))*cos(3*b*log(c)) + sin(4*b*log(c))*sin(3*b*log(c)))*x*sin(3*b*log(x^n) + 3*a) - ((b*c
os(b*log(c))*sin(4*b*log(c)) - b*cos(4*b*log(c))*sin(b*log(c)))*n - cos(4*b*log(c))*cos(b*log(c)) - sin(4*b*lo
g(c))*sin(b*log(c)))*x*sin(b*log(x^n) + a))*sin(4*b*log(x^n) + 4*a) - (2*((b*cos(3*b*log(c))*cos(2*b*log(c)) +
b*sin(3*b*log(c))*sin(2*b*log(c)))*n + cos(2*b*log(c))*sin(3*b*log(c)) - cos(3*b*log(c))*sin(2*b*log(c)))*x*c
os(2*b*log(x^n) + 2*a) + 2*((b*cos(2*b*log(c))*sin(3*b*log(c)) - b*cos(3*b*log(c))*sin(2*b*log(c)))*n - cos(3*
b*log(c))*cos(2*b*log(c)) - sin(3*b*log(c))*sin(2*b*log(c)))*x*sin(2*b*log(x^n) + 2*a) + (b*n*cos(3*b*log(c))
+ sin(3*b*log(c)))*x)*sin(3*b*log(x^n) + 3*a) - 2*(((b*cos(2*b*log(c))*cos(b*log(c)) + b*sin(2*b*log(c))*sin(b
*log(c)))*n + cos(b*log(c))*sin(2*b*log(c)) - cos(2*b*log(c))*sin(b*log(c)))*x*cos(b*log(x^n) + a) + ((b*cos(b
*log(c))*sin(2*b*log(c)) - b*cos(2*b*log(c))*sin(b*log(c)))*n - cos(2*b*log(c))*cos(b*log(c)) - sin(2*b*log(c)
)*sin(b*log(c)))*x*sin(b*log(x^n) + a))*sin(2*b*log(x^n) + 2*a))/((cos(4*b*log(c))^2 + sin(4*b*log(c))^2)*cos(
4*b*log(x^n) + 4*a)^2 + 4*(cos(2*b*log(c))^2 + sin(2*b*log(c))^2)*cos(2*b*log(x^n) + 2*a)^2 + (cos(4*b*log(c))
^2 + sin(4*b*log(c))^2)*sin(4*b*log(x^n) + 4*a)^2 + 4*(cos(2*b*log(c))^2 + sin(2*b*log(c))^2)*sin(2*b*log(x^n)
+ 2*a)^2 + 2*(2*(cos(4*b*log(c))*cos(2*b*log(c)) + sin(4*b*log(c))*sin(2*b*log(c)))*cos(2*b*log(x^n) + 2*a) +
2*(cos(2*b*log(c))*sin(4*b*log(c)) - cos(4*b*log(c))*sin(2*b*log(c)))*sin(2*b*log(x^n) + 2*a) + cos(4*b*log(c
)))*cos(4*b*log(x^n) + 4*a) + 4*cos(2*b*log(c))*cos(2*b*log(x^n) + 2*a) - 2*(2*(cos(2*b*log(c))*sin(4*b*log(c)
) - cos(4*b*log(c))*sin(2*b*log(c)))*cos(2*b*log(x^n) + 2*a) - 2*(cos(4*b*log(c))*cos(2*b*log(c)) + sin(4*b*lo
g(c))*sin(2*b*log(c)))*sin(2*b*log(x^n) + 2*a) + sin(4*b*log(c)))*sin(4*b*log(x^n) + 4*a) - 4*sin(2*b*log(c))*
sin(2*b*log(x^n) + 2*a) + 1)

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mupad [B]  time = 3.42, size = 87, normalized size = 2.12 \[ \frac {2\,x\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,1{}\mathrm {i}}\,\left (-1+b\,n\,1{}\mathrm {i}\right )-2\,x\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,1{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}\,\left (1+b\,n\,1{}\mathrm {i}\right )}{{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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