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*Log[c*x^n]]) + b*n*x*Sec[a + b*Log[c*x^n]]*Tan[a + b*Log[c*x^n]]
________________________________________________________________________________________
Rubi [C] time = 0.132896, 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 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]
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])
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*}
Mathematica [A] time = 0.655003, 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]])
________________________________________________________________________________________
Maple [C] time = 0.48, size = 537, normalized size = 13.1 \begin{align*} \text{result too large to display} \end{align*}
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*x/(exp(I*(-I*b*Pi*csgn(I*c*x^n)^3+I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*b*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*b
*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)+2*b*ln(x^n)+2*b*ln(c)+2*a))+1)^2*(((x^n)^(I*b))^3*(c^(I*b))^3*b*n*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)-(x^n)^(I*b)*c^(I*b)*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*((x^n)^(I*b))^3*(c^(I*b))^3*exp(3/2*b*Pi*csgn(I*c*x^n)^3)*exp(-3/2*b*P
i*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)*c
sgn(I*x^n))*exp(3*I*a)-I*(x^n)^(I*b)*c^(I*b)*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))
________________________________________________________________________________________
Maxima [B] time = 2.87452, size = 2290, normalized size = 55.85 \begin{align*} \text{result too large to display} \end{align*}
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)
________________________________________________________________________________________
Fricas [A] time = 0.474285, size = 146, normalized size = 3.56 \begin{align*} \frac{b n x \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - x \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{\cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2}} \end{align*}
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
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}
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)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
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)