3.279 \(\int x^m \sec ^2(a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=102 \[ \frac {4 e^{2 i a} x^{m+1} \left (c x^n\right )^{2 i b} \, _2F_1\left (2,-\frac {i (m+1)-2 b n}{2 b n};-\frac {i (m+1)-4 b n}{2 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{2 i b n+m+1} \]

[Out]

4*exp(2*I*a)*x^(1+m)*(c*x^n)^(2*I*b)*hypergeom([2, 1/2*(-I*(1+m)+2*b*n)/b/n],[1/2*(-I*(1+m)+4*b*n)/b/n],-exp(2
*I*a)*(c*x^n)^(2*I*b))/(1+m+2*I*b*n)

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Rubi [A]  time = 0.08, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4509, 4505, 364} \[ \frac {4 e^{2 i a} x^{m+1} \left (c x^n\right )^{2 i b} \, _2F_1\left (2,-\frac {i (m+1)-2 b n}{2 b n};-\frac {i (m+1)-4 b n}{2 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{2 i b n+m+1} \]

Antiderivative was successfully verified.

[In]

Int[x^m*Sec[a + b*Log[c*x^n]]^2,x]

[Out]

(4*E^((2*I)*a)*x^(1 + m)*(c*x^n)^((2*I)*b)*Hypergeometric2F1[2, -(I*(1 + m) - 2*b*n)/(2*b*n), -(I*(1 + m) - 4*
b*n)/(2*b*n), -(E^((2*I)*a)*(c*x^n)^((2*I)*b))])/(1 + m + (2*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 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 4509

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

Rubi steps

\begin {align*} \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {\left (x^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \operatorname {Subst}\left (\int x^{-1+\frac {1+m}{n}} \sec ^2(a+b \log (x)) \, dx,x,c x^n\right )}{n}\\ &=\frac {\left (4 e^{2 i a} x^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \operatorname {Subst}\left (\int \frac {x^{-1+2 i b+\frac {1+m}{n}}}{\left (1+e^{2 i a} x^{2 i b}\right )^2} \, dx,x,c x^n\right )}{n}\\ &=\frac {4 e^{2 i a} x^{1+m} \left (c x^n\right )^{2 i b} \, _2F_1\left (2,-\frac {i (1+m)-2 b n}{2 b n};-\frac {i (1+m)-4 b n}{2 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+m+2 i b n}\\ \end {align*}

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Mathematica [B]  time = 17.18, size = 482, normalized size = 4.73 \[ \frac {x^{m+1} \sin (b n \log (x)) \sec \left (a+b \left (\log \left (c x^n\right )-n \log (x)\right )\right ) \sec \left (a+b \left (\log \left (c x^n\right )-n \log (x)\right )+b n \log (x)\right )}{b n}-\frac {(m+1) \sec \left (a+b \left (\log \left (c x^n\right )-n \log (x)\right )\right ) \left (\frac {x^{m+1} \sin (b n \log (x)) \sec \left (a+b \log \left (c x^n\right )\right )}{m+1}-\frac {i \cos \left (a+b \left (\log \left (c x^n\right )-n \log (x)\right )\right ) \exp \left (-\frac {(2 m+1) \left (a+b \left (\log \left (c x^n\right )-n \log (x)\right )\right )}{b n}\right ) \left ((2 i b n+m+1) \left (-\exp \left (\frac {2 a m+a+b (2 m+1) \left (\log \left (c x^n\right )-n \log (x)\right )+b (m+1) n \log (x)}{b n}\right )\right ) \, _2F_1\left (1,-\frac {i (m+1)}{2 b n};1-\frac {i (m+1)}{2 b n};-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )+(m+1) \exp \left (\frac {a (2 i b n+2 m+1)}{b n}+\frac {(2 i b n+2 m+1) \left (\log \left (c x^n\right )-n \log (x)\right )}{n}+\log (x) (2 i b n+m+1)\right ) \, _2F_1\left (1,-\frac {i (m+2 i b n+1)}{2 b n};-\frac {i (m+4 i b n+1)}{2 b n};-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )-i (2 i b n+m+1) \tan \left (a+b \log \left (c x^n\right )\right ) \exp \left (\frac {2 a m+a+b (2 m+1) \left (\log \left (c x^n\right )-n \log (x)\right )+b (m+1) n \log (x)}{b n}\right )\right )}{(m+1) (2 i b n+m+1)}\right )}{b n} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^m*Sec[a + b*Log[c*x^n]]^2,x]

[Out]

(x^(1 + m)*Sec[a + b*(-(n*Log[x]) + Log[c*x^n])]*Sec[a + b*n*Log[x] + b*(-(n*Log[x]) + Log[c*x^n])]*Sin[b*n*Lo
g[x]])/(b*n) - ((1 + m)*Sec[a + b*(-(n*Log[x]) + Log[c*x^n])]*((x^(1 + m)*Sec[a + b*Log[c*x^n]]*Sin[b*n*Log[x]
])/(1 + m) - (I*Cos[a + b*(-(n*Log[x]) + Log[c*x^n])]*(-(E^((a + 2*a*m + b*(1 + m)*n*Log[x] + b*(1 + 2*m)*(-(n
*Log[x]) + Log[c*x^n]))/(b*n))*(1 + m + (2*I)*b*n)*Hypergeometric2F1[1, ((-1/2*I)*(1 + m))/(b*n), 1 - ((I/2)*(
1 + m))/(b*n), -E^((2*I)*(a + b*Log[c*x^n]))]) + E^((a*(1 + 2*m + (2*I)*b*n))/(b*n) + (1 + m + (2*I)*b*n)*Log[
x] + ((1 + 2*m + (2*I)*b*n)*(-(n*Log[x]) + Log[c*x^n]))/n)*(1 + m)*Hypergeometric2F1[1, ((-1/2*I)*(1 + m + (2*
I)*b*n))/(b*n), ((-1/2*I)*(1 + m + (4*I)*b*n))/(b*n), -E^((2*I)*(a + b*Log[c*x^n]))] - I*E^((a + 2*a*m + b*(1
+ m)*n*Log[x] + b*(1 + 2*m)*(-(n*Log[x]) + Log[c*x^n]))/(b*n))*(1 + m + (2*I)*b*n)*Tan[a + b*Log[c*x^n]]))/(E^
(((1 + 2*m)*(a + b*(-(n*Log[x]) + Log[c*x^n])))/(b*n))*(1 + m)*(1 + m + (2*I)*b*n))))/(b*n)

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fricas [F]  time = 1.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^m*sec(b*log(c*x^n) + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m*sec(b*log(c*x^n) + a)^2, x)

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maple [F]  time = 1.53, size = 0, normalized size = 0.00 \[ \int x^{m} \left (\sec ^{2}\left (a +b \ln \left (c \,x^{n}\right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*sec(a+b*ln(c*x^n))^2,x)

[Out]

int(x^m*sec(a+b*ln(c*x^n))^2,x)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^m}{{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/cos(a + b*log(c*x^n))^2,x)

[Out]

int(x^m/cos(a + b*log(c*x^n))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*sec(a+b*ln(c*x**n))**2,x)

[Out]

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

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