3.3.44 \(\int x \sec ^2(a+b \log (c x^n)) \, dx\) [244]

Optimal. Leaf size=79 \[ \frac {2 e^{2 i a} x^2 \left (c x^n\right )^{2 i b} \, _2F_1\left (2,1-\frac {i}{b n};2-\frac {i}{b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+i b n} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4605, 4601, 371} \begin {gather*} \frac {2 e^{2 i a} x^2 \left (c x^n\right )^{2 i b} \, _2F_1\left (2,1-\frac {i}{b n};2-\frac {i}{b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+i b n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sec[a + b*Log[c*x^n]]^2,x]

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 4601

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 4605

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

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Mathematica [A]
time = 5.63, size = 149, normalized size = 1.89 \begin {gather*} \frac {x^2 \left (e^{2 i a} \left (c x^n\right )^{2 i b} \, _2F_1\left (1,1-\frac {i}{b n};2-\frac {i}{b n};-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )+(-i+b n) \left (-i \, _2F_1\left (1,-\frac {i}{b n};1-\frac {i}{b n};-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )+\tan \left (a+b \log \left (c x^n\right )\right )\right )\right )}{b n (-i+b n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int x \left (\sec ^{2}\left (a +b \ln \left (c \,x^{n}\right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(x^2*cos(2*b*log(x^n) + 2*a)*sin(2*b*log(c)) + x^2*cos(2*b*log(c))*sin(2*b*log(x^n) + 2*a) - 2*(2*b^2*n^2*co
s(2*b*log(c))*cos(2*b*log(x^n) + 2*a) - 2*b^2*n^2*sin(2*b*log(c))*sin(2*b*log(x^n) + 2*a) + (b^2*cos(2*b*log(c
))^2 + b^2*sin(2*b*log(c))^2)*n^2*cos(2*b*log(x^n) + 2*a)^2 + (b^2*cos(2*b*log(c))^2 + b^2*sin(2*b*log(c))^2)*
n^2*sin(2*b*log(x^n) + 2*a)^2 + b^2*n^2)*integrate((x*cos(2*b*log(x^n) + 2*a)*sin(2*b*log(c)) + x*cos(2*b*log(
c))*sin(2*b*log(x^n) + 2*a))/(2*b^2*n^2*cos(2*b*log(c))*cos(2*b*log(x^n) + 2*a) - 2*b^2*n^2*sin(2*b*log(c))*si
n(2*b*log(x^n) + 2*a) + (b^2*cos(2*b*log(c))^2 + b^2*sin(2*b*log(c))^2)*n^2*cos(2*b*log(x^n) + 2*a)^2 + (b^2*c
os(2*b*log(c))^2 + b^2*sin(2*b*log(c))^2)*n^2*sin(2*b*log(x^n) + 2*a)^2 + b^2*n^2), x))/(2*b*n*cos(2*b*log(c))
*cos(2*b*log(x^n) + 2*a) + (b*cos(2*b*log(c))^2 + b*sin(2*b*log(c))^2)*n*cos(2*b*log(x^n) + 2*a)^2 - 2*b*n*sin
(2*b*log(c))*sin(2*b*log(x^n) + 2*a) + (b*cos(2*b*log(c))^2 + b*sin(2*b*log(c))^2)*n*sin(2*b*log(x^n) + 2*a)^2
+ b*n)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/cos(a + b*log(c*x^n))^2,x)

[Out]

int(x/cos(a + b*log(c*x^n))^2, x)

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