3.285 \(\int \frac{x^m}{\sec ^{\frac{3}{2}}(a+b \log (c x^n))} \, dx\)

Optimal. Leaf size=130 \[ \frac{2 x^{m+1} \text{Hypergeometric2F1}\left (-\frac{3}{2},-\frac{3 b n+2 i m+2 i}{4 b n},-\frac{-b n+2 i m+2 i}{4 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(-3 i b n+2 m+2) \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \sec ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )} \]

[Out]

(2*x^(1 + m)*Hypergeometric2F1[-3/2, -(2*I + (2*I)*m + 3*b*n)/(4*b*n), -(2*I + (2*I)*m - b*n)/(4*b*n), -(E^((2
*I)*a)*(c*x^n)^((2*I)*b))])/((2 + 2*m - (3*I)*b*n)*(1 + E^((2*I)*a)*(c*x^n)^((2*I)*b))^(3/2)*Sec[a + b*Log[c*x
^n]]^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0959615, antiderivative size = 126, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {4509, 4507, 364} \[ \frac{2 x^{m+1} \, _2F_1\left (-\frac{3}{2},\frac{1}{4} \left (-\frac{2 i (m+1)}{b n}-3\right );-\frac{2 i m-b n+2 i}{4 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(-3 i b n+2 m+2) \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \sec ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(1 + m)*Hypergeometric2F1[-3/2, (-3 - ((2*I)*(1 + m))/(b*n))/4, -(2*I + (2*I)*m - b*n)/(4*b*n), -(E^((2*I
)*a)*(c*x^n)^((2*I)*b))])/((2 + 2*m - (3*I)*b*n)*(1 + E^((2*I)*a)*(c*x^n)^((2*I)*b))^(3/2)*Sec[a + b*Log[c*x^n
]]^(3/2))

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])

Rule 4507

Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(Sec[d*(a + b*Log[x])]^p*(1
+ E^(2*I*a*d)*x^(2*I*b*d))^p)/x^(I*b*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, p}, 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 \frac{x^m}{\sec ^{\frac{3}{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 \frac{x^{-1+\frac{1+m}{n}}}{\sec ^{\frac{3}{2}}(a+b \log (x))} \, dx,x,c x^n\right )}{n}\\ &=\frac{\left (x^{1+m} \left (c x^n\right )^{\frac{3 i b}{2}-\frac{1+m}{n}}\right ) \operatorname{Subst}\left (\int x^{-1-\frac{3 i b}{2}+\frac{1+m}{n}} \left (1+e^{2 i a} x^{2 i b}\right )^{3/2} \, dx,x,c x^n\right )}{n \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \sec ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )}\\ &=\frac{2 x^{1+m} \, _2F_1\left (-\frac{3}{2},\frac{1}{4} \left (-3-\frac{2 i (1+m)}{b n}\right );-\frac{2 i+2 i m-b n}{4 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2+2 m-3 i b n) \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \sec ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )}\\ \end{align*}

Mathematica [A]  time = 2.51573, size = 202, normalized size = 1.55 \[ \frac{2 x^{m+1} \left (3 b^2 n^2 \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sec ^2\left (a+b \log \left (c x^n\right )\right ) \text{Hypergeometric2F1}\left (1,-\frac{-3 b n+2 i m+2 i}{4 b n},-\frac{-5 b n+2 i m+2 i}{4 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )+(i b n+2 m+2) \left (3 b n \tan \left (a+b \log \left (c x^n\right )\right )+2 m+2\right )\right )}{(i b n+2 m+2) (-3 i b n+2 m+2) (3 i b n+2 m+2) \sec ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*x^(1 + m)*(3*b^2*n^2*(1 + E^((2*I)*a)*(c*x^n)^((2*I)*b))*Hypergeometric2F1[1, -(2*I + (2*I)*m - 3*b*n)/(4*b
*n), -(2*I + (2*I)*m - 5*b*n)/(4*b*n), -E^((2*I)*(a + b*Log[c*x^n]))]*Sec[a + b*Log[c*x^n]]^2 + (2 + 2*m + I*b
*n)*(2 + 2*m + 3*b*n*Tan[a + b*Log[c*x^n]])))/((2 + 2*m + I*b*n)*(2 + 2*m - (3*I)*b*n)*(2 + 2*m + (3*I)*b*n)*S
ec[a + b*Log[c*x^n]]^(3/2))

________________________________________________________________________________________

Maple [F]  time = 0.288, size = 0, normalized size = 0. \begin{align*} \int{{x}^{m} \left ( \sec \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{\sec \left (b \log \left (c x^{n}\right ) + a\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m/sec(a+b*ln(c*x**n))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out