\(\int \frac {x^m}{\sqrt {\cos (a+b \log (c x^n))}} \, dx\) [129]

Optimal result

Integrand size = 19, antiderivative size = 130 \[ \int \frac {x^m}{\sqrt {\cos \left (a+b \log \left (c x^n\right )\right )}} \, dx=\frac {2 x^{1+m} \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {2 i+2 i m-b n}{4 b n},-\frac {2 i+2 i m-5 b n}{4 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2+2 m+i b n) \sqrt {\cos \left (a+b \log \left (c x^n\right )\right )}} \]

[Out]

2*x^(1+m)*hypergeom([1/2, 1/4*(-2*I-2*I*m+b*n)/b/n],[1/4*(-2*I-2*I*m+5*b*n)/b/n],-exp(2*I*a)*(c*x^n)^(2*I*b))*
(1+exp(2*I*a)*(c*x^n)^(2*I*b))^(1/2)/(2+2*m+I*b*n)/cos(a+b*ln(c*x^n))^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4582, 4580, 371} \[ \int \frac {x^m}{\sqrt {\cos \left (a+b \log \left (c x^n\right )\right )}} \, dx=\frac {2 x^{m+1} \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {2 i m-b n+2 i}{4 b n},-\frac {2 i m-5 b n+2 i}{4 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(i b n+2 m+2) \sqrt {\cos \left (a+b \log \left (c x^n\right )\right )}} \]

[In]

Int[x^m/Sqrt[Cos[a + b*Log[c*x^n]]],x]

[Out]

(2*x^(1 + m)*Sqrt[1 + E^((2*I)*a)*(c*x^n)^((2*I)*b)]*Hypergeometric2F1[1/2, -1/4*(2*I + (2*I)*m - b*n)/(b*n),
-1/4*(2*I + (2*I)*m - 5*b*n)/(b*n), -(E^((2*I)*a)*(c*x^n)^((2*I)*b))])/((2 + 2*m + I*b*n)*Sqrt[Cos[a + b*Log[c
*x^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 4580

Int[Cos[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_)*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[Cos[d*(a + b*Log[x])]^p*(x^(
I*b*d*p)/(1 + E^(2*I*a*d)*x^(2*I*b*d))^p), Int[(e*x)^m*((1 + E^(2*I*a*d)*x^(2*I*b*d))^p/x^(I*b*d*p)), x], x] /
; FreeQ[{a, b, d, e, m, p}, x] &&  !IntegerQ[p]

Rule 4582

Int[Cos[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[(e*x)^(m + 1)
/(e*n*(c*x^n)^((m + 1)/n)), Subst[Int[x^((m + 1)/n - 1)*Cos[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*} \text {integral}& = \frac {\left (x^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {1+m}{n}}}{\sqrt {\cos (a+b \log (x))}} \, dx,x,c x^n\right )}{n} \\ & = \frac {\left (x^{1+m} \left (c x^n\right )^{-\frac {i b}{2}-\frac {1+m}{n}} \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}}\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {i b}{2}+\frac {1+m}{n}}}{\sqrt {1+e^{2 i a} x^{2 i b}}} \, dx,x,c x^n\right )}{n \sqrt {\cos \left (a+b \log \left (c x^n\right )\right )}} \\ & = \frac {2 x^{1+m} \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {2 i+2 i m-b n}{4 b n},-\frac {2 i+2 i m-5 b n}{4 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2+2 m+i b n) \sqrt {\cos \left (a+b \log \left (c x^n\right )\right )}} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 0.69 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.92 \[ \int \frac {x^m}{\sqrt {\cos \left (a+b \log \left (c x^n\right )\right )}} \, dx=\frac {2 \left (1+e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right ) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,-\frac {2 i+2 i m-3 b n}{4 b n},-\frac {2 i+2 i m-5 b n}{4 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )}{(2+2 m+i b n) \sqrt {\cos \left (a+b \log \left (c x^n\right )\right )}} \]

[In]

Integrate[x^m/Sqrt[Cos[a + b*Log[c*x^n]]],x]

[Out]

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

Maple [F]

\[\int \frac {x^{m}}{\sqrt {\cos \left (a +b \ln \left (c \,x^{n}\right )\right )}}d x\]

[In]

int(x^m/cos(a+b*ln(c*x^n))^(1/2),x)

[Out]

int(x^m/cos(a+b*ln(c*x^n))^(1/2),x)

Fricas [F(-2)]

Exception generated. \[ \int \frac {x^m}{\sqrt {\cos \left (a+b \log \left (c x^n\right )\right )}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^m/cos(a+b*log(c*x^n))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

\[ \int \frac {x^m}{\sqrt {\cos \left (a+b \log \left (c x^n\right )\right )}} \, dx=\int \frac {x^{m}}{\sqrt {\cos {\left (a + b \log {\left (c x^{n} \right )} \right )}}}\, dx \]

[In]

integrate(x**m/cos(a+b*ln(c*x**n))**(1/2),x)

[Out]

Integral(x**m/sqrt(cos(a + b*log(c*x**n))), x)

Maxima [F]

\[ \int \frac {x^m}{\sqrt {\cos \left (a+b \log \left (c x^n\right )\right )}} \, dx=\int { \frac {x^{m}}{\sqrt {\cos \left (b \log \left (c x^{n}\right ) + a\right )}} \,d x } \]

[In]

integrate(x^m/cos(a+b*log(c*x^n))^(1/2),x, algorithm="maxima")

[Out]

integrate(x^m/sqrt(cos(b*log(c*x^n) + a)), x)

Giac [F]

\[ \int \frac {x^m}{\sqrt {\cos \left (a+b \log \left (c x^n\right )\right )}} \, dx=\int { \frac {x^{m}}{\sqrt {\cos \left (b \log \left (c x^{n}\right ) + a\right )}} \,d x } \]

[In]

integrate(x^m/cos(a+b*log(c*x^n))^(1/2),x, algorithm="giac")

[Out]

integrate(x^m/sqrt(cos(b*log(c*x^n) + a)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^m}{\sqrt {\cos \left (a+b \log \left (c x^n\right )\right )}} \, dx=\int \frac {x^m}{\sqrt {\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}} \,d x \]

[In]

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

[Out]

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