3.1.56 \(\int \frac {\sqrt {\sin (a+b \log (c x^n))}}{x^2} \, dx\) [56]

Optimal. Leaf size=111 \[ -\frac {2 \, _2F_1\left (-\frac {1}{2},\frac {1}{4} \left (-1+\frac {2 i}{b n}\right );\frac {1}{4} \left (3+\frac {2 i}{b n}\right );e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}}{(2+i b n) x \sqrt {1-e^{2 i a} \left (c x^n\right )^{2 i b}}} \]

[Out]

-2*hypergeom([-1/2, -1/4+1/2*I/b/n],[3/4+1/2*I/b/n],exp(2*I*a)*(c*x^n)^(2*I*b))*sin(a+b*ln(c*x^n))^(1/2)/(2+I*
b*n)/x/(1-exp(2*I*a)*(c*x^n)^(2*I*b))^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 111, normalized size of antiderivative = 1.00, 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 = {4581, 4579, 371} \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {1}{2},\frac {1}{4} \left (\frac {2 i}{b n}-1\right );\frac {1}{4} \left (3+\frac {2 i}{b n}\right );e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}}{x (2+i b n) \sqrt {1-e^{2 i a} \left (c x^n\right )^{2 i b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sin[a + b*Log[c*x^n]]]/x^2,x]

[Out]

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

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 4579

Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_), x_Symbol] :> Dist[Sin[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 4581

Int[((e_.)*(x_))^(m_.)*Sin[((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)*Sin[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 \frac {\sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}}{x^2} \, dx &=\frac {\left (c x^n\right )^{\frac {1}{n}} \text {Subst}\left (\int x^{-1-\frac {1}{n}} \sqrt {\sin (a+b \log (x))} \, dx,x,c x^n\right )}{n x}\\ &=\frac {\left (\left (c x^n\right )^{\frac {i b}{2}+\frac {1}{n}} \sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}\right ) \text {Subst}\left (\int x^{-1-\frac {i b}{2}-\frac {1}{n}} \sqrt {1-e^{2 i a} x^{2 i b}} \, dx,x,c x^n\right )}{n x \sqrt {1-e^{2 i a} \left (c x^n\right )^{2 i b}}}\\ &=-\frac {2 \, _2F_1\left (-\frac {1}{2},\frac {1}{4} \left (-1+\frac {2 i}{b n}\right );\frac {1}{4} \left (3+\frac {2 i}{b n}\right );e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}}{(2+i b n) x \sqrt {1-e^{2 i a} \left (c x^n\right )^{2 i b}}}\\ \end {align*}

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Mathematica [A]
time = 6.79, size = 149, normalized size = 1.34 \begin {gather*} \frac {i \sqrt {2} \sqrt {-i e^{-i a} \left (c x^n\right )^{-i b} \left (-1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )} \, _2F_1\left (-\frac {1}{2},-\frac {1}{4}+\frac {i}{2 b n};\frac {3}{4}+\frac {i}{2 b n};e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(-2 i+b n) x \sqrt {1-e^{2 i a} \left (c x^n\right )^{2 i b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Sin[a + b*Log[c*x^n]]]/x^2,x]

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {\sin }\left (a +b \ln \left (c \,x^{n}\right )\right )}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(sin(a+b*ln(c*x^n))^(1/2)/x^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(sin(a+b*log(c*x^n))^(1/2)/x^2,x, algorithm="maxima")

[Out]

integrate(sqrt(sin(b*log(c*x^n) + a))/x^2, x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(sin(a + b*log(c*x**n)))/x**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(sin(a+b*log(c*x^n))^(1/2)/x^2,x, algorithm="giac")

[Out]

integrate(sqrt(sin(b*log(c*x^n) + a))/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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