Integrand size = 17, antiderivative size = 111 \[ \int x \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 x^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4} \left (-3-\frac {4 i}{b n}\right ),\frac {1}{4} \left (1-\frac {4 i}{b n}\right ),e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{(4-3 i b n) \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2}} \] Output:
2*x^2*hypergeom([-3/2, -3/4-I/b/n],[1/4-I/b/n],exp(2*I*a)*(c*x^n)^(2*I*b)) *sin(a+b*ln(c*x^n))^(3/2)/(4-3*I*b*n)/(1-exp(2*I*a)*(c*x^n)^(2*I*b))^(3/2)
Time = 1.11 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.96 \[ \int x \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {6 i b^2 \sqrt {2-2 e^{2 i \left (a+b \log \left (c x^n\right )\right )}} n^2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4}-\frac {i}{b n},\frac {5}{4}-\frac {i}{b n},e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )}{\sqrt {-i e^{-i \left (a+b \log \left (c x^n\right )\right )} \left (-1+e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )} (-4 i+b n) (-4 i+3 b n) (4 i+3 b n)}+\frac {2 x^2 \sqrt {\sin \left (a+b \log \left (c x^n\right )\right )} \left (-3 b n \cos \left (a+b \log \left (c x^n\right )\right )+4 \sin \left (a+b \log \left (c x^n\right )\right )\right )}{16+9 b^2 n^2} \] Input:
Integrate[x*Sin[a + b*Log[c*x^n]]^(3/2),x]
Output:
((-6*I)*b^2*Sqrt[2 - 2*E^((2*I)*(a + b*Log[c*x^n]))]*n^2*x^2*Hypergeometri c2F1[1/2, 1/4 - I/(b*n), 5/4 - I/(b*n), E^((2*I)*(a + b*Log[c*x^n]))])/(Sq rt[((-I)*(-1 + E^((2*I)*(a + b*Log[c*x^n]))))/E^(I*(a + b*Log[c*x^n]))]*(- 4*I + b*n)*(-4*I + 3*b*n)*(4*I + 3*b*n)) + (2*x^2*Sqrt[Sin[a + b*Log[c*x^n ]]]*(-3*b*n*Cos[a + b*Log[c*x^n]] + 4*Sin[a + b*Log[c*x^n]]))/(16 + 9*b^2* n^2)
Time = 0.31 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4996, 4994, 888}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 4996 |
\(\displaystyle \frac {x^2 \left (c x^n\right )^{-2/n} \int \left (c x^n\right )^{\frac {2}{n}-1} \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 4994 |
\(\displaystyle \frac {x^2 \left (c x^n\right )^{-\frac {2}{n}+\frac {3 i b}{2}} \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \int \left (c x^n\right )^{-\frac {3 i b}{2}+\frac {2}{n}-1} \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2}d\left (c x^n\right )}{n \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {2 x^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4} \left (-3-\frac {4 i}{b n}\right ),\frac {1}{4} \left (1-\frac {4 i}{b n}\right ),e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{(4-3 i b n) \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2}}\) |
Input:
Int[x*Sin[a + b*Log[c*x^n]]^(3/2),x]
Output:
(2*x^2*Hypergeometric2F1[-3/2, (-3 - (4*I)/(b*n))/4, (1 - (4*I)/(b*n))/4, E^((2*I)*a)*(c*x^n)^((2*I)*b)]*Sin[a + b*Log[c*x^n]]^(3/2))/((4 - (3*I)*b* n)*(1 - E^((2*I)*a)*(c*x^n)^((2*I)*b))^(3/2))
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_), x_Symbol] : > Simp[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] /; Fr eeQ[{a, b, d, e, m, p}, x] && !IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_ .), x_Symbol] :> Simp[(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])
\[\int x {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}^{\frac {3}{2}}d x\]
Input:
int(x*sin(a+b*ln(c*x^n))^(3/2),x)
Output:
int(x*sin(a+b*ln(c*x^n))^(3/2),x)
Exception generated. \[ \int x \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x*sin(a+b*log(c*x^n))^(3/2),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
Timed out. \[ \int x \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \] Input:
integrate(x*sin(a+b*ln(c*x**n))**(3/2),x)
Output:
Timed out
\[ \int x \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x \sin \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(x*sin(a+b*log(c*x^n))^(3/2),x, algorithm="maxima")
Output:
integrate(x*sin(b*log(c*x^n) + a)^(3/2), x)
\[ \int x \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x \sin \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(x*sin(a+b*log(c*x^n))^(3/2),x, algorithm="giac")
Output:
integrate(x*sin(b*log(c*x^n) + a)^(3/2), x)
Timed out. \[ \int x \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\int x\,{\sin \left (a+b\,\ln \left (c\,x^n\right )\right )}^{3/2} \,d x \] Input:
int(x*sin(a + b*log(c*x^n))^(3/2),x)
Output:
int(x*sin(a + b*log(c*x^n))^(3/2), x)
\[ \int x \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {\sqrt {\sin \left (\mathrm {log}\left (x^{n} c \right ) b +a \right )}\, \sin \left (\mathrm {log}\left (x^{n} c \right ) b +a \right ) x^{2}}{2}-\frac {3 \left (\int \sqrt {\sin \left (\mathrm {log}\left (x^{n} c \right ) b +a \right )}\, \cos \left (\mathrm {log}\left (x^{n} c \right ) b +a \right ) x d x \right ) b n}{4} \] Input:
int(x*sin(a+b*log(c*x^n))^(3/2),x)
Output:
(2*sqrt(sin(log(x**n*c)*b + a))*sin(log(x**n*c)*b + a)*x**2 - 3*int(sqrt(s in(log(x**n*c)*b + a))*cos(log(x**n*c)*b + a)*x,x)*b*n)/4