\(\int (e x)^m \sin ^{\frac {3}{2}}(d (a+b \log (c x^n))) \, dx\) [74]

Optimal result

Integrand size = 23, antiderivative size = 150 \[ \int (e x)^m \sin ^{\frac {3}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {2 (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {2 i+2 i m+3 b d n}{4 b d n},-\frac {2 i+2 i m-b d n}{4 b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d}\right ) \sin ^{\frac {3}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (2+2 m-3 i b d n) \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4581, 4579, 371} \[ \int (e x)^m \sin ^{\frac {3}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {2 (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4} \left (-\frac {2 i (m+1)}{b d n}-3\right ),-\frac {2 i m-b d n+2 i}{4 b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d}\right ) \sin ^{\frac {3}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (-3 i b d n+2 m+2) \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{3/2}} \]

[In]

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

[Out]

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

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*} \text {integral}& = \frac {\left ((e x)^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \text {Subst}\left (\int x^{-1+\frac {1+m}{n}} \sin ^{\frac {3}{2}}(d (a+b \log (x))) \, dx,x,c x^n\right )}{e n} \\ & = \frac {\left ((e x)^{1+m} \left (c x^n\right )^{\frac {3 i b d}{2}-\frac {1+m}{n}} \sin ^{\frac {3}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )\right ) \text {Subst}\left (\int x^{-1-\frac {3 i b d}{2}+\frac {1+m}{n}} \left (1-e^{2 i a d} x^{2 i b d}\right )^{3/2} \, dx,x,c x^n\right )}{e n \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{3/2}} \\ & = \frac {2 (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4} \left (-3-\frac {2 i (1+m)}{b d n}\right ),-\frac {2 i+2 i m-b d n}{4 b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d}\right ) \sin ^{\frac {3}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (2+2 m-3 i b d n) \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.56 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.71 \[ \int (e x)^m \sin ^{\frac {3}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {2 (e x)^m \left (\frac {3 b^2 d^2 \sqrt {2-2 e^{2 i d \left (a+b \log \left (c x^n\right )\right )}} n^2 x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {-2 i-2 i m+b d n}{4 b d n},-\frac {2 i+2 i m-5 b d n}{4 b d n},e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )}{\sqrt {-i e^{-i d \left (a+b \log \left (c x^n\right )\right )} \left (-1+e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )} (2+2 m+i b d n)}+x \sqrt {\sin \left (d \left (a+b \log \left (c x^n\right )\right )\right )} \left (-3 b d n \cos \left (d \left (a+b \log \left (c x^n\right )\right )\right )+2 (1+m) \sin \left (d \left (a+b \log \left (c x^n\right )\right )\right )\right )\right )}{4+8 m+4 m^2+9 b^2 d^2 n^2} \]

[In]

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

[Out]

(2*(e*x)^m*((3*b^2*d^2*Sqrt[2 - 2*E^((2*I)*d*(a + b*Log[c*x^n]))]*n^2*x*Hypergeometric2F1[1/2, (-2*I - (2*I)*m
+ b*d*n)/(4*b*d*n), -1/4*(2*I + (2*I)*m - 5*b*d*n)/(b*d*n), E^((2*I)*d*(a + b*Log[c*x^n]))])/(Sqrt[((-I)*(-1
+ E^((2*I)*d*(a + b*Log[c*x^n]))))/E^(I*d*(a + b*Log[c*x^n]))]*(2 + 2*m + I*b*d*n)) + x*Sqrt[Sin[d*(a + b*Log[
c*x^n])]]*(-3*b*d*n*Cos[d*(a + b*Log[c*x^n])] + 2*(1 + m)*Sin[d*(a + b*Log[c*x^n])])))/(4 + 8*m + 4*m^2 + 9*b^
2*d^2*n^2)

Maple [F]

\[\int \left (e x \right )^{m} {\sin \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{\frac {3}{2}}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int (e x)^m \sin ^{\frac {3}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (e x)^m \sin ^{\frac {3}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int (e x)^m \sin ^{\frac {3}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \sin \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{\frac {3}{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (e x)^m \sin ^{\frac {3}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \sin \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{\frac {3}{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (e x)^m \sin ^{\frac {3}{2}}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int {\sin \left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^{3/2}\,{\left (e\,x\right )}^m \,d x \]

[In]

int(sin(d*(a + b*log(c*x^n)))^(3/2)*(e*x)^m,x)

[Out]

int(sin(d*(a + b*log(c*x^n)))^(3/2)*(e*x)^m, x)