\(\int (e x)^{-1+n} (b \sin (c+d x^n))^p \, dx\) [131]

Optimal result

Integrand size = 20, antiderivative size = 92 \[ \int (e x)^{-1+n} \left (b \sin \left (c+d x^n\right )\right )^p \, dx=\frac {x^{-n} (e x)^n \cos \left (c+d x^n\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+p}{2},\frac {3+p}{2},\sin ^2\left (c+d x^n\right )\right ) \left (b \sin \left (c+d x^n\right )\right )^{1+p}}{b d e n (1+p) \sqrt {\cos ^2\left (c+d x^n\right )}} \] Output:

(e*x)^n*cos(c+d*x^n)*hypergeom([1/2, 1/2*p+1/2],[3/2+1/2*p],sin(c+d*x^n)^2 
)*(b*sin(c+d*x^n))^(p+1)/b/d/e/n/(p+1)/(x^n)/(cos(c+d*x^n)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.96 \[ \int (e x)^{-1+n} \left (b \sin \left (c+d x^n\right )\right )^p \, dx=\frac {x^{1-n} (e x)^{-1+n} \sqrt {\cos ^2\left (c+d x^n\right )} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+p}{2},\frac {3+p}{2},\sin ^2\left (c+d x^n\right )\right ) \left (b \sin \left (c+d x^n\right )\right )^p \tan \left (c+d x^n\right )}{d n (1+p)} \] Input:

Integrate[(e*x)^(-1 + n)*(b*Sin[c + d*x^n])^p,x]
 

Output:

(x^(1 - n)*(e*x)^(-1 + n)*Sqrt[Cos[c + d*x^n]^2]*Hypergeometric2F1[1/2, (1 
 + p)/2, (3 + p)/2, Sin[c + d*x^n]^2]*(b*Sin[c + d*x^n])^p*Tan[c + d*x^n]) 
/(d*n*(1 + p))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3862, 3860, 3042, 3122}

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.

\(\displaystyle \int (e x)^{n-1} \left (b \sin \left (c+d x^n\right )\right )^p \, dx\)

\(\Big \downarrow \) 3862

\(\displaystyle \frac {x^{-n} (e x)^n \int x^{n-1} \left (b \sin \left (d x^n+c\right )\right )^pdx}{e}\)

\(\Big \downarrow \) 3860

\(\displaystyle \frac {x^{-n} (e x)^n \int \left (b \sin \left (d x^n+c\right )\right )^pdx^n}{e n}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {x^{-n} (e x)^n \int \left (b \sin \left (d x^n+c\right )\right )^pdx^n}{e n}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {x^{-n} (e x)^n \cos \left (c+d x^n\right ) \left (b \sin \left (c+d x^n\right )\right )^{p+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {p+1}{2},\frac {p+3}{2},\sin ^2\left (d x^n+c\right )\right )}{b d e n (p+1) \sqrt {\cos ^2\left (c+d x^n\right )}}\)

Input:

Int[(e*x)^(-1 + n)*(b*Sin[c + d*x^n])^p,x]
 

Output:

((e*x)^n*Cos[c + d*x^n]*Hypergeometric2F1[1/2, (1 + p)/2, (3 + p)/2, Sin[c 
 + d*x^n]^2]*(b*Sin[c + d*x^n])^(1 + p))/(b*d*e*n*(1 + p)*x^n*Sqrt[Cos[c + 
 d*x^n]^2])
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 3860

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^ 
p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simplify[ 
(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[ 
(m + 1)/n], 0]))
 

rule 3862

Int[((e_)*(x_))^(m_)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_ 
Symbol] :> Simp[e^IntPart[m]*((e*x)^FracPart[m]/x^FracPart[m])   Int[x^m*(a 
 + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && Int 
egerQ[Simplify[(m + 1)/n]]
 

Maple [F]

\[\int \left (e x \right )^{-1+n} {\left (b \sin \left (c +d \,x^{n}\right )\right )}^{p}d x\]

Input:

int((e*x)^(-1+n)*(b*sin(c+d*x^n))^p,x)
 

Output:

int((e*x)^(-1+n)*(b*sin(c+d*x^n))^p,x)
 

Fricas [F]

\[ \int (e x)^{-1+n} \left (b \sin \left (c+d x^n\right )\right )^p \, dx=\int { \left (e x\right )^{n - 1} \left (b \sin \left (d x^{n} + c\right )\right )^{p} \,d x } \] Input:

integrate((e*x)^(-1+n)*(b*sin(c+d*x^n))^p,x, algorithm="fricas")
 

Output:

integral((e*x)^(n - 1)*(b*sin(d*x^n + c))^p, x)
 

Sympy [F]

\[ \int (e x)^{-1+n} \left (b \sin \left (c+d x^n\right )\right )^p \, dx=\int \left (b \sin {\left (c + d x^{n} \right )}\right )^{p} \left (e x\right )^{n - 1}\, dx \] Input:

integrate((e*x)**(-1+n)*(b*sin(c+d*x**n))**p,x)
 

Output:

Integral((b*sin(c + d*x**n))**p*(e*x)**(n - 1), x)
 

Maxima [F]

\[ \int (e x)^{-1+n} \left (b \sin \left (c+d x^n\right )\right )^p \, dx=\int { \left (e x\right )^{n - 1} \left (b \sin \left (d x^{n} + c\right )\right )^{p} \,d x } \] Input:

integrate((e*x)^(-1+n)*(b*sin(c+d*x^n))^p,x, algorithm="maxima")
 

Output:

integrate((e*x)^(n - 1)*(b*sin(d*x^n + c))^p, x)
 

Giac [F]

\[ \int (e x)^{-1+n} \left (b \sin \left (c+d x^n\right )\right )^p \, dx=\int { \left (e x\right )^{n - 1} \left (b \sin \left (d x^{n} + c\right )\right )^{p} \,d x } \] Input:

integrate((e*x)^(-1+n)*(b*sin(c+d*x^n))^p,x, algorithm="giac")
 

Output:

integrate((e*x)^(n - 1)*(b*sin(d*x^n + c))^p, x)
 

Mupad [F(-1)]

Timed out. \[ \int (e x)^{-1+n} \left (b \sin \left (c+d x^n\right )\right )^p \, dx=\int {\left (b\,\sin \left (c+d\,x^n\right )\right )}^p\,{\left (e\,x\right )}^{n-1} \,d x \] Input:

int((b*sin(c + d*x^n))^p*(e*x)^(n - 1),x)
 

Output:

int((b*sin(c + d*x^n))^p*(e*x)^(n - 1), x)
 

Reduce [F]

\[ \int (e x)^{-1+n} \left (b \sin \left (c+d x^n\right )\right )^p \, dx=\frac {e^{n} b^{p} \left (\int \frac {x^{n} \sin \left (x^{n} d +c \right )^{p}}{x}d x \right )}{e} \] Input:

int((e*x)^(-1+n)*(b*sin(c+d*x^n))^p,x)
 

Output:

(e**n*b**p*int((x**n*sin(x**n*d + c)**p)/x,x))/e