Integrand size = 12, antiderivative size = 77 \[ \int \left (c \sin ^4(a+b x)\right )^p \, dx=\frac {\cos (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1+4 p),\frac {1}{2} (3+4 p),\sin ^2(a+b x)\right ) \sin (a+b x) \left (c \sin ^4(a+b x)\right )^p}{b (1+4 p) \sqrt {\cos ^2(a+b x)}} \] Output:
cos(b*x+a)*hypergeom([1/2, 1/2+2*p],[3/2+2*p],sin(b*x+a)^2)*sin(b*x+a)*(c* sin(b*x+a)^4)^p/b/(1+4*p)/(cos(b*x+a)^2)^(1/2)
Time = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.84 \[ \int \left (c \sin ^4(a+b x)\right )^p \, dx=\frac {\sqrt {\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}+2 p,\frac {3}{2}+2 p,\sin ^2(a+b x)\right ) \left (c \sin ^4(a+b x)\right )^p \tan (a+b x)}{b+4 b p} \] Input:
Integrate[(c*Sin[a + b*x]^4)^p,x]
Output:
(Sqrt[Cos[a + b*x]^2]*Hypergeometric2F1[1/2, 1/2 + 2*p, 3/2 + 2*p, Sin[a + b*x]^2]*(c*Sin[a + b*x]^4)^p*Tan[a + b*x])/(b + 4*b*p)
Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3686, 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.
\(\displaystyle \int \left (c \sin ^4(a+b x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (c \sin (a+b x)^4\right )^pdx\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle \sin ^{-4 p}(a+b x) \left (c \sin ^4(a+b x)\right )^p \int \sin ^{4 p}(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin ^{-4 p}(a+b x) \left (c \sin ^4(a+b x)\right )^p \int \sin (a+b x)^{4 p}dx\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\sin (a+b x) \cos (a+b x) \left (c \sin ^4(a+b x)\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (4 p+1),\frac {1}{2} (4 p+3),\sin ^2(a+b x)\right )}{b (4 p+1) \sqrt {\cos ^2(a+b x)}}\) |
Input:
Int[(c*Sin[a + b*x]^4)^p,x]
Output:
(Cos[a + b*x]*Hypergeometric2F1[1/2, (1 + 4*p)/2, (3 + 4*p)/2, Sin[a + b*x ]^2]*Sin[a + b*x]*(c*Sin[a + b*x]^4)^p)/(b*(1 + 4*p)*Sqrt[Cos[a + b*x]^2])
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]
Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^ n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Si n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / ; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
\[\int \left (c \sin \left (b x +a \right )^{4}\right )^{p}d x\]
Input:
int((c*sin(b*x+a)^4)^p,x)
Output:
int((c*sin(b*x+a)^4)^p,x)
\[ \int \left (c \sin ^4(a+b x)\right )^p \, dx=\int { \left (c \sin \left (b x + a\right )^{4}\right )^{p} \,d x } \] Input:
integrate((c*sin(b*x+a)^4)^p,x, algorithm="fricas")
Output:
integral((c*cos(b*x + a)^4 - 2*c*cos(b*x + a)^2 + c)^p, x)
\[ \int \left (c \sin ^4(a+b x)\right )^p \, dx=\int \left (c \sin ^{4}{\left (a + b x \right )}\right )^{p}\, dx \] Input:
integrate((c*sin(b*x+a)**4)**p,x)
Output:
Integral((c*sin(a + b*x)**4)**p, x)
\[ \int \left (c \sin ^4(a+b x)\right )^p \, dx=\int { \left (c \sin \left (b x + a\right )^{4}\right )^{p} \,d x } \] Input:
integrate((c*sin(b*x+a)^4)^p,x, algorithm="maxima")
Output:
integrate((c*sin(b*x + a)^4)^p, x)
\[ \int \left (c \sin ^4(a+b x)\right )^p \, dx=\int { \left (c \sin \left (b x + a\right )^{4}\right )^{p} \,d x } \] Input:
integrate((c*sin(b*x+a)^4)^p,x, algorithm="giac")
Output:
integrate((c*sin(b*x + a)^4)^p, x)
Timed out. \[ \int \left (c \sin ^4(a+b x)\right )^p \, dx=\int {\left (c\,{\sin \left (a+b\,x\right )}^4\right )}^p \,d x \] Input:
int((c*sin(a + b*x)^4)^p,x)
Output:
int((c*sin(a + b*x)^4)^p, x)
\[ \int \left (c \sin ^4(a+b x)\right )^p \, dx=c^{p} \left (\int \sin \left (b x +a \right )^{4 p}d x \right ) \] Input:
int((c*sin(b*x+a)^4)^p,x)
Output:
c**p*int(sin(a + b*x)**(4*p),x)