Integrand size = 21, antiderivative size = 80 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \, dx=-\frac {2 \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3+2 n),\frac {1}{4} (7+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (3+2 n) \sqrt {\sin ^2(c+d x)}} \] Output:
-2*cos(d*x+c)^(3/2)*(b*cos(d*x+c))^n*hypergeom([1/2, 3/4+1/2*n],[7/4+1/2*n ],cos(d*x+c)^2)*sin(d*x+c)/d/(3+2*n)/(sin(d*x+c)^2)^(1/2)
Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \, dx=-\frac {\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^n \csc (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} \left (\frac {3}{2}+n\right ),\frac {1}{2} \left (\frac {7}{2}+n\right ),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}}{d \left (\frac {3}{2}+n\right )} \] Input:
Integrate[Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^n,x]
Output:
-((Cos[c + d*x]^(3/2)*(b*Cos[c + d*x])^n*Csc[c + d*x]*Hypergeometric2F1[1/ 2, (3/2 + n)/2, (7/2 + n)/2, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2])/(d*(3/2 + n)))
Time = 0.25 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2034, 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.
\(\displaystyle \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \, dx\) |
\(\Big \downarrow \) 2034 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \cos ^{n+\frac {1}{2}}(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {1}{2}}dx\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle -\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n+3),\frac {1}{4} (2 n+7),\cos ^2(c+d x)\right )}{d (2 n+3) \sqrt {\sin ^2(c+d x)}}\) |
Input:
Int[Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^n,x]
Output:
(-2*Cos[c + d*x]^(3/2)*(b*Cos[c + d*x])^n*Hypergeometric2F1[1/2, (3 + 2*n) /4, (7 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(3 + 2*n)*Sqrt[Sin[c + d *x]^2])
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[b^IntPart [n]*((b*v)^FracPart[n]/(a^IntPart[n]*(a*v)^FracPart[n])) Int[(a*v)^(m + n )*Fx, x], x] /; FreeQ[{a, b, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[m + n]
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]
\[\int \sqrt {\cos \left (d x +c \right )}\, \left (\cos \left (d x +c \right ) b \right )^{n}d x\]
Input:
int(cos(d*x+c)^(1/2)*(cos(d*x+c)*b)^n,x)
Output:
int(cos(d*x+c)^(1/2)*(cos(d*x+c)*b)^n,x)
\[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{n} \sqrt {\cos \left (d x + c\right )} \,d x } \] Input:
integrate(cos(d*x+c)^(1/2)*(b*cos(d*x+c))^n,x, algorithm="fricas")
Output:
integral((b*cos(d*x + c))^n*sqrt(cos(d*x + c)), x)
\[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \, dx=\int \left (b \cos {\left (c + d x \right )}\right )^{n} \sqrt {\cos {\left (c + d x \right )}}\, dx \] Input:
integrate(cos(d*x+c)**(1/2)*(b*cos(d*x+c))**n,x)
Output:
Integral((b*cos(c + d*x))**n*sqrt(cos(c + d*x)), x)
\[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{n} \sqrt {\cos \left (d x + c\right )} \,d x } \] Input:
integrate(cos(d*x+c)^(1/2)*(b*cos(d*x+c))^n,x, algorithm="maxima")
Output:
integrate((b*cos(d*x + c))^n*sqrt(cos(d*x + c)), x)
\[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{n} \sqrt {\cos \left (d x + c\right )} \,d x } \] Input:
integrate(cos(d*x+c)^(1/2)*(b*cos(d*x+c))^n,x, algorithm="giac")
Output:
integrate((b*cos(d*x + c))^n*sqrt(cos(d*x + c)), x)
Timed out. \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \, dx=\int \sqrt {\cos \left (c+d\,x\right )}\,{\left (b\,\cos \left (c+d\,x\right )\right )}^n \,d x \] Input:
int(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^n,x)
Output:
int(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^n, x)
\[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \, dx=b^{n} \left (\int \cos \left (d x +c \right )^{n +\frac {1}{2}}d x \right ) \] Input:
int(cos(d*x+c)^(1/2)*(b*cos(d*x+c))^n,x)
Output:
b**n*int(cos(c + d*x)**((2*n + 1)/2),x)