Integrand size = 31, antiderivative size = 163 \[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=-\frac {2 A \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1+2 n),\frac {1}{4} (5+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1+2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3+2 n),\frac {1}{4} (7+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (3+2 n) \sqrt {\sin ^2(c+d x)}} \]
-2*B*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)-2*A*(b*cos(d*x +c))^n*hypergeom([1/2, 1/4+1/2*n],[5/4+1/2*n],cos(d*x+c)^2)*sin(d*x+c)*cos (d*x+c)^(1/2)/d/(1+2*n)/(sin(d*x+c)^2)^(1/2)
Time = 0.17 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.85 \[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=-\frac {2 \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \csc (c+d x) \left (A (3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1+2 n),\frac {1}{4} (5+2 n),\cos ^2(c+d x)\right )+B (1+2 n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3+2 n),\frac {1}{4} (7+2 n),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (1+2 n) (3+2 n)} \]
(-2*Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^n*Csc[c + d*x]*(A*(3 + 2*n)*Hyperg eometric2F1[1/2, (1 + 2*n)/4, (5 + 2*n)/4, Cos[c + d*x]^2] + B*(1 + 2*n)*C os[c + d*x]*Hypergeometric2F1[1/2, (3 + 2*n)/4, (7 + 2*n)/4, Cos[c + d*x]^ 2])*Sqrt[Sin[c + d*x]^2])/(d*(1 + 2*n)*(3 + 2*n))
Time = 0.42 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2034, 3042, 3227, 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.
| \(\displaystyle \int \frac {(A+B \cos (c+d x)) (b \cos (c+d x))^n}{\sqrt {\cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 2034 |
| \(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \cos ^{n-\frac {1}{2}}(c+d x) (A+B \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {1}{2}} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (A \int \cos ^{n-\frac {1}{2}}(c+d x)dx+B \int \cos ^{n+\frac {1}{2}}(c+d x)dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (A \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {1}{2}}dx+B \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {1}{2}}dx\right )\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (-\frac {2 A \sin (c+d x) \cos ^{n+\frac {1}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n+1),\frac {1}{4} (2 n+5),\cos ^2(c+d x)\right )}{d (2 n+1) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \sin (c+d x) \cos ^{n+\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n+3),\frac {1}{4} (2 n+7),\cos ^2(c+d x)\right )}{d (2 n+3) \sqrt {\sin ^2(c+d x)}}\right )\) |
((b*Cos[c + d*x])^n*((-2*A*Cos[c + d*x]^(1/2 + n)*Hypergeometric2F1[1/2, ( 1 + 2*n)/4, (5 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 + 2*n)*Sqrt[S in[c + d*x]^2]) - (2*B*Cos[c + d*x]^(3/2 + 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])))/Cos[c + d*x]^n
3.10.22.3.1 Defintions of rubi rules used
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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
\[\int \frac {\left (\cos \left (d x +c \right ) b \right )^{n} \left (A +B \cos \left (d x +c \right )\right )}{\sqrt {\cos \left (d x +c \right )}}d x\]
\[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\int \frac {\left (b \cos {\left (c + d x \right )}\right )^{n} \left (A + B \cos {\left (c + d x \right )}\right )}{\sqrt {\cos {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{\sqrt {\cos \left (c+d\,x\right )}} \,d x \]