Integrand size = 40, antiderivative size = 163 \[ \int \frac {(b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 B (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-5+2 n),\frac {1}{4} (-1+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (5-2 n) \cos ^{\frac {5}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}}+\frac {2 C (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3+2 n),\frac {1}{4} (1+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (3-2 n) \cos ^{\frac {3}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}} \]
2*B*(b*cos(d*x+c))^n*hypergeom([1/2, -5/4+1/2*n],[-1/4+1/2*n],cos(d*x+c)^2 )*sin(d*x+c)/d/(5-2*n)/cos(d*x+c)^(5/2)/(sin(d*x+c)^2)^(1/2)+2*C*(b*cos(d* x+c))^n*hypergeom([1/2, -3/4+1/2*n],[1/4+1/2*n],cos(d*x+c)^2)*sin(d*x+c)/d /(3-2*n)/cos(d*x+c)^(3/2)/(sin(d*x+c)^2)^(1/2)
Time = 0.15 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.85 \[ \int \frac {(b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {2 (b \cos (c+d x))^n \csc (c+d x) \left (B (-3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-5+2 n),\frac {1}{4} (-1+2 n),\cos ^2(c+d x)\right )+C (-5+2 n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3+2 n),\frac {1}{4} (1+2 n),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (-5+2 n) (-3+2 n) \cos ^{\frac {5}{2}}(c+d x)} \]
(-2*(b*Cos[c + d*x])^n*Csc[c + d*x]*(B*(-3 + 2*n)*Hypergeometric2F1[1/2, ( -5 + 2*n)/4, (-1 + 2*n)/4, Cos[c + d*x]^2] + C*(-5 + 2*n)*Cos[c + d*x]*Hyp ergeometric2F1[1/2, (-3 + 2*n)/4, (1 + 2*n)/4, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(-5 + 2*n)*(-3 + 2*n)*Cos[c + d*x]^(5/2))
Time = 0.49 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {2034, 3042, 3489, 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.
\(\displaystyle \int \frac {(b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2034 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \cos ^{n-\frac {9}{2}}(c+d x) \left (C \cos ^2(c+d x)+B \cos (c+d x)\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {9}{2}} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 3489 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \cos ^{n-\frac {7}{2}}(c+d x) (B+C \cos (c+d x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {7}{2}} \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (B \int \cos ^{n-\frac {7}{2}}(c+d x)dx+C \int \cos ^{n-\frac {5}{2}}(c+d x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (B \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {7}{2}}dx+C \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {5}{2}}dx\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {2 B \sin (c+d x) \cos ^{n-\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n-5),\frac {1}{4} (2 n-1),\cos ^2(c+d x)\right )}{d (5-2 n) \sqrt {\sin ^2(c+d x)}}+\frac {2 C \sin (c+d x) \cos ^{n-\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n-3),\frac {1}{4} (2 n+1),\cos ^2(c+d x)\right )}{d (3-2 n) \sqrt {\sin ^2(c+d x)}}\right )\) |
((b*Cos[c + d*x])^n*((2*B*Cos[c + d*x]^(-5/2 + n)*Hypergeometric2F1[1/2, ( -5 + 2*n)/4, (-1 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(5 - 2*n)*Sqrt [Sin[c + d*x]^2]) + (2*C*Cos[c + d*x]^(-3/2 + n)*Hypergeometric2F1[1/2, (- 3 + 2*n)/4, (1 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(3 - 2*n)*Sqrt[S in[c + d*x]^2])))/Cos[c + d*x]^n
3.3.31.3.1 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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b Int[(b*Sin[e + f* x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, m}, x]
\[\int \frac {\left (\cos \left (d x +c \right ) b \right )^{n} \left (B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{\cos \left (d x +c \right )^{\frac {9}{2}}}d x\]
\[ \int \frac {(b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]
integrate((b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2), x, algorithm="fricas")
Timed out. \[ \int \frac {(b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {(b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]
integrate((b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2), x, algorithm="maxima")
\[ \int \frac {(b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]
integrate((b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2), x, algorithm="giac")
Timed out. \[ \int \frac {(b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )}{{\cos \left (c+d\,x\right )}^{9/2}} \,d x \]