Integrand size = 33, antiderivative size = 142 \[ \int \frac {(b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {2 C (b \cos (c+d x))^n \sin (c+d x)}{d (5-2 n) \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (A (5-2 n)+C (7-2 n)) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-7+2 n),\frac {1}{4} (-3+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (5-2 n) (7-2 n) \cos ^{\frac {7}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}} \]
-2*C*(b*cos(d*x+c))^n*sin(d*x+c)/d/(5-2*n)/cos(d*x+c)^(7/2)+2*(A*(5-2*n)+C *(7-2*n))*(b*cos(d*x+c))^n*hypergeom([1/2, -7/4+1/2*n],[-3/4+1/2*n],cos(d* x+c)^2)*sin(d*x+c)/d/(4*n^2-24*n+35)/cos(d*x+c)^(7/2)/(sin(d*x+c)^2)^(1/2)
Time = 0.14 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.99 \[ \int \frac {(b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {2 (b \cos (c+d x))^n \csc (c+d x) \left (A (-3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-7+2 n),\frac {1}{4} (-3+2 n),\cos ^2(c+d x)\right )+C (-7+2 n) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3+2 n),\frac {1}{4} (1+2 n),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (-7+2 n) (-3+2 n) \cos ^{\frac {7}{2}}(c+d x)} \]
(-2*(b*Cos[c + d*x])^n*Csc[c + d*x]*(A*(-3 + 2*n)*Hypergeometric2F1[1/2, ( -7 + 2*n)/4, (-3 + 2*n)/4, Cos[c + d*x]^2] + C*(-7 + 2*n)*Cos[c + d*x]^2*H ypergeometric2F1[1/2, (-3 + 2*n)/4, (1 + 2*n)/4, Cos[c + d*x]^2])*Sqrt[Sin [c + d*x]^2])/(d*(-7 + 2*n)*(-3 + 2*n)*Cos[c + d*x]^(7/2))
Time = 0.46 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {2034, 3042, 3493, 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.
\(\displaystyle \int \frac {\left (A+C \cos ^2(c+d x)\right ) (b \cos (c+d x))^n}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2034 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \cos ^{n-\frac {9}{2}}(c+d x) \left (C \cos ^2(c+d x)+A\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {9}{2}} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx\) |
\(\Big \downarrow \) 3493 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {(A (5-2 n)+C (7-2 n)) \int \cos ^{n-\frac {9}{2}}(c+d x)dx}{5-2 n}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {7}{2}}(c+d x)}{d (5-2 n)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {(A (5-2 n)+C (7-2 n)) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {9}{2}}dx}{5-2 n}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {7}{2}}(c+d x)}{d (5-2 n)}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {2 (A (5-2 n)+C (7-2 n)) \sin (c+d x) \cos ^{n-\frac {7}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n-7),\frac {1}{4} (2 n-3),\cos ^2(c+d x)\right )}{d (5-2 n) (7-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {7}{2}}(c+d x)}{d (5-2 n)}\right )\) |
((b*Cos[c + d*x])^n*((-2*C*Cos[c + d*x]^(-7/2 + n)*Sin[c + d*x])/(d*(5 - 2 *n)) + (2*(A*(5 - 2*n) + C*(7 - 2*n))*Cos[c + d*x]^(-7/2 + n)*Hypergeometr ic2F1[1/2, (-7 + 2*n)/4, (-3 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(5 - 2*n)*(7 - 2*n)*Sqrt[Sin[c + d*x]^2])))/Cos[c + d*x]^n
3.2.97.3.1 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_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*( x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f *(m + 2))), x] + Simp[(A*(m + 2) + C*(m + 1))/(m + 2) Int[(b*Sin[e + f*x] )^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && !LtQ[m, -1]
\[\int \frac {\left (\cos \left (d x +c \right ) b \right )^{n} \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{\cos \left (d x +c \right )^{\frac {9}{2}}}d x\]
\[ \int \frac {(b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]
Timed out. \[ \int \frac {(b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {(b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]
\[ \int \frac {(b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]
Timed out. \[ \int \frac {(b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^n}{{\cos \left (c+d\,x\right )}^{9/2}} \,d x \]