Integrand size = 41, antiderivative size = 217 \[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 C (b \cos (c+d x))^n \sin (c+d x)}{d (1+2 n) \sqrt {\cos (c+d x)}}+\frac {2 (A-C (1-2 n)+2 A n) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-1+2 n),\frac {1}{4} (3+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d \left (1-4 n^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {2 B \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1+2 n),\frac {1}{4} (5+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1+2 n) \sqrt {\sin ^2(c+d x)}} \]
2*C*(b*cos(d*x+c))^n*sin(d*x+c)/d/(1+2*n)/cos(d*x+c)^(1/2)+2*(A-C*(1-2*n)+ 2*A*n)*(b*cos(d*x+c))^n*hypergeom([1/2, -1/4+1/2*n],[3/4+1/2*n],cos(d*x+c) ^2)*sin(d*x+c)/d/(-4*n^2+1)/cos(d*x+c)^(1/2)/(sin(d*x+c)^2)^(1/2)-2*B*(b*c os(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 = 0.31 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.76 \[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 (b \cos (c+d x))^n \csc (c+d x) \left (-\left ((A+2 A n+C (-1+2 n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-1+2 n),\frac {1}{4} (3+2 n),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )+(-1+2 n) \left (C \sin ^2(c+d x)-B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1+2 n),\frac {1}{4} (5+2 n),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )\right )}{d \left (-1+4 n^2\right ) \sqrt {\cos (c+d x)}} \]
(2*(b*Cos[c + d*x])^n*Csc[c + d*x]*(-((A + 2*A*n + C*(-1 + 2*n))*Hypergeom etric2F1[1/2, (-1 + 2*n)/4, (3 + 2*n)/4, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x] ^2]) + (-1 + 2*n)*(C*Sin[c + d*x]^2 - B*Cos[c + d*x]*Hypergeometric2F1[1/2 , (1 + 2*n)/4, (5 + 2*n)/4, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2])))/(d*(-1 + 4*n^2)*Sqrt[Cos[c + d*x]])
Time = 0.63 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {2034, 3042, 3502, 27, 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.
\(\displaystyle \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2034 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \cos ^{n-\frac {3}{2}}(c+d x) \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {3}{2}} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {2 \int \frac {1}{2} \cos ^{n-\frac {3}{2}}(c+d x) (2 n A+A-C (1-2 n)+B (2 n+1) \cos (c+d x))dx}{2 n+1}+\frac {2 C \sin (c+d x) \cos ^{n-\frac {1}{2}}(c+d x)}{d (2 n+1)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {\int \cos ^{n-\frac {3}{2}}(c+d x) (2 n A+A-C (1-2 n)+B (2 n+1) \cos (c+d x))dx}{2 n+1}+\frac {2 C \sin (c+d x) \cos ^{n-\frac {1}{2}}(c+d x)}{d (2 n+1)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {3}{2}} \left (2 n A+A-C (1-2 n)+B (2 n+1) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 n+1}+\frac {2 C \sin (c+d x) \cos ^{n-\frac {1}{2}}(c+d x)}{d (2 n+1)}\right )\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {(2 A n+A-C (1-2 n)) \int \cos ^{n-\frac {3}{2}}(c+d x)dx+B (2 n+1) \int \cos ^{n-\frac {1}{2}}(c+d x)dx}{2 n+1}+\frac {2 C \sin (c+d x) \cos ^{n-\frac {1}{2}}(c+d x)}{d (2 n+1)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {(2 A n+A-C (1-2 n)) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {3}{2}}dx+B (2 n+1) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {1}{2}}dx}{2 n+1}+\frac {2 C \sin (c+d x) \cos ^{n-\frac {1}{2}}(c+d x)}{d (2 n+1)}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {\frac {2 (2 A n+A-C (1-2 n)) \sin (c+d x) \cos ^{n-\frac {1}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n-1),\frac {1}{4} (2 n+3),\cos ^2(c+d x)\right )}{d (1-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \sin (c+d x) \cos ^{n+\frac {1}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n+1),\frac {1}{4} (2 n+5),\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}}{2 n+1}+\frac {2 C \sin (c+d x) \cos ^{n-\frac {1}{2}}(c+d x)}{d (2 n+1)}\right )\) |
((b*Cos[c + d*x])^n*((2*C*Cos[c + d*x]^(-1/2 + n)*Sin[c + d*x])/(d*(1 + 2* n)) + ((2*(A - C*(1 - 2*n) + 2*A*n)*Cos[c + d*x]^(-1/2 + n)*Hypergeometric 2F1[1/2, (-1 + 2*n)/4, (3 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 - 2*n)*Sqrt[Sin[c + d*x]^2]) - (2*B*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*Sqrt[Sin[ c + d*x]^2]))/(1 + 2*n)))/Cos[c + d*x]^n
3.4.80.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
\[\int \frac {\left (\cos \left (d x +c \right ) b \right )^{n} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{\cos \left (d x +c \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2 ),x, algorithm="fricas")
\[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (b \cos {\left (c + d x \right )}\right )^{n} \left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right )}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
Integral((b*cos(c + d*x))**n*(A + B*cos(c + d*x) + C*cos(c + d*x)**2)/cos( c + d*x)**(3/2), x)
\[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2 ),x, algorithm="maxima")
\[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2 ),x, algorithm="giac")
Timed out. \[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^{3/2}} \,d x \]