Integrand size = 41, antiderivative size = 223 \[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 C (b \cos (c+d x))^n \sin (c+d x)}{d (3-2 n) \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (A (3-2 n)+C (5-2 n)) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-5+2 n),\frac {1}{4} (-1+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (3-2 n) (5-2 n) \cos ^{\frac {5}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}}+\frac {2 B (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3+2 n),\frac {1}{4} (1+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (3-2 n) \cos ^{\frac {3}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}} \] Output:
-2*C*(b*cos(d*x+c))^n*sin(d*x+c)/d/(3-2*n)/cos(d*x+c)^(5/2)+2*(A*(3-2*n)+C *(5-2*n))*(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/(3-2*n)/(5-2*n)/cos(d*x+c)^(5/2)/(sin(d*x+c)^2)^(1/2) +2*B*(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 = 0.51 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.78 \[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 (b \cos (c+d x))^n \csc (c+d x) \left (-\left ((C (-5+2 n)+A (-3+2 n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-5+2 n),\frac {1}{4} (-1+2 n),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )+(-5+2 n) \left (C \sin ^2(c+d x)-B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3+2 n),\frac {1}{4} (1+2 n),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )\right )}{d (-5+2 n) (-3+2 n) \cos ^{\frac {5}{2}}(c+d x)} \] Input:
Integrate[((b*Cos[c + d*x])^n*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos [c + d*x]^(7/2),x]
Output:
(2*(b*Cos[c + d*x])^n*Csc[c + d*x]*(-((C*(-5 + 2*n) + A*(-3 + 2*n))*Hyperg eometric2F1[1/2, (-5 + 2*n)/4, (-1 + 2*n)/4, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2]) + (-5 + 2*n)*(C*Sin[c + d*x]^2 - B*Cos[c + d*x]*Hypergeometric2F1 [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 = 0.64 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.96, 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.
\(\displaystyle \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2034 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \cos ^{n-\frac {7}{2}}(c+d x) \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {7}{2}} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (-\frac {2 \int -\frac {1}{2} \cos ^{n-\frac {7}{2}}(c+d x) \left (2 A \left (\frac {3}{2}-n\right )+2 C \left (\frac {5}{2}-n\right )+B (3-2 n) \cos (c+d x)\right )dx}{3-2 n}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {5}{2}}(c+d x)}{d (3-2 n)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {\int \cos ^{n-\frac {7}{2}}(c+d x) (A (3-2 n)+B \cos (c+d x) (3-2 n)+C (5-2 n))dx}{3-2 n}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {5}{2}}(c+d x)}{d (3-2 n)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {7}{2}} \left (A (3-2 n)+B \sin \left (c+d x+\frac {\pi }{2}\right ) (3-2 n)+C (5-2 n)\right )dx}{3-2 n}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {5}{2}}(c+d x)}{d (3-2 n)}\right )\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {(A (3-2 n)+C (5-2 n)) \int \cos ^{n-\frac {7}{2}}(c+d x)dx+B (3-2 n) \int \cos ^{n-\frac {5}{2}}(c+d x)dx}{3-2 n}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {5}{2}}(c+d x)}{d (3-2 n)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {(A (3-2 n)+C (5-2 n)) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {7}{2}}dx+B (3-2 n) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {5}{2}}dx}{3-2 n}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {5}{2}}(c+d x)}{d (3-2 n)}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {\frac {2 (A (3-2 n)+C (5-2 n)) \sin (c+d x) \cos ^{n-\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n-5),\frac {1}{4} (2 n-1),\cos ^2(c+d x)\right )}{d (5-2 n) \sqrt {\sin ^2(c+d x)}}+\frac {2 B \sin (c+d x) \cos ^{n-\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n-3),\frac {1}{4} (2 n+1),\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}}{3-2 n}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {5}{2}}(c+d x)}{d (3-2 n)}\right )\) |
Input:
Int[((b*Cos[c + d*x])^n*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d *x]^(7/2),x]
Output:
((b*Cos[c + d*x])^n*((-2*C*Cos[c + d*x]^(-5/2 + n)*Sin[c + d*x])/(d*(3 - 2 *n)) + ((2*(A*(3 - 2*n) + C*(5 - 2*n))*Cos[c + d*x]^(-5/2 + n)*Hypergeomet ric2F1[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*B*Cos[c + d*x]^(-3/2 + n)*Hypergeometr ic2F1[1/2, (-3 + 2*n)/4, (1 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*Sqr t[Sin[c + d*x]^2]))/(3 - 2*n)))/Cos[c + d*x]^n
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]
\[\int \frac {\left (b \cos \left (d x +c \right )\right )^{n} \left (A +B \cos \left (d x +c \right )+C \cos \left (d x +c \right )^{2}\right )}{\cos \left (d x +c \right )^{\frac {7}{2}}}d x\]
Input:
int((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x)
Output:
int((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x)
\[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2 ),x, algorithm="fricas")
Output:
integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^n/cos(d* x + c)^(7/2), x)
Timed out. \[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((b*cos(d*x+c))**n*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**( 7/2),x)
Output:
Timed out
\[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2 ),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^n/cos(d *x + c)^(7/2), x)
\[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2 ),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^n/cos(d *x + c)^(7/2), x)
Timed out. \[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^{7/2}} \,d x \] Input:
int(((b*cos(c + d*x))^n*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d *x)^(7/2),x)
Output:
int(((b*cos(c + d*x))^n*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d *x)^(7/2), x)
\[ \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=b^{n} \left (\left (\int \frac {\cos \left (d x +c \right )^{n +\frac {1}{2}}}{\cos \left (d x +c \right )^{4}}d x \right ) a +\left (\int \frac {\cos \left (d x +c \right )^{n +\frac {1}{2}}}{\cos \left (d x +c \right )^{3}}d x \right ) b +\left (\int \frac {\cos \left (d x +c \right )^{n +\frac {1}{2}}}{\cos \left (d x +c \right )^{2}}d x \right ) c \right ) \] Input:
int((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x)
Output:
b**n*(int(cos(c + d*x)**((2*n + 1)/2)/cos(c + d*x)**4,x)*a + int(cos(c + d *x)**((2*n + 1)/2)/cos(c + d*x)**3,x)*b + int(cos(c + d*x)**((2*n + 1)/2)/ cos(c + d*x)**2,x)*c)