Integrand size = 31, antiderivative size = 163 \[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 A (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3+2 n),\frac {1}{4} (1+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (3-2 n) \cos ^{\frac {3}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}}+\frac {2 B (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-1+2 n),\frac {1}{4} (3+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1-2 n) \sqrt {\cos (c+d x)} \sqrt {\sin ^2(c+d x)}} \] Output:
2*A*(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)+2*B*(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/ (1-2*n)/cos(d*x+c)^(1/2)/(sin(d*x+c)^2)^(1/2)
Time = 0.17 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.85 \[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 (b \cos (c+d x))^n \csc (c+d x) \left (A (-1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3+2 n),\frac {1}{4} (1+2 n),\cos ^2(c+d x)\right )+B (-3+2 n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-1+2 n),\frac {1}{4} (3+2 n),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (-3+2 n) (-1+2 n) \cos ^{\frac {3}{2}}(c+d x)} \] Input:
Integrate[((b*Cos[c + d*x])^n*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(5/2),x]
Output:
(-2*(b*Cos[c + d*x])^n*Csc[c + d*x]*(A*(-1 + 2*n)*Hypergeometric2F1[1/2, ( -3 + 2*n)/4, (1 + 2*n)/4, Cos[c + d*x]^2] + B*(-3 + 2*n)*Cos[c + d*x]*Hype rgeometric2F1[1/2, (-1 + 2*n)/4, (3 + 2*n)/4, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(-3 + 2*n)*(-1 + 2*n)*Cos[c + d*x]^(3/2))
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.
\(\displaystyle \int \frac {(A+B \cos (c+d x)) (b \cos (c+d x))^n}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2034 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \cos ^{n-\frac {5}{2}}(c+d x) (A+B \cos (c+d x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {5}{2}} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (A \int \cos ^{n-\frac {5}{2}}(c+d x)dx+B \int \cos ^{n-\frac {3}{2}}(c+d x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (A \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {5}{2}}dx+B \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {3}{2}}dx\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {2 A \sin (c+d x) \cos ^{n-\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n-3),\frac {1}{4} (2 n+1),\cos ^2(c+d x)\right )}{d (3-2 n) \sqrt {\sin ^2(c+d x)}}+\frac {2 B \sin (c+d x) \cos ^{n-\frac {1}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n-1),\frac {1}{4} (2 n+3),\cos ^2(c+d x)\right )}{d (1-2 n) \sqrt {\sin ^2(c+d x)}}\right )\) |
Input:
Int[((b*Cos[c + d*x])^n*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(5/2),x]
Output:
((b*Cos[c + d*x])^n*((2*A*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[ Sin[c + d*x]^2]) + (2*B*Cos[c + d*x]^(-1/2 + n)*Hypergeometric2F1[1/2, (-1 + 2*n)/4, (3 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 - 2*n)*Sqrt[Si n[c + d*x]^2])))/Cos[c + d*x]^n
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]
\[\int \frac {\left (\cos \left (d x +c \right ) b \right )^{n} \left (A +B \cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )^{\frac {5}{2}}}d x\]
Input:
int((cos(d*x+c)*b)^n*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x)
Output:
int((cos(d*x+c)*b)^n*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x)
\[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm= "fricas")
Output:
integral((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n/cos(d*x + c)^(5/2), x)
Timed out. \[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((b*cos(d*x+c))**n*(A+B*cos(d*x+c))/cos(d*x+c)**(5/2),x)
Output:
Timed out
\[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm= "maxima")
Output:
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n/cos(d*x + c)^(5/2), x)
\[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm= "giac")
Output:
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n/cos(d*x + c)^(5/2), x)
Timed out. \[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\cos \left (c+d\,x\right )}^{5/2}} \,d x \] Input:
int(((b*cos(c + d*x))^n*(A + B*cos(c + d*x)))/cos(c + d*x)^(5/2),x)
Output:
int(((b*cos(c + d*x))^n*(A + B*cos(c + d*x)))/cos(c + d*x)^(5/2), x)
\[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=b^{n} \left (\left (\int \frac {\cos \left (d x +c \right )^{n +\frac {1}{2}}}{\cos \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\cos \left (d x +c \right )^{n +\frac {1}{2}}}{\cos \left (d x +c \right )^{2}}d x \right ) b \right ) \] Input:
int((b*cos(d*x+c))^n*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x)
Output:
b**n*(int(cos(c + d*x)**((2*n + 1)/2)/cos(c + d*x)**3,x)*a + int(cos(c + d *x)**((2*n + 1)/2)/cos(c + d*x)**2,x)*b)