Integrand size = 31, antiderivative size = 163 \[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 A (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-1+2 n),\frac {1}{4} (3+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1-2 n) \sqrt {\cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {2 B \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1+2 n),\frac {1}{4} (5+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1+2 n) \sqrt {\sin ^2(c+d x)}} \]
2*A*(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)-2*B*(b*cos(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 = 0.21 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.82 \[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 (b \cos (c+d x))^n \csc (c+d x) \left (A (1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-1+2 n),\frac {1}{4} (3+2 n),\cos ^2(c+d x)\right )+B (-1+2 n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1+2 n),\frac {1}{4} (5+2 n),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d \left (-1+4 n^2\right ) \sqrt {\cos (c+d x)}} \]
(-2*(b*Cos[c + d*x])^n*Csc[c + d*x]*(A*(1 + 2*n)*Hypergeometric2F1[1/2, (- 1 + 2*n)/4, (3 + 2*n)/4, Cos[c + d*x]^2] + B*(-1 + 2*n)*Cos[c + d*x]*Hyper geometric2F1[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 = 0.42 (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.
| \(\displaystyle \int \frac {(A+B \cos (c+d x)) (b \cos (c+d x))^n}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 2034 |
| \(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \cos ^{n-\frac {3}{2}}(c+d x) (A+B \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {3}{2}} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (A \int \cos ^{n-\frac {3}{2}}(c+d x)dx+B \int \cos ^{n-\frac {1}{2}}(c+d x)dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (A \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {3}{2}}dx+B \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {1}{2}}dx\right )\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {2 A \sin (c+d x) \cos ^{n-\frac {1}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n-1),\frac {1}{4} (2 n+3),\cos ^2(c+d x)\right )}{d (1-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \sin (c+d x) \cos ^{n+\frac {1}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n+1),\frac {1}{4} (2 n+5),\cos ^2(c+d x)\right )}{d (2 n+1) \sqrt {\sin ^2(c+d x)}}\right )\) |
((b*Cos[c + d*x])^n*((2*A*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[ 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*(1 + 2*n)*Sqrt[Sin[ c + d*x]^2])))/Cos[c + d*x]^n
3.10.23.3.1 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]
\[\int \frac {\left (\cos \left (d x +c \right ) b \right )^{n} \left (A +B \cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (b \cos {\left (c + d x \right )}\right )^{n} \left (A + B \cos {\left (c + d x \right )}\right )}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\cos \left (c+d\,x\right )}^{3/2}} \,d x \]