Integrand size = 33, antiderivative size = 140 \[ \int \frac {(b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 C (b \cos (c+d x))^n \sin (c+d x)}{d (1-2 n) \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (A+C (3-2 n)-2 A n) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3+2 n),\frac {1}{4} (1+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1-2 n) (3-2 n) \cos ^{\frac {3}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}} \] Output:
-2*C*(b*cos(d*x+c))^n*sin(d*x+c)/d/(1-2*n)/cos(d*x+c)^(3/2)+2*(A+C*(3-2*n) -2*A*n)*(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/(1-2*n)/(3-2*n)/cos(d*x+c)^(3/2)/(sin(d*x+c)^2)^(1/2)
Time = 0.20 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00 \[ \int \frac {(b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 (b \cos (c+d x))^n \csc (c+d x) \left (A (1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3+2 n),\frac {1}{4} (1+2 n),\cos ^2(c+d x)\right )+C (-3+2 n) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1+2 n),\frac {1}{4} (5+2 n),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (-3+2 n) (1+2 n) \cos ^{\frac {3}{2}}(c+d x)} \] Input:
Integrate[((b*Cos[c + d*x])^n*(A + C*Cos[c + d*x]^2))/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] + C*(-3 + 2*n)*Cos[c + d*x]^2*Hyp ergeometric2F1[1/2, (1 + 2*n)/4, (5 + 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 = 0.46 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.05, 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.
\(\displaystyle \int \frac {\left (A+C \cos ^2(c+d x)\right ) (b \cos (c+d x))^n}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2034 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \cos ^{n-\frac {5}{2}}(c+d x) \left (C \cos ^2(c+d x)+A\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {5}{2}} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx\) |
\(\Big \downarrow \) 3493 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {(A (1-2 n)+C (3-2 n)) \int \cos ^{n-\frac {5}{2}}(c+d x)dx}{1-2 n}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {3}{2}}(c+d x)}{d (1-2 n)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {(A (1-2 n)+C (3-2 n)) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {5}{2}}dx}{1-2 n}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {3}{2}}(c+d x)}{d (1-2 n)}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \cos ^{-n}(c+d x) (b \cos (c+d x))^n \left (\frac {2 (A (1-2 n)+C (3-2 n)) \sin (c+d x) \cos ^{n-\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n-3),\frac {1}{4} (2 n+1),\cos ^2(c+d x)\right )}{d (1-2 n) (3-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \cos ^{n-\frac {3}{2}}(c+d x)}{d (1-2 n)}\right )\) |
Input:
Int[((b*Cos[c + d*x])^n*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(5/2),x]
Output:
((b*Cos[c + d*x])^n*((-2*C*Cos[c + d*x]^(-3/2 + n)*Sin[c + d*x])/(d*(1 - 2 *n)) + (2*(A*(1 - 2*n) + C*(3 - 2*n))*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*(1 - 2*n)*(3 - 2*n)*Sqrt[Sin[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_.)*((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]
\[\int \frac {\left (b \cos \left (d x +c \right )\right )^{n} \left (A +C \cos \left (d x +c \right )^{2}\right )}{\cos \left (d x +c \right )^{\frac {5}{2}}}d x\]
Input:
int((b*cos(d*x+c))^n*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x)
Output:
int((b*cos(d*x+c))^n*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x)
\[ \int \frac {(b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*cos(d*x+c))^n*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorith m="fricas")
Output:
integral((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^n/cos(d*x + c)^(5/2), x)
Timed out. \[ \int \frac {(b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((b*cos(d*x+c))**n*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(5/2),x)
Output:
Timed out
\[ \int \frac {(b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*cos(d*x+c))^n*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorith m="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^n/cos(d*x + c)^(5/2), x)
\[ \int \frac {(b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*cos(d*x+c))^n*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorith m="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^n/cos(d*x + c)^(5/2), x)
Timed out. \[ \int \frac {(b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^n}{{\cos \left (c+d\,x\right )}^{5/2}} \,d x \] Input:
int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^n)/cos(c + d*x)^(5/2),x)
Output:
int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^n)/cos(c + d*x)^(5/2), x)
\[ \int \frac {(b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=b^{n} \left (\left (\int \frac {\cos \left (d x +c \right )^{n +\frac {1}{2}}}{\cos \left (d x +c \right )}d x \right ) c +\left (\int \frac {\cos \left (d x +c \right )^{n +\frac {1}{2}}}{\cos \left (d x +c \right )^{3}}d x \right ) a \right ) \] Input:
int((b*cos(d*x+c))^n*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x)
Output:
b**n*(int(cos(c + d*x)**((2*n + 1)/2)/cos(c + d*x),x)*c + int(cos(c + d*x) **((2*n + 1)/2)/cos(c + d*x)**3,x)*a)