Integrand size = 19, antiderivative size = 75 \[ \int \cos ^2(c+d x) (b \sec (c+d x))^n \, dx=-\frac {b^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-n}{2},\frac {5-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-3+n} \sin (c+d x)}{d (3-n) \sqrt {\sin ^2(c+d x)}} \] Output:
-b^3*hypergeom([1/2, 3/2-1/2*n],[5/2-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^( -3+n)*sin(d*x+c)/d/(3-n)/(sin(d*x+c)^2)^(1/2)
Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int \cos ^2(c+d x) (b \sec (c+d x))^n \, dx=\frac {\cos ^2(c+d x) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2+n),\frac {n}{2},\sec ^2(c+d x)\right ) (b \sec (c+d x))^n \sqrt {-\tan ^2(c+d x)}}{d (-2+n)} \] Input:
Integrate[Cos[c + d*x]^2*(b*Sec[c + d*x])^n,x]
Output:
(Cos[c + d*x]^2*Cot[c + d*x]*Hypergeometric2F1[1/2, (-2 + n)/2, n/2, Sec[c + d*x]^2]*(b*Sec[c + d*x])^n*Sqrt[-Tan[c + d*x]^2])/(d*(-2 + n))
Time = 0.35 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 2030, 4259, 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.
\(\displaystyle \int \cos ^2(c+d x) (b \sec (c+d x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^n}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b^2 \int \left (b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{n-2}dx\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle b^2 \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\cos (c+d x)}{b}\right )^{2-n}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{2-n}dx\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle -\frac {b^3 \sin (c+d x) (b \sec (c+d x))^{n-3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-n}{2},\frac {5-n}{2},\cos ^2(c+d x)\right )}{d (3-n) \sqrt {\sin ^2(c+d x)}}\) |
Input:
Int[Cos[c + d*x]^2*(b*Sec[c + d*x])^n,x]
Output:
-((b^3*Hypergeometric2F1[1/2, (3 - n)/2, (5 - n)/2, Cos[c + d*x]^2]*(b*Sec [c + d*x])^(-3 + n)*Sin[c + d*x])/(d*(3 - n)*Sqrt[Sin[c + d*x]^2]))
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^(n - 1)*((Sin[c + d*x]/b)^(n - 1) Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
\[\int \cos \left (d x +c \right )^{2} \left (b \sec \left (d x +c \right )\right )^{n}d x\]
Input:
int(cos(d*x+c)^2*(b*sec(d*x+c))^n,x)
Output:
int(cos(d*x+c)^2*(b*sec(d*x+c))^n,x)
\[ \int \cos ^2(c+d x) (b \sec (c+d x))^n \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{2} \,d x } \] Input:
integrate(cos(d*x+c)^2*(b*sec(d*x+c))^n,x, algorithm="fricas")
Output:
integral((b*sec(d*x + c))^n*cos(d*x + c)^2, x)
\[ \int \cos ^2(c+d x) (b \sec (c+d x))^n \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{n} \cos ^{2}{\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)**2*(b*sec(d*x+c))**n,x)
Output:
Integral((b*sec(c + d*x))**n*cos(c + d*x)**2, x)
\[ \int \cos ^2(c+d x) (b \sec (c+d x))^n \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{2} \,d x } \] Input:
integrate(cos(d*x+c)^2*(b*sec(d*x+c))^n,x, algorithm="maxima")
Output:
integrate((b*sec(d*x + c))^n*cos(d*x + c)^2, x)
\[ \int \cos ^2(c+d x) (b \sec (c+d x))^n \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{2} \,d x } \] Input:
integrate(cos(d*x+c)^2*(b*sec(d*x+c))^n,x, algorithm="giac")
Output:
integrate((b*sec(d*x + c))^n*cos(d*x + c)^2, x)
Timed out. \[ \int \cos ^2(c+d x) (b \sec (c+d x))^n \, dx=\int {\cos \left (c+d\,x\right )}^2\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n \,d x \] Input:
int(cos(c + d*x)^2*(b/cos(c + d*x))^n,x)
Output:
int(cos(c + d*x)^2*(b/cos(c + d*x))^n, x)
\[ \int \cos ^2(c+d x) (b \sec (c+d x))^n \, dx=b^{n} \left (\int \sec \left (d x +c \right )^{n} \cos \left (d x +c \right )^{2}d x \right ) \] Input:
int(cos(d*x+c)^2*(b*sec(d*x+c))^n,x)
Output:
b**n*int(sec(c + d*x)**n*cos(c + d*x)**2,x)