Integrand size = 19, antiderivative size = 70 \[ \int (b \cos (c+d x))^n \sec ^4(c+d x) \, dx=\frac {b^3 (b \cos (c+d x))^{-3+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-3+n),\frac {1}{2} (-1+n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (3-n) \sqrt {\sin ^2(c+d x)}} \] Output:
b^3*(b*cos(d*x+c))^(-3+n)*hypergeom([1/2, -3/2+1/2*n],[-1/2+1/2*n],cos(d*x +c)^2)*sin(d*x+c)/d/(3-n)/(sin(d*x+c)^2)^(1/2)
Time = 0.06 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.03 \[ \int (b \cos (c+d x))^n \sec ^4(c+d x) \, dx=-\frac {(b \cos (c+d x))^n \csc (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-3+n),\frac {1}{2} (-1+n),\cos ^2(c+d x)\right ) \sec ^3(c+d x) \sqrt {\sin ^2(c+d x)}}{d (-3+n)} \] Input:
Integrate[(b*Cos[c + d*x])^n*Sec[c + d*x]^4,x]
Output:
-(((b*Cos[c + d*x])^n*Csc[c + d*x]*Hypergeometric2F1[1/2, (-3 + n)/2, (-1 + n)/2, Cos[c + d*x]^2]*Sec[c + d*x]^3*Sqrt[Sin[c + d*x]^2])/(d*(-3 + n)))
Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 2030, 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.
\(\displaystyle \int \sec ^4(c+d x) (b \cos (c+d x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^n}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b^4 \int \left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{n-4}dx\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {b^3 \sin (c+d x) (b \cos (c+d x))^{n-3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n-3}{2},\frac {n-1}{2},\cos ^2(c+d x)\right )}{d (3-n) \sqrt {\sin ^2(c+d x)}}\) |
Input:
Int[(b*Cos[c + d*x])^n*Sec[c + d*x]^4,x]
Output:
(b^3*(b*Cos[c + d*x])^(-3 + n)*Hypergeometric2F1[1/2, (-3 + n)/2, (-1 + n) /2, Cos[c + d*x]^2]*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]
\[\int \left (\cos \left (d x +c \right ) b \right )^{n} \sec \left (d x +c \right )^{4}d x\]
Input:
int((cos(d*x+c)*b)^n*sec(d*x+c)^4,x)
Output:
int((cos(d*x+c)*b)^n*sec(d*x+c)^4,x)
\[ \int (b \cos (c+d x))^n \sec ^4(c+d x) \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{4} \,d x } \] Input:
integrate((b*cos(d*x+c))^n*sec(d*x+c)^4,x, algorithm="fricas")
Output:
integral((b*cos(d*x + c))^n*sec(d*x + c)^4, x)
\[ \int (b \cos (c+d x))^n \sec ^4(c+d x) \, dx=\int \left (b \cos {\left (c + d x \right )}\right )^{n} \sec ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate((b*cos(d*x+c))**n*sec(d*x+c)**4,x)
Output:
Integral((b*cos(c + d*x))**n*sec(c + d*x)**4, x)
\[ \int (b \cos (c+d x))^n \sec ^4(c+d x) \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{4} \,d x } \] Input:
integrate((b*cos(d*x+c))^n*sec(d*x+c)^4,x, algorithm="maxima")
Output:
integrate((b*cos(d*x + c))^n*sec(d*x + c)^4, x)
\[ \int (b \cos (c+d x))^n \sec ^4(c+d x) \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{4} \,d x } \] Input:
integrate((b*cos(d*x+c))^n*sec(d*x+c)^4,x, algorithm="giac")
Output:
integrate((b*cos(d*x + c))^n*sec(d*x + c)^4, x)
Timed out. \[ \int (b \cos (c+d x))^n \sec ^4(c+d x) \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^n}{{\cos \left (c+d\,x\right )}^4} \,d x \] Input:
int((b*cos(c + d*x))^n/cos(c + d*x)^4,x)
Output:
int((b*cos(c + d*x))^n/cos(c + d*x)^4, x)
\[ \int (b \cos (c+d x))^n \sec ^4(c+d x) \, dx=b^{n} \left (\int \cos \left (d x +c \right )^{n} \sec \left (d x +c \right )^{4}d x \right ) \] Input:
int((b*cos(d*x+c))^n*sec(d*x+c)^4,x)
Output:
b**n*int(cos(c + d*x)**n*sec(c + d*x)**4,x)