Integrand size = 17, antiderivative size = 48 \[ \int \sec (a+b x) (c \sin (a+b x))^m \, dx=\frac {\operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},\sin ^2(a+b x)\right ) (c \sin (a+b x))^{1+m}}{b c (1+m)} \] Output:
hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],sin(b*x+a)^2)*(c*sin(b*x+a))^(1+m)/b/ c/(1+m)
Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.06 \[ \int \sec (a+b x) (c \sin (a+b x))^m \, dx=\frac {\operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},1+\frac {1+m}{2},\sin ^2(a+b x)\right ) \sin (a+b x) (c \sin (a+b x))^m}{b (1+m)} \] Input:
Integrate[Sec[a + b*x]*(c*Sin[a + b*x])^m,x]
Output:
(Hypergeometric2F1[1, (1 + m)/2, 1 + (1 + m)/2, Sin[a + b*x]^2]*Sin[a + b* x]*(c*Sin[a + b*x])^m)/(b*(1 + m))
Time = 0.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3042, 3044, 27, 278}
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.
\(\displaystyle \int \sec (a+b x) (c \sin (a+b x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c \sin (a+b x))^m}{\cos (a+b x)}dx\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {\int \frac {c^2 (c \sin (a+b x))^m}{c^2-c^2 \sin ^2(a+b x)}d(c \sin (a+b x))}{b c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {c \int \frac {(c \sin (a+b x))^m}{c^2-c^2 \sin ^2(a+b x)}d(c \sin (a+b x))}{b}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {(c \sin (a+b x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},\sin ^2(a+b x)\right )}{b c (m+1)}\) |
Input:
Int[Sec[a + b*x]*(c*Sin[a + b*x])^m,x]
Output:
(Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, Sin[a + b*x]^2]*(c*Sin[a + b*x ])^(1 + m))/(b*c*(1 + m))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[1/(a*f) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a *Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
\[\int \sec \left (b x +a \right ) \left (c \sin \left (b x +a \right )\right )^{m}d x\]
Input:
int(sec(b*x+a)*(c*sin(b*x+a))^m,x)
Output:
int(sec(b*x+a)*(c*sin(b*x+a))^m,x)
\[ \int \sec (a+b x) (c \sin (a+b x))^m \, dx=\int { \left (c \sin \left (b x + a\right )\right )^{m} \sec \left (b x + a\right ) \,d x } \] Input:
integrate(sec(b*x+a)*(c*sin(b*x+a))^m,x, algorithm="fricas")
Output:
integral((c*sin(b*x + a))^m*sec(b*x + a), x)
\[ \int \sec (a+b x) (c \sin (a+b x))^m \, dx=\int \left (c \sin {\left (a + b x \right )}\right )^{m} \sec {\left (a + b x \right )}\, dx \] Input:
integrate(sec(b*x+a)*(c*sin(b*x+a))**m,x)
Output:
Integral((c*sin(a + b*x))**m*sec(a + b*x), x)
\[ \int \sec (a+b x) (c \sin (a+b x))^m \, dx=\int { \left (c \sin \left (b x + a\right )\right )^{m} \sec \left (b x + a\right ) \,d x } \] Input:
integrate(sec(b*x+a)*(c*sin(b*x+a))^m,x, algorithm="maxima")
Output:
integrate((c*sin(b*x + a))^m*sec(b*x + a), x)
\[ \int \sec (a+b x) (c \sin (a+b x))^m \, dx=\int { \left (c \sin \left (b x + a\right )\right )^{m} \sec \left (b x + a\right ) \,d x } \] Input:
integrate(sec(b*x+a)*(c*sin(b*x+a))^m,x, algorithm="giac")
Output:
integrate((c*sin(b*x + a))^m*sec(b*x + a), x)
Timed out. \[ \int \sec (a+b x) (c \sin (a+b x))^m \, dx=\int \frac {{\left (c\,\sin \left (a+b\,x\right )\right )}^m}{\cos \left (a+b\,x\right )} \,d x \] Input:
int((c*sin(a + b*x))^m/cos(a + b*x),x)
Output:
int((c*sin(a + b*x))^m/cos(a + b*x), x)
\[ \int \sec (a+b x) (c \sin (a+b x))^m \, dx=c^{m} \left (\int \sin \left (b x +a \right )^{m} \sec \left (b x +a \right )d x \right ) \] Input:
int(sec(b*x+a)*(c*sin(b*x+a))^m,x)
Output:
c**m*int(sin(a + b*x)**m*sec(a + b*x),x)