Integrand size = 32, antiderivative size = 92 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{(c-c \sec (e+f x))^2} \, dx=-\frac {2^{\frac {1}{2}+m} a \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{2}-m,-\frac {1}{2},\frac {1}{2} (1-\sec (e+f x))\right ) (1+\sec (e+f x))^{\frac {1}{2}-m} (a+a \sec (e+f x))^{-1+m} \tan (e+f x)}{3 f (c-c \sec (e+f x))^2} \]
-1/3*2^(1/2+m)*a*hypergeom([-3/2, 1/2-m],[-1/2],1/2-1/2*sec(f*x+e))*(1+sec (f*x+e))^(1/2-m)*(a+a*sec(f*x+e))^(-1+m)*tan(f*x+e)/f/(c-c*sec(f*x+e))^2
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{(c-c \sec (e+f x))^2} \, dx=\int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{(c-c \sec (e+f x))^2} \, dx \]
Time = 0.31 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3042, 4449, 80, 79}
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.
\(\displaystyle \int \frac {\sec (e+f x) (a \sec (e+f x)+a)^m}{(c-c \sec (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^m}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4449 |
\(\displaystyle -\frac {a c \tan (e+f x) \int \frac {(\sec (e+f x) a+a)^{m-\frac {1}{2}}}{(c-c \sec (e+f x))^{5/2}}d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle -\frac {a c 2^{m-\frac {1}{2}} \tan (e+f x) (\sec (e+f x)+1)^{\frac {1}{2}-m} (a \sec (e+f x)+a)^{m-1} \int \frac {\left (\frac {1}{2} \sec (e+f x)+\frac {1}{2}\right )^{m-\frac {1}{2}}}{(c-c \sec (e+f x))^{5/2}}d\sec (e+f x)}{f \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {a 2^{m+\frac {1}{2}} \tan (e+f x) (\sec (e+f x)+1)^{\frac {1}{2}-m} (a \sec (e+f x)+a)^{m-1} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{2}-m,-\frac {1}{2},\frac {1}{2} (1-\sec (e+f x))\right )}{3 f (c-c \sec (e+f x))^2}\) |
-1/3*(2^(1/2 + m)*a*Hypergeometric2F1[-3/2, 1/2 - m, -1/2, (1 - Sec[e + f* x])/2]*(1 + Sec[e + f*x])^(1/2 - m)*(a + a*Sec[e + f*x])^(-1 + m)*Tan[e + f*x])/(f*(c - c*Sec[e + f*x])^2)
3.2.55.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) ^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) ), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(c sc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[a*c*(Cot[e + f *x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]])) Subst[Int[(a + b*x)^(m - 1/2)*(c + d*x)^(n - 1/2), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
\[\int \frac {\sec \left (f x +e \right ) \left (a +a \sec \left (f x +e \right )\right )^{m}}{\left (c -c \sec \left (f x +e \right )\right )^{2}}d x\]
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{(c-c \sec (e+f x))^2} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )}{{\left (c \sec \left (f x + e\right ) - c\right )}^{2}} \,d x } \]
integral((a*sec(f*x + e) + a)^m*sec(f*x + e)/(c^2*sec(f*x + e)^2 - 2*c^2*s ec(f*x + e) + c^2), x)
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{(c-c \sec (e+f x))^2} \, dx=\frac {\int \frac {\left (a \sec {\left (e + f x \right )} + a\right )^{m} \sec {\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} - 2 \sec {\left (e + f x \right )} + 1}\, dx}{c^{2}} \]
Integral((a*sec(e + f*x) + a)**m*sec(e + f*x)/(sec(e + f*x)**2 - 2*sec(e + f*x) + 1), x)/c**2
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{(c-c \sec (e+f x))^2} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )}{{\left (c \sec \left (f x + e\right ) - c\right )}^{2}} \,d x } \]
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{(c-c \sec (e+f x))^2} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )}{{\left (c \sec \left (f x + e\right ) - c\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{(c-c \sec (e+f x))^2} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m}{\cos \left (e+f\,x\right )\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^2} \,d x \]