Integrand size = 34, antiderivative size = 180 \[ \int \frac {(g \sec (e+f x))^p (c-c \sec (e+f x))}{a+a \sec (e+f x)} \, dx=-\frac {c g (1-2 p) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-p}{2},\frac {3-p}{2},\cos ^2(e+f x)\right ) (g \sec (e+f x))^{-1+p} \sin (e+f x)}{a f (1-p) \sqrt {\sin ^2(e+f x)}}+\frac {2 c \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {p}{2},\frac {2-p}{2},\cos ^2(e+f x)\right ) (g \sec (e+f x))^p \sin (e+f x)}{a f \sqrt {\sin ^2(e+f x)}}-\frac {2 c (g \sec (e+f x))^p \tan (e+f x)}{f (a+a \sec (e+f x))} \] Output:
-c*g*(1-2*p)*hypergeom([1/2, 1/2-1/2*p],[3/2-1/2*p],cos(f*x+e)^2)*(g*sec(f *x+e))^(-1+p)*sin(f*x+e)/a/f/(1-p)/(sin(f*x+e)^2)^(1/2)+2*c*hypergeom([1/2 , -1/2*p],[1-1/2*p],cos(f*x+e)^2)*(g*sec(f*x+e))^p*sin(f*x+e)/a/f/(sin(f*x +e)^2)^(1/2)-2*c*(g*sec(f*x+e))^p*tan(f*x+e)/f/(a+a*sec(f*x+e))
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 14.60 (sec) , antiderivative size = 3396, normalized size of antiderivative = 18.87 \[ \int \frac {(g \sec (e+f x))^p (c-c \sec (e+f x))}{a+a \sec (e+f x)} \, dx=\text {Result too large to show} \] Input:
Integrate[((g*Sec[e + f*x])^p*(c - c*Sec[e + f*x]))/(a + a*Sec[e + f*x]),x ]
Output:
(-6*c*Sec[e + f*x]^p*(g*Sec[e + f*x])^p*Tan[(e + f*x)/2]^3*(-((AppellF1[1/ 2, p, 1 - p, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2 ]^2)/(3*AppellF1[1/2, p, 1 - p, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2] ^2] + 2*((-1 + p)*AppellF1[3/2, p, 2 - p, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + p*AppellF1[3/2, 1 + p, 1 - p, 5/2, Tan[(e + f*x)/2]^2, -Ta n[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)) + AppellF1[1/2, p, -p, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]/(3*AppellF1[1/2, p, -p, 3/2, Tan[(e + f *x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*p*(AppellF1[3/2, p, 1 - p, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + AppellF1[3/2, 1 + p, -p, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)))/(a*f*(3*Sec[(e + f *x)/2]^2*Sec[e + f*x]^p*(-((AppellF1[1/2, p, 1 - p, 3/2, Tan[(e + f*x)/2]^ 2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2)/(3*AppellF1[1/2, p, 1 - p, 3/2 , Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-1 + p)*AppellF1[3/2, p, 2 - p, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + p*AppellF1[3/2, 1 + p, 1 - p, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2] ^2)) + AppellF1[1/2, p, -p, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]/ (3*AppellF1[1/2, p, -p, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2* p*(AppellF1[3/2, p, 1 - p, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + AppellF1[3/2, 1 + p, -p, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*T an[(e + f*x)/2]^2)) + 6*p*Sec[e + f*x]^(1 + p)*Sin[e + f*x]*Tan[(e + f*...
Time = 0.73 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3042, 4508, 3042, 4274, 3042, 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.
\(\displaystyle \int \frac {(c-c \sec (e+f x)) (g \sec (e+f x))^p}{a \sec (e+f x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right ) \left (g \csc \left (e+f x+\frac {\pi }{2}\right )\right )^p}{a \csc \left (e+f x+\frac {\pi }{2}\right )+a}dx\) |
\(\Big \downarrow \) 4508 |
\(\displaystyle \frac {\int (g \sec (e+f x))^p (a c (1-2 p)+2 a c p \sec (e+f x))dx}{a^2}-\frac {2 c \tan (e+f x) (g \sec (e+f x))^p}{f (a \sec (e+f x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (g \csc \left (e+f x+\frac {\pi }{2}\right )\right )^p \left (a c (1-2 p)+2 a c p \csc \left (e+f x+\frac {\pi }{2}\right )\right )dx}{a^2}-\frac {2 c \tan (e+f x) (g \sec (e+f x))^p}{f (a \sec (e+f x)+a)}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {a c (1-2 p) \int (g \sec (e+f x))^pdx+\frac {2 a c p \int (g \sec (e+f x))^{p+1}dx}{g}}{a^2}-\frac {2 c \tan (e+f x) (g \sec (e+f x))^p}{f (a \sec (e+f x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a c (1-2 p) \int \left (g \csc \left (e+f x+\frac {\pi }{2}\right )\right )^pdx+\frac {2 a c p \int \left (g \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{p+1}dx}{g}}{a^2}-\frac {2 c \tan (e+f x) (g \sec (e+f x))^p}{f (a \sec (e+f x)+a)}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \frac {\frac {2 a c p \left (\frac {\cos (e+f x)}{g}\right )^p (g \sec (e+f x))^p \int \left (\frac {\cos (e+f x)}{g}\right )^{-p-1}dx}{g}+a c (1-2 p) \left (\frac {\cos (e+f x)}{g}\right )^p (g \sec (e+f x))^p \int \left (\frac {\cos (e+f x)}{g}\right )^{-p}dx}{a^2}-\frac {2 c \tan (e+f x) (g \sec (e+f x))^p}{f (a \sec (e+f x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 a c p \left (\frac {\cos (e+f x)}{g}\right )^p (g \sec (e+f x))^p \int \left (\frac {\sin \left (e+f x+\frac {\pi }{2}\right )}{g}\right )^{-p-1}dx}{g}+a c (1-2 p) \left (\frac {\cos (e+f x)}{g}\right )^p (g \sec (e+f x))^p \int \left (\frac {\sin \left (e+f x+\frac {\pi }{2}\right )}{g}\right )^{-p}dx}{a^2}-\frac {2 c \tan (e+f x) (g \sec (e+f x))^p}{f (a \sec (e+f x)+a)}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\frac {2 a c \sin (e+f x) (g \sec (e+f x))^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {p}{2},\frac {2-p}{2},\cos ^2(e+f x)\right )}{f \sqrt {\sin ^2(e+f x)}}-\frac {a c g (1-2 p) \sin (e+f x) (g \sec (e+f x))^{p-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-p}{2},\frac {3-p}{2},\cos ^2(e+f x)\right )}{f (1-p) \sqrt {\sin ^2(e+f x)}}}{a^2}-\frac {2 c \tan (e+f x) (g \sec (e+f x))^p}{f (a \sec (e+f x)+a)}\) |
Input:
Int[((g*Sec[e + f*x])^p*(c - c*Sec[e + f*x]))/(a + a*Sec[e + f*x]),x]
Output:
(-((a*c*g*(1 - 2*p)*Hypergeometric2F1[1/2, (1 - p)/2, (3 - p)/2, Cos[e + f *x]^2]*(g*Sec[e + f*x])^(-1 + p)*Sin[e + f*x])/(f*(1 - p)*Sqrt[Sin[e + f*x ]^2])) + (2*a*c*Hypergeometric2F1[1/2, -1/2*p, (2 - p)/2, Cos[e + f*x]^2]* (g*Sec[e + f*x])^p*Sin[e + f*x])/(f*Sqrt[Sin[e + f*x]^2]))/a^2 - (2*c*(g*S ec[e + f*x])^p*Tan[e + f*x])/(f*(a + a*Sec[e + f*x]))
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[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B , 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0]
\[\int \frac {\left (g \sec \left (f x +e \right )\right )^{p} \left (c -c \sec \left (f x +e \right )\right )}{a +a \sec \left (f x +e \right )}d x\]
Input:
int((g*sec(f*x+e))^p*(c-c*sec(f*x+e))/(a+a*sec(f*x+e)),x)
Output:
int((g*sec(f*x+e))^p*(c-c*sec(f*x+e))/(a+a*sec(f*x+e)),x)
\[ \int \frac {(g \sec (e+f x))^p (c-c \sec (e+f x))}{a+a \sec (e+f x)} \, dx=\int { -\frac {{\left (c \sec \left (f x + e\right ) - c\right )} \left (g \sec \left (f x + e\right )\right )^{p}}{a \sec \left (f x + e\right ) + a} \,d x } \] Input:
integrate((g*sec(f*x+e))^p*(c-c*sec(f*x+e))/(a+a*sec(f*x+e)),x, algorithm= "fricas")
Output:
integral(-(c*sec(f*x + e) - c)*(g*sec(f*x + e))^p/(a*sec(f*x + e) + a), x)
\[ \int \frac {(g \sec (e+f x))^p (c-c \sec (e+f x))}{a+a \sec (e+f x)} \, dx=- \frac {c \left (\int \left (- \frac {\left (g \sec {\left (e + f x \right )}\right )^{p}}{\sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {\left (g \sec {\left (e + f x \right )}\right )^{p} \sec {\left (e + f x \right )}}{\sec {\left (e + f x \right )} + 1}\, dx\right )}{a} \] Input:
integrate((g*sec(f*x+e))**p*(c-c*sec(f*x+e))/(a+a*sec(f*x+e)),x)
Output:
-c*(Integral(-(g*sec(e + f*x))**p/(sec(e + f*x) + 1), x) + Integral((g*sec (e + f*x))**p*sec(e + f*x)/(sec(e + f*x) + 1), x))/a
\[ \int \frac {(g \sec (e+f x))^p (c-c \sec (e+f x))}{a+a \sec (e+f x)} \, dx=\int { -\frac {{\left (c \sec \left (f x + e\right ) - c\right )} \left (g \sec \left (f x + e\right )\right )^{p}}{a \sec \left (f x + e\right ) + a} \,d x } \] Input:
integrate((g*sec(f*x+e))^p*(c-c*sec(f*x+e))/(a+a*sec(f*x+e)),x, algorithm= "maxima")
Output:
-integrate((c*sec(f*x + e) - c)*(g*sec(f*x + e))^p/(a*sec(f*x + e) + a), x )
Exception generated. \[ \int \frac {(g \sec (e+f x))^p (c-c \sec (e+f x))}{a+a \sec (e+f x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((g*sec(f*x+e))^p*(c-c*sec(f*x+e))/(a+a*sec(f*x+e)),x, algorithm= "giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{1,[0,1,2,0]%%%}+%%%{1,[0,1,0,0]%%%} / %%%{2,[0,0,0,1]%%%} Error: Ba
Timed out. \[ \int \frac {(g \sec (e+f x))^p (c-c \sec (e+f x))}{a+a \sec (e+f x)} \, dx=\int \frac {\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )\,{\left (\frac {g}{\cos \left (e+f\,x\right )}\right )}^p}{a+\frac {a}{\cos \left (e+f\,x\right )}} \,d x \] Input:
int(((c - c/cos(e + f*x))*(g/cos(e + f*x))^p)/(a + a/cos(e + f*x)),x)
Output:
int(((c - c/cos(e + f*x))*(g/cos(e + f*x))^p)/(a + a/cos(e + f*x)), x)
\[ \int \frac {(g \sec (e+f x))^p (c-c \sec (e+f x))}{a+a \sec (e+f x)} \, dx=\frac {g^{p} c \left (\int \frac {\sec \left (f x +e \right )^{p}}{\sec \left (f x +e \right )+1}d x -\left (\int \frac {\sec \left (f x +e \right )^{p} \sec \left (f x +e \right )}{\sec \left (f x +e \right )+1}d x \right )\right )}{a} \] Input:
int((g*sec(f*x+e))^p*(c-c*sec(f*x+e))/(a+a*sec(f*x+e)),x)
Output:
(g**p*c*(int(sec(e + f*x)**p/(sec(e + f*x) + 1),x) - int((sec(e + f*x)**p* sec(e + f*x))/(sec(e + f*x) + 1),x)))/a