Integrand size = 33, antiderivative size = 299 \[ \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{a+b \cos (e+f x)} \, dx=-\frac {(A b-a B) \operatorname {AppellF1}\left (\frac {1}{2},\frac {m}{2},1,\frac {3}{2},\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \cos ^2(e+f x)^{m/2} (c \sec (e+f x))^{1+m} \sin (e+f x)}{\left (a^2-b^2\right ) c f}+\frac {a (A b-a B) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1+m}{2},1,\frac {3}{2},\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos ^2(e+f x)^{\frac {1+m}{2}} (c \sec (e+f x))^{1+m} \sin (e+f x)}{b \left (a^2-b^2\right ) c f}-\frac {B c \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right ) (c \sec (e+f x))^{-1+m} \sin (e+f x)}{b f (1-m) \sqrt {\sin ^2(e+f x)}} \]
-(A*b-B*a)*AppellF1(1/2,1/2*m,1,3/2,sin(f*x+e)^2,-b^2*sin(f*x+e)^2/(a^2-b^ 2))*cos(f*x+e)*(cos(f*x+e)^2)^(1/2*m)*(c*sec(f*x+e))^(1+m)*sin(f*x+e)/(a^2 -b^2)/c/f+a*(A*b-B*a)*AppellF1(1/2,1/2+1/2*m,1,3/2,sin(f*x+e)^2,-b^2*sin(f *x+e)^2/(a^2-b^2))*(cos(f*x+e)^2)^(1/2+1/2*m)*(c*sec(f*x+e))^(1+m)*sin(f*x +e)/b/(a^2-b^2)/c/f-B*c*hypergeom([1/2, -1/2*m+1/2],[3/2-1/2*m],cos(f*x+e) ^2)*(c*sec(f*x+e))^(-1+m)*sin(f*x+e)/b/f/(1-m)/(sin(f*x+e)^2)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(10630\) vs. \(2(299)=598\).
Time = 32.02 (sec) , antiderivative size = 10630, normalized size of antiderivative = 35.55 \[ \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{a+b \cos (e+f x)} \, dx=\text {Result too large to show} \]
Time = 1.37 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 3439, 3042, 4526, 3042, 4259, 3042, 3122, 4356, 3042, 3302, 3042, 3668, 333}
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.
\(\displaystyle \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{a+b \cos (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (A+B \sin \left (e+f x+\frac {\pi }{2}\right )\right ) \left (c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^m}{a+b \sin \left (e+f x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3439 |
\(\displaystyle \int \frac {(A \sec (e+f x)+B) (c \sec (e+f x))^m}{a \sec (e+f x)+b}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (A \csc \left (e+f x+\frac {\pi }{2}\right )+B\right ) \left (c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^m}{a \csc \left (e+f x+\frac {\pi }{2}\right )+b}dx\) |
\(\Big \downarrow \) 4526 |
\(\displaystyle \frac {(A b-a B) \int \frac {(c \sec (e+f x))^{m+1}}{b+a \sec (e+f x)}dx}{b c}+\frac {B \int (c \sec (e+f x))^mdx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A b-a B) \int \frac {\left (c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{m+1}}{b+a \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{b c}+\frac {B \int \left (c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^mdx}{b}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \frac {(A b-a B) \int \frac {\left (c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{m+1}}{b+a \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{b c}+\frac {B \left (\frac {\cos (e+f x)}{c}\right )^m (c \sec (e+f x))^m \int \left (\frac {\cos (e+f x)}{c}\right )^{-m}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A b-a B) \int \frac {\left (c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{m+1}}{b+a \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{b c}+\frac {B \left (\frac {\cos (e+f x)}{c}\right )^m (c \sec (e+f x))^m \int \left (\frac {\sin \left (e+f x+\frac {\pi }{2}\right )}{c}\right )^{-m}dx}{b}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {(A b-a B) \int \frac {\left (c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{m+1}}{b+a \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{b c}-\frac {B c \sin (e+f x) (c \sec (e+f x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right )}{b f (1-m) \sqrt {\sin ^2(e+f x)}}\) |
\(\Big \downarrow \) 4356 |
\(\displaystyle \frac {(A b-a B) \cos ^{m+1}(e+f x) (c \sec (e+f x))^{m+1} \int \frac {\cos ^{-m}(e+f x)}{a+b \cos (e+f x)}dx}{b c}-\frac {B c \sin (e+f x) (c \sec (e+f x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right )}{b f (1-m) \sqrt {\sin ^2(e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A b-a B) \cos ^{m+1}(e+f x) (c \sec (e+f x))^{m+1} \int \frac {\sin \left (e+f x+\frac {\pi }{2}\right )^{-m}}{a+b \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{b c}-\frac {B c \sin (e+f x) (c \sec (e+f x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right )}{b f (1-m) \sqrt {\sin ^2(e+f x)}}\) |
\(\Big \downarrow \) 3302 |
\(\displaystyle \frac {(A b-a B) \cos ^{m+1}(e+f x) (c \sec (e+f x))^{m+1} \left (a \int \frac {\cos ^{-m}(e+f x)}{a^2-b^2 \cos ^2(e+f x)}dx-b \int \frac {\cos ^{1-m}(e+f x)}{a^2-b^2 \cos ^2(e+f x)}dx\right )}{b c}-\frac {B c \sin (e+f x) (c \sec (e+f x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right )}{b f (1-m) \sqrt {\sin ^2(e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A b-a B) \cos ^{m+1}(e+f x) (c \sec (e+f x))^{m+1} \left (a \int \frac {\sin \left (e+f x+\frac {\pi }{2}\right )^{-m}}{a^2-b^2 \sin \left (e+f x+\frac {\pi }{2}\right )^2}dx-b \int \frac {\sin \left (e+f x+\frac {\pi }{2}\right )^{1-m}}{a^2-b^2 \sin \left (e+f x+\frac {\pi }{2}\right )^2}dx\right )}{b c}-\frac {B c \sin (e+f x) (c \sec (e+f x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right )}{b f (1-m) \sqrt {\sin ^2(e+f x)}}\) |
\(\Big \downarrow \) 3668 |
\(\displaystyle \frac {(A b-a B) \cos ^{m+1}(e+f x) (c \sec (e+f x))^{m+1} \left (\frac {a \cos ^{-m-1}(e+f x) \cos ^2(e+f x)^{\frac {m+1}{2}} \int \frac {\left (1-\sin ^2(e+f x)\right )^{\frac {1}{2} (-m-1)}}{a^2-b^2+b^2 \sin ^2(e+f x)}d\sin (e+f x)}{f}-\frac {b \cos ^{-m}(e+f x) \cos ^2(e+f x)^{m/2} \int \frac {\left (1-\sin ^2(e+f x)\right )^{-m/2}}{a^2-b^2+b^2 \sin ^2(e+f x)}d\sin (e+f x)}{f}\right )}{b c}-\frac {B c \sin (e+f x) (c \sec (e+f x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right )}{b f (1-m) \sqrt {\sin ^2(e+f x)}}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle \frac {(A b-a B) \cos ^{m+1}(e+f x) (c \sec (e+f x))^{m+1} \left (\frac {a \sin (e+f x) \cos ^{-m-1}(e+f x) \cos ^2(e+f x)^{\frac {m+1}{2}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {m+1}{2},1,\frac {3}{2},\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )}-\frac {b \sin (e+f x) \cos ^{-m}(e+f x) \cos ^2(e+f x)^{m/2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {m}{2},1,\frac {3}{2},\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )}\right )}{b c}-\frac {B c \sin (e+f x) (c \sec (e+f x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right )}{b f (1-m) \sqrt {\sin ^2(e+f x)}}\) |
-((B*c*Hypergeometric2F1[1/2, (1 - m)/2, (3 - m)/2, Cos[e + f*x]^2]*(c*Sec [e + f*x])^(-1 + m)*Sin[e + f*x])/(b*f*(1 - m)*Sqrt[Sin[e + f*x]^2])) + (( A*b - a*B)*Cos[e + f*x]^(1 + m)*(c*Sec[e + f*x])^(1 + m)*(-((b*AppellF1[1/ 2, m/2, 1, 3/2, Sin[e + f*x]^2, -((b^2*Sin[e + f*x]^2)/(a^2 - b^2))]*(Cos[ e + f*x]^2)^(m/2)*Sin[e + f*x])/((a^2 - b^2)*f*Cos[e + f*x]^m)) + (a*Appel lF1[1/2, (1 + m)/2, 1, 3/2, Sin[e + f*x]^2, -((b^2*Sin[e + f*x]^2)/(a^2 - b^2))]*Cos[e + f*x]^(-1 - m)*(Cos[e + f*x]^2)^((1 + m)/2)*Sin[e + f*x])/(( a^2 - b^2)*f)))/(b*c)
3.7.40.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
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[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[a Int[(d*Sin[e + f*x])^n/(a^2 - b^2*Sin[e + f*x] ^2), x], x] - Simp[b/d Int[(d*Sin[e + f*x])^(n + 1)/(a^2 - b^2*Sin[e + f* x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* (x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim p[g^(m + n) Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[( -ff)*d^(2*IntPart[(m - 1)/2] + 1)*((d*Sin[e + f*x])^(2*FracPart[(m - 1)/2]) /(f*(Sin[e + f*x]^2)^FracPart[(m - 1)/2])) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && !IntegerQ[m]
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_))^(m_.), x_Symbol] :> Simp[Sin[e + f*x]^n*(d*Csc[e + f*x])^n Int[(b + a*Sin[e + f*x])^m/Sin[e + f*x]^(m + n), x], x] /; FreeQ[{a, b, d, e, f, n} , x] && NeQ[a^2 - b^2, 0] && IntegerQ[m]
Int[((csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[A/a Int[ (d*Csc[e + f*x])^n, x], x] - Simp[(A*b - a*B)/(a*d) Int[(d*Csc[e + f*x])^ (n + 1)/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
\[\int \frac {\left (A +\cos \left (f x +e \right ) B \right ) \left (c \sec \left (f x +e \right )\right )^{m}}{a +b \cos \left (f x +e \right )}d x\]
\[ \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{a+b \cos (e+f x)} \, dx=\int { \frac {{\left (B \cos \left (f x + e\right ) + A\right )} \left (c \sec \left (f x + e\right )\right )^{m}}{b \cos \left (f x + e\right ) + a} \,d x } \]
\[ \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{a+b \cos (e+f x)} \, dx=\int \frac {\left (c \sec {\left (e + f x \right )}\right )^{m} \left (A + B \cos {\left (e + f x \right )}\right )}{a + b \cos {\left (e + f x \right )}}\, dx \]
\[ \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{a+b \cos (e+f x)} \, dx=\int { \frac {{\left (B \cos \left (f x + e\right ) + A\right )} \left (c \sec \left (f x + e\right )\right )^{m}}{b \cos \left (f x + e\right ) + a} \,d x } \]
Exception generated. \[ \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{a+b \cos (e+f x)} \, dx=\text {Exception raised: TypeError} \]
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,[0,1,0,0]%%%} / %%%{1,[0,0,1,0]%%%}+%%%{-1,[0,0,0,1]%%% } Error:
Timed out. \[ \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{a+b \cos (e+f x)} \, dx=\int \frac {{\left (\frac {c}{\cos \left (e+f\,x\right )}\right )}^m\,\left (A+B\,\cos \left (e+f\,x\right )\right )}{a+b\,\cos \left (e+f\,x\right )} \,d x \]