Integrand size = 21, antiderivative size = 294 \[ \int \frac {\cos ^m(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {b^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-1-m),2,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^{1+m}(c+d x) \cos ^2(c+d x)^{\frac {1}{2} (-1-m)} \sin (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac {a^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},2,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^{-1+m}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} \sin (c+d x)}{\left (a^2-b^2\right )^2 d}-\frac {2 a b \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},2,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \sin (c+d x)}{\left (a^2-b^2\right )^2 d} \] Output:
b^2*AppellF1(1/2,-1/2-1/2*m,2,3/2,sin(d*x+c)^2,-b^2*sin(d*x+c)^2/(a^2-b^2) )*cos(d*x+c)^(1+m)*(cos(d*x+c)^2)^(-1/2-1/2*m)*sin(d*x+c)/(a^2-b^2)^2/d+a^ 2*AppellF1(1/2,1/2-1/2*m,2,3/2,sin(d*x+c)^2,-b^2*sin(d*x+c)^2/(a^2-b^2))*c os(d*x+c)^(-1+m)*(cos(d*x+c)^2)^(1/2-1/2*m)*sin(d*x+c)/(a^2-b^2)^2/d-2*a*b *AppellF1(1/2,-1/2*m,2,3/2,sin(d*x+c)^2,-b^2*sin(d*x+c)^2/(a^2-b^2))*cos(d *x+c)^m*sin(d*x+c)/(a^2-b^2)^2/d/((cos(d*x+c)^2)^(1/2*m))
Leaf count is larger than twice the leaf count of optimal. \(7214\) vs. \(2(294)=588\).
Time = 29.15 (sec) , antiderivative size = 7214, normalized size of antiderivative = 24.54 \[ \int \frac {\cos ^m(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Result too large to show} \] Input:
Integrate[Cos[c + d*x]^m/(a + b*Cos[c + d*x])^2,x]
Output:
Result too large to show
Time = 0.68 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3303, 2009}
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.
\(\displaystyle \int \frac {\cos ^m(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^m}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3303 |
\(\displaystyle \int \left (-\frac {2 a b \cos ^{m+1}(c+d x)}{\left (a^2-b^2 \cos ^2(c+d x)\right )^2}+\frac {b^2 \cos ^{m+2}(c+d x)}{\left (b^2 \cos ^2(c+d x)-a^2\right )^2}+\frac {a^2 \cos ^m(c+d x)}{\left (a^2-b^2 \cos ^2(c+d x)\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x) \cos ^2(c+d x)^{\frac {1}{2} (-m-1)} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-m-1),2,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )^2}+\frac {a^2 \sin (c+d x) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},2,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )^2}-\frac {2 a b \sin (c+d x) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},2,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )^2}\) |
Input:
Int[Cos[c + d*x]^m/(a + b*Cos[c + d*x])^2,x]
Output:
(b^2*AppellF1[1/2, (-1 - m)/2, 2, 3/2, Sin[c + d*x]^2, -((b^2*Sin[c + d*x] ^2)/(a^2 - b^2))]*Cos[c + d*x]^(1 + m)*(Cos[c + d*x]^2)^((-1 - m)/2)*Sin[c + d*x])/((a^2 - b^2)^2*d) + (a^2*AppellF1[1/2, (1 - m)/2, 2, 3/2, Sin[c + d*x]^2, -((b^2*Sin[c + d*x]^2)/(a^2 - b^2))]*Cos[c + d*x]^(-1 + m)*(Cos[c + d*x]^2)^((1 - m)/2)*Sin[c + d*x])/((a^2 - b^2)^2*d) - (2*a*b*AppellF1[1 /2, -1/2*m, 2, 3/2, Sin[c + d*x]^2, -((b^2*Sin[c + d*x]^2)/(a^2 - b^2))]*C os[c + d*x]^m*Sin[c + d*x])/((a^2 - b^2)^2*d*(Cos[c + d*x]^2)^(m/2))
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(1/((a - b*sin[ e + f*x])^m/(a^2 - b^2*sin[e + f*x]^2)^m)), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0] && ILtQ[m, -1]
\[\int \frac {\cos \left (d x +c \right )^{m}}{\left (a +\cos \left (d x +c \right ) b \right )^{2}}d x\]
Input:
int(cos(d*x+c)^m/(a+cos(d*x+c)*b)^2,x)
Output:
int(cos(d*x+c)^m/(a+cos(d*x+c)*b)^2,x)
\[ \int \frac {\cos ^m(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {\cos \left (d x + c\right )^{m}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(cos(d*x+c)^m/(a+b*cos(d*x+c))^2,x, algorithm="fricas")
Output:
integral(cos(d*x + c)^m/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2), x )
Timed out. \[ \int \frac {\cos ^m(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**m/(a+b*cos(d*x+c))**2,x)
Output:
Timed out
\[ \int \frac {\cos ^m(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {\cos \left (d x + c\right )^{m}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(cos(d*x+c)^m/(a+b*cos(d*x+c))^2,x, algorithm="maxima")
Output:
integrate(cos(d*x + c)^m/(b*cos(d*x + c) + a)^2, x)
\[ \int \frac {\cos ^m(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {\cos \left (d x + c\right )^{m}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(cos(d*x+c)^m/(a+b*cos(d*x+c))^2,x, algorithm="giac")
Output:
integrate(cos(d*x + c)^m/(b*cos(d*x + c) + a)^2, x)
Timed out. \[ \int \frac {\cos ^m(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^m}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \] Input:
int(cos(c + d*x)^m/(a + b*cos(c + d*x))^2,x)
Output:
int(cos(c + d*x)^m/(a + b*cos(c + d*x))^2, x)
\[ \int \frac {\cos ^m(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int \frac {\cos \left (d x +c \right )^{m}}{\cos \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) a b +a^{2}}d x \] Input:
int(cos(d*x+c)^m/(a+b*cos(d*x+c))^2,x)
Output:
int(cos(c + d*x)**m/(cos(c + d*x)**2*b**2 + 2*cos(c + d*x)*a*b + a**2),x)