Integrand size = 35, antiderivative size = 449 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{7/2}} \, dx=-\frac {2 \left (3 a^3 b B+29 a b^3 B+2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 \left (a^2-b^2\right )^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (3 a^2 b B+5 b^3 B+2 a^3 C-2 a b^2 (4 A+5 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{5 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}+\frac {2 \left (3 a^2 b B+5 b^3 B+2 a^3 C-2 a b^2 (4 A+5 C)\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (3 a^3 b B+29 a b^3 B+2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 C)\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^3 d \sqrt {a+b \cos (c+d x)}} \]
-2/5*(A*b^2-a*(B*b-C*a))*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(5/2)+2 /15*(3*B*a^2*b+5*B*b^3+2*a^3*C-2*a*b^2*(4*A+5*C))*sin(d*x+c)/b/(a^2-b^2)^2 /d/(a+b*cos(d*x+c))^(3/2)+2/15*(3*B*a^3*b+29*B*a*b^3+2*a^4*C-3*b^4*(3*A+5* C)-a^2*b^2*(23*A+19*C))*sin(d*x+c)/b/(a^2-b^2)^3/d/(a+b*cos(d*x+c))^(1/2)- 2/15*(3*B*a^3*b+29*B*a*b^3+2*a^4*C-3*b^4*(3*A+5*C)-a^2*b^2*(23*A+19*C))*(c os(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c) ,2^(1/2)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/b^2/(a^2-b^2)^3/d/((a+b*c os(d*x+c))/(a+b))^(1/2)+2/15*(3*B*a^2*b+5*B*b^3+2*a^3*C-2*a*b^2*(4*A+5*C)) *(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2 *c),2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/b^2/(a^2-b^2)^ 2/d/(a+b*cos(d*x+c))^(1/2)
Time = 4.75 (sec) , antiderivative size = 433, normalized size of antiderivative = 0.96 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{7/2}} \, dx=\frac {2 \left (\frac {\left (\frac {a+b \cos (c+d x)}{a+b}\right )^{5/2} \left (b^2 \left (-27 a^2 b B-5 b^3 B+a^3 (15 A+7 C)+a b^2 (17 A+25 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+\left (-3 a^3 b B-29 a b^3 B-2 a^4 C+3 b^4 (3 A+5 C)+a^2 b^2 (23 A+19 C)\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )\right )}{(a-b)^3 (a+b)}+\frac {b \left (68 a^4 A b^2+13 a^2 A b^4+15 A b^6-18 a^5 b B-53 a^3 b^3 B-25 a b^5 B-2 a^6 C+48 a^4 b^2 C+35 a^2 b^4 C+15 b^6 C+2 b \left (-9 a^4 b B-60 a^2 b^3 B+5 b^5 B-6 a^5 C+10 a b^4 (A+2 C)+2 a^3 b^2 (27 A+25 C)\right ) \cos (c+d x)+b^2 \left (-3 a^3 b B-29 a b^3 B-2 a^4 C+3 b^4 (3 A+5 C)+a^2 b^2 (23 A+19 C)\right ) \cos (2 (c+d x))\right ) \sin (c+d x)}{2 \left (-a^2+b^2\right )^3}\right )}{15 b^2 d (a+b \cos (c+d x))^{5/2}} \]
(2*((((a + b*Cos[c + d*x])/(a + b))^(5/2)*(b^2*(-27*a^2*b*B - 5*b^3*B + a^ 3*(15*A + 7*C) + a*b^2*(17*A + 25*C))*EllipticF[(c + d*x)/2, (2*b)/(a + b) ] + (-3*a^3*b*B - 29*a*b^3*B - 2*a^4*C + 3*b^4*(3*A + 5*C) + a^2*b^2*(23*A + 19*C))*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)])))/((a - b)^3*(a + b)) + (b*(68*a^4*A*b^2 + 13*a^ 2*A*b^4 + 15*A*b^6 - 18*a^5*b*B - 53*a^3*b^3*B - 25*a*b^5*B - 2*a^6*C + 48 *a^4*b^2*C + 35*a^2*b^4*C + 15*b^6*C + 2*b*(-9*a^4*b*B - 60*a^2*b^3*B + 5* b^5*B - 6*a^5*C + 10*a*b^4*(A + 2*C) + 2*a^3*b^2*(27*A + 25*C))*Cos[c + d* x] + b^2*(-3*a^3*b*B - 29*a*b^3*B - 2*a^4*C + 3*b^4*(3*A + 5*C) + a^2*b^2* (23*A + 19*C))*Cos[2*(c + d*x)])*Sin[c + d*x])/(2*(-a^2 + b^2)^3)))/(15*b^ 2*d*(a + b*Cos[c + d*x])^(5/2))
Time = 2.22 (sec) , antiderivative size = 472, normalized size of antiderivative = 1.05, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.514, Rules used = {3042, 3500, 27, 3042, 3233, 27, 3042, 3233, 27, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}
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.
\(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx\) |
\(\Big \downarrow \) 3500 |
\(\displaystyle -\frac {2 \int \frac {5 b (b B-a (A+C))+\left (-2 C a^2-3 b B a+3 A b^2+5 b^2 C\right ) \cos (c+d x)}{2 (a+b \cos (c+d x))^{5/2}}dx}{5 b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {5 b (b B-a (A+C))+\left (-2 C a^2-3 b B a+3 A b^2+5 b^2 C\right ) \cos (c+d x)}{(a+b \cos (c+d x))^{5/2}}dx}{5 b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {5 b (b B-a (A+C))+\left (-2 C a^2-3 b B a+3 A b^2+5 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{5 b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3233 |
\(\displaystyle -\frac {-\frac {2 \int -\frac {3 b \left (-\left ((5 A+3 C) a^2\right )+8 b B a-b^2 (3 A+5 C)\right )-\left (2 C a^3+3 b B a^2-2 b^2 (4 A+5 C) a+5 b^3 B\right ) \cos (c+d x)}{2 (a+b \cos (c+d x))^{3/2}}dx}{3 \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right )}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{5 b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {\int \frac {3 b \left (-\left ((5 A+3 C) a^2\right )+8 b B a-b^2 (3 A+5 C)\right )-\left (2 C a^3+3 b B a^2-2 b^2 (4 A+5 C) a+5 b^3 B\right ) \cos (c+d x)}{(a+b \cos (c+d x))^{3/2}}dx}{3 \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right )}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{5 b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\int \frac {3 b \left (-\left ((5 A+3 C) a^2\right )+8 b B a-b^2 (3 A+5 C)\right )+\left (-2 C a^3-3 b B a^2+2 b^2 (4 A+5 C) a-5 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right )}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{5 b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3233 |
\(\displaystyle -\frac {\frac {-\frac {2 \int -\frac {b \left (-\left ((15 A+7 C) a^3\right )+27 b B a^2-b^2 (17 A+25 C) a+5 b^3 B\right )+\left (2 C a^4+3 b B a^3-b^2 (23 A+19 C) a^2+29 b^3 B a-3 b^4 (3 A+5 C)\right ) \cos (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}-\frac {2 \sin (c+d x) \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right )}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right )}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{5 b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {\frac {\int \frac {b \left (-\left ((15 A+7 C) a^3\right )+27 b B a^2-b^2 (17 A+25 C) a+5 b^3 B\right )+\left (2 C a^4+3 b B a^3-b^2 (23 A+19 C) a^2+29 b^3 B a-3 b^4 (3 A+5 C)\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}-\frac {2 \sin (c+d x) \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right )}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right )}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{5 b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\frac {\int \frac {b \left (-\left ((15 A+7 C) a^3\right )+27 b B a^2-b^2 (17 A+25 C) a+5 b^3 B\right )+\left (2 C a^4+3 b B a^3-b^2 (23 A+19 C) a^2+29 b^3 B a-3 b^4 (3 A+5 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {2 \sin (c+d x) \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right )}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right )}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{5 b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3231 |
\(\displaystyle -\frac {\frac {\frac {\frac {\left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right ) \int \sqrt {a+b \cos (c+d x)}dx}{b}-\frac {\left (a^2-b^2\right ) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}}{a^2-b^2}-\frac {2 \sin (c+d x) \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right )}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right )}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{5 b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\frac {\frac {\left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (a^2-b^2\right ) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{a^2-b^2}-\frac {2 \sin (c+d x) \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right )}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right )}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{5 b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3134 |
\(\displaystyle -\frac {\frac {\frac {\frac {\left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{a^2-b^2}-\frac {2 \sin (c+d x) \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right )}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right )}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{5 b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\frac {\frac {\left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{a^2-b^2}-\frac {2 \sin (c+d x) \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right )}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right )}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{5 b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3132 |
\(\displaystyle -\frac {\frac {\frac {\frac {2 \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{a^2-b^2}-\frac {2 \sin (c+d x) \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right )}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right )}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{5 b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3142 |
\(\displaystyle -\frac {\frac {\frac {\frac {2 \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}}{a^2-b^2}-\frac {2 \sin (c+d x) \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right )}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right )}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{5 b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\frac {\frac {2 \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}}{a^2-b^2}-\frac {2 \sin (c+d x) \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right )}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right )}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{5 b \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3140 |
\(\displaystyle -\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}-\frac {\frac {\frac {\frac {2 \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \cos (c+d x)}}}{a^2-b^2}-\frac {2 \sin (c+d x) \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right )}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 \sin (c+d x) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right )}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{5 b \left (a^2-b^2\right )}\) |
(-2*(A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(5*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(5/2)) - ((-2*(3*a^2*b*B + 5*b^3*B + 2*a^3*C - 2*a*b^2*(4*A + 5*C) )*Sin[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) + (((2*(3*a^3 *b*B + 29*a*b^3*B + 2*a^4*C - 3*b^4*(3*A + 5*C) - a^2*b^2*(23*A + 19*C))*S qrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(b*d*Sqrt[( a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*(3*a^2*b*B + 5*b^3*B + 2*a^ 3*C - 2*a*b^2*(4*A + 5*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(b*d*Sqrt[a + b*Cos[c + d*x]]))/(a^2 - b^2) - ( 2*(3*a^3*b*B + 29*a*b^3*B + 2*a^4*C - 3*b^4*(3*A + 5*C) - a^2*b^2*(23*A + 19*C))*Sin[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]))/(3*(a^2 - b ^2)))/(5*b*(a^2 - b^2))
3.11.60.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)] Int[Sqrt[a/(a + b) + ( b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 , 0] && !GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]] Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && !GtQ[a + b, 0]
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b Int[1/Sqrt[a + b*Sin[e + f*x ]], x], x] + Simp[d/b Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a, b , c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(1319\) vs. \(2(483)=966\).
Time = 12.15 (sec) , antiderivative size = 1320, normalized size of antiderivative = 2.94
method | result | size |
default | \(\text {Expression too large to display}\) | \(1320\) |
parts | \(\text {Expression too large to display}\) | \(2962\) |
-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*C/b^2/s in(1/2*d*x+1/2*c)^2/(2*b*sin(1/2*d*x+1/2*c)^2-a-b)/(a^2-b^2)*(-2*b*sin(1/2 *d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*b*sin(1/2*d*x+1/2*c)^2* cos(1/2*d*x+1/2*c)+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/ 2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)) *a-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a- b))^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)))+2*(B*b-2*C*a )/b^2*(1/6/b/(a-b)/(a+b)*cos(1/2*d*x+1/2*c)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+ b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2+1/2/b*(a-b))^2+8/3*b* sin(1/2*d*x+1/2*c)^2/(a-b)^2/(a+b)^2*cos(1/2*d*x+1/2*c)*a/(-(-2*cos(1/2*d* x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+(3*a-b)/(3*a^3+3*a^2*b-3*a*b ^2-3*b^3)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a- b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*Ell ipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-4/3*a/(a-b)/(a+b)^2*(sin(1/2 *d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*si n(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d* x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/ 2))))+2*(A*b^2-B*a*b+C*a^2)/b^2*(1/20/b^2/(a-b)/(a+b)*cos(1/2*d*x+1/2*c)*( -2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1 /2*c)^2+1/2/b*(a-b))^3+4/15*a/b/(a-b)^2/(a+b)^2*cos(1/2*d*x+1/2*c)*(-2*...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.25 (sec) , antiderivative size = 1813, normalized size of antiderivative = 4.04 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{7/2}} \, dx=\text {Too large to display} \]
1/45*(6*(C*a^6*b^2 + 9*B*a^5*b^3 - (34*A + 25*C)*a^4*b^4 + 25*B*a^3*b^5 + (5*A - 8*C)*a^2*b^6 - 2*B*a*b^7 - 3*A*b^8 + (2*C*a^4*b^4 + 3*B*a^3*b^5 - ( 23*A + 19*C)*a^2*b^6 + 29*B*a*b^7 - 3*(3*A + 5*C)*b^8)*cos(d*x + c)^2 + (6 *C*a^5*b^3 + 9*B*a^4*b^4 - 2*(27*A + 25*C)*a^3*b^5 + 60*B*a^2*b^6 - 10*(A + 2*C)*a*b^7 - 5*B*b^8)*cos(d*x + c))*sqrt(b*cos(d*x + c) + a)*sin(d*x + c ) - (sqrt(2)*(4*I*C*a^5*b^3 + 6*I*B*a^4*b^4 - I*(A + 17*C)*a^3*b^5 - 23*I* B*a^2*b^6 + 3*I*(11*A + 15*C)*a*b^7 - 15*I*B*b^8)*cos(d*x + c)^3 - 3*sqrt( 2)*(-4*I*C*a^6*b^2 - 6*I*B*a^5*b^3 + I*(A + 17*C)*a^4*b^4 + 23*I*B*a^3*b^5 - 3*I*(11*A + 15*C)*a^2*b^6 + 15*I*B*a*b^7)*cos(d*x + c)^2 - 3*sqrt(2)*(- 4*I*C*a^7*b - 6*I*B*a^6*b^2 + I*(A + 17*C)*a^5*b^3 + 23*I*B*a^4*b^4 - 3*I* (11*A + 15*C)*a^3*b^5 + 15*I*B*a^2*b^6)*cos(d*x + c) + sqrt(2)*(4*I*C*a^8 + 6*I*B*a^7*b - I*(A + 17*C)*a^6*b^2 - 23*I*B*a^5*b^3 + 3*I*(11*A + 15*C)* a^4*b^4 - 15*I*B*a^3*b^5))*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 - 3*b^2) /b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b) - (sqrt(2)*(-4*I*C*a^5*b^3 - 6*I*B*a^4*b^4 + I*(A + 17*C)*a^ 3*b^5 + 23*I*B*a^2*b^6 - 3*I*(11*A + 15*C)*a*b^7 + 15*I*B*b^8)*cos(d*x + c )^3 - 3*sqrt(2)*(4*I*C*a^6*b^2 + 6*I*B*a^5*b^3 - I*(A + 17*C)*a^4*b^4 - 23 *I*B*a^3*b^5 + 3*I*(11*A + 15*C)*a^2*b^6 - 15*I*B*a*b^7)*cos(d*x + c)^2 - 3*sqrt(2)*(4*I*C*a^7*b + 6*I*B*a^6*b^2 - I*(A + 17*C)*a^5*b^3 - 23*I*B*a^4 *b^4 + 3*I*(11*A + 15*C)*a^3*b^5 - 15*I*B*a^2*b^6)*cos(d*x + c) + sqrt(...
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{7/2}} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{7/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{7/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{7/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \]