Integrand size = 25, antiderivative size = 512 \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\frac {5 \left (3 a^2-2 b^2\right ) e^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{7/2} \left (-a^2+b^2\right )^{3/4} d}+\frac {5 \left (3 a^2-2 b^2\right ) e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{7/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {15 a e^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{4 b^4 d \sqrt {e \sin (c+d x)}}+\frac {5 a \left (3 a^2-2 b^2\right ) e^4 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{8 b^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {5 a \left (3 a^2-2 b^2\right ) e^4 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{8 b^4 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {5 e^3 (3 a+2 b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2} \]
5/8*(3*a^2-2*b^2)*e^(7/2)*arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^( 1/4)/e^(1/2))/b^(7/2)/(-a^2+b^2)^(3/4)/d+5/8*(3*a^2-2*b^2)*e^(7/2)*arctanh (b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(7/2)/(-a^2+b^2) ^(3/4)/d+1/2*e*(e*sin(d*x+c))^(5/2)/b/d/(a+b*cos(d*x+c))^2+15/4*a*e^4*(sin (1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/ 2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/b^4/d/(e*sin(d*x+c))^(1/2)-5 /8*a*(3*a^2-2*b^2)*e^4*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*P i+1/2*d*x)*EllipticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*b/(b-(-a^2+b^2)^(1/2)),2 ^(1/2))*sin(d*x+c)^(1/2)/b^4/d/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(e*sin(d*x+c)) ^(1/2)-5/8*a*(3*a^2-2*b^2)*e^4*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2 *c+1/4*Pi+1/2*d*x)*EllipticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*b/(b+(-a^2+b^2)^ (1/2)),2^(1/2))*sin(d*x+c)^(1/2)/b^4/d/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(e*sin (d*x+c))^(1/2)+5/4*e^3*(3*a+2*b*cos(d*x+c))*(e*sin(d*x+c))^(1/2)/b^3/d/(a+ b*cos(d*x+c))
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 10.94 (sec) , antiderivative size = 946, normalized size of antiderivative = 1.85 \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\frac {\csc ^3(c+d x) (e \sin (c+d x))^{7/2} \left (7 a^2+2 b^2+9 a b \cos (c+d x)+\frac {(a+b \cos (c+d x)) (-6 b-7 a \cos (c+d x)+4 b \cos (2 (c+d x))) \left (8 (a+b)-5 (3 a+2 b) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}+(3 a-2 b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{\frac {1}{9} \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sin (c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \left (-45 (3 a+2 b) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+18 (3 a-2 b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )+\frac {9 (3 a+2 b) \left (2 (a-b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},1,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}+9 (3 a-2 b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )-\frac {5 (3 a-2 b) \left (2 (a-b) \operatorname {AppellF1}\left (\frac {9}{4},\frac {1}{2},2,\frac {13}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) \operatorname {AppellF1}\left (\frac {9}{4},\frac {3}{2},1,\frac {13}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+\cos (c+d x) \left (8 (a+b)-5 (3 a+2 b) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}+(3 a-2 b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}\right )}{4 b^3 d (a+b \cos (c+d x))^2} \]
(Csc[c + d*x]^3*(e*Sin[c + d*x])^(7/2)*(7*a^2 + 2*b^2 + 9*a*b*Cos[c + d*x] + ((a + b*Cos[c + d*x])*(-6*b - 7*a*Cos[c + d*x] + 4*b*Cos[2*(c + d*x)])* (8*(a + b) - 5*(3*a + 2*b)*AppellF1[1/4, 1/2, 1, 5/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)]*Sqrt[Sec[(c + d*x)/2]^2] + (3*a - 2*b)*AppellF1[5/4, 1/2, 1, 9/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d* x)/2]^2)/(a + b)]*Sqrt[Sec[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2))/((Sqrt[Sec [(c + d*x)/2]^2]*Sin[c + d*x]*Tan[(c + d*x)/2]*(-45*(3*a + 2*b)*AppellF1[1 /4, 1/2, 1, 5/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b )] + 18*(3*a - 2*b)*AppellF1[5/4, 1/2, 1, 9/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)]*Sec[(c + d*x)/2]^2 + (9*(3*a + 2*b)*(2*(a - b)*AppellF1[5/4, 1/2, 2, 9/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d *x)/2]^2)/(a + b)] + (a + b)*AppellF1[5/4, 3/2, 1, 9/4, -Tan[(c + d*x)/2]^ 2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)])*Sec[(c + d*x)/2]^2)/(a + b) + 9 *(3*a - 2*b)*AppellF1[5/4, 1/2, 1, 9/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan [(c + d*x)/2]^2)/(a + b)]*Tan[(c + d*x)/2]^2 - (5*(3*a - 2*b)*(2*(a - b)*A ppellF1[9/4, 1/2, 2, 13/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2] ^2)/(a + b)] + (a + b)*AppellF1[9/4, 3/2, 1, 13/4, -Tan[(c + d*x)/2]^2, (( -a + b)*Tan[(c + d*x)/2]^2)/(a + b)])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]^ 2)/(a + b)))/9 + Cos[c + d*x]*(8*(a + b) - 5*(3*a + 2*b)*AppellF1[1/4, 1/2 , 1, 5/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)]*S...
Time = 2.23 (sec) , antiderivative size = 483, normalized size of antiderivative = 0.94, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.840, Rules used = {3042, 3172, 25, 3042, 3342, 27, 3042, 3346, 3042, 3121, 3042, 3120, 3181, 266, 756, 218, 221, 3042, 3286, 3042, 3284}
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.
\(\displaystyle \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{7/2}}{\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\) |
\(\Big \downarrow \) 3172 |
\(\displaystyle \frac {5 e^2 \int -\frac {\cos (c+d x) (e \sin (c+d x))^{3/2}}{(a+b \cos (c+d x))^2}dx}{4 b}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \int \frac {\cos (c+d x) (e \sin (c+d x))^{3/2}}{(a+b \cos (c+d x))^2}dx}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \int \frac {\left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{4 b}\) |
\(\Big \downarrow \) 3342 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (-\frac {e^2 \int -\frac {2 b+3 a \cos (c+d x)}{2 (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \int \frac {2 b+3 a \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \int \frac {2 b-3 a \sin \left (c+d x-\frac {\pi }{2}\right )}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
\(\Big \downarrow \) 3346 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {3 a \int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{b}-\frac {\left (3 a^2-2 b^2\right ) \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {3 a \int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{b}-\frac {\left (3 a^2-2 b^2\right ) \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {3 a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{b \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {3 a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{b \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
\(\Big \downarrow \) 3181 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b^2 \sin ^2(c+d x) e^2+\left (a^2-b^2\right ) e^2\right )}d(e \sin (c+d x))}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {2 b e \int \frac {1}{b^2 e^4 \sin ^4(c+d x)+\left (a^2-b^2\right ) e^2}d\sqrt {e \sin (c+d x)}}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\int \frac {1}{b e^2 \sin ^2(c+d x)+\sqrt {b^2-a^2} e}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
\(\Big \downarrow \) 3286 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {5 e^2 \left (\frac {e^2 \left (\frac {6 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (3 a^2-2 b^2\right ) \left (-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}+\frac {a \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {b^2-a^2} \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {b^2-a^2} \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}}\right )}{b}\right )}{2 b^2}-\frac {e \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{b^2 d (a+b \cos (c+d x))}\right )}{4 b}\) |
(e*(e*Sin[c + d*x])^(5/2))/(2*b*d*(a + b*Cos[c + d*x])^2) - (5*e^2*(-((e*( 3*a + 2*b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]])/(b^2*d*(a + b*Cos[c + d*x])) ) + (e^2*((6*a*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(b*d*S qrt[e*Sin[c + d*x]]) - ((3*a^2 - 2*b^2)*((-2*b*e*(-1/2*ArcTan[(Sqrt[b]*Sqr t[e]*Sin[c + d*x])/(-a^2 + b^2)^(1/4)]/(Sqrt[b]*(-a^2 + b^2)^(3/4)*e^(3/2) ) - ArcTanh[(Sqrt[b]*Sqrt[e]*Sin[c + d*x])/(-a^2 + b^2)^(1/4)]/(2*Sqrt[b]* (-a^2 + b^2)^(3/4)*e^(3/2))))/d + (a*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2 ]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(Sqrt[-a^2 + b^2]*(b - Sqrt [-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) - (a*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(Sqrt[-a^2 + b^2]*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]])))/b))/(2*b^2)))/(4*b)
3.1.81.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + 1))), x] + Simp[g^2*((p - 1)/(b*(m + 1))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; Fre eQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && I ntegersQ[2*m, 2*p]
Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)* (x_)])), x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Simp[-a/(2*q) Int[1/( Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (Simp[b*(g/f) Subst[ Int[1/(Sqrt[x]*(g^2*(a^2 - b^2) + b^2*x^2)), x], x, g*Cos[e + f*x]], x] - S imp[a/(2*q) Int[1/(Sqrt[g*Cos[e + f*x]]*(q - b*Cos[e + f*x])), x], x])] / ; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt [c + d*Sin[e + f*x]] Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + d))*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] && NeQ[c^2 - d^2, 0] && !GtQ[c + d, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*(g*C os[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d*p + b*d*(m + 1)*Sin[e + f*x])/(b^2*f*(m + 1)*(m + p + 1))), x] + Simp[g^2*(( p - 1)/(b^2*(m + 1)*(m + p + 1))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin [e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Sin[e + f*x ], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && Lt Q[m, -1] && GtQ[p, 1] && NeQ[m + p + 1, 0] && IntegerQ[2*m]
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((c_.) + (d_.)*sin[(e_.) + (f_.)* (x_)]))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d/b Int [(g*Cos[e + f*x])^p, x], x] + Simp[(b*c - a*d)/b Int[(g*Cos[e + f*x])^p/( a + b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(2588\) vs. \(2(536)=1072\).
Time = 98.09 (sec) , antiderivative size = 2589, normalized size of antiderivative = 5.06
(2*e^3*b*((e*sin(d*x+c))^(1/2)/b^4-e^2/b^4*(-1/8*(e*sin(d*x+c))^(1/2)*e^2* (-11*a^2*b^2*cos(d*x+c)^2+2*b^4*cos(d*x+c)^2+7*a^4+2*a^2*b^2)/(-b^2*cos(d* x+c)^2*e^2+a^2*e^2)^2+5/64*(3*a^2-2*b^2)*(e^2*(a^2-b^2)/b^2)^(1/4)/(a^2*e^ 2-b^2*e^2)*2^(1/2)*(ln((e*sin(d*x+c)+(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+ c))^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2))/(e*sin(d*x+c)-(e^2*(a^2-b^2)/ b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2)))+2*arct an(2^(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)+1)+2*arctan(2^(1 /2)/(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)-1))))-(cos(d*x+c)^2*e*s in(d*x+c))^(1/2)*e^4*a*(-3/b^4*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2) *sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*EllipticF((1-sin(d*x+c ))^(1/2),1/2*2^(1/2))+(-10*a^2+6*b^2)/b^4*(-1/2/(-a^2+b^2)^(1/2)/b*(1-sin( d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin( d*x+c))^(1/2)/(1-(-a^2+b^2)^(1/2)/b)*EllipticPi((1-sin(d*x+c))^(1/2),1/(1- (-a^2+b^2)^(1/2)/b),1/2*2^(1/2))+1/2/(-a^2+b^2)^(1/2)/b*(1-sin(d*x+c))^(1/ 2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin(d*x+c))^(1/ 2)/(1+(-a^2+b^2)^(1/2)/b)*EllipticPi((1-sin(d*x+c))^(1/2),1/(1+(-a^2+b^2)^ (1/2)/b),1/2*2^(1/2)))+1/b^4*(11*a^4-14*a^2*b^2+3*b^4)*(1/2*b^2/e/a^2/(a^2 -b^2)*(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(-b^2*cos(d*x+c)^2+a^2)+1/4/a^2/(a ^2-b^2)*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos( d*x+c)^2*e*sin(d*x+c))^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2)...
Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]
\[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
\[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\int \frac {{\left (e\,\sin \left (c+d\,x\right )\right )}^{7/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]