Integrand size = 25, antiderivative size = 529 \[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \cos (c+d x))^3} \, dx=-\frac {\left (3 a^2+2 b^2\right ) \sqrt {e} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 \sqrt {b} \left (-a^2+b^2\right )^{9/4} d}+\frac {\left (3 a^2+2 b^2\right ) \sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 \sqrt {b} \left (-a^2+b^2\right )^{9/4} d}+\frac {a \left (3 a^2+2 b^2\right ) e \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{8 b \left (a^2-b^2\right )^2 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {a \left (3 a^2+2 b^2\right ) e \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{8 b \left (a^2-b^2\right )^2 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {5 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d \sqrt {\sin (c+d x)}}-\frac {b (e \sin (c+d x))^{3/2}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {5 a b (e \sin (c+d x))^{3/2}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))} \]
-1/2*b*(e*sin(d*x+c))^(3/2)/(a^2-b^2)/d/e/(a+b*cos(d*x+c))^2-5/4*a*b*(e*si n(d*x+c))^(3/2)/(a^2-b^2)^2/d/e/(a+b*cos(d*x+c))-1/8*(3*a^2+2*b^2)*arctan( b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))*e^(1/2)/(-a^2+b^2)^ (9/4)/d/b^(1/2)+1/8*(3*a^2+2*b^2)*arctanh(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a ^2+b^2)^(1/4)/e^(1/2))*e^(1/2)/(-a^2+b^2)^(9/4)/d/b^(1/2)-1/8*a*(3*a^2+2*b ^2)*e*(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)*Ellipt icPi(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/(a^2-b^2)^2/d/(b-(-a^2+b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-1/8*a*(3 *a^2+2*b^2)*e*(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))*s in(d*x+c)^(1/2)/b/(a^2-b^2)^2/d/(b+(-a^2+b^2)^(1/2))/(e*sin(d*x+c))^(1/2)- 5/4*a*(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)*Ellipt icE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2/d/ sin(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 12.24 (sec) , antiderivative size = 748, normalized size of antiderivative = 1.41 \[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \cos (c+d x))^3} \, dx=\frac {\sqrt {e \sin (c+d x)} \left (-\frac {2 b \left (7 a^2-2 b^2+5 a b \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {\cos (c+d x) \left (a+b \sqrt {\cos ^2(c+d x)}\right ) \left (\frac {5 a \sec (c+d x) \left (3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )\right )+8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)\right )}{\sqrt {b} \left (-a^2+b^2\right )}+\frac {48 \left (4 a^2+b^2\right ) \left (\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+i b \sin (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+i b \sin (c+d x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}+\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)}{3 \left (a^2-b^2\right )}\right )}{\sqrt {\cos ^2(c+d x)}}\right )}{12 (a-b)^2 (a+b)^2 (a+b \cos (c+d x)) \sqrt {\sin (c+d x)}}\right )}{8 d} \]
(Sqrt[e*Sin[c + d*x]]*((-2*b*(7*a^2 - 2*b^2 + 5*a*b*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(a + b*Cos[c + d*x])^2) + (Cos[c + d*x]*(a + b*Sqrt[C os[c + d*x]^2])*((5*a*Sec[c + d*x]*(3*Sqrt[2]*a*(a^2 - b^2)^(3/4)*(2*ArcTa n[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] - Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + b*Sin[c + d*x]] + Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + b*Sin[c + d*x]]) + 8*b^(5/2)*AppellF1[3/4, -1/2, 1, 7/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sin[c + d*x]^(3/2)))/(Sqrt[b] *(-a^2 + b^2)) + (48*(4*a^2 + b^2)*(((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)*Sq rt[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt [b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[c + d*x]] + Log [Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[c + d*x]]))/(Sqrt[b]*(-a^2 + b^2)^(1/4)) + (a*AppellF1[3/4, 1/2, 1, 7/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sin[c + d*x]^( 3/2))/(3*(a^2 - b^2))))/Sqrt[Cos[c + d*x]^2]))/(12*(a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])*Sqrt[Sin[c + d*x]])))/(8*d)
Time = 2.27 (sec) , antiderivative size = 477, normalized size of antiderivative = 0.90, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.840, Rules used = {3042, 3173, 27, 3042, 3343, 27, 3042, 3346, 3042, 3121, 3042, 3119, 3180, 266, 827, 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.
\(\displaystyle \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \cos (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\) |
\(\Big \downarrow \) 3173 |
\(\displaystyle -\frac {\int -\frac {(4 a-b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{2 (a+b \cos (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(4 a-b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{(a+b \cos (c+d x))^2}dx}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sqrt {-e \cos \left (c+d x+\frac {\pi }{2}\right )} \left (4 a-b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3343 |
\(\displaystyle \frac {-\frac {\int -\frac {\left (8 a^2+5 b \cos (c+d x) a+2 b^2\right ) \sqrt {e \sin (c+d x)}}{2 (a+b \cos (c+d x))}dx}{a^2-b^2}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {\left (2 \left (4 a^2+b^2\right )+5 a b \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)}dx}{2 \left (a^2-b^2\right )}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {\sqrt {-e \cos \left (c+d x+\frac {\pi }{2}\right )} \left (2 \left (4 a^2+b^2\right )+5 a b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 \left (a^2-b^2\right )}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3346 |
\(\displaystyle \frac {\frac {\left (3 a^2+2 b^2\right ) \int \frac {\sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)}dx+5 a \int \sqrt {e \sin (c+d x)}dx}{2 \left (a^2-b^2\right )}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\left (3 a^2+2 b^2\right ) \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{a-b \sin \left (c+d x-\frac {\pi }{2}\right )}dx+5 a \int \sqrt {e \sin (c+d x)}dx}{2 \left (a^2-b^2\right )}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {\frac {\left (3 a^2+2 b^2\right ) \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{a-b \sin \left (c+d x-\frac {\pi }{2}\right )}dx+\frac {5 a \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{\sqrt {\sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\left (3 a^2+2 b^2\right ) \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{a-b \sin \left (c+d x-\frac {\pi }{2}\right )}dx+\frac {5 a \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{\sqrt {\sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {\left (3 a^2+2 b^2\right ) \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{a-b \sin \left (c+d x-\frac {\pi }{2}\right )}dx+\frac {10 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3180 |
\(\displaystyle \frac {\frac {\left (3 a^2+2 b^2\right ) \left (-\frac {b e \int \frac {\sqrt {e \sin (c+d x)}}{b^2 \sin ^2(c+d x) e^2+\left (a^2-b^2\right ) e^2}d(e \sin (c+d x))}{d}-\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )+\frac {10 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {\frac {\left (3 a^2+2 b^2\right ) \left (-\frac {2 b e \int \frac {e^2 \sin ^2(c+d x)}{b^2 e^4 \sin ^4(c+d x)+\left (a^2-b^2\right ) e^2}d\sqrt {e \sin (c+d x)}}{d}-\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )+\frac {10 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {\frac {\left (3 a^2+2 b^2\right ) \left (-\frac {2 b e \left (\frac {\int \frac {1}{b e^2 \sin ^2(c+d x)+\sqrt {b^2-a^2} e}d\sqrt {e \sin (c+d x)}}{2 b}-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 b}\right )}{d}-\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )+\frac {10 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\left (3 a^2+2 b^2\right ) \left (-\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 b}\right )}{d}-\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )+\frac {10 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\left (3 a^2+2 b^2\right ) \left (-\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}-\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\right )+\frac {10 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\left (3 a^2+2 b^2\right ) \left (-\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}-\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\right )+\frac {10 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3286 |
\(\displaystyle \frac {\frac {\left (3 a^2+2 b^2\right ) \left (-\frac {a e \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b \sqrt {e \sin (c+d x)}}+\frac {a e \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\right )+\frac {10 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\left (3 a^2+2 b^2\right ) \left (-\frac {a e \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b \sqrt {e \sin (c+d x)}}+\frac {a e \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\right )+\frac {10 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {\frac {\left (3 a^2+2 b^2\right ) \left (-\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}+\frac {a e \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}+\frac {a e \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}}\right )+\frac {10 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {5 a b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
-1/2*(b*(e*Sin[c + d*x])^(3/2))/((a^2 - b^2)*d*e*(a + b*Cos[c + d*x])^2) + ((-5*a*b*(e*Sin[c + d*x])^(3/2))/((a^2 - b^2)*d*e*(a + b*Cos[c + d*x])) + ((10*a*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(d*Sqrt[Sin [c + d*x]]) + (3*a^2 + 2*b^2)*((-2*b*e*(ArcTan[(Sqrt[b]*Sqrt[e]*Sin[c + d* x])/(-a^2 + b^2)^(1/4)]/(2*b^(3/2)*(-a^2 + b^2)^(1/4)*Sqrt[e]) - ArcTanh[( Sqrt[b]*Sqrt[e]*Sin[c + d*x])/(-a^2 + b^2)^(1/4)]/(2*b^(3/2)*(-a^2 + b^2)^ (1/4)*Sqrt[e])))/d + (a*e*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - Pi /2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(b*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin [c + d*x]]) + (a*e*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d* x)/2, 2]*Sqrt[Sin[c + d*x]])/(b*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d* x]])))/(2*(a^2 - b^2)))/(4*(a^2 - b^2))
3.1.84.3.1 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[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m + 1)/(f*g*(a^2 - b^2)*(m + 1))), x] + Simp[1/((a^2 - b^2)*(m + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*(a*(m + 1) - b*(m + p + 2)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b ^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*p]
Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_ )]), x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Simp[a*(g/(2*b)) Int[1/(S qrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (-Simp[a*(g/(2*b)) In t[1/(Sqrt[g*Cos[e + f*x]]*(q - b*Cos[e + f*x])), x], x] + Simp[b*(g/f) Su bst[Int[Sqrt[x]/(g^2*(a^2 - b^2) + b^2*x^2), x], x, g*Cos[e + f*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[(-(b*c - a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m + 1)/(f*g*(a^2 - b^2)*(m + 1))), x] + Simp[1/((a^2 - b^2)*(m + 1)) Int[(g*Cos[e + f*x])^p *(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ [a^2 - b^2, 0] && LtQ[m, -1] && 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. \(2364\) vs. \(2(553)=1106\).
Time = 7.73 (sec) , antiderivative size = 2365, normalized size of antiderivative = 4.47
(2*e^3*b*(-1/8/(a^4-2*a^2*b^2+b^4)*(e*sin(d*x+c))^(3/2)*(-3*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) ^2-1/64*(3*a^2+2*b^2)/(a^4-2*a^2*b^2+b^4)/e^2/b^2/(e^2*(a^2-b^2)/b^2)^(1/4 )*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*arctan(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*sin(d*x+c)) ^(1/2)*e*a*(3/2*b^2/e/a^2/(a^2-b^2)*sin(d*x+c)*(cos(d*x+c)^2*e*sin(d*x+c)) ^(1/2)/(-b^2*cos(d*x+c)^2+a^2)-3/2/a^2/(a^2-b^2)*(1-sin(d*x+c))^(1/2)*(2*s in(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*Elli pticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))+3/4/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))-9/8/(a^2-b^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)/(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))+3/4/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)/(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))-9/8/(a^2-b^2)/b^2*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2...
Timed out. \[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {\sqrt {e \sin \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
\[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {\sqrt {e \sin \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \cos (c+d x))^3} \, dx=\int \frac {\sqrt {e\,\sin \left (c+d\,x\right )}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]