Integrand size = 25, antiderivative size = 445 \[ \int \frac {1}{(a+b \cos (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=-\frac {3 a \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 \left (-a^2+b^2\right )^{7/4} d \sqrt {e}}-\frac {3 a \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 \left (-a^2+b^2\right )^{7/4} d \sqrt {e}}-\frac {\operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b^2\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 a^2 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 a^2 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 \left (a^2-b^2\right ) \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}-\frac {b \sqrt {e \sin (c+d x)}}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x))} \]
-3/2*a*arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))*b^(1/ 2)/(-a^2+b^2)^(7/4)/d/e^(1/2)-3/2*a*arctanh(b^(1/2)*(e*sin(d*x+c))^(1/2)/( -a^2+b^2)^(1/4)/e^(1/2))*b^(1/2)/(-a^2+b^2)^(7/4)/d/e^(1/2)+(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)/(a^2-b^2)/d/(e*sin(d*x+c))^(1/2)-3/2*a ^2*(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)*EllipticP i(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)/(a^2-b^2)/d/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(e*sin(d*x+c))^(1/2)-3/2*a^2 *(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)/(a^2-b^2)/d/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(e*sin(d*x+c))^(1/2)-b*(e*sin( d*x+c))^(1/2)/(a^2-b^2)/d/e/(a+b*cos(d*x+c))
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 9.67 (sec) , antiderivative size = 1182, normalized size of antiderivative = 2.66 \[ \int \frac {1}{(a+b \cos (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=-\frac {b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}+\frac {\sqrt {\sin (c+d x)} \left (-\frac {2 b \cos ^2(c+d x) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right ) \left (\frac {a \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 )}{4 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4}}+\frac {5 b \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sqrt {\sin (c+d x)} \sqrt {1-\sin ^2(c+d x)}}{\left (-5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )+2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},-\frac {1}{2},2,\frac {9}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )+\left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )\right ) \sin ^2(c+d x)\right ) \left (a^2+b^2 \left (-1+\sin ^2(c+d x)\right )\right )}\right )}{(a+b \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac {4 a \cos (c+d x) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right ) \left (-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {b} \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 )}{\left (-a^2+b^2\right )^{3/4}}+\frac {5 a \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sqrt {\sin (c+d x)}}{\sqrt {1-\sin ^2(c+d x)} \left (5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )-2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )+\left (-a^2+b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},1,\frac {9}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )\right ) \sin ^2(c+d x)\right ) \left (a^2+b^2 \left (-1+\sin ^2(c+d x)\right )\right )}\right )}{(a+b \cos (c+d x)) \sqrt {1-\sin ^2(c+d x)}}\right )}{2 (a-b) (a+b) d \sqrt {e \sin (c+d x)}} \]
-((b*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x] ])) + (Sqrt[Sin[c + d*x]]*((-2*b*Cos[c + d*x]^2*(a + b*Sqrt[1 - Sin[c + d* x]^2])*((a*(-2*ArcTan[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]]))/(4*Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)) + (5*b*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2 , (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sqrt[Sin[c + d*x]]*Sqrt[1 - Sin[c + d *x]^2])/((-5*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (b^2* Sin[c + d*x]^2)/(-a^2 + b^2)] + 2*(2*b^2*AppellF1[5/4, -1/2, 2, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)] + (a^2 - b^2)*AppellF1[5/4, 1/2, 1, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)])*Sin[c + d *x]^2)*(a^2 + b^2*(-1 + Sin[c + d*x]^2)))))/((a + b*Cos[c + d*x])*(1 - Sin [c + d*x]^2)) + (4*a*Cos[c + d*x]*(a + b*Sqrt[1 - Sin[c + d*x]^2])*(((-1/8 + I/8)*Sqrt[b]*(2*ArcTan[1 - ((1 + I)*Sqrt[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)*Sqr t[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]]))/(-a^2 + ...
Time = 1.91 (sec) , antiderivative size = 423, normalized size of antiderivative = 0.95, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.720, Rules used = {3042, 3173, 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.
\(\displaystyle \int \frac {1}{\sqrt {e \sin (c+d x)} (a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3173 |
\(\displaystyle -\frac {\int -\frac {2 a-b \cos (c+d x)}{2 (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{a^2-b^2}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {2 a-b \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{2 \left (a^2-b^2\right )}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {2 a+b \sin \left (c+d x-\frac {\pi }{2}\right )}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{2 \left (a^2-b^2\right )}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3346 |
\(\displaystyle \frac {3 a \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx-\int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{2 \left (a^2-b^2\right )}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 a \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx-\int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{2 \left (a^2-b^2\right )}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {3 a \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx-\frac {\sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{\sqrt {e \sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 a \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx-\frac {\sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{\sqrt {e \sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {3 a \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3181 |
\(\displaystyle \frac {3 a \left (-\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b^2 \sin ^2(c+d x) e^2+\left (a^2-b^2\right ) e^2\right )}d(e \sin (c+d x))}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {3 a \left (-\frac {2 b e \int \frac {1}{b^2 e^4 \sin ^4(c+d x)+\left (a^2-b^2\right ) e^2}d\sqrt {e \sin (c+d x)}}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {3 a \left (-\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\int \frac {1}{b e^2 \sin ^2(c+d x)+\sqrt {b^2-a^2} e}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {3 a \left (-\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {3 a \left (-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 a \left (-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3286 |
\(\displaystyle \frac {3 a \left (-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 a \left (-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {3 a \left (-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}+\frac {a \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {b^2-a^2} \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {b^2-a^2} \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}}\right )-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{2 \left (a^2-b^2\right )}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
-((b*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)*d*e*(a + b*Cos[c + d*x]))) + ((-2* EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(d*Sqrt[e*Sin[c + d*x ]]) + 3*a*((-2*b*e*(-1/2*ArcTan[(Sqrt[b]*Sqrt[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]*S in[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]*Sqr t[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*S in[c + d*x]])))/(2*(a^2 - b^2))
3.1.74.3.1 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[(-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[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_)*((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. \(1279\) vs. \(2(476)=952\).
Time = 4.31 (sec) , antiderivative size = 1280, normalized size of antiderivative = 2.88
(-4*a*b*e^3*(1/4*(e*sin(d*x+c))^(1/2)/(a^2*e^2-b^2*e^2)/(-b^2*cos(d*x+c)^2 *e^2+a^2*e^2)+3/32/(a^2*e^2-b^2*e^2)^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)*(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))+2*a^2*(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))-5 /8/(a^2-b^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/ 4/a^2/(a^2-b^2)/(-a^2+b^2)^(1/2)*b*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2...
\[ \int \frac {1}{(a+b \cos (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sqrt {e \sin \left (d x + c\right )}} \,d x } \]
integral(sqrt(e*sin(d*x + c))/((b^2*e*cos(d*x + c)^2 + 2*a*b*e*cos(d*x + c ) + a^2*e)*sin(d*x + c)), x)
\[ \int \frac {1}{(a+b \cos (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\int \frac {1}{\sqrt {e \sin {\left (c + d x \right )}} \left (a + b \cos {\left (c + d x \right )}\right )^{2}}\, dx \]
Timed out. \[ \int \frac {1}{(a+b \cos (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+b \cos (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sqrt {e \sin \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+b \cos (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\int \frac {1}{\sqrt {e\,\sin \left (c+d\,x\right )}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]