Integrand size = 31, antiderivative size = 323 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {\left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 a b d}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}-\frac {\left (4 a^2+57 b^2\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{15 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {a \left (4 a^2+11 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{15 b^2 d \sqrt {a+b \sin (c+d x)}}+\frac {b \operatorname {EllipticPi}\left (2,\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{d \sqrt {a+b \sin (c+d x)}} \]
-2/5*cos(d*x+c)*(a+b*sin(d*x+c))^(3/2)/b/d-cot(d*x+c)*(a+b*sin(d*x+c))^(3/ 2)/a/d+1/15*(4*a^2+15*b^2)*cos(d*x+c)*(a+b*sin(d*x+c))^(1/2)/a/b/d+1/15*(4 *a^2+57*b^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) *EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*(a+b*sin(d*x +c))^(1/2)/b^2/d/((a+b*sin(d*x+c))/(a+b))^(1/2)-1/15*a*(4*a^2+11*b^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)*(b/(a+b))^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/2 )/b^2/d/(a+b*sin(d*x+c))^(1/2)-b*(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),2,2^(1/2)*(b/(a+ b))^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/2)/d/(a+b*sin(d*x+c))^(1/2)
Result contains complex when optimal does not.
Time = 2.67 (sec) , antiderivative size = 422, normalized size of antiderivative = 1.31 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {\frac {2 i \left (4 a^2+57 b^2\right ) \left (-2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (-2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right ),\frac {a+b}{a-b}\right )+b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right ) \sec (c+d x) \sqrt {-\frac {b (-1+\sin (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sin (c+d x))}{a-b}}}{a b^3 \sqrt {-\frac {1}{a+b}}}+\frac {184 a \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{\sqrt {a+b \sin (c+d x)}}+\frac {2 \left (4 a^2+27 b^2\right ) \operatorname {EllipticPi}\left (2,\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{b \sqrt {a+b \sin (c+d x)}}-\frac {4 \sqrt {a+b \sin (c+d x)} (2 a \cos (c+d x)+3 b (5 \cot (c+d x)+\sin (2 (c+d x))))}{b}}{60 d} \]
(((2*I)*(4*a^2 + 57*b^2)*(-2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt[-(a + b)^( -1)]*Sqrt[a + b*Sin[c + d*x]]], (a + b)/(a - b)] + b*(-2*a*EllipticF[I*Arc Sinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Sin[c + d*x]]], (a + b)/(a - b)] + b*E llipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Sin[c + d*x] ]], (a + b)/(a - b)]))*Sec[c + d*x]*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b) )]*Sqrt[-((b*(1 + Sin[c + d*x]))/(a - b))])/(a*b^3*Sqrt[-(a + b)^(-1)]) + (184*a*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/Sqrt[a + b*Sin[c + d*x]] + (2*(4*a^2 + 27*b^2)*EllipticPi [2, (-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b )])/(b*Sqrt[a + b*Sin[c + d*x]]) - (4*Sqrt[a + b*Sin[c + d*x]]*(2*a*Cos[c + d*x] + 3*b*(5*Cot[c + d*x] + Sin[2*(c + d*x)])))/b)/(60*d)
Time = 2.40 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.05, number of steps used = 21, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.677, Rules used = {3042, 3373, 27, 3042, 3528, 27, 3042, 3538, 25, 3042, 3134, 3042, 3132, 3481, 3042, 3142, 3042, 3140, 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.
\(\displaystyle \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^4 \sqrt {a+b \sin (c+d x)}}{\sin (c+d x)^2}dx\) |
\(\Big \downarrow \) 3373 |
\(\displaystyle -\frac {2 \int -\frac {1}{4} \csc (c+d x) \sqrt {a+b \sin (c+d x)} \left (5 b^2-14 a \sin (c+d x) b-\left (4 a^2+15 b^2\right ) \sin ^2(c+d x)\right )dx}{5 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \csc (c+d x) \sqrt {a+b \sin (c+d x)} \left (5 b^2-14 a \sin (c+d x) b-\left (4 a^2+15 b^2\right ) \sin ^2(c+d x)\right )dx}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sqrt {a+b \sin (c+d x)} \left (5 b^2-14 a \sin (c+d x) b-\left (4 a^2+15 b^2\right ) \sin (c+d x)^2\right )}{\sin (c+d x)}dx}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 3528 |
\(\displaystyle \frac {\frac {2}{3} \int \frac {\csc (c+d x) \left (-46 b \sin (c+d x) a^2+15 b^2 a-\left (4 a^2+57 b^2\right ) \sin ^2(c+d x) a\right )}{2 \sqrt {a+b \sin (c+d x)}}dx+\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{3} \int \frac {\csc (c+d x) \left (-46 b \sin (c+d x) a^2+15 b^2 a-\left (4 a^2+57 b^2\right ) \sin ^2(c+d x) a\right )}{\sqrt {a+b \sin (c+d x)}}dx+\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} \int \frac {-46 b \sin (c+d x) a^2+15 b^2 a-\left (4 a^2+57 b^2\right ) \sin (c+d x)^2 a}{\sin (c+d x) \sqrt {a+b \sin (c+d x)}}dx+\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {\frac {1}{3} \left (-\frac {a \left (4 a^2+57 b^2\right ) \int \sqrt {a+b \sin (c+d x)}dx}{b}-\frac {\int -\frac {\csc (c+d x) \left (15 a b^3+a^2 \left (4 a^2+11 b^2\right ) \sin (c+d x)\right )}{\sqrt {a+b \sin (c+d x)}}dx}{b}\right )+\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {\int \frac {\csc (c+d x) \left (15 a b^3+a^2 \left (4 a^2+11 b^2\right ) \sin (c+d x)\right )}{\sqrt {a+b \sin (c+d x)}}dx}{b}-\frac {a \left (4 a^2+57 b^2\right ) \int \sqrt {a+b \sin (c+d x)}dx}{b}\right )+\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {\int \frac {15 a b^3+a^2 \left (4 a^2+11 b^2\right ) \sin (c+d x)}{\sin (c+d x) \sqrt {a+b \sin (c+d x)}}dx}{b}-\frac {a \left (4 a^2+57 b^2\right ) \int \sqrt {a+b \sin (c+d x)}dx}{b}\right )+\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 3134 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {\int \frac {15 a b^3+a^2 \left (4 a^2+11 b^2\right ) \sin (c+d x)}{\sin (c+d x) \sqrt {a+b \sin (c+d x)}}dx}{b}-\frac {a \left (4 a^2+57 b^2\right ) \sqrt {a+b \sin (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}\right )+\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {\int \frac {15 a b^3+a^2 \left (4 a^2+11 b^2\right ) \sin (c+d x)}{\sin (c+d x) \sqrt {a+b \sin (c+d x)}}dx}{b}-\frac {a \left (4 a^2+57 b^2\right ) \sqrt {a+b \sin (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}\right )+\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 3132 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {\int \frac {15 a b^3+a^2 \left (4 a^2+11 b^2\right ) \sin (c+d x)}{\sin (c+d x) \sqrt {a+b \sin (c+d x)}}dx}{b}-\frac {2 a \left (4 a^2+57 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}\right )+\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {a^2 \left (4 a^2+11 b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx+15 a b^3 \int \frac {\csc (c+d x)}{\sqrt {a+b \sin (c+d x)}}dx}{b}-\frac {2 a \left (4 a^2+57 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}\right )+\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {a^2 \left (4 a^2+11 b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx+15 a b^3 \int \frac {1}{\sin (c+d x) \sqrt {a+b \sin (c+d x)}}dx}{b}-\frac {2 a \left (4 a^2+57 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}\right )+\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 3142 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {\frac {a^2 \left (4 a^2+11 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{\sqrt {a+b \sin (c+d x)}}+15 a b^3 \int \frac {1}{\sin (c+d x) \sqrt {a+b \sin (c+d x)}}dx}{b}-\frac {2 a \left (4 a^2+57 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}\right )+\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {\frac {a^2 \left (4 a^2+11 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{\sqrt {a+b \sin (c+d x)}}+15 a b^3 \int \frac {1}{\sin (c+d x) \sqrt {a+b \sin (c+d x)}}dx}{b}-\frac {2 a \left (4 a^2+57 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}\right )+\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 3140 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {15 a b^3 \int \frac {1}{\sin (c+d x) \sqrt {a+b \sin (c+d x)}}dx+\frac {2 a^2 \left (4 a^2+11 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \sin (c+d x)}}}{b}-\frac {2 a \left (4 a^2+57 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}\right )+\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 3286 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {\frac {15 a b^3 \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {\csc (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{\sqrt {a+b \sin (c+d x)}}+\frac {2 a^2 \left (4 a^2+11 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \sin (c+d x)}}}{b}-\frac {2 a \left (4 a^2+57 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}\right )+\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {\frac {15 a b^3 \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sin (c+d x) \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{\sqrt {a+b \sin (c+d x)}}+\frac {2 a^2 \left (4 a^2+11 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \sin (c+d x)}}}{b}-\frac {2 a \left (4 a^2+57 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}\right )+\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {\frac {2 \left (4 a^2+15 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}+\frac {1}{3} \left (\frac {\frac {2 a^2 \left (4 a^2+11 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \sin (c+d x)}}+\frac {30 a b^3 \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \sin (c+d x)}}}{b}-\frac {2 a \left (4 a^2+57 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}\right )}{10 a b}-\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^{3/2}}{a d}\) |
(-2*Cos[c + d*x]*(a + b*Sin[c + d*x])^(3/2))/(5*b*d) - (Cot[c + d*x]*(a + b*Sin[c + d*x])^(3/2))/(a*d) + ((2*(4*a^2 + 15*b^2)*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(3*d) + ((-2*a*(4*a^2 + 57*b^2)*EllipticE[(c - Pi/2 + d*x )/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(b*d*Sqrt[(a + b*Sin[c + d*x ])/(a + b)]) + ((2*a^2*(4*a^2 + 11*b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b )/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(d*Sqrt[a + b*Sin[c + d*x]] ) + (30*a*b^3*EllipticPi[2, (c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b *Sin[c + d*x])/(a + b)])/(d*Sqrt[a + b*Sin[c + d*x]]))/b)/3)/(10*a*b)
3.12.46.3.1 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[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_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[Cos[e + f*x]*(a + b* Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x])^(n + 1)/(a*d*f*(n + 1))), x] + (-Si mp[Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x])^(n + 2)/(b*d ^2*f*(m + n + 4))), x] + Simp[1/(a*b*d*(n + 1)*(m + n + 4)) Int[(a + b*Si n[e + f*x])^m*(d*Sin[e + f*x])^(n + 1)*Simp[a^2*(n + 1)*(n + 2) - b^2*(m + n + 2)*(m + n + 4) + a*b*(m + 3)*Sin[e + f*x] - (a^2*(n + 1)*(n + 3) - b^2* (m + n + 3)*(m + n + 4))*Sin[e + f*x]^2, x], x], x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n]) && ! m < -1 && LtQ[n, -1] && NeQ[m + n + 4, 0]
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
Time = 1.40 (sec) , antiderivative size = 657, normalized size of antiderivative = 2.03
method | result | size |
default | \(-\frac {-6 a \,b^{4} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )-8 a^{2} b^{3} \left (\cos ^{4}\left (d x +c \right )\right )+\left (2 a^{3} b^{2}+21 a \,b^{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+23 a^{2} b^{3} \left (\cos ^{2}\left (d x +c \right )\right )-\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \left (4 E\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{5}+53 E\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3} b^{2}-57 E\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{4}-4 F\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{4} b -42 F\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3} b^{2}-11 F\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} b^{3}+57 F\left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{4}-15 \Pi \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \frac {a -b}{a}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{4}+15 \Pi \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \frac {a -b}{a}, \sqrt {\frac {a -b}{a +b}}\right ) b^{5}\right ) \sin \left (d x +c \right )}{15 a \,b^{3} \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {a +b \sin \left (d x +c \right )}\, d}\) | \(657\) |
-1/15*(-6*a*b^4*cos(d*x+c)^4*sin(d*x+c)-8*a^2*b^3*cos(d*x+c)^4+(2*a^3*b^2+ 21*a*b^4)*cos(d*x+c)^2*sin(d*x+c)+23*a^2*b^3*cos(d*x+c)^2-(b/(a-b)*sin(d*x +c)+a/(a-b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c )+b/(a+b))^(1/2)*(4*EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a +b))^(1/2))*a^5+53*EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+ b))^(1/2))*a^3*b^2-57*EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/ (a+b))^(1/2))*a*b^4-4*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/ (a+b))^(1/2))*a^4*b-42*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b) /(a+b))^(1/2))*a^3*b^2-11*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a -b)/(a+b))^(1/2))*a^2*b^3+57*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2), ((a-b)/(a+b))^(1/2))*a*b^4-15*EllipticPi((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2 ),(a-b)/a,((a-b)/(a+b))^(1/2))*a*b^4+15*EllipticPi((b/(a-b)*sin(d*x+c)+a/( a-b))^(1/2),(a-b)/a,((a-b)/(a+b))^(1/2))*b^5)*sin(d*x+c))/a/b^3/sin(d*x+c) /cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d
Timed out. \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\text {Timed out} \]
\[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\int \sqrt {a + b \sin {\left (c + d x \right )}} \cos ^{2}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx \]
\[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\int { \sqrt {b \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{2} \cot \left (d x + c\right )^{2} \,d x } \]
Timed out. \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^2\,{\mathrm {cot}\left (c+d\,x\right )}^2\,\sqrt {a+b\,\sin \left (c+d\,x\right )} \,d x \]