Integrand size = 25, antiderivative size = 838 \[ \int \frac {1}{(a+b \sec (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=-\frac {3 b^3 \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{2 a^{3/2} \left (a^2-b^2\right )^{7/4} d \sqrt {e}}-\frac {2 b \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{a^{3/2} \left (a^2-b^2\right )^{3/4} d \sqrt {e}}-\frac {3 b^3 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{2 a^{3/2} \left (a^2-b^2\right )^{7/4} d \sqrt {e}}-\frac {2 b \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{a^{3/2} \left (a^2-b^2\right )^{3/4} d \sqrt {e}}+\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{a^2 d \sqrt {e \sin (c+d x)}}+\frac {b^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{a^2 \left (a^2-b^2\right ) d \sqrt {e \sin (c+d x)}}+\frac {2 b^2 \operatorname {EllipticPi}\left (\frac {2 a}{a-\sqrt {a^2-b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{a^2 \left (a^2-b^2-a \sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 b^4 \operatorname {EllipticPi}\left (\frac {2 a}{a-\sqrt {a^2-b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 a^2 \left (a^2-b^2\right ) \left (a^2-b^2-a \sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {2 b^2 \operatorname {EllipticPi}\left (\frac {2 a}{a+\sqrt {a^2-b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{a^2 \left (a^2-b^2+a \sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 b^4 \operatorname {EllipticPi}\left (\frac {2 a}{a+\sqrt {a^2-b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 a^2 \left (a^2-b^2\right ) \left (a^2-b^2+a \sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {b^2 \sqrt {e \sin (c+d x)}}{a \left (a^2-b^2\right ) d e (b+a \cos (c+d x))} \]
-3/2*b^3*arctan(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))/a^(3 /2)/(a^2-b^2)^(7/4)/d/e^(1/2)-2*b*arctan(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2 -b^2)^(1/4)/e^(1/2))/a^(3/2)/(a^2-b^2)^(3/4)/d/e^(1/2)-3/2*b^3*arctanh(a^( 1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))/a^(3/2)/(a^2-b^2)^(7/4) /d/e^(1/2)-2*b*arctanh(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2 ))/a^(3/2)/(a^2-b^2)^(3/4)/d/e^(1/2)-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))*si n(d*x+c)^(1/2)/a^2/d/(e*sin(d*x+c))^(1/2)-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))*sin(d*x+c)^(1/2)/a^2/(a^2-b^2)/d/(e*sin(d*x+c))^(1/2)-2*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)*EllipticPi(cos(1/2*c+ 1/4*Pi+1/2*d*x),2*a/(a-(a^2-b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/a^2/d/(a ^2-b^2-a*(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-3/2*b^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*a/(a-(a^2-b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/a^2/(a^2-b^2)/d/(a ^2-b^2-a*(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-2*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)*EllipticPi(cos(1/2*c+1/4*Pi+1/2*d *x),2*a/(a+(a^2-b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/a^2/d/(a^2-b^2+a*(a^ 2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-3/2*b^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*a...
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 14.65 (sec) , antiderivative size = 1246, normalized size of antiderivative = 1.49 \[ \int \frac {1}{(a+b \sec (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\frac {(b+a \cos (c+d x))^2 \sec ^2(c+d x) \sqrt {\sin (c+d x)} \left (\frac {2 \left (-2 a^2+b^2\right ) \cos ^2(c+d x) \left (b+a \sqrt {1-\sin ^2(c+d x)}\right ) \left (\frac {b \left (-2 \arctan \left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )\right )}{4 \sqrt {2} \sqrt {a} \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 {a^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 {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )+2 \left (2 a^2 \operatorname {AppellF1}\left (\frac {5}{4},-\frac {1}{2},2,\frac {9}{4},\sin ^2(c+d x),\frac {a^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 {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )\right ) \sin ^2(c+d x)\right ) \left (b^2+a^2 \left (-1+\sin ^2(c+d x)\right )\right )}\right )}{(b+a \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac {4 a b \cos (c+d x) \left (b+a \sqrt {1-\sin ^2(c+d x)}\right ) \left (-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {a} \left (2 \arctan \left (1-\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+\log \left (\sqrt {a^2-b^2}-(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+i a \sin (c+d x)\right )-\log \left (\sqrt {a^2-b^2}+(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+i a \sin (c+d x)\right )\right )}{\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 {a^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 {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )+2 \left (2 a^2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},\sin ^2(c+d x),\frac {a^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 {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )\right ) \sin ^2(c+d x)\right ) \left (b^2+a^2 \left (-1+\sin ^2(c+d x)\right )\right )}\right )}{(b+a \cos (c+d x)) \sqrt {1-\sin ^2(c+d x)}}\right )}{2 a (-a+b) (a+b) d (a+b \sec (c+d x))^2 \sqrt {e \sin (c+d x)}}+\frac {b^2 (b+a \cos (c+d x)) \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2 \sqrt {e \sin (c+d x)}} \]
((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*Sqrt[Sin[c + d*x]]*((2*(-2*a^2 + b^ 2)*Cos[c + d*x]^2*(b + a*Sqrt[1 - Sin[c + d*x]^2])*((b*(-2*ArcTan[1 - (Sqr t[2]*Sqrt[a]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[ 2]*Sqrt[a]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] - Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + a*Sin[c + d*x]] + Log[Sqrt[-a^2 + b^2] + Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d *x]] + a*Sin[c + d*x]]))/(4*Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(3/4)) - (5*a*(a^ 2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (a^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, (a^2*Sin[c + d*x]^2)/(a^2 - b^ 2)] + 2*(2*a^2*AppellF1[5/4, -1/2, 2, 9/4, Sin[c + d*x]^2, (a^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, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)])*Sin[c + d*x]^2)*(b^2 + a^2*(-1 + Sin [c + d*x]^2)))))/((b + a*Cos[c + d*x])*(1 - Sin[c + d*x]^2)) + (4*a*b*Cos[ c + d*x]*(b + a*Sqrt[1 - Sin[c + d*x]^2])*(((-1/8 + I/8)*Sqrt[a]*(2*ArcTan [1 - ((1 + I)*Sqrt[a]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[a]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] + Log[Sqrt[a^2 - b^2] - (1 + I)*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*a*Sin[c + d*x]] - Log[Sqrt[a^2 - b^2] + (1 + I)*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Sin[ c + d*x]] + I*a*Sin[c + d*x]]))/(a^2 - b^2)^(3/4) + (5*b*(a^2 - b^2)*Ap...
Time = 2.24 (sec) , antiderivative size = 838, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4360, 3042, 3391, 2009}
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.
| \(\displaystyle \int \frac {1}{\sqrt {e \sin (c+d x)} (a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{\sqrt {e \sin (c+d x)} (-a \cos (c+d x)-b)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a \sin \left (c+d x-\frac {\pi }{2}\right )-b\right )^2}dx\) |
| \(\Big \downarrow \) 3391 |
| \(\displaystyle \int \left (\frac {b^2}{a^2 \sqrt {e \sin (c+d x)} (-a \cos (c+d x)-b)^2}+\frac {2 b}{a^2 \sqrt {e \sin (c+d x)} (-a \cos (c+d x)-b)}+\frac {1}{a^2 \sqrt {e \sin (c+d x)}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 \operatorname {EllipticPi}\left (\frac {2 a}{a-\sqrt {a^2-b^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} b^4}{2 a^2 \left (a^2-b^2\right ) \left (a^2-\sqrt {a^2-b^2} a-b^2\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 \operatorname {EllipticPi}\left (\frac {2 a}{a+\sqrt {a^2-b^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} b^4}{2 a^2 \left (a^2-b^2\right ) \left (a^2+\sqrt {a^2-b^2} a-b^2\right ) d \sqrt {e \sin (c+d x)}}-\frac {3 \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b^3}{2 a^{3/2} \left (a^2-b^2\right )^{7/4} d \sqrt {e}}-\frac {3 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b^3}{2 a^{3/2} \left (a^2-b^2\right )^{7/4} d \sqrt {e}}+\frac {\sqrt {e \sin (c+d x)} b^2}{a \left (a^2-b^2\right ) d e (b+a \cos (c+d x))}+\frac {\operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} b^2}{a^2 \left (a^2-b^2\right ) d \sqrt {e \sin (c+d x)}}+\frac {2 \operatorname {EllipticPi}\left (\frac {2 a}{a-\sqrt {a^2-b^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} b^2}{a^2 \left (a^2-\sqrt {a^2-b^2} a-b^2\right ) d \sqrt {e \sin (c+d x)}}+\frac {2 \operatorname {EllipticPi}\left (\frac {2 a}{a+\sqrt {a^2-b^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} b^2}{a^2 \left (a^2+\sqrt {a^2-b^2} a-b^2\right ) d \sqrt {e \sin (c+d x)}}-\frac {2 \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b}{a^{3/2} \left (a^2-b^2\right )^{3/4} d \sqrt {e}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b}{a^{3/2} \left (a^2-b^2\right )^{3/4} d \sqrt {e}}+\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)}}{a^2 d \sqrt {e \sin (c+d x)}}\) |
(-3*b^3*ArcTan[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])] )/(2*a^(3/2)*(a^2 - b^2)^(7/4)*d*Sqrt[e]) - (2*b*ArcTan[(Sqrt[a]*Sqrt[e*Si n[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/(a^(3/2)*(a^2 - b^2)^(3/4)*d*Sq rt[e]) - (3*b^3*ArcTanh[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)* Sqrt[e])])/(2*a^(3/2)*(a^2 - b^2)^(7/4)*d*Sqrt[e]) - (2*b*ArcTanh[(Sqrt[a] *Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/(a^(3/2)*(a^2 - b^2)^ (3/4)*d*Sqrt[e]) + (2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]]) /(a^2*d*Sqrt[e*Sin[c + d*x]]) + (b^2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt [Sin[c + d*x]])/(a^2*(a^2 - b^2)*d*Sqrt[e*Sin[c + d*x]]) + (2*b^2*Elliptic Pi[(2*a)/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]]) /(a^2*(a^2 - b^2 - a*Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) + (3*b^4*Ell ipticPi[(2*a)/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d *x]])/(2*a^2*(a^2 - b^2)*(a^2 - b^2 - a*Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) + (2*b^2*EllipticPi[(2*a)/(a + Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2 , 2]*Sqrt[Sin[c + d*x]])/(a^2*(a^2 - b^2 + a*Sqrt[a^2 - b^2])*d*Sqrt[e*Sin [c + d*x]]) + (3*b^4*EllipticPi[(2*a)/(a + Sqrt[a^2 - b^2]), (c - Pi/2 + d *x)/2, 2]*Sqrt[Sin[c + d*x]])/(2*a^2*(a^2 - b^2)*(a^2 - b^2 + a*Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) + (b^2*Sqrt[e*Sin[c + d*x]])/(a*(a^2 - b^2) *d*e*(b + a*Cos[c + d*x]))
3.3.46.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[m] && (G tQ[m, 0] || IntegerQ[n])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 23.71 (sec) , antiderivative size = 1239, normalized size of antiderivative = 1.48
(4*a*b*e*(1/4*b^2/a^2/(a^2-b^2)*(e*sin(d*x+c))^(1/2)/(-a^2*e^2*cos(d*x+c)^ 2+b^2*e^2)+1/16*(4*a^2-b^2)/a^2/(a^2-b^2)*(e^2*(a^2-b^2)/a^2)^(1/4)/(-a^2* e^2+b^2*e^2)*(ln(((e*sin(d*x+c))^(1/2)+(e^2*(a^2-b^2)/a^2)^(1/4))/((e*sin( d*x+c))^(1/2)-(e^2*(a^2-b^2)/a^2)^(1/4)))+2*arctan((e*sin(d*x+c))^(1/2)/(e ^2*(a^2-b^2)/a^2)^(1/4))))+(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*(-1/a^2*(-sin (d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*s in(d*x+c))^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+2/a^2*b^4*(- 1/2*a^2/e/b^2/(a^2-b^2)*(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(-cos(d*x+c)^2*a ^2+b^2)-1/4/b^2/(a^2-b^2)*(-sin(d*x+c)+1)^(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((-sin(d*x+c)+1)^ (1/2),1/2*2^(1/2))-1/4/b^2/(a^2-b^2)^(3/2)*a*(-sin(d*x+c)+1)^(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)/a)*EllipticPi((-sin(d*x+c)+1)^(1/2),1/(1-(a^2-b^2)^(1/2)/a),1/ 2*2^(1/2))+5/8/(a^2-b^2)^(3/2)/a*(-sin(d*x+c)+1)^(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)/ a)*EllipticPi((-sin(d*x+c)+1)^(1/2),1/(1-(a^2-b^2)^(1/2)/a),1/2*2^(1/2))+1 /4/b^2/(a^2-b^2)^(3/2)*a*(-sin(d*x+c)+1)^(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)/a)*Ellip ticPi((-sin(d*x+c)+1)^(1/2),1/(1+(a^2-b^2)^(1/2)/a),1/2*2^(1/2))-5/8/(a^2- b^2)^(3/2)/a*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1...
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+b \sec (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\int \frac {1}{\sqrt {e \sin {\left (c + d x \right )}} \left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+b \sec (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {e \sin \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{\sqrt {e\,\sin \left (c+d\,x\right )}\,{\left (b+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]