Integrand size = 25, antiderivative size = 473 \[ \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \cos (c+d x))^2} \, dx=-\frac {7 a \left (-a^2+b^2\right )^{3/4} e^{9/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{9/2} d}+\frac {7 a \left (-a^2+b^2\right )^{3/4} e^{9/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{9/2} d}-\frac {7 a^2 \left (a^2-b^2\right ) e^5 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 b^5 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {7 a^2 \left (a^2-b^2\right ) e^5 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 b^5 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {7 \left (5 a^2-3 b^2\right ) e^4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 b^4 d \sqrt {\sin (c+d x)}}-\frac {7 e^3 (5 a-3 b \cos (c+d x)) (e \sin (c+d x))^{3/2}}{15 b^3 d}+\frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))} \] Output:
-7/2*a*(-a^2+b^2)^(3/4)*e^(9/2)*arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+ b^2)^(1/4)/e^(1/2))/b^(9/2)/d+7/2*a*(-a^2+b^2)^(3/4)*e^(9/2)*arctanh(b^(1/ 2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(9/2)/d+7/2*a^2*(a^2-b ^2)*e^5*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^5/(b-(-a^2+b^2)^(1/2))/d/(e*sin(d*x+c))^(1/2)+7/2* a^2*(a^2-b^2)*e^5*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^5/(b+(-a^2+b^2)^(1/2))/d/(e*sin(d*x+c))^ (1/2)-7/5*(5*a^2-3*b^2)*e^4*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*( e*sin(d*x+c))^(1/2)/b^4/d/sin(d*x+c)^(1/2)-7/15*e^3*(5*a-3*b*cos(d*x+c))*( e*sin(d*x+c))^(3/2)/b^3/d+e*(e*sin(d*x+c))^(7/2)/b/d/(a+b*cos(d*x+c))
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 17.29 (sec) , antiderivative size = 835, normalized size of antiderivative = 1.77 \[ \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \cos (c+d x))^2} \, dx=\frac {7 (e \sin (c+d x))^{9/2} \left (\frac {\left (5 a^2-3 b^2\right ) \cos ^2(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 ) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right )}{12 b^{3/2} \left (-a^2+b^2\right ) (a+b \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac {4 a b \cos (c+d x) \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 ) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right )}{(a+b \cos (c+d x)) \sqrt {1-\sin ^2(c+d x)}}\right )}{10 b^3 d \sin ^{\frac {9}{2}}(c+d x)}+\frac {\csc ^4(c+d x) (e \sin (c+d x))^{9/2} \left (-\frac {4 a \sin (c+d x)}{3 b^3}+\frac {-a^2 \sin (c+d x)+b^2 \sin (c+d x)}{b^3 (a+b \cos (c+d x))}+\frac {\sin (2 (c+d x))}{5 b^2}\right )}{d} \] Input:
Integrate[(e*Sin[c + d*x])^(9/2)/(a + b*Cos[c + d*x])^2,x]
Output:
(7*(e*Sin[c + d*x])^(9/2)*(((5*a^2 - 3*b^2)*Cos[c + d*x]^2*(3*Sqrt[2]*a*(a ^2 - b^2)^(3/4)*(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[Si n[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))*(a + b*Sqrt[1 - Sin[c + d*x]^2]))/(12*b^(3/2)*(-a^2 + b^2)* (a + b*Cos[c + d*x])*(1 - Sin[c + d*x]^2)) + (4*a*b*Cos[c + d*x]*(((1/8 + I/8)*(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)*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)))*(a + b*Sqrt[1 - Sin[c + d*x]^2]))/((a + b*Cos[c + d*x])*Sqrt[1 - Sin[c + d*x]^2])))/(10*b^3*d*S in[c + d*x]^(9/2)) + (Csc[c + d*x]^4*(e*Sin[c + d*x])^(9/2)*((-4*a*Sin[c + d*x])/(3*b^3) + (-(a^2*Sin[c + d*x]) + b^2*Sin[c + d*x])/(b^3*(a + b*Cos[ c + d*x])) + Sin[2*(c + d*x)]/(5*b^2)))/d
Time = 2.10 (sec) , antiderivative size = 460, normalized size of antiderivative = 0.97, 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, 3344, 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.
\(\displaystyle \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{9/2}}{\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3172 |
\(\displaystyle \frac {7 e^2 \int -\frac {\cos (c+d x) (e \sin (c+d x))^{5/2}}{a+b \cos (c+d x)}dx}{2 b}+\frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \int \frac {\cos (c+d x) (e \sin (c+d x))^{5/2}}{a+b \cos (c+d x)}dx}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \int \frac {\left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}\) |
\(\Big \downarrow \) 3344 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e^2 \int -\frac {\left (2 a b+\left (5 a^2-3 b^2\right ) \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{2 (a+b \cos (c+d x))}dx}{5 b^2}+\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}\right )}{2 b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}-\frac {e^2 \int \frac {\left (2 a b+\left (5 a^2-3 b^2\right ) \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)}dx}{5 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}-\frac {e^2 \int \frac {\sqrt {-e \cos \left (c+d x+\frac {\pi }{2}\right )} \left (2 a b+\left (5 a^2-3 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3346 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}-\frac {e^2 \left (\frac {\left (5 a^2-3 b^2\right ) \int \sqrt {e \sin (c+d x)}dx}{b}-\frac {5 a \left (a^2-b^2\right ) \int \frac {\sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)}dx}{b}\right )}{5 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}-\frac {e^2 \left (\frac {\left (5 a^2-3 b^2\right ) \int \sqrt {e \sin (c+d x)}dx}{b}-\frac {5 a \left (a^2-b^2\right ) \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{a-b \sin \left (c+d x-\frac {\pi }{2}\right )}dx}{b}\right )}{5 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}-\frac {e^2 \left (\frac {\left (5 a^2-3 b^2\right ) \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{b \sqrt {\sin (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{a-b \sin \left (c+d x-\frac {\pi }{2}\right )}dx}{b}\right )}{5 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}-\frac {e^2 \left (\frac {\left (5 a^2-3 b^2\right ) \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{b \sqrt {\sin (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{a-b \sin \left (c+d x-\frac {\pi }{2}\right )}dx}{b}\right )}{5 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{b d \sqrt {\sin (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{a-b \sin \left (c+d x-\frac {\pi }{2}\right )}dx}{b}\right )}{5 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3180 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{b d \sqrt {\sin (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {b e \int \frac {\sqrt {e \sin (c+d x)}}{b^2 \sin ^2(c+d x) e^2+\left (a^2-b^2\right ) e^2}d(e \sin (c+d x))}{d}-\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{5 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{b d \sqrt {\sin (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {2 b e \int \frac {e^2 \sin ^2(c+d x)}{b^2 e^4 \sin ^4(c+d x)+\left (a^2-b^2\right ) e^2}d\sqrt {e \sin (c+d x)}}{d}-\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{5 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{b d \sqrt {\sin (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {2 b e \left (\frac {\int \frac {1}{b e^2 \sin ^2(c+d x)+\sqrt {b^2-a^2} e}d\sqrt {e \sin (c+d x)}}{2 b}-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 b}\right )}{d}-\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{5 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{b d \sqrt {\sin (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 b}\right )}{d}-\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{5 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{b d \sqrt {\sin (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}-\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\right )}{b}\right )}{5 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{b d \sqrt {\sin (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}-\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\right )}{b}\right )}{5 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3286 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{b d \sqrt {\sin (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {a e \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b \sqrt {e \sin (c+d x)}}+\frac {a e \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\right )}{b}\right )}{5 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{b d \sqrt {\sin (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {a e \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b \sqrt {e \sin (c+d x)}}+\frac {a e \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\right )}{b}\right )}{5 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {e (e \sin (c+d x))^{7/2}}{b d (a+b \cos (c+d x))}-\frac {7 e^2 \left (\frac {2 e (e \sin (c+d x))^{3/2} (5 a-3 b \cos (c+d x))}{15 b^2 d}-\frac {e^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{b d \sqrt {\sin (c+d x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}+\frac {a e \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}+\frac {a e \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}}\right )}{b}\right )}{5 b^2}\right )}{2 b}\) |
Input:
Int[(e*Sin[c + d*x])^(9/2)/(a + b*Cos[c + d*x])^2,x]
Output:
(e*(e*Sin[c + d*x])^(7/2))/(b*d*(a + b*Cos[c + d*x])) - (7*e^2*((2*e*(5*a - 3*b*Cos[c + d*x])*(e*Sin[c + d*x])^(3/2))/(15*b^2*d) - (e^2*((2*(5*a^2 - 3*b^2)*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(b*d*Sqrt[S in[c + d*x]]) - (5*a*(a^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]) - ArcTan h[(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]])))/b))/(5*b^2)))/(2*b)
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[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[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[g*(g* Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d* p + b*d*(m + p)*Sin[e + f*x])/(b^2*f*(m + p)*(m + p + 1))), x] + Simp[g^2*( (p - 1)/(b^2*(m + p)*(m + p + 1))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Si n[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1) - d*(a^ 2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && 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. \(1627\) vs. \(2(409)=818\).
Time = 16.41 (sec) , antiderivative size = 1628, normalized size of antiderivative = 3.44
Input:
int((e*sin(d*x+c))^(9/2)/(a+cos(d*x+c)*b)^2,x,method=_RETURNVERBOSE)
Output:
(-4*e^3*a*b*(1/3*(e*sin(d*x+c))^(3/2)/b^4-e^2/b^4*((-1/4*a^2+1/4*b^2)*(e*s in(d*x+c))^(3/2)/(-b^2*cos(d*x+c)^2*e^2+a^2*e^2)+1/8*(7/4*a^2-7/4*b^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^5*(-1/5/b^4/(cos(d*x+c)^2*e*sin(d*x+c)) ^(1/2)*(30*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*El lipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2-16*(1-sin(d*x+c))^(1/2)*(2+2 *sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^( 1/2))*b^2-15*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)* EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2+8*(1-sin(d*x+c))^(1/2)*(2+ 2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^ (1/2))*b^2+2*b^2*cos(d*x+c)^4-2*cos(d*x+c)^2*b^2)-(5*a^4-6*a^2*b^2+b^4)/b^ 4*(-1/2/b^2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(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/b^2*(1-sin(d*x+c)) ^(1/2)*(2+2*sin(d*x+c))^(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^...
Timed out. \[ \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate((e*sin(d*x+c))^(9/2)/(a+b*cos(d*x+c))^2,x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate((e*sin(d*x+c))**(9/2)/(a+b*cos(d*x+c))**2,x)
Output:
Timed out
\[ \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {9}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate((e*sin(d*x+c))^(9/2)/(a+b*cos(d*x+c))^2,x, algorithm="maxima")
Output:
integrate((e*sin(d*x + c))^(9/2)/(b*cos(d*x + c) + a)^2, x)
\[ \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {9}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate((e*sin(d*x+c))^(9/2)/(a+b*cos(d*x+c))^2,x, algorithm="giac")
Output:
integrate((e*sin(d*x + c))^(9/2)/(b*cos(d*x + c) + a)^2, x)
Timed out. \[ \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \cos (c+d x))^2} \, dx=\int \frac {{\left (e\,\sin \left (c+d\,x\right )\right )}^{9/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \] Input:
int((e*sin(c + d*x))^(9/2)/(a + b*cos(c + d*x))^2,x)
Output:
int((e*sin(c + d*x))^(9/2)/(a + b*cos(c + d*x))^2, x)
\[ \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \cos (c+d x))^2} \, dx=\sqrt {e}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) a b +a^{2}}d x \right ) e^{4} \] Input:
int((e*sin(d*x+c))^(9/2)/(a+b*cos(d*x+c))^2,x)
Output:
sqrt(e)*int((sqrt(sin(c + d*x))*sin(c + d*x)**4)/(cos(c + d*x)**2*b**2 + 2 *cos(c + d*x)*a*b + a**2),x)*e**4