Integrand size = 25, antiderivative size = 487 \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^2} \, dx=\frac {5 a \sqrt [4]{-a^2+b^2} e^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{7/2} d}+\frac {5 a \sqrt [4]{-a^2+b^2} e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{7/2} d}+\frac {5 \left (3 a^2-b^2\right ) e^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 b^4 d \sqrt {e \sin (c+d x)}}-\frac {5 a^2 \left (a^2-b^2\right ) e^4 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 b^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}-\frac {5 a^2 \left (a^2-b^2\right ) e^4 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 b^4 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}-\frac {5 e^3 (3 a-b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{3 b^3 d}+\frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))} \]
5/2*a*(-a^2+b^2)^(1/4)*e^(7/2)*arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b ^2)^(1/4)/e^(1/2))/b^(7/2)/d+5/2*a*(-a^2+b^2)^(1/4)*e^(7/2)*arctanh(b^(1/2 )*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(7/2)/d+e*(e*sin(d*x+c) )^(5/2)/b/d/(a+b*cos(d*x+c))-5/3*(3*a^2-b^2)*e^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)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2 ^(1/2))*sin(d*x+c)^(1/2)/b^4/d/(e*sin(d*x+c))^(1/2)+5/2*a^2*(a^2-b^2)*e^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(c os(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^4/d/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(e*sin(d*x+c))^(1/2)+5/2*a^2*(a^2-b^2 )*e^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)*Ellipt icPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*sin(d*x+c )^(1/2)/b^4/d/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(e*sin(d*x+c))^(1/2)-5/3*e^3*(3 *a-b*cos(d*x+c))*(e*sin(d*x+c))^(1/2)/b^3/d
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 13.65 (sec) , antiderivative size = 1956, normalized size of antiderivative = 4.02 \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \]
(((2*Cos[c + d*x])/(3*b^2) + (-a^2 + b^2)/(b^3*(a + b*Cos[c + d*x])))*Csc[ c + d*x]^3*(e*Sin[c + d*x])^(7/2))/d + ((e*Sin[c + d*x])^(7/2)*((2*(3*a^2 - 5*b^2)*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)) + (8*a*b*C os[c + d*x]*(a + b*Sqrt[1 - Sin[c + d*x]^2])*(((-1/8 + I/8)*Sqrt[b]*(2*Arc Tan[1 - ((1 + I)*Sqrt[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTa n[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/...
Time = 2.25 (sec) , antiderivative size = 479, normalized size of antiderivative = 0.98, 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, 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.
\(\displaystyle \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{7/2}}{\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3172 |
\(\displaystyle \frac {5 e^2 \int -\frac {\cos (c+d x) (e \sin (c+d x))^{3/2}}{a+b \cos (c+d x)}dx}{2 b}+\frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \int \frac {\cos (c+d x) (e \sin (c+d x))^{3/2}}{a+b \cos (c+d x)}dx}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \int \frac {\left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}\) |
\(\Big \downarrow \) 3344 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e^2 \int -\frac {2 a b+\left (3 a^2-b^2\right ) \cos (c+d x)}{2 (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{3 b^2}+\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}\right )}{2 b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}-\frac {e^2 \int \frac {2 a b+\left (3 a^2-b^2\right ) \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{3 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}-\frac {e^2 \int \frac {2 a b-\left (3 a^2-b^2\right ) \sin \left (c+d x-\frac {\pi }{2}\right )}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{3 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3346 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}-\frac {e^2 \left (\frac {\left (3 a^2-b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{b}-\frac {3 a \left (a^2-b^2\right ) \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{b}\right )}{3 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}-\frac {e^2 \left (\frac {\left (3 a^2-b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{b}-\frac {3 a \left (a^2-b^2\right ) \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}\right )}{3 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}-\frac {e^2 \left (\frac {\left (3 a^2-b^2\right ) \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{b \sqrt {e \sin (c+d x)}}-\frac {3 a \left (a^2-b^2\right ) \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}\right )}{3 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}-\frac {e^2 \left (\frac {\left (3 a^2-b^2\right ) \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{b \sqrt {e \sin (c+d x)}}-\frac {3 a \left (a^2-b^2\right ) \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}\right )}{3 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}-\frac {e^2 \left (\frac {2 \left (3 a^2-b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {3 a \left (a^2-b^2\right ) \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}\right )}{3 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3181 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}-\frac {e^2 \left (\frac {2 \left (3 a^2-b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {3 a \left (a^2-b^2\right ) \left (-\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b^2 \sin ^2(c+d x) e^2+\left (a^2-b^2\right ) e^2\right )}d(e \sin (c+d x))}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{3 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}-\frac {e^2 \left (\frac {2 \left (3 a^2-b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {3 a \left (a^2-b^2\right ) \left (-\frac {2 b e \int \frac {1}{b^2 e^4 \sin ^4(c+d x)+\left (a^2-b^2\right ) e^2}d\sqrt {e \sin (c+d x)}}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{3 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}-\frac {e^2 \left (\frac {2 \left (3 a^2-b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {3 a \left (a^2-b^2\right ) \left (-\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\int \frac {1}{b e^2 \sin ^2(c+d x)+\sqrt {b^2-a^2} e}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{3 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}-\frac {e^2 \left (\frac {2 \left (3 a^2-b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {3 a \left (a^2-b^2\right ) \left (-\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{3 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}-\frac {e^2 \left (\frac {2 \left (3 a^2-b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {3 a \left (a^2-b^2\right ) \left (-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{3 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}-\frac {e^2 \left (\frac {2 \left (3 a^2-b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {3 a \left (a^2-b^2\right ) \left (-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{3 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3286 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}-\frac {e^2 \left (\frac {2 \left (3 a^2-b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {3 a \left (a^2-b^2\right ) \left (-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{3 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}-\frac {e^2 \left (\frac {2 \left (3 a^2-b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {3 a \left (a^2-b^2\right ) \left (-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{3 b^2}\right )}{2 b}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {e (e \sin (c+d x))^{5/2}}{b d (a+b \cos (c+d x))}-\frac {5 e^2 \left (\frac {2 e \sqrt {e \sin (c+d x)} (3 a-b \cos (c+d x))}{3 b^2 d}-\frac {e^2 \left (\frac {2 \left (3 a^2-b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {3 a \left (a^2-b^2\right ) \left (-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}+\frac {a \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {b^2-a^2} \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {b^2-a^2} \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}}\right )}{b}\right )}{3 b^2}\right )}{2 b}\) |
(e*(e*Sin[c + d*x])^(5/2))/(b*d*(a + b*Cos[c + d*x])) - (5*e^2*((2*e*(3*a - b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]])/(3*b^2*d) - (e^2*((2*(3*a^2 - b^2) *EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(b*d*Sqrt[e*Sin[c + d*x]]) - (3*a*(a^2 - b^2)*((-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[(S qrt[b]*Sqrt[e]*Sin[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]*Sqrt[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*Sin[c + d*x]])))/b))/(3*b^2)))/(2*b)
3.1.70.3.1 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[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[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_)*((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. \(1500\) vs. \(2(513)=1026\).
Time = 17.46 (sec) , antiderivative size = 1501, normalized size of antiderivative = 3.08
(-4*e^3*a*b*((e*sin(d*x+c))^(1/2)/b^4-e^2/b^4*((-1/4*a^2+1/4*b^2)*(e*sin(d *x+c))^(1/2)/(-b^2*cos(d*x+c)^2*e^2+a^2*e^2)+5/32*(a^2-b^2)*(e^2*(a^2-b^2) /b^2)^(1/4)/(a^2*e^2-b^2*e^2)*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^4*(-1/3/b^4/(cos(d*x+c)^2*e*sin(d*x+ c))^(1/2)*(9*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)* EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2-4*(1-sin(d*x+c))^(1/2)*(2* sin(d*x+c)+2)^(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)^2*sin(d*x+c))-1/b^4*(5*a^4-6*a^2*b^2+b^4)*(-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)*EllipticP i((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)/(co s(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*(a^4-2*a^2*b^2+b^ 4)/b^4*(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*co s(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)...
Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^2} \, dx=\int \frac {{\left (e\,\sin \left (c+d\,x\right )\right )}^{7/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]