Integrand size = 31, antiderivative size = 413 \[ \int \frac {(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {a \left (-a^2+b^2\right )^{3/4} g^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} f}+\frac {a \left (-a^2+b^2\right )^{3/4} g^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} f}-\frac {2 \left (5 a^2-3 b^2\right ) g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b^3 f \sqrt {\cos (e+f x)}}+\frac {a^2 \left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^2 \left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f} \]
-a*(-a^2+b^2)^(3/4)*g^(5/2)*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2) ^(1/4)/g^(1/2))/b^(7/2)/f+a*(-a^2+b^2)^(3/4)*g^(5/2)*arctanh(b^(1/2)*(g*co s(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(7/2)/f-2/15*g*(g*cos(f*x+e))^ (3/2)*(5*a-3*b*sin(f*x+e))/b^2/f+a^2*(a^2-b^2)*g^3*(cos(1/2*f*x+1/2*e)^2)^ (1/2)/cos(1/2*f*x+1/2*e)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b-(-a^2+b^2)^( 1/2)),2^(1/2))*cos(f*x+e)^(1/2)/b^4/f/(b-(-a^2+b^2)^(1/2))/(g*cos(f*x+e))^ (1/2)+a^2*(a^2-b^2)*g^3*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*El lipticPi(sin(1/2*f*x+1/2*e),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*cos(f*x+e)^( 1/2)/b^4/f/(b+(-a^2+b^2)^(1/2))/(g*cos(f*x+e))^(1/2)-2/5*(5*a^2-3*b^2)*g^2 *(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*f*x+1/2 *e),2^(1/2))*(g*cos(f*x+e))^(1/2)/b^3/f/cos(f*x+e)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 17.14 (sec) , antiderivative size = 737, normalized size of antiderivative = 1.78 \[ \int \frac {(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\frac {(g \cos (e+f x))^{5/2} \left (\frac {2 \cos ^{\frac {3}{2}}(e+f x) (-5 a+3 b \sin (e+f x))}{3 b^2}+\frac {\left (5 a^2-3 b^2\right ) \left (8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \left (a+b \sqrt {\sin ^2(e+f x)}\right )}{12 b^{7/2} \left (-a^2+b^2\right ) (a+b \sin (e+f x))}+\frac {4 a \left (\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)}{3 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}\right ) \sin (e+f x) \left (a+b \sqrt {\sin ^2(e+f x)}\right )}{b \sqrt {\sin ^2(e+f x)} (a+b \sin (e+f x))}\right )}{5 f \cos ^{\frac {5}{2}}(e+f x)} \]
((g*Cos[e + f*x])^(5/2)*((2*Cos[e + f*x]^(3/2)*(-5*a + 3*b*Sin[e + f*x]))/ (3*b^2) + ((5*a^2 - 3*b^2)*(8*b^(5/2)*AppellF1[3/4, -1/2, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Cos[e + f*x]^(3/2) + 3*Sqrt[2]* a*(a^2 - b^2)^(3/4)*(2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^ 2 - b^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] - Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqr t[Cos[e + f*x]] + b*Cos[e + f*x]] + Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b]* (a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Cos[e + f*x]]))*(a + b*Sqrt[Sin[e + f*x]^2]))/(12*b^(7/2)*(-a^2 + b^2)*(a + b*Sin[e + f*x])) + (4*a*((a*App ellF1[3/4, 1/2, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] *Cos[e + f*x]^(3/2))/(3*(a^2 - b^2)) + ((1/8 + I/8)*(2*ArcTan[1 - ((1 + I) *Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*S qrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]] + Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x ]] + I*b*Cos[e + f*x]]))/(Sqrt[b]*(-a^2 + b^2)^(1/4)))*Sin[e + f*x]*(a + b *Sqrt[Sin[e + f*x]^2]))/(b*Sqrt[Sin[e + f*x]^2]*(a + b*Sin[e + f*x]))))/(5 *f*Cos[e + f*x]^(5/2))
Time = 1.72 (sec) , antiderivative size = 402, normalized size of antiderivative = 0.97, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.581, Rules used = {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.
| \(\displaystyle \int \frac {\sin (e+f x) (g \cos (e+f x))^{5/2}}{a+b \sin (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x) (g \cos (e+f x))^{5/2}}{a+b \sin (e+f x)}dx\) |
| \(\Big \downarrow \) 3344 |
| \(\displaystyle \frac {2 g^2 \int -\frac {\sqrt {g \cos (e+f x)} \left (2 a b+\left (5 a^2-3 b^2\right ) \sin (e+f x)\right )}{2 (a+b \sin (e+f x))}dx}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {g^2 \int \frac {\sqrt {g \cos (e+f x)} \left (2 a b+\left (5 a^2-3 b^2\right ) \sin (e+f x)\right )}{a+b \sin (e+f x)}dx}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {g^2 \int \frac {\sqrt {g \cos (e+f x)} \left (2 a b+\left (5 a^2-3 b^2\right ) \sin (e+f x)\right )}{a+b \sin (e+f x)}dx}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
| \(\Big \downarrow \) 3346 |
| \(\displaystyle -\frac {g^2 \left (\frac {\left (5 a^2-3 b^2\right ) \int \sqrt {g \cos (e+f x)}dx}{b}-\frac {5 a \left (a^2-b^2\right ) \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx}{b}\right )}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {g^2 \left (\frac {\left (5 a^2-3 b^2\right ) \int \sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{b}-\frac {5 a \left (a^2-b^2\right ) \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx}{b}\right )}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle -\frac {g^2 \left (\frac {\left (5 a^2-3 b^2\right ) \sqrt {g \cos (e+f x)} \int \sqrt {\cos (e+f x)}dx}{b \sqrt {\cos (e+f x)}}-\frac {5 a \left (a^2-b^2\right ) \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx}{b}\right )}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {g^2 \left (\frac {\left (5 a^2-3 b^2\right ) \sqrt {g \cos (e+f x)} \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{b \sqrt {\cos (e+f x)}}-\frac {5 a \left (a^2-b^2\right ) \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx}{b}\right )}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {g^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {5 a \left (a^2-b^2\right ) \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx}{b}\right )}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
| \(\Big \downarrow \) 3180 |
| \(\displaystyle -\frac {g^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {5 a \left (a^2-b^2\right ) \left (\frac {b g \int \frac {\sqrt {g \cos (e+f x)}}{b^2 \cos ^2(e+f x) g^2+\left (a^2-b^2\right ) g^2}d(g \cos (e+f x))}{f}-\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {g^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {5 a \left (a^2-b^2\right ) \left (\frac {2 b g \int \frac {g^2 \cos ^2(e+f x)}{b^2 g^4 \cos ^4(e+f x)+\left (a^2-b^2\right ) g^2}d\sqrt {g \cos (e+f x)}}{f}-\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle -\frac {g^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {5 a \left (a^2-b^2\right ) \left (\frac {2 b g \left (\frac {\int \frac {1}{b g^2 \cos ^2(e+f x)+\sqrt {b^2-a^2} g}d\sqrt {g \cos (e+f x)}}{2 b}-\frac {\int \frac {1}{\sqrt {b^2-a^2} g-b g^2 \cos ^2(e+f x)}d\sqrt {g \cos (e+f x)}}{2 b}\right )}{f}-\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {g^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {5 a \left (a^2-b^2\right ) \left (\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\int \frac {1}{\sqrt {b^2-a^2} g-b g^2 \cos ^2(e+f x)}d\sqrt {g \cos (e+f x)}}{2 b}\right )}{f}-\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {g^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b}+\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{f}\right )}{b}\right )}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {g^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {a g \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )} \left (b \sin \left (e+f x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 b}+\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{f}\right )}{b}\right )}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle -\frac {g^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {a g \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b \sqrt {g \cos (e+f x)}}+\frac {a g \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b \sqrt {g \cos (e+f x)}}+\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{f}\right )}{b}\right )}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {g^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {5 a \left (a^2-b^2\right ) \left (-\frac {a g \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{2 b \sqrt {g \cos (e+f x)}}+\frac {a g \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )} \left (b \sin \left (e+f x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 b \sqrt {g \cos (e+f x)}}+\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{f}\right )}{b}\right )}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle -\frac {g^2 \left (\frac {2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {5 a \left (a^2-b^2\right ) \left (\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{f}+\frac {a g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}+\frac {a g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}\right )}{b}\right )}{5 b^2}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}\) |
-1/5*(g^2*((2*(5*a^2 - 3*b^2)*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(b*f*Sqrt[Cos[e + f*x]]) - (5*a*(a^2 - b^2)*((2*b*g*(ArcTan[(Sqrt[b]*S qrt[g]*Cos[e + f*x])/(-a^2 + b^2)^(1/4)]/(2*b^(3/2)*(-a^2 + b^2)^(1/4)*Sqr t[g]) - ArcTanh[(Sqrt[b]*Sqrt[g]*Cos[e + f*x])/(-a^2 + b^2)^(1/4)]/(2*b^(3 /2)*(-a^2 + b^2)^(1/4)*Sqrt[g])))/f + (a*g*Sqrt[Cos[e + f*x]]*EllipticPi[( 2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b*(b - Sqrt[-a^2 + b^2])*f* Sqrt[g*Cos[e + f*x]]) + (a*g*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt [-a^2 + b^2]), (e + f*x)/2, 2])/(b*(b + Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]])))/b))/b^2 - (2*g*(g*Cos[e + f*x])^(3/2)*(5*a - 3*b*Sin[e + f*x]))/ (15*b^2*f)
3.14.85.3.1 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[(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[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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 7.29 (sec) , antiderivative size = 1533, normalized size of antiderivative = 3.71
(-4*g^3*a*(1/6/b^2*(-2*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*sin(1/2*f*x+1/2 *e)^2+(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2))/g-1/16/b^4*(a^2-b^2)/(g^2*(a^2- b^2)/b^2)^(1/4)*2^(1/2)*(ln((2*g*cos(1/2*f*x+1/2*e)^2-g-(g^2*(a^2-b^2)/b^2 )^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/ 2))/(2*g*cos(1/2*f*x+1/2*e)^2-g+(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x +1/2*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/2)))+2*arctan((2^(1/2)*( 2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/4))/(g^2*(a^2-b^2 )/b^2)^(1/4))+2*arctan((2^(1/2)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)-(g^2*(a ^2-b^2)/b^2)^(1/4))/(g^2*(a^2-b^2)/b^2)^(1/4))))-1/12*(g*(2*cos(1/2*f*x+1/ 2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*g^3*(64*cos(1/2*f*x+1/2*e)^5*sin(1/2 *f*x+1/2*e)^2*a^2*b^4-64*cos(1/2*f*x+1/2*e)^5*a^2*b^4-96*cos(1/2*f*x+1/2*e )^3*sin(1/2*f*x+1/2*e)^2*a^2*b^4-48*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))* (sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*sin(1/2*f*x+ 1/2*e)^2*a^4*b^2+64*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))*(sin(1/2*f*x+1/2 *e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*sin(1/2*f*x+1/2*e)^2*a^2*b^4 -48*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1- 2*cos(1/2*f*x+1/2*e)^2)^(1/2)*sin(1/2*f*x+1/2*e)^2*a^2*b^4+96*cos(1/2*f*x+ 1/2*e)^3*a^2*b^4+32*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)^2*a^2*b^4+48*Ell ipticF(cos(1/2*f*x+1/2*e),2^(1/2))*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1 /2*f*x+1/2*e)^2)^(1/2)*a^4*b^2-64*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))...
Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]
\[ \int \frac {(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} \sin \left (f x + e\right )}{b \sin \left (f x + e\right ) + a} \,d x } \]
\[ \int \frac {(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} \sin \left (f x + e\right )}{b \sin \left (f x + e\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\int \frac {\sin \left (e+f\,x\right )\,{\left (g\,\cos \left (e+f\,x\right )\right )}^{5/2}}{a+b\,\sin \left (e+f\,x\right )} \,d x \]