Integrand size = 25, antiderivative size = 124 \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=-\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))}{3 d e (e \sin (c+d x))^{3/2}}+\frac {2 \left (a^2-2 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 d e^2 \sqrt {e \sin (c+d x)}}-\frac {2 a b \sqrt {e \sin (c+d x)}}{3 d e^3} \]
-2/3*(b+a*cos(d*x+c))*(a+b*cos(d*x+c))/d/e/(e*sin(d*x+c))^(3/2)-2/3*(a^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)*Ellipt icF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/d/e^2/(e*sin(d*x+c ))^(1/2)-2/3*a*b*(e*sin(d*x+c))^(1/2)/d/e^3
Time = 1.00 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.61 \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=-\frac {2 \left (2 a b+\left (a^2+b^2\right ) \cos (c+d x)+\left (a^2-2 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right ) \sin ^{\frac {3}{2}}(c+d x)\right )}{3 d e (e \sin (c+d x))^{3/2}} \]
(-2*(2*a*b + (a^2 + b^2)*Cos[c + d*x] + (a^2 - 2*b^2)*EllipticF[(-2*c + Pi - 2*d*x)/4, 2]*Sin[c + d*x]^(3/2)))/(3*d*e*(e*Sin[c + d*x])^(3/2))
Time = 0.56 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3170, 27, 3042, 3148, 3042, 3121, 3042, 3120}
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.
| \(\displaystyle \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )^2}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3170 |
| \(\displaystyle -\frac {2 \int -\frac {a^2-b \cos (c+d x) a-2 b^2}{2 \sqrt {e \sin (c+d x)}}dx}{3 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{3 d e (e \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a^2-b \cos (c+d x) a-2 b^2}{\sqrt {e \sin (c+d x)}}dx}{3 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{3 d e (e \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a^2+b \sin \left (c+d x-\frac {\pi }{2}\right ) a-2 b^2}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}dx}{3 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{3 d e (e \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {\left (a^2-2 b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}}dx-\frac {2 a b \sqrt {e \sin (c+d x)}}{d e}}{3 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{3 d e (e \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2-2 b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}}dx-\frac {2 a b \sqrt {e \sin (c+d x)}}{d e}}{3 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{3 d e (e \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {\frac {\left (a^2-2 b^2\right ) \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{\sqrt {e \sin (c+d x)}}-\frac {2 a b \sqrt {e \sin (c+d x)}}{d e}}{3 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{3 d e (e \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (a^2-2 b^2\right ) \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{\sqrt {e \sin (c+d x)}}-\frac {2 a b \sqrt {e \sin (c+d x)}}{d e}}{3 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{3 d e (e \sin (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {2 \left (a^2-2 b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}-\frac {2 a b \sqrt {e \sin (c+d x)}}{d e}}{3 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{3 d e (e \sin (c+d x))^{3/2}}\) |
(-2*(b + a*Cos[c + d*x])*(a + b*Cos[c + d*x]))/(3*d*e*(e*Sin[c + d*x])^(3/ 2)) + ((2*(a^2 - 2*b^2)*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x] ])/(d*Sqrt[e*Sin[c + d*x]]) - (2*a*b*Sqrt[e*Sin[c + d*x]])/(d*e))/(3*e^2)
3.1.47.3.1 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[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 _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-(g*Cos[e + f*x])^(p + 1))*(a + b*Sin[e + f*x ])^(m - 1)*((b + a*Sin[e + f*x])/(f*g*(p + 1))), x] + Simp[1/(g^2*(p + 1)) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(p + 2) + a*b*(m + p + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g }, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (IntegersQ[2*m, 2* p] || IntegerQ[m])
Time = 3.37 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.63
| method | result | size |
| default | \(\frac {-\frac {4 a b}{3 e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) a^{2}-2 b^{2} \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+2 a^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+2 b^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(202\) |
| parts | \(-\frac {a^{2} \left (\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{3}\left (d x +c \right )\right )+2 \sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 b^{2} \left (\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+\sin ^{3}\left (d x +c \right )-\sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {4 a b}{3 \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}} e d}\) | \(234\) |
(-4/3*a*b/e/(e*sin(d*x+c))^(3/2)-1/3/e^2*((1-sin(d*x+c))^(1/2)*(2*sin(d*x+ c)+2)^(1/2)*sin(d*x+c)^(5/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a ^2-2*b^2*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(5/2)*Elli pticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))+2*a^2*cos(d*x+c)^2*sin(d*x+c)+2*b^ 2*cos(d*x+c)^2*sin(d*x+c))/sin(d*x+c)^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.42 \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\frac {{\left (\sqrt {2} {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - \sqrt {2} {\left (a^{2} - 2 \, b^{2}\right )}\right )} \sqrt {-i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - \sqrt {2} {\left (a^{2} - 2 \, b^{2}\right )}\right )} \sqrt {i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (2 \, a b + {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {e \sin \left (d x + c\right )}}{3 \, {\left (d e^{3} \cos \left (d x + c\right )^{2} - d e^{3}\right )}} \]
1/3*((sqrt(2)*(a^2 - 2*b^2)*cos(d*x + c)^2 - sqrt(2)*(a^2 - 2*b^2))*sqrt(- I*e)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*( a^2 - 2*b^2)*cos(d*x + c)^2 - sqrt(2)*(a^2 - 2*b^2))*sqrt(I*e)*weierstrass PInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(2*a*b + (a^2 + b^2)*cos (d*x + c))*sqrt(e*sin(d*x + c)))/(d*e^3*cos(d*x + c)^2 - d*e^3)
\[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\int \frac {\left (a + b \cos {\left (c + d x \right )}\right )^{2}}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]