Integrand size = 25, antiderivative size = 120 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^4} \, dx=\frac {10 e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 a^4 d \sqrt {e \cos (c+d x)}}-\frac {4 e (e \cos (c+d x))^{5/2}}{7 a d (a+a \sin (c+d x))^3}+\frac {20 e^3 \sqrt {e \cos (c+d x)}}{21 d \left (a^4+a^4 \sin (c+d x)\right )} \]
-4/7*e*(e*cos(d*x+c))^(5/2)/a/d/(a+a*sin(d*x+c))^3+10/21*e^4*(cos(1/2*d*x+ 1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*c os(d*x+c)^(1/2)/a^4/d/(e*cos(d*x+c))^(1/2)+20/21*e^3*(e*cos(d*x+c))^(1/2)/ d/(a^4+a^4*sin(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.55 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {(e \cos (c+d x))^{9/2} \operatorname {Hypergeometric2F1}\left (\frac {9}{4},\frac {11}{4},\frac {13}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{9\ 2^{3/4} a^4 d e (1+\sin (c+d x))^{9/4}} \]
-1/9*((e*Cos[c + d*x])^(9/2)*Hypergeometric2F1[9/4, 11/4, 13/4, (1 - Sin[c + d*x])/2])/(2^(3/4)*a^4*d*e*(1 + Sin[c + d*x])^(9/4))
Time = 0.53 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3159, 3042, 3159, 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.
| \(\displaystyle \int \frac {(e \cos (c+d x))^{7/2}}{(a \sin (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(e \cos (c+d x))^{7/2}}{(a \sin (c+d x)+a)^4}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle -\frac {5 e^2 \int \frac {(e \cos (c+d x))^{3/2}}{(\sin (c+d x) a+a)^2}dx}{7 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{7 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {5 e^2 \int \frac {(e \cos (c+d x))^{3/2}}{(\sin (c+d x) a+a)^2}dx}{7 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{7 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle -\frac {5 e^2 \left (-\frac {e^2 \int \frac {1}{\sqrt {e \cos (c+d x)}}dx}{3 a^2}-\frac {4 e \sqrt {e \cos (c+d x)}}{3 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{7 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{7 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {5 e^2 \left (-\frac {e^2 \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a^2}-\frac {4 e \sqrt {e \cos (c+d x)}}{3 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{7 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{7 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle -\frac {5 e^2 \left (-\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 a^2 \sqrt {e \cos (c+d x)}}-\frac {4 e \sqrt {e \cos (c+d x)}}{3 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{7 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{7 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {5 e^2 \left (-\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a^2 \sqrt {e \cos (c+d x)}}-\frac {4 e \sqrt {e \cos (c+d x)}}{3 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{7 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{7 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {5 e^2 \left (-\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d \sqrt {e \cos (c+d x)}}-\frac {4 e \sqrt {e \cos (c+d x)}}{3 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{7 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{7 a d (a \sin (c+d x)+a)^3}\) |
(-4*e*(e*Cos[c + d*x])^(5/2))/(7*a*d*(a + a*Sin[c + d*x])^3) - (5*e^2*((-2 *e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*a^2*d*Sqrt[e*Cos[c + d*x]]) - (4*e*Sqrt[e*Cos[c + d*x]])/(3*d*(a^2 + a^2*Sin[c + d*x]))))/(7*a ^2)
3.3.67.3.1 Defintions of rubi rules used
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[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f *x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 ))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & & NeQ[2*m + p + 1, 0] && !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
Leaf count of result is larger than twice the leaf count of optimal. \(400\) vs. \(2(132)=264\).
Time = 6.54 (sec) , antiderivative size = 401, normalized size of antiderivative = 3.34
\[\frac {2 \left (-40 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+128 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+60 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-128 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-30 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-112 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-16 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+112 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e^{4}}{21 \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) a^{4} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\]
2/21/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^ 2-1)/a^4/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(-40*(sin( 1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d *x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^6+128*sin(1/2*d*x+1/2*c)^6*cos(1/2 *d*x+1/2*c)+60*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2 ^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-128*sin(1/2* d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-30*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF( cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1 /2*c)^2-112*sin(1/2*d*x+1/2*c)^5-16*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c )+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elliptic F(cos(1/2*d*x+1/2*c),2^(1/2))+112*sin(1/2*d*x+1/2*c)^3-4*sin(1/2*d*x+1/2*c ))*e^4/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.57 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^4} \, dx=\frac {5 \, {\left (-i \, \sqrt {2} e^{3} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} e^{3} \sin \left (d x + c\right ) + 2 i \, \sqrt {2} e^{3}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (i \, \sqrt {2} e^{3} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} e^{3} \sin \left (d x + c\right ) - 2 i \, \sqrt {2} e^{3}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 8 \, {\left (4 \, e^{3} \sin \left (d x + c\right ) + e^{3}\right )} \sqrt {e \cos \left (d x + c\right )}}{21 \, {\left (a^{4} d \cos \left (d x + c\right )^{2} - 2 \, a^{4} d \sin \left (d x + c\right ) - 2 \, a^{4} d\right )}} \]
1/21*(5*(-I*sqrt(2)*e^3*cos(d*x + c)^2 + 2*I*sqrt(2)*e^3*sin(d*x + c) + 2* I*sqrt(2)*e^3)*sqrt(e)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(I*sqrt(2)*e^3*cos(d*x + c)^2 - 2*I*sqrt(2)*e^3*sin(d*x + c) - 2*I*sqrt(2)*e^3)*sqrt(e)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d *x + c)) - 8*(4*e^3*sin(d*x + c) + e^3)*sqrt(e*cos(d*x + c)))/(a^4*d*cos(d *x + c)^2 - 2*a^4*d*sin(d*x + c) - 2*a^4*d)
Timed out. \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^4} \, dx=\text {Timed out} \]
\[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
\[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^4} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \]