Integrand size = 25, antiderivative size = 112 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^2} \, dx=\frac {10 e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d \sqrt {e \cos (c+d x)}}+\frac {10 e^3 \sqrt {e \cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2+a^2 \sin (c+d x)\right )} \]
4*e*(e*cos(d*x+c))^(5/2)/d/(a^2+a^2*sin(d*x+c))+10/3*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))*cos(d *x+c)^(1/2)/a^2/d/(e*cos(d*x+c))^(1/2)+10/3*e^3*sin(d*x+c)*(e*cos(d*x+c))^ (1/2)/a^2/d
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 = 0.59 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 \sqrt [4]{2} (e \cos (c+d x))^{9/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {9}{4},\frac {13}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{9 a^2 d e (1+\sin (c+d x))^{9/4}} \]
(-2*2^(1/4)*(e*Cos[c + d*x])^(9/2)*Hypergeometric2F1[3/4, 9/4, 13/4, (1 - Sin[c + d*x])/2])/(9*a^2*d*e*(1 + Sin[c + d*x])^(9/4))
Time = 0.48 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.01, 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, 3115, 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.
\(\displaystyle \int \frac {(e \cos (c+d x))^{7/2}}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(e \cos (c+d x))^{7/2}}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3159 |
\(\displaystyle \frac {5 e^2 \int (e \cos (c+d x))^{3/2}dx}{a^2}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 e^2 \int \left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx}{a^2}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {5 e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \cos (c+d x)}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {5 e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {5 e^2 \left (\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\) |
(4*e*(e*Cos[c + d*x])^(5/2))/(d*(a^2 + a^2*Sin[c + d*x])) + (5*e^2*((2*e^2 *Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[e*Cos[c + d*x]]) + (2*e*Sqrt[e*Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/a^2
3.3.45.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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]
Time = 18.70 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.38
method | result | size |
default | \(\frac {2 e^{4} \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) | \(155\) |
2/3/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^4*(4*sin( 1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*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)*Ellipt icF(cos(1/2*d*x+1/2*c),2^(1/2))-12*sin(1/2*d*x+1/2*c)^3+6*sin(1/2*d*x+1/2* c))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.79 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^2} \, dx=\frac {-5 i \, \sqrt {2} e^{\frac {7}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} e^{\frac {7}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (e^{3} \sin \left (d x + c\right ) - 6 \, e^{3}\right )} \sqrt {e \cos \left (d x + c\right )}}{3 \, a^{2} d} \]
1/3*(-5*I*sqrt(2)*e^(7/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin( d*x + c)) + 5*I*sqrt(2)*e^(7/2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(e^3*sin(d*x + c) - 6*e^3)*sqrt(e*cos(d*x + c)))/(a^2* d)
Timed out. \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^2} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]