Integrand size = 25, antiderivative size = 107 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^3} \, dx=-\frac {10 e^3 \sqrt {e \cos (c+d x)}}{3 a^3 d}-\frac {10 e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^3 d \sqrt {e \cos (c+d x)}}-\frac {4 e (e \cos (c+d x))^{5/2}}{3 a d (a+a \sin (c+d x))^2} \]
-4/3*e*(e*cos(d*x+c))^(5/2)/a/d/(a+a*sin(d*x+c))^2-10/3*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))*co s(d*x+c)^(1/2)/a^3/d/(e*cos(d*x+c))^(1/2)-10/3*e^3*(e*cos(d*x+c))^(1/2)/a^ 3/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.62 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^3} \, dx=-\frac {\sqrt [4]{2} (e \cos (c+d x))^{9/2} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {9}{4},\frac {13}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{9 a^3 d e (1+\sin (c+d x))^{9/4}} \]
-1/9*(2^(1/4)*(e*Cos[c + d*x])^(9/2)*Hypergeometric2F1[7/4, 9/4, 13/4, (1 - Sin[c + d*x])/2])/(a^3*d*e*(1 + Sin[c + d*x])^(9/4))
Time = 0.52 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.05, 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, 3161, 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.
\(\displaystyle \int \frac {(e \cos (c+d x))^{7/2}}{(a \sin (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(e \cos (c+d x))^{7/2}}{(a \sin (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3159 |
\(\displaystyle -\frac {5 e^2 \int \frac {(e \cos (c+d x))^{3/2}}{\sin (c+d x) a+a}dx}{3 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{3 a d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {5 e^2 \int \frac {(e \cos (c+d x))^{3/2}}{\sin (c+d x) a+a}dx}{3 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{3 a d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3161 |
\(\displaystyle -\frac {5 e^2 \left (\frac {e^2 \int \frac {1}{\sqrt {e \cos (c+d x)}}dx}{a}+\frac {2 e \sqrt {e \cos (c+d x)}}{a d}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{3 a d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {5 e^2 \left (\frac {e^2 \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}+\frac {2 e \sqrt {e \cos (c+d x)}}{a d}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{3 a d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle -\frac {5 e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a \sqrt {e \cos (c+d x)}}+\frac {2 e \sqrt {e \cos (c+d x)}}{a d}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{3 a d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {5 e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \sqrt {e \cos (c+d x)}}+\frac {2 e \sqrt {e \cos (c+d x)}}{a d}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{3 a d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {5 e^2 \left (\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d \sqrt {e \cos (c+d x)}}+\frac {2 e \sqrt {e \cos (c+d x)}}{a d}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{3 a d (a \sin (c+d x)+a)^2}\) |
(-5*e^2*((2*e*Sqrt[e*Cos[c + d*x]])/(a*d) + (2*e^2*Sqrt[Cos[c + d*x]]*Elli pticF[(c + d*x)/2, 2])/(a*d*Sqrt[e*Cos[c + d*x]])))/(3*a^2) - (4*e*(e*Cos[ c + d*x])^(5/2))/(3*a*d*(a + a*Sin[c + d*x])^2)
3.3.57.3.1 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]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[g*((g*Cos[e + f*x])^(p - 1)/(b*f*(p - 1))), x] + Si mp[g^2/a Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g}, x ] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]
Time = 175.13 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.05
method | result | size |
default | \(-\frac {2 \left (-10 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e^{4}}{3 \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) | \(219\) |
-2/3/(2*sin(1/2*d*x+1/2*c)^2-1)/a^3/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2 *c)^2*e+e)^(1/2)*(-10*(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)^2-12*si n(1/2*d*x+1/2*c)^5+8*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)*EllipticF(cos(1/2*d*x+1 /2*c),2^(1/2))+12*sin(1/2*d*x+1/2*c)^3-7*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 = 143, normalized size of antiderivative = 1.34 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^3} \, dx=-\frac {5 \, {\left (-i \, \sqrt {2} e^{3} \sin \left (d x + c\right ) - i \, \sqrt {2} e^{3}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (i \, \sqrt {2} e^{3} \sin \left (d x + c\right ) + i \, \sqrt {2} e^{3}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (3 \, e^{3} \sin \left (d x + c\right ) + 7 \, e^{3}\right )} \sqrt {e \cos \left (d x + c\right )}}{3 \, {\left (a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \]
-1/3*(5*(-I*sqrt(2)*e^3*sin(d*x + c) - I*sqrt(2)*e^3)*sqrt(e)*weierstrassP Inverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(I*sqrt(2)*e^3*sin(d*x + c) + I*sqrt(2)*e^3)*sqrt(e)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*s in(d*x + c)) + 2*(3*e^3*sin(d*x + c) + 7*e^3)*sqrt(e*cos(d*x + c)))/(a^3*d *sin(d*x + c) + a^3*d)
Timed out. \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]
\[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
\[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^3} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3} \,d x \]