Integrand size = 27, antiderivative size = 115 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2}{5 d e (e \cos (c+d x))^{3/2} \sqrt {a+a \sin (c+d x)}}-\frac {8 \sqrt {a+a \sin (c+d x)}}{5 a d e (e \cos (c+d x))^{3/2}}+\frac {16 (a+a \sin (c+d x))^{3/2}}{15 a^2 d e (e \cos (c+d x))^{3/2}} \]
16/15*(a+a*sin(d*x+c))^(3/2)/a^2/d/e/(e*cos(d*x+c))^(3/2)-2/5/d/e/(e*cos(d *x+c))^(3/2)/(a+a*sin(d*x+c))^(1/2)-8/5*(a+a*sin(d*x+c))^(1/2)/a/d/e/(e*co s(d*x+c))^(3/2)
Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.49 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}} \, dx=\frac {2 \left (-7+4 \sin (c+d x)+8 \sin ^2(c+d x)\right )}{15 d e (e \cos (c+d x))^{3/2} \sqrt {a (1+\sin (c+d x))}} \]
(2*(-7 + 4*Sin[c + d*x] + 8*Sin[c + d*x]^2))/(15*d*e*(e*Cos[c + d*x])^(3/2 )*Sqrt[a*(1 + Sin[c + d*x])])
Time = 0.52 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 3151, 3042, 3151, 3042, 3150}
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.
\(\displaystyle \int \frac {1}{\sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3151 |
\(\displaystyle \frac {4 \int \frac {\sqrt {\sin (c+d x) a+a}}{(e \cos (c+d x))^{5/2}}dx}{5 a}-\frac {2}{5 d e \sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 \int \frac {\sqrt {\sin (c+d x) a+a}}{(e \cos (c+d x))^{5/2}}dx}{5 a}-\frac {2}{5 d e \sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3151 |
\(\displaystyle \frac {4 \left (\frac {2 \int \frac {(\sin (c+d x) a+a)^{3/2}}{(e \cos (c+d x))^{5/2}}dx}{a}-\frac {2 \sqrt {a \sin (c+d x)+a}}{d e (e \cos (c+d x))^{3/2}}\right )}{5 a}-\frac {2}{5 d e \sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 \left (\frac {2 \int \frac {(\sin (c+d x) a+a)^{3/2}}{(e \cos (c+d x))^{5/2}}dx}{a}-\frac {2 \sqrt {a \sin (c+d x)+a}}{d e (e \cos (c+d x))^{3/2}}\right )}{5 a}-\frac {2}{5 d e \sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3150 |
\(\displaystyle \frac {4 \left (\frac {4 (a \sin (c+d x)+a)^{3/2}}{3 a d e (e \cos (c+d x))^{3/2}}-\frac {2 \sqrt {a \sin (c+d x)+a}}{d e (e \cos (c+d x))^{3/2}}\right )}{5 a}-\frac {2}{5 d e \sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^{3/2}}\) |
-2/(5*d*e*(e*Cos[c + d*x])^(3/2)*Sqrt[a + a*Sin[c + d*x]]) + (4*((-2*Sqrt[ a + a*Sin[c + d*x]])/(d*e*(e*Cos[c + d*x])^(3/2)) + (4*(a + a*Sin[c + d*x] )^(3/2))/(3*a*d*e*(e*Cos[c + d*x])^(3/2))))/(5*a)
3.4.3.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^m/(a*f*g*m)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[Simplify[m + p + 1], 0] && !ILtQ[p, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^m/(a*f*g*Simplify[2*m + p + 1])), x] + Simp[Simplify[m + p + 1]/(a*Simpl ify[2*m + p + 1]) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x] , x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simpli fy[m + p + 1], 0] && NeQ[2*m + p + 1, 0] && !IGtQ[m, 0]
Time = 2.82 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.47
method | result | size |
default | \(\frac {-\frac {16 \cos \left (d x +c \right )}{15}+\frac {8 \tan \left (d x +c \right )}{15}+\frac {2 \sec \left (d x +c \right )}{15}}{d \,e^{2} \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, \sqrt {e \cos \left (d x +c \right )}}\) | \(54\) |
2/15/d/e^2/(a*(1+sin(d*x+c)))^(1/2)/(e*cos(d*x+c))^(1/2)*(-8*cos(d*x+c)+4* tan(d*x+c)+sec(d*x+c))
Time = 0.28 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2 \, \sqrt {e \cos \left (d x + c\right )} {\left (8 \, \cos \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) - 1\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{15 \, {\left (a d e^{3} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a d e^{3} \cos \left (d x + c\right )^{2}\right )}} \]
-2/15*sqrt(e*cos(d*x + c))*(8*cos(d*x + c)^2 - 4*sin(d*x + c) - 1)*sqrt(a* sin(d*x + c) + a)/(a*d*e^3*cos(d*x + c)^2*sin(d*x + c) + a*d*e^3*cos(d*x + c)^2)
\[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {1}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \left (e \cos {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (97) = 194\).
Time = 0.31 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.50 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2 \, {\left (7 \, \sqrt {a} \sqrt {e} - \frac {8 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {25 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {25 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {8 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {7 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{15 \, {\left (a e^{3} + \frac {3 \, a e^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a e^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a e^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}}} \]
-2/15*(7*sqrt(a)*sqrt(e) - 8*sqrt(a)*sqrt(e)*sin(d*x + c)/(cos(d*x + c) + 1) - 25*sqrt(a)*sqrt(e)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 25*sqrt(a)*s qrt(e)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 8*sqrt(a)*sqrt(e)*sin(d*x + c )^5/(cos(d*x + c) + 1)^5 - 7*sqrt(a)*sqrt(e)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/((a*e^3 + 3*a*e^3*sin( d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*e^3*sin(d*x + c)^4/(cos(d*x + c) + 1 )^4 + a*e^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2))
Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}} \, dx=\text {Timed out} \]
Time = 6.61 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}} \, dx=-\frac {8\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}\,\left (8\,\cos \left (c+d\,x\right )+6\,\cos \left (3\,c+3\,d\,x\right )-\sin \left (2\,c+2\,d\,x\right )+2\,\sin \left (4\,c+4\,d\,x\right )\right )}{15\,a\,d\,e^2\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\left (4\,\sin \left (c+d\,x\right )+4\,\cos \left (2\,c+2\,d\,x\right )-\cos \left (4\,c+4\,d\,x\right )+4\,\sin \left (3\,c+3\,d\,x\right )+5\right )} \]