Integrand size = 27, antiderivative size = 218 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a+a \sin (c+d x))^{3/2}}-\frac {2 e^{5/2} \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d \left (a^3+a^3 \cos (c+d x)+a^3 \sin (c+d x)\right )}-\frac {2 e^{5/2} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d \left (a^3+a^3 \cos (c+d x)+a^3 \sin (c+d x)\right )} \]
-4/3*e*(e*cos(d*x+c))^(3/2)/a/d/(a+a*sin(d*x+c))^(3/2)-2*e^(5/2)*arcsinh(( e*cos(d*x+c))^(1/2)/e^(1/2))*(1+cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2)/d /(a^3+a^3*cos(d*x+c)+a^3*sin(d*x+c))-2*e^(5/2)*arctan(sin(d*x+c)*e^(1/2)/( e*cos(d*x+c))^(1/2)/(1+cos(d*x+c))^(1/2))*(1+cos(d*x+c))^(1/2)*(a+a*sin(d* x+c))^(1/2)/d/(a^3+a^3*cos(d*x+c)+a^3*sin(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.37 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {\sqrt [4]{2} (e \cos (c+d x))^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {7}{4},\frac {11}{4},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt {a (1+\sin (c+d x))}}{7 a^3 d e (1+\sin (c+d x))^{9/4}} \]
-1/7*(2^(1/4)*(e*Cos[c + d*x])^(7/2)*Hypergeometric2F1[7/4, 7/4, 11/4, (1 - Sin[c + d*x])/2]*Sqrt[a*(1 + Sin[c + d*x])])/(a^3*d*e*(1 + Sin[c + d*x]) ^(9/4))
Time = 0.81 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {3042, 3159, 3042, 3163, 3042, 25, 3254, 216, 3312, 63, 222}
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.
| \(\displaystyle \int \frac {(e \cos (c+d x))^{5/2}}{(a \sin (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(e \cos (c+d x))^{5/2}}{(a \sin (c+d x)+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle -\frac {e^2 \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {\sin (c+d x) a+a}}dx}{a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {\sin (c+d x) a+a}}dx}{a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3163 |
| \(\displaystyle -\frac {e^2 \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sqrt {\cos (c+d x)+1}}{\sqrt {e \cos (c+d x)}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}-\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {\cos (c+d x)+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}-\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int -\frac {\cos \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {e^2 \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}+\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {e \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \sqrt {\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3254 |
| \(\displaystyle -\frac {e^2 \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {e \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \sqrt {\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}-\frac {2 e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x)}{\cos (c+d x)+1}+1}d\left (-\frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {\cos (c+d x)+1}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {e^2 \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {e \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \sqrt {\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3312 |
| \(\displaystyle -\frac {e^2 \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {1}{\sqrt {e \cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{d (a \sin (c+d x)+a \cos (c+d x)+a)}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 63 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {1}{\sqrt {\cos (c+d x)+1}}d\sqrt {e \cos (c+d x)}}{d (a \sin (c+d x)+a \cos (c+d x)+a)}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle -\frac {e^2 \left (\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a \sin (c+d x)+a)^{3/2}}\) |
(-4*e*(e*Cos[c + d*x])^(3/2))/(3*a*d*(a + a*Sin[c + d*x])^(3/2)) - (e^2*(( 2*Sqrt[e]*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*Sqrt[1 + Cos[c + d*x]]*Sqr t[a + a*Sin[c + d*x]])/(d*(a + a*Cos[c + d*x] + a*Sin[c + d*x])) + (2*Sqrt [e]*ArcTan[(Sqrt[e]*Sin[c + d*x])/(Sqrt[e*Cos[c + d*x]]*Sqrt[1 + Cos[c + d *x]])]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*(a + a*Cos[c + d*x] + a*Sin[c + d*x]))))/a^2
3.4.15.3.1 Defintions of rubi rules used
Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2/b S ubst[Int[1/Sqrt[c + d*(x^2/b)], x], x, Sqrt[b*x]], x] /; FreeQ[{b, c, d}, x ] && GtQ[c, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
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[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[g*Sqrt[1 + Cos[e + f*x]]*(Sqrt[a + b*Sin[e + f*x ]]/(a + a*Cos[e + f*x] + b*Sin[e + f*x])) Int[Sqrt[1 + Cos[e + f*x]]/Sqrt [g*Cos[e + f*x]], x], x] - Simp[g*Sqrt[1 + Cos[e + f*x]]*(Sqrt[a + b*Sin[e + f*x]]/(b + b*Cos[e + f*x] + a*Sin[e + f*x])) Int[Sin[e + f*x]/(Sqrt[g*C os[e + f*x]]*Sqrt[1 + Cos[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, g}, x] & & EqQ[a^2 - b^2, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b + d*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]))], x ] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 3.32 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.32
| method | result | size |
| default | \(-\frac {2 \left (3 \sin \left (d x +c \right ) \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-3 \sin \left (d x +c \right ) \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+2 \cos \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-2 \sin \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3 \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-3 \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+2 \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {e \cos \left (d x +c \right )}\, e^{2}}{3 d \left (1+\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) a^{2} \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {a \left (1+\sin \left (d x +c \right )\right )}}\) | \(287\) |
-2/3/d*(3*sin(d*x+c)*arctan((-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-3*sin(d*x+ c)*arctanh(sin(d*x+c)/(1+cos(d*x+c))/(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+2 *cos(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-2*sin(d*x+c)*(-cos(d*x+c)/( 1+cos(d*x+c)))^(1/2)+3*arctan((-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-3*arctan h(sin(d*x+c)/(1+cos(d*x+c))/(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+2*(-cos(d* x+c)/(1+cos(d*x+c)))^(1/2))*(e*cos(d*x+c))^(1/2)*e^2/(1+cos(d*x+c)+sin(d*x +c))/a^2/(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/(a*(1+sin(d*x+c)))^(1/2)
Result contains complex when optimal does not.
Time = 0.53 (sec) , antiderivative size = 1334, normalized size of antiderivative = 6.12 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
-1/6*(3*(a^3*d*cos(d*x + c)^2 - a^3*d*cos(d*x + c) - 2*a^3*d - (a^3*d*cos( d*x + c) + 2*a^3*d)*sin(d*x + c))*(-e^10/(a^10*d^4))^(1/4)*log((2*(e^7*sin (d*x + c) + (a^5*d^2*e^2*cos(d*x + c) + a^5*d^2*e^2)*sqrt(-e^10/(a^10*d^4) ))*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a) + (2*a^8*d^3*cos(d*x + c) ^2 + a^8*d^3*cos(d*x + c) - a^8*d^3*sin(d*x + c) - a^8*d^3)*(-e^10/(a^10*d ^4))^(3/4) + (a^3*d*e^5*cos(d*x + c) + a^3*d*e^5 + (2*a^3*d*e^5*cos(d*x + c) + a^3*d*e^5)*sin(d*x + c))*(-e^10/(a^10*d^4))^(1/4))/(cos(d*x + c) + si n(d*x + c) + 1)) - 3*(a^3*d*cos(d*x + c)^2 - a^3*d*cos(d*x + c) - 2*a^3*d - (a^3*d*cos(d*x + c) + 2*a^3*d)*sin(d*x + c))*(-e^10/(a^10*d^4))^(1/4)*lo g((2*(e^7*sin(d*x + c) + (a^5*d^2*e^2*cos(d*x + c) + a^5*d^2*e^2)*sqrt(-e^ 10/(a^10*d^4)))*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a) - (2*a^8*d^3 *cos(d*x + c)^2 + a^8*d^3*cos(d*x + c) - a^8*d^3*sin(d*x + c) - a^8*d^3)*( -e^10/(a^10*d^4))^(3/4) - (a^3*d*e^5*cos(d*x + c) + a^3*d*e^5 + (2*a^3*d*e ^5*cos(d*x + c) + a^3*d*e^5)*sin(d*x + c))*(-e^10/(a^10*d^4))^(1/4))/(cos( d*x + c) + sin(d*x + c) + 1)) - 3*(I*a^3*d*cos(d*x + c)^2 - I*a^3*d*cos(d* x + c) - 2*I*a^3*d + (-I*a^3*d*cos(d*x + c) - 2*I*a^3*d)*sin(d*x + c))*(-e ^10/(a^10*d^4))^(1/4)*log((2*(e^7*sin(d*x + c) - (a^5*d^2*e^2*cos(d*x + c) + a^5*d^2*e^2)*sqrt(-e^10/(a^10*d^4)))*sqrt(e*cos(d*x + c))*sqrt(a*sin(d* x + c) + a) + (2*I*a^8*d^3*cos(d*x + c)^2 + I*a^8*d^3*cos(d*x + c) - I*a^8 *d^3*sin(d*x + c) - I*a^8*d^3)*(-e^10/(a^10*d^4))^(3/4) + (-I*a^3*d*e^5...
Timed out. \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]