Integrand size = 27, antiderivative size = 204 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{5/2}} \, dx=\frac {2 a^2 \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d e^{5/2} (1+\cos (c+d x)+\sin (c+d x))}-\frac {2 a^2 \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d e^{5/2} (1+\cos (c+d x)+\sin (c+d x))}+\frac {4 a (a+a \sin (c+d x))^{3/2}}{3 d e (e \cos (c+d x))^{3/2}} \]
4/3*a*(a+a*sin(d*x+c))^(3/2)/d/e/(e*cos(d*x+c))^(3/2)+2*a^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/e^(5 /2)/(1+cos(d*x+c)+sin(d*x+c))-2*a^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/e^(5/2)/(1+cos(d*x+c)+sin(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.12 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.38 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{5/2}} \, dx=\frac {4\ 2^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {3}{4},\frac {1}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (a (1+\sin (c+d x)))^{5/2}}{3 d e (e \cos (c+d x))^{3/2} (1+\sin (c+d x))^{7/4}} \]
(4*2^(3/4)*Hypergeometric2F1[-3/4, -3/4, 1/4, (1 - Sin[c + d*x])/2]*(a*(1 + Sin[c + d*x]))^(5/2))/(3*d*e*(e*Cos[c + d*x])^(3/2)*(1 + Sin[c + d*x])^( 7/4))
Time = 0.80 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {3042, 3155, 3042, 3156, 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.
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^{5/2}}{(e \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^{5/2}}{(e \cos (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3155 |
| \(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}-\frac {a^2 \int \frac {\sqrt {\sin (c+d x) a+a}}{\sqrt {e \cos (c+d x)}}dx}{e^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}-\frac {a^2 \int \frac {\sqrt {\sin (c+d x) a+a}}{\sqrt {e \cos (c+d x)}}dx}{e^2}\) |
| \(\Big \downarrow \) 3156 |
| \(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}-\frac {a^2 \left (\frac {\sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sqrt {\cos (c+d x)+1}}{\sqrt {e \cos (c+d x)}}dx}{\sin (c+d x)+\cos (c+d x)+1}+\frac {\sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {\cos (c+d x)+1}}dx}{\sin (c+d x)+\cos (c+d x)+1}\right )}{e^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}-\frac {a^2 \left (\frac {\sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int -\frac {\cos \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx}{\sin (c+d x)+\cos (c+d x)+1}+\frac {\sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sin (c+d x)+\cos (c+d x)+1}\right )}{e^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}-\frac {a^2 \left (\frac {\sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sin (c+d x)+\cos (c+d x)+1}-\frac {\sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {e \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \sqrt {\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+1}}dx}{\sin (c+d x)+\cos (c+d x)+1}\right )}{e^2}\) |
| \(\Big \downarrow \) 3254 |
| \(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}-\frac {a^2 \left (-\frac {\sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {e \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \sqrt {\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+1}}dx}{\sin (c+d x)+\cos (c+d x)+1}-\frac {2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x)}{\cos (c+d x)+1}+1}d\left (-\frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {\cos (c+d x)+1}}\right )}{d (\sin (c+d x)+\cos (c+d x)+1)}\right )}{e^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}-\frac {a^2 \left (\frac {2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d \sqrt {e} (\sin (c+d x)+\cos (c+d x)+1)}-\frac {\sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {e \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \sqrt {\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+1}}dx}{\sin (c+d x)+\cos (c+d x)+1}\right )}{e^2}\) |
| \(\Big \downarrow \) 3312 |
| \(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}-\frac {a^2 \left (\frac {2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d \sqrt {e} (\sin (c+d x)+\cos (c+d x)+1)}-\frac {\sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {1}{\sqrt {e \cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{d (\sin (c+d x)+\cos (c+d x)+1)}\right )}{e^2}\) |
| \(\Big \downarrow \) 63 |
| \(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}-\frac {a^2 \left (\frac {2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d \sqrt {e} (\sin (c+d x)+\cos (c+d x)+1)}-\frac {2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {1}{\sqrt {\cos (c+d x)+1}}d\sqrt {e \cos (c+d x)}}{d e (\sin (c+d x)+\cos (c+d x)+1)}\right )}{e^2}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}-\frac {a^2 \left (\frac {2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d \sqrt {e} (\sin (c+d x)+\cos (c+d x)+1)}-\frac {2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e} (\sin (c+d x)+\cos (c+d x)+1)}\right )}{e^2}\) |
(4*a*(a + a*Sin[c + d*x])^(3/2))/(3*d*e*(e*Cos[c + d*x])^(3/2)) - (a^2*((- 2*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a* Sin[c + d*x]])/(d*Sqrt[e]*(1 + Cos[c + d*x] + Sin[c + d*x])) + (2*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*Sqrt[e]*(1 + Cos[c + d*x] + Sin[c + d*x]))))/e^2
3.3.93.3.1 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*b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(p + 1))), x] + Simp[b^2*((2*m + p - 1)/(g^2*(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] && GtQ[m, 1] && LtQ[p, -1] && Int egersQ[2*m, 2*p]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[cos[(e_.) + (f_.)*(x_)] *(g_.)], x_Symbol] :> Simp[a*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[b*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[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 = 2.94 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.42
| method | result | size |
| default | \(\frac {2 \left (3 \sin \left (d x +c \right ) \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+3 \sin \left (d x +c \right ) \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-2 \cos \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-2 \sin \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-3 \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-3 \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-2 \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, a^{2}}{3 d \left (-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1\right ) \sqrt {e \cos \left (d x +c \right )}\, \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, e^{2}}\) | \(289\) |
2/3/d*(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))+3*sin(d*x+c)*arctan((-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*arctanh(sin(d*x+c)/(1+cos(d*x+c))/(-cos(d*x+c)/(1+co s(d*x+c)))^(1/2))-3*arctan((-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-2*(-cos(d*x +c)/(1+cos(d*x+c)))^(1/2))*(a*(1+sin(d*x+c)))^(1/2)*a^2/(-cos(d*x+c)+sin(d *x+c)-1)/(e*cos(d*x+c))^(1/2)/(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/e^2
Result contains complex when optimal does not.
Time = 0.54 (sec) , antiderivative size = 1120, normalized size of antiderivative = 5.49 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
-1/6*(8*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a)*a^2 + 3*(I*d*e^3*sin (d*x + c) - I*d*e^3)*(-a^10/(d^4*e^10))^(1/4)*log(-(2*(a^7*sin(d*x + c) + (a^2*d^2*e^5*cos(d*x + c) + a^2*d^2*e^5)*sqrt(-a^10/(d^4*e^10)))*sqrt(e*co s(d*x + c))*sqrt(a*sin(d*x + c) + a) - (I*d^3*e^8*cos(d*x + c) + I*d^3*e^8 + (2*I*d^3*e^8*cos(d*x + c) + I*d^3*e^8)*sin(d*x + c))*(-a^10/(d^4*e^10)) ^(3/4) - (-2*I*a^5*d*e^3*cos(d*x + c)^2 - I*a^5*d*e^3*cos(d*x + c) + I*a^5 *d*e^3*sin(d*x + c) + I*a^5*d*e^3)*(-a^10/(d^4*e^10))^(1/4))/(cos(d*x + c) + sin(d*x + c) + 1)) + 3*(-I*d*e^3*sin(d*x + c) + I*d*e^3)*(-a^10/(d^4*e^ 10))^(1/4)*log(-(2*(a^7*sin(d*x + c) + (a^2*d^2*e^5*cos(d*x + c) + a^2*d^2 *e^5)*sqrt(-a^10/(d^4*e^10)))*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a ) - (-I*d^3*e^8*cos(d*x + c) - I*d^3*e^8 + (-2*I*d^3*e^8*cos(d*x + c) - I* d^3*e^8)*sin(d*x + c))*(-a^10/(d^4*e^10))^(3/4) - (2*I*a^5*d*e^3*cos(d*x + c)^2 + I*a^5*d*e^3*cos(d*x + c) - I*a^5*d*e^3*sin(d*x + c) - I*a^5*d*e^3) *(-a^10/(d^4*e^10))^(1/4))/(cos(d*x + c) + sin(d*x + c) + 1)) + 3*(d*e^3*s in(d*x + c) - d*e^3)*(-a^10/(d^4*e^10))^(1/4)*log(-(2*(a^7*sin(d*x + c) - (a^2*d^2*e^5*cos(d*x + c) + a^2*d^2*e^5)*sqrt(-a^10/(d^4*e^10)))*sqrt(e*co s(d*x + c))*sqrt(a*sin(d*x + c) + a) + (d^3*e^8*cos(d*x + c) + d^3*e^8 + ( 2*d^3*e^8*cos(d*x + c) + d^3*e^8)*sin(d*x + c))*(-a^10/(d^4*e^10))^(3/4) + (2*a^5*d*e^3*cos(d*x + c)^2 + a^5*d*e^3*cos(d*x + c) - a^5*d*e^3*sin(d*x + c) - a^5*d*e^3)*(-a^10/(d^4*e^10))^(1/4))/(cos(d*x + c) + sin(d*x + c...
Timed out. \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]