Integrand size = 25, antiderivative size = 171 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {7 \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)}}+\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {a+a \cos (c+d x)}} \] Output:
7/4*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/a^(1/2)/d-2^(1/2)*ar ctan(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2 ))/a^(1/2)/d-1/4*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/2* cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)
Time = 0.31 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\left (\arcsin \left (\sqrt {1-\cos (c+d x)}\right )+8 \arcsin \left (\sqrt {\cos (c+d x)}\right )-4 \sqrt {2} \arctan \left (\frac {\sqrt {\cos (c+d x)}}{\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}}\right )-2 \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)+\sqrt {-((-1+\cos (c+d x)) \cos (c+d x))}\right ) \sin (c+d x)}{4 d \sqrt {1-\cos (c+d x)} \sqrt {a (1+\cos (c+d x))}} \] Input:
Integrate[Cos[c + d*x]^(5/2)/Sqrt[a + a*Cos[c + d*x]],x]
Output:
-1/4*((ArcSin[Sqrt[1 - Cos[c + d*x]]] + 8*ArcSin[Sqrt[Cos[c + d*x]]] - 4*S qrt[2]*ArcTan[Sqrt[Cos[c + d*x]]/Sqrt[Sin[(c + d*x)/2]^2]] - 2*Sqrt[1 - Co s[c + d*x]]*Cos[c + d*x]^(3/2) + Sqrt[-((-1 + Cos[c + d*x])*Cos[c + d*x])] )*Sin[c + d*x])/(d*Sqrt[1 - Cos[c + d*x]]*Sqrt[a*(1 + Cos[c + d*x])])
Time = 1.01 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.08, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 3257, 25, 3042, 3462, 27, 3042, 3461, 3042, 3253, 223, 3261, 218}
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.
\(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a \cos (c+d x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}{\sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
\(\Big \downarrow \) 3257 |
\(\displaystyle \frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\int -\frac {\sqrt {\cos (c+d x)} (3 a-a \cos (c+d x))}{\sqrt {\cos (c+d x) a+a}}dx}{4 a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\sqrt {\cos (c+d x)} (3 a-a \cos (c+d x))}{\sqrt {\cos (c+d x) a+a}}dx}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (3 a-a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 3462 |
\(\displaystyle \frac {\frac {\int -\frac {a^2-7 a^2 \cos (c+d x)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}-\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {-\frac {\int \frac {a^2-7 a^2 \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a}-\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {\int \frac {a^2-7 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}-\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 3461 |
\(\displaystyle \frac {-\frac {8 a^2 \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx-7 a \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx}{2 a}-\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {8 a^2 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-7 a \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}-\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 3253 |
\(\displaystyle \frac {-\frac {8 a^2 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {14 a \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{2 a}-\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {-\frac {8 a^2 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {14 a^{3/2} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}-\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 3261 |
\(\displaystyle \frac {-\frac {-\frac {16 a^3 \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {14 a^{3/2} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}-\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {-\frac {\frac {8 \sqrt {2} a^{3/2} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {14 a^{3/2} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}-\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\) |
Input:
Int[Cos[c + d*x]^(5/2)/Sqrt[a + a*Cos[c + d*x]],x]
Output:
(Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) + (-1/2*( (-14*a^(3/2)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (8*Sqrt[2]*a^(3/2)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x ]]*Sqrt[a + a*Cos[c + d*x]])])/d)/a - (a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/ (d*Sqrt[a + a*Cos[c + d*x]]))/(4*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[-2/f Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E qQ[a^2 - b^2, 0] && EqQ[d, a/b]
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_. ) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*d*Cos[e + f*x]*((c + d*Sin[e + f*x]) ^(n - 1)/(f*(2*n - 1)*Sqrt[a + b*Sin[e + f*x]])), x] - Simp[1/(b*(2*n - 1)) Int[((c + d*Sin[e + f*x])^(n - 2)/Sqrt[a + b*Sin[e + f*x]])*Simp[a*c*d - b*(2*d^2*(n - 1) + c^2*(2*n - 1)) + d*(a*d - b*c*(4*n - 3))*Sin[e + f*x], x], 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] && GtQ[n, 1] && IntegerQ[2*n]
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f) Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S in[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[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Sim p[(A*b - a*B)/b Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) , x], x] + Simp[B/b Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]] , x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[ a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Sin[e + f*x])^m*(c + d*S in[e + f*x])^(n - 1)*Simp[A*b*c*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Time = 9.42 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.89
method | result | size |
default | \(\frac {\left (7 \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\sin \left (2 d x +2 c \right )-\sin \left (d x +c \right )\right )+8 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}}{8 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a}\) | \(153\) |
Input:
int(cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/8/d*(7*2^(1/2)*arctan(tan(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+2^(1 /2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(sin(2*d*x+2*c)-sin(d*x+c))+8*arcsin (cot(d*x+c)-csc(d*x+c)))*cos(d*x+c)^(1/2)*2^(1/2)*(a*(cos(d*x+c)+1))^(1/2) /(cos(d*x+c)+1)/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/a
Time = 0.18 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {a \cos \left (d x + c\right ) + a} {\left (2 \, \cos \left (d x + c\right ) - 1\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 7 \, \sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )}\right ) - \frac {4 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a}}\right )}{\sqrt {a}}}{4 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:
integrate(cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")
Output:
1/4*(sqrt(a*cos(d*x + c) + a)*(2*cos(d*x + c) - 1)*sqrt(cos(d*x + c))*sin( d*x + c) + 7*sqrt(a)*(cos(d*x + c) + 1)*arctan(sqrt(a*cos(d*x + c) + a)*sq rt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))) - 4*sqrt(2)*(a*cos(d*x + c) + a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c)/((cos(d*x + c)^2 + cos(d*x + c))*sqrt( a)))/sqrt(a))/(a*d*cos(d*x + c) + a*d)
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(5/2)/(a+a*cos(d*x+c))**(1/2),x)
Output:
Timed out
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {a \cos \left (d x + c\right ) + a}} \,d x } \] Input:
integrate(cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate(cos(d*x + c)^(5/2)/sqrt(a*cos(d*x + c) + a), x)
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}}{\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \] Input:
int(cos(c + d*x)^(5/2)/(a + a*cos(c + d*x))^(1/2),x)
Output:
int(cos(c + d*x)^(5/2)/(a + a*cos(c + d*x))^(1/2), x)
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right )+1}d x \right )}{a} \] Input:
int(cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x)
Output:
(sqrt(a)*int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)**2)/( cos(c + d*x) + 1),x))/a