\(\int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{9/2}} \, dx\) [260]

Optimal result

Integrand size = 25, antiderivative size = 217 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{9/2}} \, dx=\frac {45 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{1024 \sqrt {2} a^{9/2} d}-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 d (a+a \cos (c+d x))^{9/2}}-\frac {5 \sqrt {\cos (c+d x)} \sin (c+d x)}{32 a d (a+a \cos (c+d x))^{7/2}}+\frac {33 \sqrt {\cos (c+d x)} \sin (c+d x)}{256 a^2 d (a+a \cos (c+d x))^{5/2}}+\frac {73 \sqrt {\cos (c+d x)} \sin (c+d x)}{1024 a^3 d (a+a \cos (c+d x))^{3/2}} \]

[Out]

-1/8*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(9/2)+45/2048*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*
x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(9/2)/d*2^(1/2)-5/32*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^(7/
2)+33/256*sin(d*x+c)*cos(d*x+c)^(1/2)/a^2/d/(a+a*cos(d*x+c))^(5/2)+73/1024*sin(d*x+c)*cos(d*x+c)^(1/2)/a^3/d/(
a+a*cos(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2844, 3056, 3057, 12, 2861, 211} \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{9/2}} \, dx=\frac {45 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{1024 \sqrt {2} a^{9/2} d}+\frac {73 \sin (c+d x) \sqrt {\cos (c+d x)}}{1024 a^3 d (a \cos (c+d x)+a)^{3/2}}+\frac {33 \sin (c+d x) \sqrt {\cos (c+d x)}}{256 a^2 d (a \cos (c+d x)+a)^{5/2}}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{8 d (a \cos (c+d x)+a)^{9/2}}-\frac {5 \sin (c+d x) \sqrt {\cos (c+d x)}}{32 a d (a \cos (c+d x)+a)^{7/2}} \]

[In]

Int[Cos[c + d*x]^(5/2)/(a + a*Cos[c + d*x])^(9/2),x]

[Out]

(45*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(1024*Sqrt[2]*a^(9/2
)*d) - (Cos[c + d*x]^(3/2)*Sin[c + d*x])/(8*d*(a + a*Cos[c + d*x])^(9/2)) - (5*Sqrt[Cos[c + d*x]]*Sin[c + d*x]
)/(32*a*d*(a + a*Cos[c + d*x])^(7/2)) + (33*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(256*a^2*d*(a + a*Cos[c + d*x])^(
5/2)) + (73*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(1024*a^3*d*(a + a*Cos[c + d*x])^(3/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2844

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Dist[1/
(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*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] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ
[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2861

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[-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*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]

Rule 3056

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^n/(a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], 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] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 3057

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 d (a+a \cos (c+d x))^{9/2}}-\frac {\int \frac {\sqrt {\cos (c+d x)} \left (\frac {3 a}{2}-6 a \cos (c+d x)\right )}{(a+a \cos (c+d x))^{7/2}} \, dx}{8 a^2} \\ & = -\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 d (a+a \cos (c+d x))^{9/2}}-\frac {5 \sqrt {\cos (c+d x)} \sin (c+d x)}{32 a d (a+a \cos (c+d x))^{7/2}}-\frac {\int \frac {\frac {15 a^2}{4}-21 a^2 \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx}{48 a^4} \\ & = -\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 d (a+a \cos (c+d x))^{9/2}}-\frac {5 \sqrt {\cos (c+d x)} \sin (c+d x)}{32 a d (a+a \cos (c+d x))^{7/2}}+\frac {33 \sqrt {\cos (c+d x)} \sin (c+d x)}{256 a^2 d (a+a \cos (c+d x))^{5/2}}-\frac {\int \frac {\frac {21 a^3}{8}-\frac {99}{4} a^3 \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}} \, dx}{192 a^6} \\ & = -\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 d (a+a \cos (c+d x))^{9/2}}-\frac {5 \sqrt {\cos (c+d x)} \sin (c+d x)}{32 a d (a+a \cos (c+d x))^{7/2}}+\frac {33 \sqrt {\cos (c+d x)} \sin (c+d x)}{256 a^2 d (a+a \cos (c+d x))^{5/2}}+\frac {73 \sqrt {\cos (c+d x)} \sin (c+d x)}{1024 a^3 d (a+a \cos (c+d x))^{3/2}}-\frac {\int -\frac {135 a^4}{16 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{384 a^8} \\ & = -\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 d (a+a \cos (c+d x))^{9/2}}-\frac {5 \sqrt {\cos (c+d x)} \sin (c+d x)}{32 a d (a+a \cos (c+d x))^{7/2}}+\frac {33 \sqrt {\cos (c+d x)} \sin (c+d x)}{256 a^2 d (a+a \cos (c+d x))^{5/2}}+\frac {73 \sqrt {\cos (c+d x)} \sin (c+d x)}{1024 a^3 d (a+a \cos (c+d x))^{3/2}}+\frac {45 \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{2048 a^4} \\ & = -\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 d (a+a \cos (c+d x))^{9/2}}-\frac {5 \sqrt {\cos (c+d x)} \sin (c+d x)}{32 a d (a+a \cos (c+d x))^{7/2}}+\frac {33 \sqrt {\cos (c+d x)} \sin (c+d x)}{256 a^2 d (a+a \cos (c+d x))^{5/2}}+\frac {73 \sqrt {\cos (c+d x)} \sin (c+d x)}{1024 a^3 d (a+a \cos (c+d x))^{3/2}}-\frac {45 \text {Subst}\left (\int \frac {1}{2 a^2+a x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{1024 a^3 d} \\ & = \frac {45 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{1024 \sqrt {2} a^{9/2} d}-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 d (a+a \cos (c+d x))^{9/2}}-\frac {5 \sqrt {\cos (c+d x)} \sin (c+d x)}{32 a d (a+a \cos (c+d x))^{7/2}}+\frac {33 \sqrt {\cos (c+d x)} \sin (c+d x)}{256 a^2 d (a+a \cos (c+d x))^{5/2}}+\frac {73 \sqrt {\cos (c+d x)} \sin (c+d x)}{1024 a^3 d (a+a \cos (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{9/2}} \, dx=\frac {\sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (5760 \text {arctanh}\left (\sqrt {-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \cos ^8\left (\frac {1}{2} (c+d x)\right )+(999+2466 \cos (c+d x)+1072 \cos (2 (c+d x))+702 \cos (3 (c+d x))+73 \cos (4 (c+d x))) \sqrt {2-2 \sec (c+d x)}\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{65536 \sqrt {2} a^4 d \sqrt {-1+\cos (c+d x)} \sqrt {a (1+\cos (c+d x))}} \]

[In]

Integrate[Cos[c + d*x]^(5/2)/(a + a*Cos[c + d*x])^(9/2),x]

[Out]

(Sec[(c + d*x)/2]^6*(5760*ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]]*Cos[(c + d*x)/2]^8 + (999 + 2466*C
os[c + d*x] + 1072*Cos[2*(c + d*x)] + 702*Cos[3*(c + d*x)] + 73*Cos[4*(c + d*x)])*Sqrt[2 - 2*Sec[c + d*x]])*Ta
n[(c + d*x)/2])/(65536*Sqrt[2]*a^4*d*Sqrt[-1 + Cos[c + d*x]]*Sqrt[a*(1 + Cos[c + d*x])])

Maple [A] (verified)

Time = 3.50 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.61

method	result	size
default	\(\frac {{\left (-\frac {\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\right )}^{\frac {5}{2}} {\left (\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1\right )}^{3} \sqrt {\frac {a}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}}\, \left (16 \left (\csc ^{7}\left (d x +c \right )\right ) \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (1-\cos \left (d x +c \right )\right )^{7}-24 \left (\csc ^{5}\left (d x +c \right )\right ) \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (1-\cos \left (d x +c \right )\right )^{5}-30 \left (\csc ^{3}\left (d x +c \right )\right ) \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (1-\cos \left (d x +c \right )\right )^{3}+83 \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-45 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}}{2048 d {\left (-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1\right )}^{\frac {5}{2}} a^{5}}\)	\(350\)

[In]

int(cos(d*x+c)^(5/2)/(a+cos(d*x+c)*a)^(9/2),x,method=_RETURNVERBOSE)

[Out]

1/2048/d*(-(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)/(csc(d*x+c)^2*(1-cos(d*x+c))^2+1))^(5/2)/(-csc(d*x+c)^2*(1-cos(d*
x+c))^2+1)^(5/2)*(csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^3*(a/(csc(d*x+c)^2*(1-cos(d*x+c))^2+1))^(1/2)*(16*csc(d*x+c
)^7*(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2)*(1-cos(d*x+c))^7-24*csc(d*x+c)^5*(-csc(d*x+c)^2*(1-cos(d*x+c))^2+
1)^(1/2)*(1-cos(d*x+c))^5-30*csc(d*x+c)^3*(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2)*(1-cos(d*x+c))^3+83*(-csc(d
*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2)*(csc(d*x+c)-cot(d*x+c))-45*arcsin(cot(d*x+c)-csc(d*x+c)))*2^(1/2)/a^5

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.14 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{9/2}} \, dx=\frac {45 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{5} + 5 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 10 \, \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) + 2 \, {\left (73 \, \cos \left (d x + c\right )^{3} + 351 \, \cos \left (d x + c\right )^{2} + 195 \, \cos \left (d x + c\right ) + 45\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2048 \, {\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \]

[In]

integrate(cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(9/2),x, algorithm="fricas")

[Out]

1/2048*(45*sqrt(2)*(cos(d*x + c)^5 + 5*cos(d*x + c)^4 + 10*cos(d*x + c)^3 + 10*cos(d*x + c)^2 + 5*cos(d*x + c)
+ 1)*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x +
c)^2 + a*cos(d*x + c))) + 2*(73*cos(d*x + c)^3 + 351*cos(d*x + c)^2 + 195*cos(d*x + c) + 45)*sqrt(a*cos(d*x +
c) + a)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^5*d*cos(d*x + c)^5 + 5*a^5*d*cos(d*x + c)^4 + 10*a^5*d*cos(d*x +
c)^3 + 10*a^5*d*cos(d*x + c)^2 + 5*a^5*d*cos(d*x + c) + a^5*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(5/2)/(a+a*cos(d*x+c))**(9/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{9/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {9}{2}}} \,d x } \]

[In]

integrate(cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(9/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^(9/2), x)

Giac [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(9/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{9/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{9/2}} \,d x \]

[In]

int(cos(c + d*x)^(5/2)/(a + a*cos(c + d*x))^(9/2),x)

[Out]

int(cos(c + d*x)^(5/2)/(a + a*cos(c + d*x))^(9/2), x)