\(\int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\) [245]

Optimal result

Integrand size = 25, antiderivative size = 174 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {2 \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}-\frac {43 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {11 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \]

[Out]

2*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d-1/4*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c
))^(5/2)-43/32*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^(5/2)/d*2^(1/2
)-11/16*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2844, 3056, 3061, 2861, 211, 2853, 222} \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {2 \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}-\frac {43 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}-\frac {11 \sin (c+d x) \sqrt {\cos (c+d x)}}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

[In]

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

[Out]

(2*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(5/2)*d) - (43*ArcTan[(Sqrt[a]*Sin[c + d*x])/(S
qrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - (Cos[c + d*x]^(3/2)*Sin[c + d*x
])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - (11*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2))

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 222

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]

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 2853

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

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 3061

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A*b - a*B)/b, Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*
x]]), x], x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {\int \frac {\sqrt {\cos (c+d x)} \left (\frac {3 a}{2}-4 a \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = -\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {11 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {\int \frac {\frac {11 a^2}{4}-8 a^2 \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4} \\ & = -\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {11 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx}{a^3}-\frac {43 \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2} \\ & = -\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {11 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {2 \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a}}} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^3 d}+\frac {43 \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 )}{16 a d} \\ & = \frac {2 \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}-\frac {43 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {11 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {\left (44 \arcsin \left (\sqrt {1-\cos (c+d x)}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+172 \arcsin \left (\sqrt {\cos (c+d x)}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )-86 \sqrt {2} \arctan \left (\frac {\sqrt {\cos (c+d x)}}{\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+15 \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)+11 \sqrt {-((-1+\cos (c+d x)) \cos (c+d x))}\right ) \sin (c+d x)}{16 d \sqrt {1-\cos (c+d x)} (a (1+\cos (c+d x)))^{5/2}} \]

[In]

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

[Out]

-1/16*((44*ArcSin[Sqrt[1 - Cos[c + d*x]]]*Cos[(c + d*x)/2]^4 + 172*ArcSin[Sqrt[Cos[c + d*x]]]*Cos[(c + d*x)/2]
^4 - 86*Sqrt[2]*ArcTan[Sqrt[Cos[c + d*x]]/Sqrt[Sin[(c + d*x)/2]^2]]*Cos[(c + d*x)/2]^4 + 15*Sqrt[1 - Cos[c + d
*x]]*Cos[c + d*x]^(3/2) + 11*Sqrt[-((-1 + Cos[c + d*x])*Cos[c + d*x])])*Sin[c + d*x])/(d*Sqrt[1 - Cos[c + d*x]
]*(a*(1 + Cos[c + d*x]))^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(329\) vs. \(2(143)=286\).

Time = 3.85 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.90

method	result	size
default	\(-\frac {{\left (-\frac {\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\right )}^{\frac {5}{2}} {\left (\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1\right )}^{3} \sqrt {\frac {a}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}}\, \left (-2 \left (\csc ^{3}\left (d x +c \right )\right ) \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (1-\cos \left (d x +c \right )\right )^{3}+32 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}\right )+13 \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-43 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}}{32 d {\left (-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1\right )}^{\frac {5}{2}} a^{3}}\)	\(330\)

[In]

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

[Out]

-1/32/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)*(-2*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+32*2^(1/2)*arctan(2^(1/2)*(-csc(d*x+c)^2*(1-cos(d
*x+c))^2+1)^(1/2)/(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)*(csc(d*x+c)-cot(d*x+c)))+13*(-csc(d*x+c)^2*(1-cos(d*x+c))^
2+1)^(1/2)*(csc(d*x+c)-cot(d*x+c))-43*arcsin(cot(d*x+c)-csc(d*x+c)))*2^(1/2)/a^3

Fricas [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.30 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {43 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (15 \, \cos \left (d x + c\right ) + 11\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 64 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

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

[Out]

1/32*(43*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*
x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - 2*sqrt(a*cos(d*x + c) + a)*(15*cos(d*x + c) + 11)*sqr
t(cos(d*x + c))*sin(d*x + c) - 64*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(a)*arctan(sqrt
(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^
2 + 3*a^3*d*cos(d*x + c) + a^3*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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