\(\int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\) [244]

Optimal result

Integrand size = 25, antiderivative size = 214 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {5 \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac {115 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {15 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {35 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}} \]

[Out]

-5*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)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+
c))^(5/2)-15/16*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)+115/32*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)+35/16*sin(d*x+c)*cos(d*x+c)^(1/2)/a^2/d/(a+a*
cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2844, 3056, 3062, 3061, 2861, 211, 2853, 222} \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {5 \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac {115 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {35 \sin (c+d x) \sqrt {\cos (c+d x)}}{16 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}-\frac {15 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

[In]

Int[Cos[c + d*x]^(7/2)/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(-5*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(5/2)*d) + (115*ArcTan[(Sqrt[a]*Sin[c + d*x])/
(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - (Cos[c + d*x]^(5/2)*Sin[c + d
*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - (15*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2
)) + (35*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*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 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]

Rule 3062

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)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*
(m + n + 1))), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[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])

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (\frac {5 a}{2}-5 a \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = -\frac {\cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {15 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {\int \frac {\sqrt {\cos (c+d x)} \left (\frac {45 a^2}{4}-\frac {35}{2} a^2 \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4} \\ & = -\frac {\cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {15 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {35 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {\int \frac {-\frac {35 a^3}{4}+20 a^3 \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^5} \\ & = -\frac {\cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {15 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {35 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {5 \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx}{2 a^3}+\frac {115 \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2} \\ & = -\frac {\cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {15 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {35 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {5 \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 {115 \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 {5 \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac {115 \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 {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {15 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {35 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.01 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\left (140 \arcsin \left (\sqrt {1-\cos (c+d x)}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+460 \arcsin \left (\sqrt {\cos (c+d x)}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )-230 \sqrt {2} \arctan \left (\frac {\sqrt {\cos (c+d x)}}{\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+55 \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)+16 \sqrt {1-\cos (c+d x)} \cos ^{\frac {5}{2}}(c+d x)+35 \sqrt {-((-1+\cos (c+d x)) \cos (c+d x))}\right ) \sin (c+d x)}{16 d \sqrt {1-\cos (c+d x)} (a (1+\cos (c+d x)))^{5/2}} \]

[In]

Integrate[Cos[c + d*x]^(7/2)/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

((140*ArcSin[Sqrt[1 - Cos[c + d*x]]]*Cos[(c + d*x)/2]^4 + 460*ArcSin[Sqrt[Cos[c + d*x]]]*Cos[(c + d*x)/2]^4 -
230*Sqrt[2]*ArcTan[Sqrt[Cos[c + d*x]]/Sqrt[Sin[(c + d*x)/2]^2]]*Cos[(c + d*x)/2]^4 + 55*Sqrt[1 - Cos[c + d*x]]
*Cos[c + d*x]^(3/2) + 16*Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(5/2) + 35*Sqrt[-((-1 + Cos[c + d*x])*Cos[c + d*x
])])*Sin[c + d*x])/(16*d*Sqrt[1 - Cos[c + d*x]]*(a*(1 + Cos[c + d*x]))^(5/2))

Maple [A] (verified)

Time = 12.74 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.61

method	result	size
default	\(\frac {\left (16 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-80 \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+55 \sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-115 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )-160 \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \cos \left (d x +c \right )+35 \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-230 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-80 \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-115 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{32 d \left (1+\cos \left (d x +c \right )\right )^{3} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{3}}\)	\(344\)

[In]

int(cos(d*x+c)^(7/2)/(a+cos(d*x+c)*a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/32/d*(16*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2*sin(d*x+c)-80*2^(1/2)*arctan(tan(d*x+c)*(cos
(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^2+55*2^(1/2)*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
-115*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^2-160*2^(1/2)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2
))*cos(d*x+c)+35*sin(d*x+c)*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-230*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*
x+c)-80*2^(1/2)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-115*arcsin(cot(d*x+c)-csc(d*x+c)))*cos(d*
x+c)^(1/2)*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))^3/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)/a^3

Fricas [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {115 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (16 \, \cos \left (d x + c\right )^{2} + 55 \, \cos \left (d x + c\right ) + 35\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 160 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

integrate(cos(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/32*(115*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)*(16*cos(d*x + c)^2 + 55*
cos(d*x + c) + 35)*sqrt(cos(d*x + c))*sin(d*x + c) - 160*(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. \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(7/2)/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {7}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(cos(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^(7/2)/(a*cos(d*x + c) + a)^(5/2), x)

Giac [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{7/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

int(cos(c + d*x)^(7/2)/(a + a*cos(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^(7/2)/(a + a*cos(c + d*x))^(5/2), x)