\(\int \frac {\cos ^2(c+d x) (A+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx\) [112]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F(-1)]
   Giac [F(-2)]
   Mupad [F(-1)]

Optimal result

Integrand size = 35, antiderivative size = 214 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(7 A+15 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(15 A+31 C) \sin (c+d x)}{5 a d \sqrt {a+a \cos (c+d x)}}+\frac {(5 A+9 C) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(5 A+13 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{10 a^2 d} \]

[Out]

-1/2*(A+C)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)-1/4*(7*A+15*C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1
/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)+1/5*(15*A+31*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)+1/10*(5*A+
9*C)*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)-1/10*(5*A+13*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/a^2/
d

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3121, 3062, 3047, 3102, 2830, 2728, 212} \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(7 A+15 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(5 A+13 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{10 a^2 d}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}+\frac {(5 A+9 C) \sin (c+d x) \cos ^2(c+d x)}{10 a d \sqrt {a \cos (c+d x)+a}}+\frac {(15 A+31 C) \sin (c+d x)}{5 a d \sqrt {a \cos (c+d x)+a}} \]

[In]

Int[(Cos[c + d*x]^2*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

-1/2*((7*A + 15*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[2]*a^(3/2)*d) - (
(A + C)*Cos[c + d*x]^3*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((15*A + 31*C)*Sin[c + d*x])/(5*a*d*Sq
rt[a + a*Cos[c + d*x]]) + ((5*A + 9*C)*Cos[c + d*x]^2*Sin[c + d*x])/(10*a*d*Sqrt[a + a*Cos[c + d*x]]) - ((5*A
+ 13*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(10*a^2*d)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2830

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

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 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])

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3121

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\cos ^2(c+d x) \left (-a (A+3 C)+\frac {1}{2} a (5 A+9 C) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(5 A+9 C) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\cos (c+d x) \left (a^2 (5 A+9 C)-\frac {3}{4} a^2 (5 A+13 C) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{5 a^3} \\ & = -\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(5 A+9 C) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {a^2 (5 A+9 C) \cos (c+d x)-\frac {3}{4} a^2 (5 A+13 C) \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{5 a^3} \\ & = -\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(5 A+9 C) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(5 A+13 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{10 a^2 d}+\frac {2 \int \frac {-\frac {3}{8} a^3 (5 A+13 C)+\frac {3}{4} a^3 (15 A+31 C) \cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{15 a^4} \\ & = -\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(15 A+31 C) \sin (c+d x)}{5 a d \sqrt {a+a \cos (c+d x)}}+\frac {(5 A+9 C) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(5 A+13 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{10 a^2 d}-\frac {(7 A+15 C) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a} \\ & = -\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(15 A+31 C) \sin (c+d x)}{5 a d \sqrt {a+a \cos (c+d x)}}+\frac {(5 A+9 C) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(5 A+13 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{10 a^2 d}+\frac {(7 A+15 C) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{2 a d} \\ & = -\frac {(7 A+15 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(15 A+31 C) \sin (c+d x)}{5 a d \sqrt {a+a \cos (c+d x)}}+\frac {(5 A+9 C) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(5 A+13 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{10 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.64 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {5 (7 A+15 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right )-\cos ^3\left (\frac {1}{2} (c+d x)\right ) (25 A+47 C+(20 A+39 C) \cos (c+d x)-2 C \cos (2 (c+d x))+C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )}{5 d (a (1+\cos (c+d x)))^{3/2} \left (-1+\sin ^2\left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

Integrate[(Cos[c + d*x]^2*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

(5*(7*A + 15*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^5 - Cos[(c + d*x)/2]^3*(25*A + 47*C + (20*A + 39*C)
*Cos[c + d*x] - 2*C*Cos[2*(c + d*x)] + C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2])/(5*d*(a*(1 + Cos[c + d*x]))^(3/2)
*(-1 + Sin[(c + d*x)/2]^2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(412\) vs. \(2(187)=374\).

Time = 6.95 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.93

method result size
default \(\frac {\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-32 C \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+32 C \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+35 A \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -40 A \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+75 C \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -80 C \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-35 A \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a +45 A \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-75 C \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a +85 C \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{20 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{\frac {5}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) \(413\)
parts \(\frac {A \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-7 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {2}\, \sqrt {a}+\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{\frac {5}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-32 \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+32 \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+75 \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -80 \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-75 \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a +85 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\right )}{20 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{\frac {5}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) \(440\)

[In]

int(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+cos(d*x+c)*a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/20/cos(1/2*d*x+1/2*c)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-32*C*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*s
in(1/2*d*x+1/2*c)^6+32*C*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4+35*A*ln(4*(a^(1/2)*(a*sin
(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))*sin(1/2*d*x+1/2*c)^2*a-40*A*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(
1/2)*sin(1/2*d*x+1/2*c)^2+75*C*ln(4*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))*sin(1/2*d*x
+1/2*c)^2*a-80*C*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2-35*A*ln(4*(a^(1/2)*(a*sin(1/2*d*x
+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))*a+45*A*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-75*C*ln(4*(a^(1/2)*(a*si
n(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))*a+85*C*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(5/2)/sin(1/
2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.02 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {5 \, \sqrt {2} {\left ({\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right ) + 7 \, A + 15 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (4 \, C \cos \left (d x + c\right )^{3} - 4 \, C \cos \left (d x + c\right )^{2} + 4 \, {\left (5 \, A + 9 \, C\right )} \cos \left (d x + c\right ) + 25 \, A + 49 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{40 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/40*(5*sqrt(2)*((7*A + 15*C)*cos(d*x + c)^2 + 2*(7*A + 15*C)*cos(d*x + c) + 7*A + 15*C)*sqrt(a)*log(-(a*cos(d
*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2
+ 2*cos(d*x + c) + 1)) + 4*(4*C*cos(d*x + c)^3 - 4*C*cos(d*x + c)^2 + 4*(5*A + 9*C)*cos(d*x + c) + 25*A + 49*C
)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**2*(A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{%%{[%%%{%%{[2475880078570760549798248448,0]:[1,0,-2]%%},
[10]%%%},0]

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

int((cos(c + d*x)^2*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^(3/2),x)

[Out]

int((cos(c + d*x)^2*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^(3/2), x)