\(\int \frac {\cos (c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx\) [412]

Optimal result

Integrand size = 41, antiderivative size = 181 \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {(3 A-7 B+11 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(3 A-9 B+13 C) \sin (c+d x)}{3 a d \sqrt {a+a \cos (c+d x)}}+\frac {(3 A-3 B+7 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d} \]

[Out]

-1/2*(A-B+C)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)+1/4*(3*A-7*B+11*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)-1/3*(3*A-9*B+13*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)+1
/6*(3*A-3*B+7*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/a^2/d

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {3120, 3047, 3102, 2830, 2728, 212} \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {(3 A-7 B+11 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(3 A-3 B+7 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{6 a^2 d}-\frac {(A-B+C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {(3 A-9 B+13 C) \sin (c+d x)}{3 a d \sqrt {a \cos (c+d x)+a}} \]

[In]

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

[Out]

((3*A - 7*B + 11*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d)
- ((A - B + C)*Cos[c + d*x]^2*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) - ((3*A - 9*B + 13*C)*Sin[c + d*x
])/(3*a*d*Sqrt[a + a*Cos[c + d*x]]) + ((3*A - 3*B + 7*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(6*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 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 3120

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + 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)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(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, B, 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-B+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\cos (c+d x) \left (2 a (B-C)+\frac {1}{2} a (3 A-3 B+7 C) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {(A-B+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {2 a (B-C) \cos (c+d x)+\frac {1}{2} a (3 A-3 B+7 C) \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {(A-B+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(3 A-3 B+7 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}+\frac {\int \frac {\frac {1}{4} a^2 (3 A-3 B+7 C)-\frac {1}{2} a^2 (3 A-9 B+13 C) \cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{3 a^3} \\ & = -\frac {(A-B+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(3 A-9 B+13 C) \sin (c+d x)}{3 a d \sqrt {a+a \cos (c+d x)}}+\frac {(3 A-3 B+7 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}+\frac {(3 A-7 B+11 C) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a} \\ & = -\frac {(A-B+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(3 A-9 B+13 C) \sin (c+d x)}{3 a d \sqrt {a+a \cos (c+d x)}}+\frac {(3 A-3 B+7 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}-\frac {(3 A-7 B+11 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 {(3 A-7 B+11 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-B+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(3 A-9 B+13 C) \sin (c+d x)}{3 a d \sqrt {a+a \cos (c+d x)}}+\frac {(3 A-3 B+7 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.72 \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {-3 (3 A-7 B+11 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right )-\cos ^3\left (\frac {1}{2} (c+d x)\right ) (-3 A+15 B-17 C+12 (B-C) \cos (c+d x)+2 C \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )}{3 d (a (1+\cos (c+d x)))^{3/2} \left (-1+\sin ^2\left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

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

[Out]

(-3*(3*A - 7*B + 11*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^5 - Cos[(c + d*x)/2]^3*(-3*A + 15*B - 17*C +
12*(B - C)*Cos[c + d*x] + 2*C*Cos[2*(c + d*x)])*Sin[(c + d*x)/2])/(3*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. \(406\) vs. \(2(158)=316\).

Time = 11.73 (sec) , antiderivative size = 407, normalized size of antiderivative = 2.25

method	result	size
default	\(\frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (16 C \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+9 A \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -21 B \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +33 C \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +24 B \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {2}\, \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-40 C \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 A \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+3 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-3 C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{12 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{\frac {5}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(407\)
parts	\(\frac {A \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (3 \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 )-\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 {B \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 (16 \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 )+8 \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 )-33 \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 -27 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+33 \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 \right )}{12 \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}\)	\(549\)

[In]

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

[Out]

1/12*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*C*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)^4+
9*A*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^2*a-21*
B*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^2*a+33*C*
2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^2*a+24*B*co
s(1/2*d*x+1/2*c)^2*2^(1/2)*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-40*C*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2
)^(1/2)*cos(1/2*d*x+1/2*c)^2-3*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+3*B*2^(1/2)*(a*sin(1/2*d*x+1/2
*c)^2)^(1/2)*a^(1/2)-3*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/cos(1/2*d*x+1/2*c)/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 = 0.29 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.20 \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {3 \, \sqrt {2} {\left ({\left (3 \, A - 7 \, B + 11 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A - 7 \, B + 11 \, C\right )} \cos \left (d x + c\right ) + 3 \, A - 7 \, B + 11 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (4 \, C \cos \left (d x + c\right )^{2} + 12 \, {\left (B - C\right )} \cos \left (d x + c\right ) - 3 \, A + 15 \, B - 19 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

1/24*(3*sqrt(2)*((3*A - 7*B + 11*C)*cos(d*x + c)^2 + 2*(3*A - 7*B + 11*C)*cos(d*x + c) + 3*A - 7*B + 11*C)*sqr
t(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)^2 + 12*(B - C)*cos(d*x + c) - 3*A + 15*B - 19*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. \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c)+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%%%{%%{[%%%{%%{[-54043195528445952,0]:[1,0,-2]%%},[6]%%%},0]
:[1,0,%%%{-

Mupad [F(-1)]

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

[In]

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

[Out]

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