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

Optimal. Leaf size=277 \[ \frac {(11 A-15 B+19 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(245 A-273 B+397 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{210 a^2 d}-\frac {(A-B+C) \sin (c+d x) \cos ^4(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}+\frac {(7 A-7 B+11 C) \sin (c+d x) \cos ^3(c+d x)}{14 a d \sqrt {a \cos (c+d x)+a}}-\frac {(35 A-63 B+67 C) \sin (c+d x) \cos ^2(c+d x)}{70 a d \sqrt {a \cos (c+d x)+a}}-\frac {(455 A-651 B+799 C) \sin (c+d x)}{105 a d \sqrt {a \cos (c+d x)+a}} \]

[Out]

-1/2*(A-B+C)*cos(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)+1/4*(11*A-15*B+19*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)-1/105*(455*A-651*B+799*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c)
)^(1/2)-1/70*(35*A-63*B+67*C)*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)+1/14*(7*A-7*B+11*C)*cos(d*x+c
)^3*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)+1/210*(245*A-273*B+397*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/a^2/d

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Rubi [A]  time = 0.87, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {3041, 2983, 2968, 3023, 2751, 2649, 206} \[ \frac {(245 A-273 B+397 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{210 a^2 d}+\frac {(11 A-15 B+19 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sin (c+d x) \cos ^4(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}+\frac {(7 A-7 B+11 C) \sin (c+d x) \cos ^3(c+d x)}{14 a d \sqrt {a \cos (c+d x)+a}}-\frac {(35 A-63 B+67 C) \sin (c+d x) \cos ^2(c+d x)}{70 a d \sqrt {a \cos (c+d x)+a}}-\frac {(455 A-651 B+799 C) \sin (c+d x)}{105 a d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((11*A - 15*B + 19*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]^4*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) - ((455*A - 651*B + 799*C)*Sin[
c + d*x])/(105*a*d*Sqrt[a + a*Cos[c + d*x]]) - ((35*A - 63*B + 67*C)*Cos[c + d*x]^2*Sin[c + d*x])/(70*a*d*Sqrt
[a + a*Cos[c + d*x]]) + ((7*A - 7*B + 11*C)*Cos[c + d*x]^3*Sin[c + d*x])/(14*a*d*Sqrt[a + a*Cos[c + d*x]]) + (
(245*A - 273*B + 397*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(210*a^2*d)

Rule 206

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

Rule 2649

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 2751

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*Sin
[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 2968

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 2983

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] && (I
ntegerQ[n] || EqQ[m + 1/2, 0])

Rule 3023

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 3041

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*} \int \frac {\cos ^3(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 {(A-B+C) \cos ^4(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\cos ^3(c+d x) \left (-2 a (A-2 B+2 C)+\frac {1}{2} a (7 A-7 B+11 C) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A-B+C) \cos ^4(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(7 A-7 B+11 C) \cos ^3(c+d x) \sin (c+d x)}{14 a d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\cos ^2(c+d x) \left (\frac {3}{2} a^2 (7 A-7 B+11 C)-\frac {1}{4} a^2 (35 A-63 B+67 C) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{7 a^3}\\ &=-\frac {(A-B+C) \cos ^4(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(35 A-63 B+67 C) \cos ^2(c+d x) \sin (c+d x)}{70 a d \sqrt {a+a \cos (c+d x)}}+\frac {(7 A-7 B+11 C) \cos ^3(c+d x) \sin (c+d x)}{14 a d \sqrt {a+a \cos (c+d x)}}+\frac {2 \int \frac {\cos (c+d x) \left (-\frac {1}{2} a^3 (35 A-63 B+67 C)+\frac {1}{8} a^3 (245 A-273 B+397 C) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{35 a^4}\\ &=-\frac {(A-B+C) \cos ^4(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(35 A-63 B+67 C) \cos ^2(c+d x) \sin (c+d x)}{70 a d \sqrt {a+a \cos (c+d x)}}+\frac {(7 A-7 B+11 C) \cos ^3(c+d x) \sin (c+d x)}{14 a d \sqrt {a+a \cos (c+d x)}}+\frac {2 \int \frac {-\frac {1}{2} a^3 (35 A-63 B+67 C) \cos (c+d x)+\frac {1}{8} a^3 (245 A-273 B+397 C) \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{35 a^4}\\ &=-\frac {(A-B+C) \cos ^4(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(35 A-63 B+67 C) \cos ^2(c+d x) \sin (c+d x)}{70 a d \sqrt {a+a \cos (c+d x)}}+\frac {(7 A-7 B+11 C) \cos ^3(c+d x) \sin (c+d x)}{14 a d \sqrt {a+a \cos (c+d x)}}+\frac {(245 A-273 B+397 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{210 a^2 d}+\frac {4 \int \frac {\frac {1}{16} a^4 (245 A-273 B+397 C)-\frac {1}{8} a^4 (455 A-651 B+799 C) \cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{105 a^5}\\ &=-\frac {(A-B+C) \cos ^4(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(455 A-651 B+799 C) \sin (c+d x)}{105 a d \sqrt {a+a \cos (c+d x)}}-\frac {(35 A-63 B+67 C) \cos ^2(c+d x) \sin (c+d x)}{70 a d \sqrt {a+a \cos (c+d x)}}+\frac {(7 A-7 B+11 C) \cos ^3(c+d x) \sin (c+d x)}{14 a d \sqrt {a+a \cos (c+d x)}}+\frac {(245 A-273 B+397 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{210 a^2 d}+\frac {(11 A-15 B+19 C) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a}\\ &=-\frac {(A-B+C) \cos ^4(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(455 A-651 B+799 C) \sin (c+d x)}{105 a d \sqrt {a+a \cos (c+d x)}}-\frac {(35 A-63 B+67 C) \cos ^2(c+d x) \sin (c+d x)}{70 a d \sqrt {a+a \cos (c+d x)}}+\frac {(7 A-7 B+11 C) \cos ^3(c+d x) \sin (c+d x)}{14 a d \sqrt {a+a \cos (c+d x)}}+\frac {(245 A-273 B+397 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{210 a^2 d}-\frac {(11 A-15 B+19 C) \operatorname {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 {(11 A-15 B+19 C) \tanh ^{-1}\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 ^4(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(455 A-651 B+799 C) \sin (c+d x)}{105 a d \sqrt {a+a \cos (c+d x)}}-\frac {(35 A-63 B+67 C) \cos ^2(c+d x) \sin (c+d x)}{70 a d \sqrt {a+a \cos (c+d x)}}+\frac {(7 A-7 B+11 C) \cos ^3(c+d x) \sin (c+d x)}{14 a d \sqrt {a+a \cos (c+d x)}}+\frac {(245 A-273 B+397 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{210 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.54, size = 180, normalized size = 0.65 \[ \frac {\frac {1}{2} \sin \left (\frac {1}{2} (c+d x)\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) (6 (140 A-273 B+277 C) \cos (c+d x)-4 (35 A-21 B+64 C) \cos (2 (c+d x))+1190 A-42 B \cos (3 (c+d x))-1974 B+18 C \cos (3 (c+d x))-15 C \cos (4 (c+d x))+2161 C)-105 (11 A-15 B+19 C) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{105 d \left (\sin ^2\left (\frac {1}{2} (c+d x)\right )-1\right ) (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-105*(11*A - 15*B + 19*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^5 + (Cos[(c + d*x)/2]^3*(1190*A - 1974*B
+ 2161*C + 6*(140*A - 273*B + 277*C)*Cos[c + d*x] - 4*(35*A - 21*B + 64*C)*Cos[2*(c + d*x)] - 42*B*Cos[3*(c +
d*x)] + 18*C*Cos[3*(c + d*x)] - 15*C*Cos[4*(c + d*x)])*Sin[(c + d*x)/2])/2)/(105*d*(a*(1 + Cos[c + d*x]))^(3/
2)*(-1 + Sin[(c + d*x)/2]^2))

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fricas [A]  time = 0.52, size = 259, normalized size = 0.94 \[ \frac {105 \, \sqrt {2} {\left ({\left (11 \, A - 15 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (11 \, A - 15 \, B + 19 \, C\right )} \cos \left (d x + c\right ) + 11 \, A - 15 \, B + 19 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (60 \, C \cos \left (d x + c\right )^{4} + 12 \, {\left (7 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{3} + 28 \, {\left (5 \, A - 3 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{2} - 12 \, {\left (35 \, A - 63 \, B + 67 \, C\right )} \cos \left (d x + c\right ) - 665 \, A + 1029 \, B - 1201 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{840 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(105*sqrt(2)*((11*A - 15*B + 19*C)*cos(d*x + c)^2 + 2*(11*A - 15*B + 19*C)*cos(d*x + c) + 11*A - 15*B +
19*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*(60*C*cos(d*x + c)^4 + 12*(7*B - 3*C)*cos(d*x + c)^3 + 2
8*(5*A - 3*B + 7*C)*cos(d*x + c)^2 - 12*(35*A - 63*B + 67*C)*cos(d*x + c) - 665*A + 1029*B - 1201*C)*sqrt(a*co
s(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)

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giac [A]  time = 3.50, size = 305, normalized size = 1.10 \[ -\frac {\frac {105 \, {\left (11 \, \sqrt {2} A - 15 \, \sqrt {2} B + 19 \, \sqrt {2} C\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {3}{2}}} + \frac {{\left ({\left ({\left ({\left (\frac {105 \, {\left (\sqrt {2} A a^{5} - \sqrt {2} B a^{5} + \sqrt {2} C a^{5}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{3}} + \frac {4 \, {\left (455 \, \sqrt {2} A a^{5} - 693 \, \sqrt {2} B a^{5} + 877 \, \sqrt {2} C a^{5}\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {14 \, {\left (305 \, \sqrt {2} A a^{5} - 453 \, \sqrt {2} B a^{5} + 517 \, \sqrt {2} C a^{5}\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {140 \, {\left (25 \, \sqrt {2} A a^{5} - 39 \, \sqrt {2} B a^{5} + 47 \, \sqrt {2} C a^{5}\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {105 \, {\left (9 \, \sqrt {2} A a^{5} - 17 \, \sqrt {2} B a^{5} + 17 \, \sqrt {2} C a^{5}\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {7}{2}}}}{420 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/420*(105*(11*sqrt(2)*A - 15*sqrt(2)*B + 19*sqrt(2)*C)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/
2*d*x + 1/2*c)^2 + a)))/a^(3/2) + ((((105*(sqrt(2)*A*a^5 - sqrt(2)*B*a^5 + sqrt(2)*C*a^5)*tan(1/2*d*x + 1/2*c)
^2/a^3 + 4*(455*sqrt(2)*A*a^5 - 693*sqrt(2)*B*a^5 + 877*sqrt(2)*C*a^5)/a^3)*tan(1/2*d*x + 1/2*c)^2 + 14*(305*s
qrt(2)*A*a^5 - 453*sqrt(2)*B*a^5 + 517*sqrt(2)*C*a^5)/a^3)*tan(1/2*d*x + 1/2*c)^2 + 140*(25*sqrt(2)*A*a^5 - 39
*sqrt(2)*B*a^5 + 47*sqrt(2)*C*a^5)/a^3)*tan(1/2*d*x + 1/2*c)^2 + 105*(9*sqrt(2)*A*a^5 - 17*sqrt(2)*B*a^5 + 17*
sqrt(2)*C*a^5)/a^3)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(7/2))/d

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maple [B]  time = 1.77, size = 577, normalized size = 2.08 \[ \frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (960 C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-96 \sqrt {2}\, \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (7 B +17 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+112 \sqrt {2}\, \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (5 A +6 B +16 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+35 \sqrt {2}\, \left (8 A \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-33 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 -48 B \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+45 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 ) a +16 C \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-57 \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 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1155 \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 A -1575 \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 B +1995 \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 C -945 A \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+1785 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-1785 C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{420 \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} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420/cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(960*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*
sin(1/2*d*x+1/2*c)^8-96*2^(1/2)*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(7*B+17*C)*sin(1/2*d*x+1/2*c)^6+112*2^(
1/2)*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*A+6*B+16*C)*sin(1/2*d*x+1/2*c)^4+35*2^(1/2)*(8*A*(a*sin(1/2*d*x
+1/2*c)^2)^(1/2)*a^(1/2)-33*A*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))*a-48*B*(a*si
n(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+45*B*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))*a+1
6*C*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-57*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+
a))*a*C)*sin(1/2*d*x+1/2*c)^2+1155*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))
*a*A-1575*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))*a*B+1995*2^(1/2)*ln(4/co
s(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))*a*C-945*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*
a^(1/2)+1785*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-1785*C*2^(1/2)*(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

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^3\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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