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

Optimal. Leaf size=236 \[ \frac {2 (21 A+19 C) \sin (c+d x) \cos ^2(c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}-\frac {2 (21 A+29 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 a d}+\frac {4 (147 A+143 C) \sin (c+d x)}{315 d \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} (A+C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 C \sin (c+d x) \cos ^4(c+d x)}{9 d \sqrt {a \cos (c+d x)+a}}-\frac {2 C \sin (c+d x) \cos ^3(c+d x)}{63 d \sqrt {a \cos (c+d x)+a}} \]

[Out]

-(A+C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)+4/315*(147*A+143*C)*si
n(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/105*(21*A+19*C)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-2/63*C*co
s(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/9*C*cos(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-2/315*(2
1*A+29*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/a/d

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Rubi [A]  time = 0.82, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3046, 2983, 2968, 3023, 2751, 2649, 206} \[ \frac {2 (21 A+19 C) \sin (c+d x) \cos ^2(c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}-\frac {2 (21 A+29 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 a d}+\frac {4 (147 A+143 C) \sin (c+d x)}{315 d \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} (A+C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 C \sin (c+d x) \cos ^4(c+d x)}{9 d \sqrt {a \cos (c+d x)+a}}-\frac {2 C \sin (c+d x) \cos ^3(c+d x)}{63 d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*(A + C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d)) + (4*(147*
A + 143*C)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (2*(21*A + 19*C)*Cos[c + d*x]^2*Sin[c + d*x])/(105
*d*Sqrt[a + a*Cos[c + d*x]]) - (2*C*Cos[c + d*x]^3*Sin[c + d*x])/(63*d*Sqrt[a + a*Cos[c + d*x]]) + (2*C*Cos[c
+ d*x]^4*Sin[c + d*x])/(9*d*Sqrt[a + a*Cos[c + d*x]]) - (2*(21*A + 29*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x]
)/(315*a*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 3046

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

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx &=\frac {2 C \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}+\frac {2 \int \frac {\cos ^3(c+d x) \left (\frac {1}{2} a (9 A+8 C)-\frac {1}{2} a C \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{9 a}\\ &=-\frac {2 C \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}+\frac {4 \int \frac {\cos ^2(c+d x) \left (-\frac {3 a^2 C}{2}+\frac {3}{4} a^2 (21 A+19 C) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{63 a^2}\\ &=\frac {2 (21 A+19 C) \cos ^2(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}-\frac {2 C \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}+\frac {8 \int \frac {\cos (c+d x) \left (\frac {3}{2} a^3 (21 A+19 C)-\frac {3}{8} a^3 (21 A+29 C) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{315 a^3}\\ &=\frac {2 (21 A+19 C) \cos ^2(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}-\frac {2 C \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}+\frac {8 \int \frac {\frac {3}{2} a^3 (21 A+19 C) \cos (c+d x)-\frac {3}{8} a^3 (21 A+29 C) \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{315 a^3}\\ &=\frac {2 (21 A+19 C) \cos ^2(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}-\frac {2 C \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (21 A+29 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 a d}+\frac {16 \int \frac {-\frac {3}{16} a^4 (21 A+29 C)+\frac {3}{8} a^4 (147 A+143 C) \cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{945 a^4}\\ &=\frac {4 (147 A+143 C) \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (21 A+19 C) \cos ^2(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}-\frac {2 C \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (21 A+29 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 a d}+(-A-C) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx\\ &=\frac {4 (147 A+143 C) \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (21 A+19 C) \cos ^2(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}-\frac {2 C \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (21 A+29 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 a d}+\frac {(2 (A+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 )}{d}\\ &=-\frac {\sqrt {2} (A+C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {4 (147 A+143 C) \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (21 A+19 C) \cos ^2(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}-\frac {2 C \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (21 A+29 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 a d}\\ \end {align*}

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Mathematica [A]  time = 0.62, size = 121, normalized size = 0.51 \[ \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (2 \sin \left (\frac {1}{2} (c+d x)\right ) (-2 (84 A+131 C) \cos (c+d x)+4 (63 A+92 C) \cos (2 (c+d x))+2436 A-10 C \cos (3 (c+d x))+35 C \cos (4 (c+d x))+2389 C)-2520 (A+C) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{1260 d \sqrt {a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*(-2520*(A + C)*ArcTanh[Sin[(c + d*x)/2]] + 2*(2436*A + 2389*C - 2*(84*A + 131*C)*Cos[c + d*x
] + 4*(63*A + 92*C)*Cos[2*(c + d*x)] - 10*C*Cos[3*(c + d*x)] + 35*C*Cos[4*(c + d*x)])*Sin[(c + d*x)/2]))/(1260
*d*Sqrt[a*(1 + Cos[c + d*x])])

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fricas [A]  time = 0.44, size = 191, normalized size = 0.81 \[ \frac {4 \, {\left (35 \, C \cos \left (d x + c\right )^{4} - 5 \, C \cos \left (d x + c\right )^{3} + 3 \, {\left (21 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (21 \, A + 29 \, C\right )} \cos \left (d x + c\right ) + 273 \, A + 257 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) + \frac {315 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right ) + {\left (A + C\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{630 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/630*(4*(35*C*cos(d*x + c)^4 - 5*C*cos(d*x + c)^3 + 3*(21*A + 19*C)*cos(d*x + c)^2 - (21*A + 29*C)*cos(d*x +
c) + 273*A + 257*C)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c) + 315*sqrt(2)*((A + C)*a*cos(d*x + c) + (A + C)*a)*l
og(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x +
c)^2 + 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c) + a*d)

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giac [A]  time = 5.88, size = 227, normalized size = 0.96 \[ \frac {\frac {315 \, {\left (\sqrt {2} A + \sqrt {2} C\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {a}} + \frac {2 \, {\left (315 \, \sqrt {2} A a^{4} + 315 \, \sqrt {2} C a^{4} + {\left (1050 \, \sqrt {2} A a^{4} + 840 \, \sqrt {2} C a^{4} + {\left (1512 \, \sqrt {2} A a^{4} + 1638 \, \sqrt {2} C a^{4} + {\left (1134 \, \sqrt {2} A a^{4} + 936 \, \sqrt {2} C a^{4} + {\left (357 \, \sqrt {2} A a^{4} + 383 \, \sqrt {2} C a^{4}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {9}{2}}}}{315 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/315*(315*(sqrt(2)*A + 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))
)/sqrt(a) + 2*(315*sqrt(2)*A*a^4 + 315*sqrt(2)*C*a^4 + (1050*sqrt(2)*A*a^4 + 840*sqrt(2)*C*a^4 + (1512*sqrt(2)
*A*a^4 + 1638*sqrt(2)*C*a^4 + (1134*sqrt(2)*A*a^4 + 936*sqrt(2)*C*a^4 + (357*sqrt(2)*A*a^4 + 383*sqrt(2)*C*a^4
)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x
+ 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(9/2))/d

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maple [A]  time = 1.17, size = 340, normalized size = 1.44 \[ \frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (1120 C \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2160 C \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+504 \sqrt {2}\, \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (A +4 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-420 \sqrt {2}\, \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (A +2 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-315 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a A -315 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a C +630 A \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+630 C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{315 a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(1/2),x)

[Out]

1/315*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(1120*C*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)
*sin(1/2*d*x+1/2*c)^8-2160*C*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^6+504*2^(1/2)*a
^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(A+4*C)*sin(1/2*d*x+1/2*c)^4-420*2^(1/2)*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2
)^(1/2)*(A+2*C)*sin(1/2*d*x+1/2*c)^2-315*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-315*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*C+630*A*2^(1/2)*(
a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+630*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(3/2)/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+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^3*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^(1/2),x)

[Out]

int((cos(c + d*x)^3*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^(1/2), 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+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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