3.516 \(\int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=263 \[ -\frac{(15 A-11 B+7 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(39 A-35 B+15 C) \sin (c+d x)}{30 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{(9 A-5 B+5 C) \sin (c+d x)}{10 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}-\frac{(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}+\frac{(147 A-95 B+75 C) \sin (c+d x)}{30 a d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}} \]

[Out]

-((15*A - 11*B + 7*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2
*Sqrt[2]*a^(3/2)*d) - ((A - B + C)*Sin[c + d*x])/(2*d*Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(3/2)) + ((9*A -
5*B + 5*C)*Sin[c + d*x])/(10*a*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) - ((39*A - 35*B + 15*C)*Sin[c +
d*x])/(30*a*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + ((147*A - 95*B + 75*C)*Sin[c + d*x])/(30*a*d*Sqr
t[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.820798, antiderivative size = 263, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {3041, 2984, 12, 2782, 205} \[ -\frac{(15 A-11 B+7 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(39 A-35 B+15 C) \sin (c+d x)}{30 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{(9 A-5 B+5 C) \sin (c+d x)}{10 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}-\frac{(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}+\frac{(147 A-95 B+75 C) \sin (c+d x)}{30 a d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + a*Cos[c + d*x])^(3/2)),x]

[Out]

-((15*A - 11*B + 7*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2
*Sqrt[2]*a^(3/2)*d) - ((A - B + C)*Sin[c + d*x])/(2*d*Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(3/2)) + ((9*A -
5*B + 5*C)*Sin[c + d*x])/(10*a*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) - ((39*A - 35*B + 15*C)*Sin[c +
d*x])/(30*a*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + ((147*A - 95*B + 75*C)*Sin[c + d*x])/(30*a*d*Sqr
t[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])

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)]

Rule 2984

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*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] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2782

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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx &=-\frac{(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\frac{1}{2} a (9 A-5 B+5 C)-a (3 A-3 B+C) \cos (c+d x)}{\cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(9 A-5 B+5 C) \sin (c+d x)}{10 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{-\frac{1}{4} a^2 (39 A-35 B+15 C)+a^2 (9 A-5 B+5 C) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{5 a^3}\\ &=-\frac{(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(9 A-5 B+5 C) \sin (c+d x)}{10 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(39 A-35 B+15 C) \sin (c+d x)}{30 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 \int \frac{\frac{1}{8} a^3 (147 A-95 B+75 C)-\frac{1}{4} a^3 (39 A-35 B+15 C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{15 a^4}\\ &=-\frac{(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(9 A-5 B+5 C) \sin (c+d x)}{10 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(39 A-35 B+15 C) \sin (c+d x)}{30 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{(147 A-95 B+75 C) \sin (c+d x)}{30 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{4 \int -\frac{15 a^4 (15 A-11 B+7 C)}{16 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{15 a^5}\\ &=-\frac{(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(9 A-5 B+5 C) \sin (c+d x)}{10 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(39 A-35 B+15 C) \sin (c+d x)}{30 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{(147 A-95 B+75 C) \sin (c+d x)}{30 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}-\frac{(15 A-11 B+7 C) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{4 a}\\ &=-\frac{(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(9 A-5 B+5 C) \sin (c+d x)}{10 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(39 A-35 B+15 C) \sin (c+d x)}{30 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{(147 A-95 B+75 C) \sin (c+d x)}{30 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{(15 A-11 B+7 C) \operatorname{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 )}{2 d}\\ &=-\frac{(15 A-11 B+7 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(9 A-5 B+5 C) \sin (c+d x)}{10 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(39 A-35 B+15 C) \sin (c+d x)}{30 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{(147 A-95 B+75 C) \sin (c+d x)}{30 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 8.11606, size = 2437, normalized size = 9.27 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + a*Cos[c + d*x])^(3/2)),x]

[Out]

(8*C*Cos[c/2 + (d*x)/2]^3*Sin[c/2 + (d*x)/2])/(5*d*(a*(1 + Cos[c + d*x]))^(3/2)*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(
5/2)) - ((A - B + C)*Cos[c/2 + (d*x)/2]^3*(1 - 2*Sin[c/2 + (d*x)/2]))/(10*d*(a*(1 + Cos[c + d*x]))^(3/2)*(1 +
Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) + ((A - B + C)*Cos[c/2 + (d*x)/2]^3*(1 + 2*Sin[c/2 + (
d*x)/2]))/(10*d*(a*(1 + Cos[c + d*x]))^(3/2)*(1 - Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) + (3
2*C*Cos[c/2 + (d*x)/2]^3*(Sin[c/2 + (d*x)/2]/(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2) + (2*Sin[c/2 + (d*x)/2])/Sqrt[
1 - 2*Sin[c/2 + (d*x)/2]^2]))/(15*d*(a*(1 + Cos[c + d*x]))^(3/2)) + ((A - B + C)*Cos[c/2 + (d*x)/2]^3*(105*Arc
Tan[(1 - 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] - (4 + 3*Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 +
(d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) + (19 + 29*Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 + (d*x)/2])*Sqrt[1
- 2*Sin[c/2 + (d*x)/2]^2]) + (67*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 - Sin[c/2 + (d*x)/2])))/(15*d*(a*(1 + Co
s[c + d*x]))^(3/2)) - ((A - B + C)*Cos[c/2 + (d*x)/2]^3*(105*ArcTan[(1 + 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[
c/2 + (d*x)/2]^2]] - (4 - 3*Sin[c/2 + (d*x)/2])/((1 + Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2))
+ (19 - 29*Sin[c/2 + (d*x)/2])/((1 + Sin[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (67*Sqrt[1 - 2*Si
n[c/2 + (d*x)/2]^2])/(1 + Sin[c/2 + (d*x)/2])))/(15*d*(a*(1 + Cos[c + d*x]))^(3/2)) + ((-A - 3*B + 7*C)*Cot[c/
2 + (d*x)/2]^3*Csc[c/2 + (d*x)/2]^4*(4725*Sin[c/2 + (d*x)/2]^2 - 48825*Sin[c/2 + (d*x)/2]^4 + 210105*Sin[c/2 +
(d*x)/2]^6 - 486630*Sin[c/2 + (d*x)/2]^8 + 655812*Sin[c/2 + (d*x)/2]^10 - 710*Hypergeometric2F1[2, 9/2, 11/2,
Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^10 - 40*Cos[(c + d*x)/2]^6*Hypergeomet
ricPFQ[{2, 2, 2, 9/2}, {1, 1, 11/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^10
- 518760*Sin[c/2 + (d*x)/2]^12 + 1770*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 +
(d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 + 226656*Sin[c/2 + (d*x)/2]^14 - 1500*Hypergeometric2F1[2, 9/2, 11/2, Sin[c
/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^14 - 42048*Sin[c/2 + (d*x)/2]^16 + 440*Hyper
geometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^16 + 4725*Ar
cTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d
*x)/2]^2)] - 56700*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^2*Sqrt
[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 291060*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/
2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^4*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 833760*ArcTa
nh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^6*Sqrt[Sin[c/2 + (d*x)/2]^2/(-
1 + 2*Sin[c/2 + (d*x)/2]^2)] + 1458000*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c
/2 + (d*x)/2]^8*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 1598400*ArcTanh[Sqrt[Sin[c/2 + (d*x
)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^10*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)
/2]^2)] + 1080000*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^12*Sqrt
[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 414720*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/
2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^14*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 69120*ArcTa
nh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^16*Sqrt[Sin[c/2 + (d*x)/2]^2/(
-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 60*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2, 2, 9/2}, {1, 11/2}, Sin[c/2 + (d*x
)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^10*(-5 + 4*Sin[c/2 + (d*x)/2]^2)))/(675*d*(a*(1 + Cos
[c + d*x]))^(3/2)*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2)*(-1 + 2*Sin[c/2 + (d*x)/2]^2))

________________________________________________________________________________________

Maple [B]  time = 0.127, size = 683, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(3/2),x)

[Out]

-1/60/d*sin(d*x+c)*(a*(1+cos(d*x+c)))^(1/2)*(225*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcsin((-1+cos(d*x+c))/si
n(d*x+c))*2^(1/2)*sin(d*x+c)*cos(d*x+c)^3-165*2^(1/2)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*B*arcsin((-
1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3+105*C*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcsin((-1+cos(d*x+c))/sin(d*x+
c))*2^(1/2)*sin(d*x+c)*cos(d*x+c)^3+450*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))
*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2-330*2^(1/2)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*B*arcsin((-1+cos(d*x
+c))/sin(d*x+c))*cos(d*x+c)^2+210*C*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/
2)*sin(d*x+c)*cos(d*x+c)^2+225*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*
sin(d*x+c)*cos(d*x+c)-165*2^(1/2)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*B*arcsin((-1+cos(d*x+c))/sin(d*
x+c))*cos(d*x+c)+105*C*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*sin(d*x+c)
*cos(d*x+c)-294*A*cos(d*x+c)^4+190*B*cos(d*x+c)^4-150*C*cos(d*x+c)^4+78*A*cos(d*x+c)^3-70*B*cos(d*x+c)^3+30*C*
cos(d*x+c)^3+240*A*cos(d*x+c)^2-160*B*cos(d*x+c)^2+120*C*cos(d*x+c)^2-48*A*cos(d*x+c)+40*B*cos(d*x+c)+24*A)/a^
2/(-1+cos(d*x+c))/(1+cos(d*x+c))^2/cos(d*x+c)^(5/2)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.14684, size = 684, normalized size = 2.6 \begin{align*} -\frac{15 \, \sqrt{2}{\left ({\left (15 \, A - 11 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (15 \, A - 11 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (15 \, A - 11 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \,{\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \,{\left ({\left (147 \, A - 95 \, B + 75 \, C\right )} \cos \left (d x + c\right )^{3} + 12 \,{\left (9 \, A - 5 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left (3 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 12 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{60 \,{\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/60*(15*sqrt(2)*((15*A - 11*B + 7*C)*cos(d*x + c)^5 + 2*(15*A - 11*B + 7*C)*cos(d*x + c)^4 + (15*A - 11*B +
7*C)*cos(d*x + c)^3)*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x +
c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))) - 2*((147*A - 95*B + 75*C)*cos(d*x + c)^3 + 12*(9*A - 5*B + 5*C)*cos(d
*x + c)^2 - 4*(3*A - 5*B)*cos(d*x + c) + 12*A)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^2*
d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(7/2)/(a+a*cos(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^(3/2)*cos(d*x + c)^(7/2)), x)