3.204 \(\int \frac{A+C \cos ^2(c+d x)}{\cos ^{\frac{9}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx\)
Optimal. Leaf size=224 \[ \frac{2 (31 A+35 C) \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}-\frac{2 (43 A+35 C) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{\sqrt{2} (A+C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 A \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{2 A \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}} \]
[Out]
(Sqrt[2]*(A + C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a
]*d) + (2*A*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)*Sqrt[a + a*Cos[c + d*x]]) - (2*A*Sin[c + d*x])/(35*d*Cos[c +
d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + (2*(31*A + 35*C)*Sin[c + d*x])/(105*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Co
s[c + d*x]]) - (2*(43*A + 35*C)*Sin[c + d*x])/(105*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])
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Rubi [A] time = 0.68914, antiderivative size = 224, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used =
{3044, 2984, 12, 2782, 205} \[ \frac{2 (31 A+35 C) \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}-\frac{2 (43 A+35 C) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{\sqrt{2} (A+C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 A \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{2 A \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(9/2)*Sqrt[a + a*Cos[c + d*x]]),x]
[Out]
(Sqrt[2]*(A + C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a
]*d) + (2*A*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)*Sqrt[a + a*Cos[c + d*x]]) - (2*A*Sin[c + d*x])/(35*d*Cos[c +
d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + (2*(31*A + 35*C)*Sin[c + d*x])/(105*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Co
s[c + d*x]]) - (2*(43*A + 35*C)*Sin[c + d*x])/(105*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])
Rule 3044
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[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n +
2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b
*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0
])
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+C \cos ^2(c+d x)}{\cos ^{\frac{9}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx &=\frac{2 A \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 \int \frac{-\frac{a A}{2}+\frac{1}{2} a (6 A+7 C) \cos (c+d x)}{\cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{7 a}\\ &=\frac{2 A \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{2 A \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{4 \int \frac{\frac{1}{4} a^2 (31 A+35 C)-a^2 A \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{35 a^2}\\ &=\frac{2 A \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{2 A \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 (31 A+35 C) \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{8 \int \frac{-\frac{1}{8} a^3 (43 A+35 C)+\frac{1}{4} a^3 (31 A+35 C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{105 a^3}\\ &=\frac{2 A \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{2 A \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 (31 A+35 C) \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{2 (43 A+35 C) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{16 \int \frac{105 a^4 (A+C)}{16 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{105 a^4}\\ &=\frac{2 A \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{2 A \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 (31 A+35 C) \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{2 (43 A+35 C) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+(A+C) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx\\ &=\frac{2 A \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{2 A \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 (31 A+35 C) \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{2 (43 A+35 C) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}-\frac{(2 a (A+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 )}{d}\\ &=\frac{\sqrt{2} (A+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 )}{\sqrt{a} d}+\frac{2 A \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{2 A \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 (31 A+35 C) \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{2 (43 A+35 C) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\\ \end{align*}
Mathematica [C] time = 10.2408, size = 2490, normalized size = 11.12 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(9/2)*Sqrt[a + a*Cos[c + d*x]]),x]
[Out]
(-2*C*Cos[c/2 + (d*x)/2]*Sin[c/2 + (d*x)/2])/(3*d*Sqrt[a*(1 + Cos[c + d*x])]*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2
)) + (2*(A + C)*Cot[c/2 + (d*x)/2]*Csc[c/2 + (d*x)/2]^8*(363825*Sin[c/2 + (d*x)/2]^2 - 4729725*Sin[c/2 + (d*x)
/2]^4 + 26785605*Sin[c/2 + (d*x)/2]^6 - 86790165*Sin[c/2 + (d*x)/2]^8 + 177677808*Sin[c/2 + (d*x)/2]^10 - 2392
83044*Sin[c/2 + (d*x)/2]^12 + 52080*Hypergeometric2F1[2, 11/2, 13/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d
*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 + 560*HypergeometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x)/
2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 + 213120160*Sin[c/2 + (d*x)/2]^14 - 168280*Hypergeom
etric2F1[2, 11/2, 13/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^14 - 2240*Hyper
geometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2
+ (d*x)/2]^14 - 121497024*Sin[c/2 + (d*x)/2]^16 + 212520*Hypergeometric2F1[2, 11/2, 13/2, Sin[c/2 + (d*x)/2]^2
/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^16 + 3360*HypergeometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 1
3/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^16 + 40125184*Sin[c/2 + (d*x)/2]^
18 - 124320*Hypergeometric2F1[2, 11/2, 13/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*
x)/2]^18 - 2240*HypergeometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 +
(d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^18 - 5840384*Sin[c/2 + (d*x)/2]^20 + 28000*Hypergeometric2F1[2, 11/2, 13/2, Si
n[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^20 + 560*HypergeometricPFQ[{2, 2, 2, 2, 1
1/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^20 + 363825*ArcT
anh[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)] - 5336100*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)] + 34636140*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)] - 131060160*
ArcTanh[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)] + 320535600*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)] - 530671680*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)] + 604296000*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)] - 468948480*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)]
+ 237726720*ArcTanh[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)] - 70963200*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 +
(d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^18*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 9461760*ArcTan
h[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^20*Sqrt[Sin[c/2 + (d*x)/2]^2/(-
1 + 2*Sin[c/2 + (d*x)/2]^2)] - 1120*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 11/2}, {1, 1, 13/2}, Sin[c/
2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12*(-6 + 5*Sin[c/2 + (d*x)/2]^2) + 280*Cos[(c
+ d*x)/2]^4*HypergeometricPFQ[{2, 2, 11/2}, {1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Si
n[c/2 + (d*x)/2]^12*(103 - 164*Sin[c/2 + (d*x)/2]^2 + 70*Sin[c/2 + (d*x)/2]^4)))/(40425*d*Sqrt[a*(1 + Cos[c +
d*x])]*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(9/2)*(-1 + 2*Sin[c/2 + (d*x)/2]^2)) + (2*C*Cos[c/2 + (d*x)/2]*((5*Sin[c/2
+ (d*x)/2])/(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2) + 2*((3*Sin[c/2 + (d*x)/2])/(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)
+ 4*(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]))))/(105*d*Sqrt[a*(1 + Cos[c + d*x])])
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Maple [B] time = 0.123, size = 554, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(a+a*cos(d*x+c))^(1/2),x)
[Out]
-1/105/d*(a*(1+cos(d*x+c)))^(1/2)*sin(d*x+c)^4*(105*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))
/sin(d*x+c))*2^(1/2)*cos(d*x+c)^4+105*C*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2
^(1/2)*cos(d*x+c)^4+315*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x
+c)^3+315*C*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)^3+315*A*(c
os(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)^2+315*C*(cos(d*x+c)/(1+c
os(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)^2+105*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5
/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)+105*C*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+c
os(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)+86*A*sin(d*x+c)*cos(d*x+c)^3+70*C*sin(d*x+c)*cos(d*x+c)^3-62*A*sin(d
*x+c)*cos(d*x+c)^2-70*C*sin(d*x+c)*cos(d*x+c)^2+6*A*sin(d*x+c)*cos(d*x+c)-30*A*sin(d*x+c))/a/(-1+cos(d*x+c))^2
/(1+cos(d*x+c))^3/cos(d*x+c)^(7/2)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 2.03486, size = 532, normalized size = 2.38 \begin{align*} -\frac{2 \,{\left ({\left (43 \, A + 35 \, C\right )} \cos \left (d x + c\right )^{3} -{\left (31 \, A + 35 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) - 15 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - \frac{105 \, \sqrt{2}{\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{5} +{\left (A + C\right )} a \cos \left (d x + c\right )^{4}\right )} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \,{\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt{a}}\right )}{\sqrt{a}}}{105 \,{\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
-1/105*(2*((43*A + 35*C)*cos(d*x + c)^3 - (31*A + 35*C)*cos(d*x + c)^2 + 3*A*cos(d*x + c) - 15*A)*sqrt(a*cos(d
*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) - 105*sqrt(2)*((A + C)*a*cos(d*x + c)^5 + (A + C)*a*cos(d*x + c)^
4)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c)/((cos(d*x + c)^2 + cos(d*x + c)
)*sqrt(a)))/sqrt(a))/(a*d*cos(d*x + c)^5 + a*d*cos(d*x + c)^4)
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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+C*cos(d*x+c)**2)/cos(d*x+c)**(9/2)/(a+a*cos(d*x+c))**(1/2),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} + A}{\sqrt{a \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )^{\frac{9}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + A)/(sqrt(a*cos(d*x + c) + a)*cos(d*x + c)^(9/2)), x)