3.1240 \(\int \frac {(A+C \cos ^2(c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=268 \[ -\frac {(15 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(9 A+5 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{10 a d \sqrt {a \cos (c+d x)+a}}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {(13 A+5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{10 a d \sqrt {a \cos (c+d x)+a}}+\frac {(49 A+25 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{10 a d \sqrt {a \cos (c+d x)+a}} \]

[Out]

-1/2*(A+C)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)-1/10*(13*A+5*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/
d/(a+a*cos(d*x+c))^(1/2)+1/10*(9*A+5*C)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)-1/4*(15*A+7*C)*
arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/
2)/a^(3/2)/d*2^(1/2)+1/10*(49*A+25*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.88, antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {4221, 3042, 2984, 12, 2782, 205} \[ -\frac {(15 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(9 A+5 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{10 a d \sqrt {a \cos (c+d x)+a}}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {(13 A+5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{10 a d \sqrt {a \cos (c+d x)+a}}+\frac {(49 A+25 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{10 a d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((15*A + 7*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[c
+ d*x]]*Sqrt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) + ((49*A + 25*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(10*a*d*S
qrt[a + a*Cos[c + d*x]]) - ((13*A + 5*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(10*a*d*Sqrt[a + a*Cos[c + d*x]]) -
((A + C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((9*A + 5*C)*Sec[c + d*x]^(5/2)*S
in[c + d*x])/(10*a*d*Sqrt[a + a*Cos[c + d*x]])

Rule 12

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

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]

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 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 3042

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[(a*(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)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(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, 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 4221

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps

\begin {align*} \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx\\ &=-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} a (9 A+5 C)-a (3 A+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+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {3}{4} a^2 (13 A+5 C)+a^2 (9 A+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 {(13 A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{8} a^3 (49 A+25 C)-\frac {3}{4} a^3 (13 A+5 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{15 a^4}\\ &=\frac {(49 A+25 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(13 A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int -\frac {15 a^4 (15 A+7 C)}{16 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{15 a^5}\\ &=\frac {(49 A+25 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(13 A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {\left ((15 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{4 a}\\ &=\frac {(49 A+25 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(13 A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {\left ((15 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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+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 ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}+\frac {(49 A+25 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(13 A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 7.80, size = 2280, normalized size = 8.51 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Cos[c/2 + (d*x)/2]^3*Sqrt[(1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1)]*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]*((4*C*Sin[c/2
+ (d*x)/2])/(5*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) - ((A + C)*(1 - 2*Sin[c/2 + (d*x)/2]))/(20*(1 + Sin[c/2 +
(d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) + ((A + C)*(1 + 2*Sin[c/2 + (d*x)/2]))/(20*(1 - Sin[c/2 + (d*x)/
2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) + (16*C*(Sin[c/2 + (d*x)/2]/(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2) + (2*Si
n[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]))/15 - ((A + C)*(-105*ArcTan[(1 - 2*Sin[c/2 + (d*x)/2])/Sqr
t[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*S
qrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 - Sin[c/2 + (d*x)/2])))/30 + ((A + C)*(-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*Sin[c/2 + (d*x)/2]^2])/(1 + Sin[c/2 + (d*x)/2])))/30 + ((-A + 7*C)*Csc[c/2 + (d*x)/2]^7*(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*HypergeometricPFQ[{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*Hy
pergeometric2F1[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 + 2266
56*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*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]^16 + 4725*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*S
in[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*Sq
rt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 833760*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)] + 1458000*Ar
cTanh[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)]]*Si
n[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*S
qrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 69120*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)] + 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)))/(1350*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2)*(-1 + 2*Sin[c/2 +
(d*x)/2]^2))))/(d*(a*(1 + Cos[c + d*x]))^(3/2))

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fricas [A]  time = 0.51, size = 214, normalized size = 0.80 \[ \frac {5 \, \sqrt {2} {\left ({\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (49 \, A + 25 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (9 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \, A \cos \left (d x + c\right ) + 4 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{20 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(5*sqrt(2)*((15*A + 7*C)*cos(d*x + c)^4 + 2*(15*A + 7*C)*cos(d*x + c)^3 + (15*A + 7*C)*cos(d*x + c)^2)*sq
rt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((49*A + 25*C)*co
s(d*x + c)^3 + 4*(9*A + 5*C)*cos(d*x + c)^2 - 4*A*cos(d*x + c) + 4*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sq
rt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.67, size = 583, normalized size = 2.18 \[ \frac {\left (75 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+35 C \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+225 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+105 C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+225 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+105 C \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+75 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+35 C \sin \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-49 A \sqrt {2}\, \left (\cos ^{4}\left (d x +c \right )\right )-25 C \sqrt {2}\, \left (\cos ^{4}\left (d x +c \right )\right )+13 A \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )+5 C \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )+40 A \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )+20 C \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )-8 A \sqrt {2}\, \cos \left (d x +c \right )+4 A \sqrt {2}\right ) \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{20 d \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{3} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20/d*(75*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+35*C
*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+225*A*arcsin((-1
+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+105*C*sin(d*x+c)*cos(d*x+c)
^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+225*A*arcsin((-1+cos(d*x+c))/sin(d*x+c
))*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+105*C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+
c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+75*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d*x+c)/(
1+cos(d*x+c)))^(5/2)+35*C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-49*A
*2^(1/2)*cos(d*x+c)^4-25*C*2^(1/2)*cos(d*x+c)^4+13*A*2^(1/2)*cos(d*x+c)^3+5*C*2^(1/2)*cos(d*x+c)^3+40*A*2^(1/2
)*cos(d*x+c)^2+20*C*2^(1/2)*cos(d*x+c)^2-8*A*2^(1/2)*cos(d*x+c)+4*A*2^(1/2))*cos(d*x+c)*sin(d*x+c)^3*(1/cos(d*
x+c))^(7/2)*(a*(1+cos(d*x+c)))^(1/2)/(-1+cos(d*x+c))^2/(1+cos(d*x+c))^3*2^(1/2)/a^2

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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((A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(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 \[ \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2))/(a + a*cos(c + d*x))^(3/2),x)

[Out]

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2))/(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((A+C*cos(d*x+c)**2)*sec(d*x+c)**(7/2)/(a+a*cos(d*x+c))**(3/2),x)

[Out]

Timed out

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