3.1241 \(\int \frac {(A+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=221 \[ \frac {(11 A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(7 A+3 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{6 a d \sqrt {a \cos (c+d x)+a}}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {(19 A+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{6 a d \sqrt {a \cos (c+d x)+a}} \]

[Out]

-1/2*(A+C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)+1/6*(7*A+3*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d/
(a+a*cos(d*x+c))^(1/2)+1/4*(11*A+3*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)-1/6*(19*A+3*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d/(a+a
*cos(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.72, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {(11 A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(7 A+3 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{6 a d \sqrt {a \cos (c+d x)+a}}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {(19 A+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{6 a d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2))/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

((11*A + 3*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) - ((19*A + 3*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(6*a*d*Sqrt
[a + a*Cos[c + d*x]]) - ((A + C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((7*A + 3
*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(6*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 {5}{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 {5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx\\ &=-\frac {(A+C) \sec ^{\frac {3}{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 (7 A+3 C)-2 a A \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(7 A+3 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 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 {1}{4} a^2 (19 A+3 C)+\frac {1}{2} a^2 (7 A+3 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{3 a^3}\\ &=-\frac {(19 A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(7 A+3 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 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 {3 a^3 (11 A+3 C)}{8 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{3 a^4}\\ &=-\frac {(19 A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(7 A+3 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}+\frac {\left ((11 A+3 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 {(19 A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(7 A+3 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}-\frac {\left ((11 A+3 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 {(11 A+3 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 {(19 A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(7 A+3 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 6.85, size = 1055, normalized size = 4.77 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2))/(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])/(3*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) - ((A + C)*(1 - 2*Sin[c/2 + (d*x)/2]))/(12*(1 + Sin[c/2 +
(d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) + ((A + C)*(1 + 2*Sin[c/2 + (d*x)/2]))/(12*(1 - Sin[c/2 + (d*x)/
2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) + (8*C*Sin[c/2 + (d*x)/2])/(3*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) - ((A
+ C)*(5*ArcTan[(1 - 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] + (1 + Sin[c/2 + (d*x)/2])/((1 - S
in[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (3*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 - Sin[c/2 + (d*
x)/2])))/2 + ((A + C)*(5*ArcTan[(1 + 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] + (1 - Sin[c/2 +
(d*x)/2])/((1 + Sin[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (3*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(
1 + Sin[c/2 + (d*x)/2])))/2 + ((A - 7*C)*Csc[c/2 + (d*x)/2]^5*(-12*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2, 2,
7/2}, {1, 9/2}, -(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*Sin[c/2 + (d*x)/2]^8 - 12*Hypergeometri
c2F1[2, 7/2, 9/2, -(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*Sin[c/2 + (d*x)/2]^8*(4 - 7*Sin[c/2 +
(d*x)/2]^2 + 3*Sin[c/2 + (d*x)/2]^4) + 7*Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*(1 - 2*Sin
[c/2 + (d*x)/2]^2)^3*(15 - 20*Sin[c/2 + (d*x)/2]^2 + 8*Sin[c/2 + (d*x)/2]^4)*((3 - 7*Sin[c/2 + (d*x)/2]^2)*Sqr
t[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))] - 3*ArcTanh[Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2
+ (d*x)/2]^2))]]*(1 - 2*Sin[c/2 + (d*x)/2]^2))))/(126*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2))))/(d*(a*(1 + Cos[c
+ d*x]))^(3/2))

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 193, normalized size = 0.87 \[ -\frac {3 \, \sqrt {2} {\left ({\left (11 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (11 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (11 \, A + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 12 \, A \cos \left (d x + c\right ) - 4 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/12*(3*sqrt(2)*((11*A + 3*C)*cos(d*x + c)^3 + 2*(11*A + 3*C)*cos(d*x + c)^2 + (11*A + 3*C)*cos(d*x + c))*sqr
t(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((19*A + 3*C)*cos(
d*x + c)^2 + 12*A*cos(d*x + c) - 4*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x
+ c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^(3/2), x)

________________________________________________________________________________________

maple [B]  time = 0.63, size = 445, normalized size = 2.01 \[ \frac {\left (33 A \sin \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+9 C \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 {3}{2}}+66 A \sin \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \cos \left (d x +c \right )+18 C \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 {3}{2}}+33 A \sin \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+9 C \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 {3}{2}}-19 A \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )-3 C \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )+7 A \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )+3 C \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )+16 A \sqrt {2}\, \cos \left (d x +c \right )-4 A \sqrt {2}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{12 d \left (-1+\cos \left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2} 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)^(5/2)/(a+a*cos(d*x+c))^(3/2),x)

[Out]

1/12/d*(33*A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2+9*C*
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)))^(3/2)+66*A*sin(d*x+c)*(
cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)+18*C*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)))^(3/2)+33*A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(
3/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+9*C*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d*x+c)/(1+cos(d
*x+c)))^(3/2)-19*A*2^(1/2)*cos(d*x+c)^3-3*C*2^(1/2)*cos(d*x+c)^3+7*A*2^(1/2)*cos(d*x+c)^2+3*C*2^(1/2)*cos(d*x+
c)^2+16*A*2^(1/2)*cos(d*x+c)-4*A*2^(1/2))*cos(d*x+c)*sin(d*x+c)*(1/cos(d*x+c))^(5/2)*(a*(1+cos(d*x+c)))^(1/2)/
(-1+cos(d*x+c))/(1+cos(d*x+c))^2*2^(1/2)/a^2

________________________________________________________________________________________

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)^(5/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(5/2))/(a + a*cos(c + d*x))^(3/2),x)

[Out]

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(5/2))/(a + a*cos(c + d*x))^(3/2), x)

________________________________________________________________________________________

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)**(5/2)/(a+a*cos(d*x+c))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________