3.1196 \(\int \frac{A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=236 \[ \frac{4 (5 A+14 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (A+3 C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(A+3 C) \sin (c+d x)}{a^2 d (\cos (c+d x)+1) \sec ^{\frac{5}{2}}(c+d x)}-\frac{5 (A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{4 (5 A+14 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac{(A+C) \sin (c+d x)}{3 d \sec ^{\frac{7}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]

[Out]

(4*(5*A + 14*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a^2*d) - (5*(A + 3*C)*Sqrt
[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) - ((A + C)*Sin[c + d*x])/(3*d*(a + a*Co
s[c + d*x])^2*Sec[c + d*x]^(7/2)) - ((A + 3*C)*Sin[c + d*x])/(a^2*d*(1 + Cos[c + d*x])*Sec[c + d*x]^(5/2)) + (
4*(5*A + 14*C)*Sin[c + d*x])/(15*a^2*d*Sec[c + d*x]^(3/2)) - (5*(A + 3*C)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c +
d*x]])

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Rubi [A]  time = 0.461364, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4221, 3042, 2977, 2748, 2635, 2641, 2639} \[ \frac{4 (5 A+14 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (A+3 C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(A+3 C) \sin (c+d x)}{a^2 d (\cos (c+d x)+1) \sec ^{\frac{5}{2}}(c+d x)}-\frac{5 (A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{4 (5 A+14 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac{(A+C) \sin (c+d x)}{3 d \sec ^{\frac{7}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(5*A + 14*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a^2*d) - (5*(A + 3*C)*Sqrt
[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) - ((A + C)*Sin[c + d*x])/(3*d*(a + a*Co
s[c + d*x])^2*Sec[c + d*x]^(7/2)) - ((A + 3*C)*Sin[c + d*x])/(a^2*d*(1 + Cos[c + d*x])*Sec[c + d*x]^(5/2)) + (
4*(5*A + 14*C)*Sin[c + d*x])/(15*a^2*d*Sec[c + d*x]^(3/2)) - (5*(A + 3*C)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c +
d*x]])

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]

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 2977

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

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x)} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\cos ^{\frac{5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx\\ &=-\frac{(A+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac{7}{2}}(c+d x)}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\cos ^{\frac{5}{2}}(c+d x) \left (-\frac{1}{2} a (A+7 C)+\frac{1}{2} a (5 A+11 C) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(A+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac{7}{2}}(c+d x)}-\frac{(A+3 C) \sin (c+d x)}{a^2 d (1+\cos (c+d x)) \sec ^{\frac{5}{2}}(c+d x)}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \cos ^{\frac{3}{2}}(c+d x) \left (-\frac{15}{2} a^2 (A+3 C)+2 a^2 (5 A+14 C) \cos (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{(A+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac{7}{2}}(c+d x)}-\frac{(A+3 C) \sin (c+d x)}{a^2 d (1+\cos (c+d x)) \sec ^{\frac{5}{2}}(c+d x)}-\frac{\left (5 (A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \cos ^{\frac{3}{2}}(c+d x) \, dx}{2 a^2}+\frac{\left (2 (5 A+14 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \cos ^{\frac{5}{2}}(c+d x) \, dx}{3 a^2}\\ &=-\frac{(A+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac{7}{2}}(c+d x)}-\frac{(A+3 C) \sin (c+d x)}{a^2 d (1+\cos (c+d x)) \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 (5 A+14 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (A+3 C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{\left (5 (A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}+\frac{\left (2 (5 A+14 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 a^2}\\ &=\frac{4 (5 A+14 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 a^2 d}-\frac{5 (A+3 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^2 d}-\frac{(A+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac{7}{2}}(c+d x)}-\frac{(A+3 C) \sin (c+d x)}{a^2 d (1+\cos (c+d x)) \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 (5 A+14 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (A+3 C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.86583, size = 813, normalized size = 3.44 \[ -\frac{4 \sqrt{2} A e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \csc \left (\frac{c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) \sec \left (\frac{c}{2}\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (\cos (c+d x) a+a)^2}-\frac{56 \sqrt{2} C e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \csc \left (\frac{c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) \sec \left (\frac{c}{2}\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{15 d (\cos (c+d x) a+a)^2}-\frac{10 A \sqrt{\cos (c+d x)} \csc \left (\frac{c}{2}\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\sec (c+d x)} \sin (c) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (\cos (c+d x) a+a)^2}-\frac{10 C \sqrt{\cos (c+d x)} \csc \left (\frac{c}{2}\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\sec (c+d x)} \sin (c) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (\cos (c+d x) a+a)^2}+\frac{\sqrt{\sec (c+d x)} \left (-\frac{2 \sec \left (\frac{c}{2}\right ) \left (A \sin \left (\frac{d x}{2}\right )+C \sin \left (\frac{d x}{2}\right )\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}-\frac{2 (A+C) \tan \left (\frac{c}{2}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}+\frac{4 \sec \left (\frac{c}{2}\right ) \left (7 A \sin \left (\frac{d x}{2}\right )+13 C \sin \left (\frac{d x}{2}\right )\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}-\frac{(20 \cos (2 c) A+60 A+151 C+73 C \cos (2 c)) \cos (d x) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right )}{10 d}-\frac{8 C \cos (2 d x) \sin (2 c)}{3 d}+\frac{2 C \cos (3 d x) \sin (3 c)}{5 d}+\frac{2 (20 A+73 C) \cos (c) \sin (d x)}{5 d}-\frac{8 C \cos (2 c) \sin (2 d x)}{3 d}+\frac{2 C \cos (3 c) \sin (3 d x)}{5 d}+\frac{4 (7 A+13 C) \tan \left (\frac{c}{2}\right )}{3 d}\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{(\cos (c+d x) a+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]
^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7
/4, -E^((2*I)*(c + d*x))])*Sec[c/2])/(3*d*E^(I*d*x)*(a + a*Cos[c + d*x])^2) - (56*Sqrt[2]*C*Sqrt[E^(I*(c + d*x
))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I
)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c
/2])/(15*d*E^(I*d*x)*(a + a*Cos[c + d*x])^2) - (10*A*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*Elliptic
F[(c + d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*Sin[c])/(3*d*(a + a*Cos[c + d*x])^2) - (10*C*Cos[c/2 + (d*x)/2]^
4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*Sin[c])/(d*(a + a*Cos[c +
d*x])^2) + (Cos[c/2 + (d*x)/2]^4*Sqrt[Sec[c + d*x]]*(-((60*A + 151*C + 20*A*Cos[2*c] + 73*C*Cos[2*c])*Cos[d*x]
*Csc[c/2]*Sec[c/2])/(10*d) - (8*C*Cos[2*d*x]*Sin[2*c])/(3*d) + (2*C*Cos[3*d*x]*Sin[3*c])/(5*d) - (2*Sec[c/2]*S
ec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(3*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(7*A*Sin[(d*x)/2
] + 13*C*Sin[(d*x)/2]))/(3*d) + (2*(20*A + 73*C)*Cos[c]*Sin[d*x])/(5*d) - (8*C*Cos[2*c]*Sin[2*d*x])/(3*d) + (2
*C*Cos[3*c]*Sin[3*d*x])/(5*d) + (4*(7*A + 13*C)*Tan[c/2])/(3*d) - (2*(A + C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3
*d)))/(a + a*Cos[c + d*x])^2

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Maple [A]  time = 1.15, size = 451, normalized size = 1.9 \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)/(a+a*cos(d*x+c))^2/sec(d*x+c)^(5/2),x)

[Out]

1/30*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-96*C*cos(1/2*d*x+1/2*c)^10+352*C*cos(1/2*d*x+1/
2*c)^8+120*A*cos(1/2*d*x+1/2*c)^6+50*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*Elliptic
F(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3+120*A*cos(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2
*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-120*C*cos(1/2*d*x+1/2*c)^6+150*C*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c
)^3+336*C*cos(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/
2*d*x+1/2*c),2^(1/2))-190*A*cos(1/2*d*x+1/2*c)^4-266*C*cos(1/2*d*x+1/2*c)^4+75*A*cos(1/2*d*x+1/2*c)^2+135*C*co
s(1/2*d*x+1/2*c)^2-5*A-5*C)/a^2/cos(1/2*d*x+1/2*c)^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(
1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C \cos \left (d x + c\right )^{2} + A}{{\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sec \left (d x + c\right )^{\frac{5}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(a+a*cos(d*x+c))**2/sec(d*x+c)**(5/2),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}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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