3.1216 \(\int (a+a \cos (c+d x))^{3/2} (A+C \cos ^2(c+d x)) \sec ^{\frac{5}{2}}(c+d x) \, dx\)

Optimal. Leaf size=181 \[ -\frac{a^2 (8 A-3 C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{3 a^{3/2} C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{d}+\frac{2 A \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d}+\frac{2 a A \sin (c+d x) \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}{d} \]

[Out]

(3*a^(3/2)*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/d
- (a^2*(8*A - 3*C)*Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*a*A*Sqrt[a + a*Cos[c +
d*x]]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d + (2*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(
3*d)

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Rubi [A]  time = 0.639784, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {4221, 3044, 2975, 2981, 2774, 216} \[ -\frac{a^2 (8 A-3 C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{3 a^{3/2} C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{d}+\frac{2 A \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d}+\frac{2 a A \sin (c+d x) \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^(3/2)*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/d
- (a^2*(8*A - 3*C)*Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*a*A*Sqrt[a + a*Cos[c +
d*x]]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d + (2*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(
3*d)

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 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 2975

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

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rule 2774

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac{5}{2}}(c+d x) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^{3/2} \left (\frac{3 a A}{2}-\frac{1}{2} a (2 A-3 C) \cos (c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{3 a}\\ &=\frac{2 a A \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{1}{4} a^2 (4 A+3 C)-\frac{1}{4} a^2 (8 A-3 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{3 a}\\ &=-\frac{a^2 (8 A-3 C) \sin (c+d x)}{3 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}+\frac{2 a A \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{1}{2} \left (3 a C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{a^2 (8 A-3 C) \sin (c+d x)}{3 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}+\frac{2 a A \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}-\frac{\left (3 a C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{a}}} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{d}\\ &=\frac{3 a^{3/2} C \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{d}-\frac{a^2 (8 A-3 C) \sin (c+d x)}{3 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}+\frac{2 a A \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.52912, size = 116, normalized size = 0.64 \[ \frac{a \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) \sqrt{a (\cos (c+d x)+1)} \left (\sin \left (\frac{1}{2} (c+d x)\right ) (20 A \cos (c+d x)+4 A+3 C \cos (2 (c+d x))+3 C)+9 \sqrt{2} C \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \cos ^{\frac{3}{2}}(c+d x)\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^(3/2)*(9*Sqrt[2]*C*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]
]*Cos[c + d*x]^(3/2) + (4*A + 3*C + 20*A*Cos[c + d*x] + 3*C*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(6*d)

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Maple [A]  time = 0.203, size = 290, normalized size = 1.6 \begin{align*} -{\frac{a\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3\,d \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 1+\cos \left ( dx+c \right ) \right ) ^{2}} \left ( 9\,C \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+18\,C \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \cos \left ( dx+c \right ) +9\,C \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) +3\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +10\,A\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +2\,A\sin \left ( dx+c \right ) \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{5}{2}}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/d*a*(9*C*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c)
)*cos(d*x+c)^2+18*C*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(
d*x+c))*cos(d*x+c)+9*C*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/c
os(d*x+c))+3*C*cos(d*x+c)^2*sin(d*x+c)+10*A*sin(d*x+c)*cos(d*x+c)+2*A*sin(d*x+c))*cos(d*x+c)*(1/cos(d*x+c))^(5
/2)*(a*(1+cos(d*x+c)))^(1/2)*sin(d*x+c)^2/(-1+cos(d*x+c))/(1+cos(d*x+c))^2

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Maxima [B]  time = 2.30544, size = 1881, normalized size = 10.39 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(6*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*a^(3/2)*sin(1/2*arctan2(si
n(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 2*(3*a*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(2
*d*x + 2*c) + (a*cos(2*d*x + 2*c)^2*sin(d*x + c) + a*sin(2*d*x + 2*c)^2*sin(d*x + c) + 2*a*cos(2*d*x + 2*c)*si
n(d*x + c) + a*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 3*(a*cos(2*d*x + 2*c)
+ a)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((a*cos(d*x + c) - a)*cos(2*d*x + 2*c)^2 + (a*
cos(d*x + c) - a)*sin(2*d*x + 2*c)^2 + 2*(a*cos(d*x + c) - a)*cos(2*d*x + 2*c) + a*cos(d*x + c) - a)*sin(1/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)
+ 1)^(1/4)*sqrt(a) + 3*((a*cos(2*d*x + 2*c)^2 + a*sin(2*d*x + 2*c)^2 + 2*a*cos(2*d*x + 2*c) + a)*arctan2(-(co
s(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2
*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(
2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) + 1) - (
a*cos(2*d*x + 2*c)^2 + a*sin(2*d*x + 2*c)^2 + 2*a*cos(2*d*x + 2*c) + a)*arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d
*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x
+ c) - cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x
+ 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1
)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) - 1) - (a*cos(2*d*x + 2*c)^2 + a*s
in(2*d*x + 2*c)^2 + 2*a*cos(2*d*x + 2*c) + a)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x +
2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2
*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) + (a*cos(2
*d*x + 2*c)^2 + a*sin(2*d*x + 2*c)^2 + 2*a*cos(2*d*x + 2*c) + a)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c
)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c
)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) +
1)) - 1))*sqrt(a))*C/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) + 16*(3*sqrt(2)*a^(3/2
)*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 2*sqrt(2)*a^(3/2)*
sin(d*x + c)^5/(cos(d*x + c) + 1)^5)*A/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(-sin(d*x + c)/(cos(d*x +
c) + 1) + 1)^(5/2)))/d

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Fricas [A]  time = 1.60115, size = 379, normalized size = 2.09 \begin{align*} -\frac{9 \,{\left (C a \cos \left (d x + c\right )^{2} + C a \cos \left (d x + c\right )\right )} \sqrt{a} \arctan \left (\frac{\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{{\left (3 \, C a \cos \left (d x + c\right )^{2} + 10 \, A a \cos \left (d x + c\right ) + 2 \, A a\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{3 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(9*(C*a*cos(d*x + c)^2 + C*a*cos(d*x + c))*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sq
rt(a)*sin(d*x + c))) - (3*C*a*cos(d*x + c)^2 + 10*A*a*cos(d*x + c) + 2*A*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x +
c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2 + d*cos(d*x + c))

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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+a*cos(d*x+c))**(3/2)*(A+C*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{\left (C \cos \left (d x + c\right )^{2} + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{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+a*cos(d*x+c))^(3/2)*(A+C*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)^(3/2)*sec(d*x + c)^(5/2), x)