3.501 \(\int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec ^{\frac{13}{2}}(c+d x) \, dx\)

Optimal. Leaf size=275 \[ \frac{2 a^2 (12 A+11 B) \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x)}{99 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (168 A+187 B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{693 d \sqrt{a \cos (c+d x)+a}}+\frac{4 a^2 (168 A+187 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{1155 d \sqrt{a \cos (c+d x)+a}}+\frac{16 a^2 (168 A+187 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3465 d \sqrt{a \cos (c+d x)+a}}+\frac{32 a^2 (168 A+187 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{3465 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a A \sin (c+d x) \sec ^{\frac{11}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{11 d} \]

[Out]

(32*a^2*(168*A + 187*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3465*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^2*(168*A +
187*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3465*d*Sqrt[a + a*Cos[c + d*x]]) + (4*a^2*(168*A + 187*B)*Sec[c + d*x
]^(5/2)*Sin[c + d*x])/(1155*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(168*A + 187*B)*Sec[c + d*x]^(7/2)*Sin[c + d*
x])/(693*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(12*A + 11*B)*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(99*d*Sqrt[a + a*
Cos[c + d*x]]) + (2*a*A*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(11/2)*Sin[c + d*x])/(11*d)

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Rubi [A]  time = 0.723155, antiderivative size = 275, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2961, 2975, 2980, 2772, 2771} \[ \frac{2 a^2 (12 A+11 B) \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x)}{99 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (168 A+187 B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{693 d \sqrt{a \cos (c+d x)+a}}+\frac{4 a^2 (168 A+187 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{1155 d \sqrt{a \cos (c+d x)+a}}+\frac{16 a^2 (168 A+187 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3465 d \sqrt{a \cos (c+d x)+a}}+\frac{32 a^2 (168 A+187 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{3465 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a A \sin (c+d x) \sec ^{\frac{11}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(13/2),x]

[Out]

(32*a^2*(168*A + 187*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3465*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^2*(168*A +
187*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3465*d*Sqrt[a + a*Cos[c + d*x]]) + (4*a^2*(168*A + 187*B)*Sec[c + d*x
]^(5/2)*Sin[c + d*x])/(1155*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(168*A + 187*B)*Sec[c + d*x]^(7/2)*Sin[c + d*
x])/(693*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(12*A + 11*B)*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(99*d*Sqrt[a + a*
Cos[c + d*x]]) + (2*a*A*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(11/2)*Sin[c + d*x])/(11*d)

Rule 2961

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Sin[e + f*x])^m*(
c + d*Sin[e + f*x])^n)/(g*Sin[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

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 2980

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((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]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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[n, -1]

Rule 2772

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), 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] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 2771

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[(-2*b^2*Cos[e + f*x])/(f*(b*c + a*d)*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]

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec ^{\frac{13}{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} (A+B \cos (c+d x))}{\cos ^{\frac{13}{2}}(c+d x)} \, dx\\ &=\frac{2 a A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{1}{11} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{1}{2} a (12 A+11 B)+\frac{1}{2} a (8 A+11 B) \cos (c+d x)\right )}{\cos ^{\frac{11}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (12 A+11 B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{99 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{1}{99} \left (a (168 A+187 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{9}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (168 A+187 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (12 A+11 B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{99 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{1}{231} \left (2 a (168 A+187 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{4 a^2 (168 A+187 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (168 A+187 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (12 A+11 B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{99 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{\left (8 a (168 A+187 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{1155}\\ &=\frac{16 a^2 (168 A+187 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3465 d \sqrt{a+a \cos (c+d x)}}+\frac{4 a^2 (168 A+187 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (168 A+187 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (12 A+11 B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{99 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{\left (16 a (168 A+187 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{3465}\\ &=\frac{32 a^2 (168 A+187 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3465 d \sqrt{a+a \cos (c+d x)}}+\frac{16 a^2 (168 A+187 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3465 d \sqrt{a+a \cos (c+d x)}}+\frac{4 a^2 (168 A+187 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (168 A+187 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (12 A+11 B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{99 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}\\ \end{align*}

Mathematica [A]  time = 0.774511, size = 146, normalized size = 0.53 \[ \frac{a \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{11}{2}}(c+d x) \sqrt{a (\cos (c+d x)+1)} ((6342 A+6193 B) \cos (c+d x)+13 (168 A+187 B) \cos (2 (c+d x))+2184 A \cos (3 (c+d x))+336 A \cos (4 (c+d x))+336 A \cos (5 (c+d x))+2478 A+2431 B \cos (3 (c+d x))+374 B \cos (4 (c+d x))+374 B \cos (5 (c+d x))+2057 B)}{3465 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(13/2),x]

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(2478*A + 2057*B + (6342*A + 6193*B)*Cos[c + d*x] + 13*(168*A + 187*B)*Cos[2*(c
+ d*x)] + 2184*A*Cos[3*(c + d*x)] + 2431*B*Cos[3*(c + d*x)] + 336*A*Cos[4*(c + d*x)] + 374*B*Cos[4*(c + d*x)]
+ 336*A*Cos[5*(c + d*x)] + 374*B*Cos[5*(c + d*x)])*Sec[c + d*x]^(11/2)*Tan[(c + d*x)/2])/(3465*d)

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Maple [A]  time = 0.777, size = 161, normalized size = 0.6 \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 2688\,A \left ( \cos \left ( dx+c \right ) \right ) ^{5}+2992\,B \left ( \cos \left ( dx+c \right ) \right ) ^{5}+1344\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1496\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1008\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+1122\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+840\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+935\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+735\,A\cos \left ( dx+c \right ) +385\,B\cos \left ( dx+c \right ) +315\,A \right ) \cos \left ( dx+c \right ) }{3465\,d\sin \left ( dx+c \right ) } \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{13}{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+cos(d*x+c)*a)^(3/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(13/2),x)

[Out]

-2/3465/d*a*(-1+cos(d*x+c))*(2688*A*cos(d*x+c)^5+2992*B*cos(d*x+c)^5+1344*A*cos(d*x+c)^4+1496*B*cos(d*x+c)^4+1
008*A*cos(d*x+c)^3+1122*B*cos(d*x+c)^3+840*A*cos(d*x+c)^2+935*B*cos(d*x+c)^2+735*A*cos(d*x+c)+385*B*cos(d*x+c)
+315*A)*cos(d*x+c)*(1/cos(d*x+c))^(13/2)*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)

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Maxima [B]  time = 2.3038, size = 961, normalized size = 3.49 \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+B*cos(d*x+c))*sec(d*x+c)^(13/2),x, algorithm="maxima")

[Out]

4/3465*(21*(165*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 495*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 1056*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 1254*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(
cos(d*x + c) + 1)^7 + 781*sqrt(2)*a^(3/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 299*sqrt(2)*a^(3/2)*sin(d*x +
c)^11/(cos(d*x + c) + 1)^11 + 46*sqrt(2)*a^(3/2)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13)*A*(sin(d*x + c)^2/(cos
(d*x + c) + 1)^2 + 1)^5/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(13/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(
13/2)*(5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*sin(d*x + c)^6/(cos
(d*x + c) + 1)^6 + 5*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 1)) + 11*(3
15*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 1155*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3
+ 2184*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 2586*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x + c)
+ 1)^7 + 1759*sqrt(2)*a^(3/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 611*sqrt(2)*a^(3/2)*sin(d*x + c)^11/(cos(
d*x + c) + 1)^11 + 94*sqrt(2)*a^(3/2)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13)*B*(sin(d*x + c)^2/(cos(d*x + c) +
1)^2 + 1)^5/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(13/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(13/2)*(5*si
n(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*sin(d*x + c)^6/(cos(d*x + c) +
1)^6 + 5*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 1)))/d

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Fricas [A]  time = 1.48124, size = 402, normalized size = 1.46 \begin{align*} \frac{2 \,{\left (16 \,{\left (168 \, A + 187 \, B\right )} a \cos \left (d x + c\right )^{5} + 8 \,{\left (168 \, A + 187 \, B\right )} a \cos \left (d x + c\right )^{4} + 6 \,{\left (168 \, A + 187 \, B\right )} a \cos \left (d x + c\right )^{3} + 5 \,{\left (168 \, A + 187 \, B\right )} a \cos \left (d x + c\right )^{2} + 35 \,{\left (21 \, A + 11 \, B\right )} a \cos \left (d x + c\right ) + 315 \, A a\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{3465 \,{\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )} \sqrt{\cos \left (d x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(13/2),x, algorithm="fricas")

[Out]

2/3465*(16*(168*A + 187*B)*a*cos(d*x + c)^5 + 8*(168*A + 187*B)*a*cos(d*x + c)^4 + 6*(168*A + 187*B)*a*cos(d*x
+ c)^3 + 5*(168*A + 187*B)*a*cos(d*x + c)^2 + 35*(21*A + 11*B)*a*cos(d*x + c) + 315*A*a)*sqrt(a*cos(d*x + c)
+ a)*sin(d*x + c)/((d*cos(d*x + c)^6 + d*cos(d*x + c)^5)*sqrt(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+B*cos(d*x+c))*sec(d*x+c)**(13/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \cos \left (d x + c\right ) + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \sec \left (d x + c\right )^{\frac{13}{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+B*cos(d*x+c))*sec(d*x+c)^(13/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^(3/2)*sec(d*x + c)^(13/2), x)