3.511 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac{13}{2}}(c+d x) \, dx\)

Optimal. Leaf size=275 \[ \frac{2 a^2 (14 A+11 B) \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{99 d}+\frac{2 a^3 (194 A+209 B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{693 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^3 (710 A+803 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{1155 d \sqrt{a \cos (c+d x)+a}}+\frac{8 a^3 (710 A+803 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3465 d \sqrt{a \cos (c+d x)+a}}+\frac{16 a^3 (710 A+803 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{3465 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a A \sin (c+d x) \sec ^{\frac{11}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d} \]

[Out]

(16*a^3*(710*A + 803*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3465*d*Sqrt[a + a*Cos[c + d*x]]) + (8*a^3*(710*A + 8
03*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3465*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(710*A + 803*B)*Sec[c + d*x]
^(5/2)*Sin[c + d*x])/(1155*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(194*A + 209*B)*Sec[c + d*x]^(7/2)*Sin[c + d*x
])/(693*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(14*A + 11*B)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(9/2)*Sin[c +
d*x])/(99*d) + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(11/2)*Sin[c + d*x])/(11*d)

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Rubi [A]  time = 0.848269, 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} \[ \frac{2 a^2 (14 A+11 B) \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{99 d}+\frac{2 a^3 (194 A+209 B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{693 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^3 (710 A+803 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{1155 d \sqrt{a \cos (c+d x)+a}}+\frac{8 a^3 (710 A+803 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3465 d \sqrt{a \cos (c+d x)+a}}+\frac{16 a^3 (710 A+803 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{3465 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a A \sin (c+d x) \sec ^{\frac{11}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(13/2),x]

[Out]

(16*a^3*(710*A + 803*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3465*d*Sqrt[a + a*Cos[c + d*x]]) + (8*a^3*(710*A + 8
03*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3465*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(710*A + 803*B)*Sec[c + d*x]
^(5/2)*Sin[c + d*x])/(1155*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(194*A + 209*B)*Sec[c + d*x]^(7/2)*Sin[c + d*x
])/(693*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(14*A + 11*B)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(9/2)*Sin[c +
d*x])/(99*d) + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*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))^{5/2} (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))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac{13}{2}}(c+d x)} \, dx\\ &=\frac{2 a A (a+a \cos (c+d x))^{3/2} \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{(a+a \cos (c+d x))^{3/2} \left (\frac{1}{2} a (14 A+11 B)+\frac{1}{2} a (6 A+11 B) \cos (c+d x)\right )}{\cos ^{\frac{11}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (14 A+11 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{1}{99} \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 (194 A+209 B)+\frac{3}{4} a^2 (46 A+55 B) \cos (c+d x)\right )}{\cos ^{\frac{9}{2}}(c+d x)} \, dx\\ &=\frac{2 a^3 (194 A+209 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 (14 A+11 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{1}{231} \left (a^2 (710 A+803 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{2 a^3 (710 A+803 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (194 A+209 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 (14 A+11 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{\left (4 a^2 (710 A+803 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{8 a^3 (710 A+803 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3465 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (710 A+803 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (194 A+209 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 (14 A+11 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{\left (8 a^2 (710 A+803 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{16 a^3 (710 A+803 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3465 d \sqrt{a+a \cos (c+d x)}}+\frac{8 a^3 (710 A+803 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3465 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (710 A+803 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (194 A+209 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 (14 A+11 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}\\ \end{align*}

Mathematica [A]  time = 1.27922, size = 147, normalized size = 0.53 \[ \frac{a^2 \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{11}{2}}(c+d x) \sqrt{a (\cos (c+d x)+1)} ((25070 A+24827 B) \cos (c+d x)+(9230 A+9284 B) \cos (2 (c+d x))+9230 A \cos (3 (c+d x))+1420 A \cos (4 (c+d x))+1420 A \cos (5 (c+d x))+9070 A+10439 B \cos (3 (c+d x))+1606 B \cos (4 (c+d x))+1606 B \cos (5 (c+d x))+7678 B)}{6930 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(13/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(9070*A + 7678*B + (25070*A + 24827*B)*Cos[c + d*x] + (9230*A + 9284*B)*Cos[2*
(c + d*x)] + 9230*A*Cos[3*(c + d*x)] + 10439*B*Cos[3*(c + d*x)] + 1420*A*Cos[4*(c + d*x)] + 1606*B*Cos[4*(c +
d*x)] + 1420*A*Cos[5*(c + d*x)] + 1606*B*Cos[5*(c + d*x)])*Sec[c + d*x]^(11/2)*Tan[(c + d*x)/2])/(6930*d)

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Maple [A]  time = 0.674, size = 163, normalized size = 0.6 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 5680\,A \left ( \cos \left ( dx+c \right ) \right ) ^{5}+6424\,B \left ( \cos \left ( dx+c \right ) \right ) ^{5}+2840\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+3212\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+2130\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+2409\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+1775\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1430\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1120\,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)^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(13/2),x)

[Out]

-2/3465/d*a^2*(-1+cos(d*x+c))*(5680*A*cos(d*x+c)^5+6424*B*cos(d*x+c)^5+2840*A*cos(d*x+c)^4+3212*B*cos(d*x+c)^4
+2130*A*cos(d*x+c)^3+2409*B*cos(d*x+c)^3+1775*A*cos(d*x+c)^2+1430*B*cos(d*x+c)^2+1120*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 = 2.4238, size = 907, normalized size = 3.3 \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))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(13/2),x, algorithm="maxima")

[Out]

8/3465*(5*(693*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 2310*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 4620*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 5478*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(
cos(d*x + c) + 1)^7 + 3575*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 1300*sqrt(2)*a^(5/2)*sin(d*x
+ c)^11/(cos(d*x + c) + 1)^11 + 200*sqrt(2)*a^(5/2)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13)*A*(sin(d*x + c)^2/(
cos(d*x + c) + 1)^2 + 1)^4/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(13/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1
)^(13/2)*(4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*sin(d*x + c)^6/(co
s(d*x + c) + 1)^6 + sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 1)) + 11*(315*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x
+ c) + 1) - 1260*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 2394*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 - 2736*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 1859*sqrt(2)*a^(5/2)*sin(d*x +
c)^9/(cos(d*x + c) + 1)^9 - 676*sqrt(2)*a^(5/2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 104*sqrt(2)*a^(5/2)*si
n(d*x + c)^13/(cos(d*x + c) + 1)^13)*B*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^4/((sin(d*x + c)/(cos(d*x + c
) + 1) + 1)^(13/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(13/2)*(4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*si
n(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + sin(d*x + c)^8/(cos(d*x + c) + 1)^
8 + 1)))/d

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Fricas [A]  time = 1.48055, size = 417, normalized size = 1.52 \begin{align*} \frac{2 \,{\left (8 \,{\left (710 \, A + 803 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} + 4 \,{\left (710 \, A + 803 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \,{\left (710 \, A + 803 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 5 \,{\left (355 \, A + 286 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 35 \,{\left (32 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right ) + 315 \, A a^{2}\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))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(13/2),x, algorithm="fricas")

[Out]

2/3465*(8*(710*A + 803*B)*a^2*cos(d*x + c)^5 + 4*(710*A + 803*B)*a^2*cos(d*x + c)^4 + 3*(710*A + 803*B)*a^2*co
s(d*x + c)^3 + 5*(355*A + 286*B)*a^2*cos(d*x + c)^2 + 35*(32*A + 11*B)*a^2*cos(d*x + c) + 315*A*a^2)*sqrt(a*co
s(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))**(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)**(13/2),x)

[Out]

Timed out

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Giac [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))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(13/2),x, algorithm="giac")

[Out]

Timed out