3.1212 \(\int (a+a \cos (c+d x))^{3/2} (A+C \cos ^2(c+d x)) \sec ^{\frac{13}{2}}(c+d x) \, dx\)

Optimal. Leaf size=266 \[ \frac{2 a^2 (28 A+33 C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{231 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (112 A+143 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{385 d \sqrt{a \cos (c+d x)+a}}+\frac{8 a^2 (112 A+143 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{1155 d \sqrt{a \cos (c+d x)+a}}+\frac{16 a^2 (112 A+143 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{1155 d \sqrt{a \cos (c+d x)+a}}+\frac{2 A \sin (c+d x) \sec ^{\frac{11}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d}+\frac{2 a A \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{33 d} \]

[Out]

(16*a^2*(112*A + 143*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(1155*d*Sqrt[a + a*Cos[c + d*x]]) + (8*a^2*(112*A + 1
43*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(1155*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(112*A + 143*C)*Sec[c + d*x]
^(5/2)*Sin[c + d*x])/(385*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(28*A + 33*C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/
(231*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*A*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(33*d) + (
2*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 = 0.836181, antiderivative size = 266, normalized size of antiderivative = 1., 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, 3044, 2975, 2980, 2772, 2771} \[ \frac{2 a^2 (28 A+33 C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{231 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (112 A+143 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{385 d \sqrt{a \cos (c+d x)+a}}+\frac{8 a^2 (112 A+143 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{1155 d \sqrt{a \cos (c+d x)+a}}+\frac{16 a^2 (112 A+143 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{1155 d \sqrt{a \cos (c+d x)+a}}+\frac{2 A \sin (c+d x) \sec ^{\frac{11}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d}+\frac{2 a A \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{33 d} \]

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]^(13/2),x]

[Out]

(16*a^2*(112*A + 143*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(1155*d*Sqrt[a + a*Cos[c + d*x]]) + (8*a^2*(112*A + 1
43*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(1155*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(112*A + 143*C)*Sec[c + d*x]
^(5/2)*Sin[c + d*x])/(385*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(28*A + 33*C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/
(231*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*A*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(33*d) + (
2*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(11/2)*Sin[c + d*x])/(11*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 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} \left (A+C \cos ^2(c+d x)\right ) \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} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac{13}{2}}(c+d x)} \, dx\\ &=\frac{2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 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 (6 A+11 C) \cos (c+d x)\right )}{\cos ^{\frac{11}{2}}(c+d x)} \, dx}{11 a}\\ &=\frac{2 a A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{33 d}+\frac{2 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 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{3}{4} a^2 (28 A+33 C)+\frac{9}{4} a^2 (8 A+11 C) \cos (c+d x)\right )}{\cos ^{\frac{9}{2}}(c+d x)} \, dx}{99 a}\\ &=\frac{2 a^2 (28 A+33 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{231 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{33 d}+\frac{2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{1}{77} \left (a (112 A+143 C) \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^2 (112 A+143 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{385 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (28 A+33 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{231 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{33 d}+\frac{2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{1}{385} \left (4 a (112 A+143 C) \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\\ &=\frac{8 a^2 (112 A+143 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (112 A+143 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{385 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (28 A+33 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{231 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{33 d}+\frac{2 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 (112 A+143 C) \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}{1155}\\ &=\frac{16 a^2 (112 A+143 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{1155 d \sqrt{a+a \cos (c+d x)}}+\frac{8 a^2 (112 A+143 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (112 A+143 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{385 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (28 A+33 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{231 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{33 d}+\frac{2 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 = 0.839492, size = 146, normalized size = 0.55 \[ \frac{a \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{11}{2}}(c+d x) \sqrt{a (\cos (c+d x)+1)} ((4228 A+4147 C) \cos (c+d x)+2 (728 A+737 C) \cos (2 (c+d x))+1456 A \cos (3 (c+d x))+224 A \cos (4 (c+d x))+224 A \cos (5 (c+d x))+1652 A+1859 C \cos (3 (c+d x))+286 C \cos (4 (c+d x))+286 C \cos (5 (c+d x))+1188 C)}{2310 d} \]

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]^(13/2),x]

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(1652*A + 1188*C + (4228*A + 4147*C)*Cos[c + d*x] + 2*(728*A + 737*C)*Cos[2*(c +
d*x)] + 1456*A*Cos[3*(c + d*x)] + 1859*C*Cos[3*(c + d*x)] + 224*A*Cos[4*(c + d*x)] + 286*C*Cos[4*(c + d*x)] +
224*A*Cos[5*(c + d*x)] + 286*C*Cos[5*(c + d*x)])*Sec[c + d*x]^(11/2)*Tan[(c + d*x)/2])/(2310*d)

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Maple [A]  time = 0.192, size = 152, normalized size = 0.6 \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 896\,A \left ( \cos \left ( dx+c \right ) \right ) ^{5}+1144\,C \left ( \cos \left ( dx+c \right ) \right ) ^{5}+448\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+572\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+336\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+429\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+280\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+165\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+245\,A\cos \left ( dx+c \right ) +105\,A \right ) \cos \left ( dx+c \right ) }{1155\,d\sin \left ( dx+c \right ) }\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{13}{2}}}} \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)^(13/2),x)

[Out]

-2/1155/d*a*(-1+cos(d*x+c))*(896*A*cos(d*x+c)^5+1144*C*cos(d*x+c)^5+448*A*cos(d*x+c)^4+572*C*cos(d*x+c)^4+336*
A*cos(d*x+c)^3+429*C*cos(d*x+c)^3+280*A*cos(d*x+c)^2+165*C*cos(d*x+c)^2+245*A*cos(d*x+c)+105*A)*cos(d*x+c)*(a*
(1+cos(d*x+c)))^(1/2)*(1/cos(d*x+c))^(13/2)/sin(d*x+c)

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Maxima [B]  time = 1.83423, size = 961, normalized size = 3.61 \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)^(13/2),x, algorithm="maxima")

[Out]

4/1155*(7*(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/(c
os(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)^(1
3/2)*(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*(10
5*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 455*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 +
868*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 962*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x + c) + 1
)^7 + 653*sqrt(2)*a^(3/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 247*sqrt(2)*a^(3/2)*sin(d*x + c)^11/(cos(d*x +
c) + 1)^11 + 38*sqrt(2)*a^(3/2)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13)*C*(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)))/d

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Fricas [A]  time = 1.49977, size = 383, normalized size = 1.44 \begin{align*} \frac{2 \,{\left (8 \,{\left (112 \, A + 143 \, C\right )} a \cos \left (d x + c\right )^{5} + 4 \,{\left (112 \, A + 143 \, C\right )} a \cos \left (d x + c\right )^{4} + 3 \,{\left (112 \, A + 143 \, C\right )} a \cos \left (d x + c\right )^{3} + 5 \,{\left (56 \, A + 33 \, C\right )} a \cos \left (d x + c\right )^{2} + 245 \, A a \cos \left (d x + c\right ) + 105 \, A a\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{1155 \,{\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+C*cos(d*x+c)^2)*sec(d*x+c)^(13/2),x, algorithm="fricas")

[Out]

2/1155*(8*(112*A + 143*C)*a*cos(d*x + c)^5 + 4*(112*A + 143*C)*a*cos(d*x + c)^4 + 3*(112*A + 143*C)*a*cos(d*x
+ c)^3 + 5*(56*A + 33*C)*a*cos(d*x + c)^2 + 245*A*a*cos(d*x + c) + 105*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+C*cos(d*x+c)**2)*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 (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{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+C*cos(d*x+c)^2)*sec(d*x+c)^(13/2),x, algorithm="giac")

[Out]

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