3.1318 \(\int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac{13}{2}}(c+d x) \, dx\)

Optimal. Leaf size=284 \[ \frac{2 a^2 (84 A+110 B+99 C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{693 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (336 A+374 B+429 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{1155 d \sqrt{a \cos (c+d x)+a}}+\frac{8 a^2 (336 A+374 B+429 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3465 d \sqrt{a \cos (c+d x)+a}}+\frac{16 a^2 (336 A+374 B+429 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3465 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a (3 A+11 B) \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{99 d}+\frac{2 A \sin (c+d x) \sec ^{\frac{11}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d} \]

[Out]

(16*a^2*(336*A + 374*B + 429*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3465*d*Sqrt[a + a*Cos[c + d*x]]) + (8*a^2*(3
36*A + 374*B + 429*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3465*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(336*A + 374
*B + 429*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(1155*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(84*A + 110*B + 99*C)*
Sec[c + d*x]^(7/2)*Sin[c + d*x])/(693*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(3*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*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 = 0.90749, antiderivative size = 284, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {4221, 3043, 2975, 2980, 2772, 2771} \[ \frac{2 a^2 (84 A+110 B+99 C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{693 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (336 A+374 B+429 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{1155 d \sqrt{a \cos (c+d x)+a}}+\frac{8 a^2 (336 A+374 B+429 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3465 d \sqrt{a \cos (c+d x)+a}}+\frac{16 a^2 (336 A+374 B+429 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3465 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a (3 A+11 B) \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{99 d}+\frac{2 A \sin (c+d x) \sec ^{\frac{11}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*a^2*(336*A + 374*B + 429*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3465*d*Sqrt[a + a*Cos[c + d*x]]) + (8*a^2*(3
36*A + 374*B + 429*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3465*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(336*A + 374
*B + 429*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(1155*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(84*A + 110*B + 99*C)*
Sec[c + d*x]^(7/2)*Sin[c + d*x])/(693*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(3*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*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 3043

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + 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 -
B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(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, B, 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+B \cos (c+d x)+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+B \cos (c+d x)+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{1}{2} a (3 A+11 B)+\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 (3 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 \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{1}{4} a^2 (84 A+110 B+99 C)+\frac{3}{4} a^2 (24 A+22 B+33 C) \cos (c+d x)\right )}{\cos ^{\frac{9}{2}}(c+d x)} \, dx}{99 a}\\ &=\frac{2 a^2 (84 A+110 B+99 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a (3 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 \cos (c+d x))^{3/2} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{1}{231} \left (a (336 A+374 B+429 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 (336 A+374 B+429 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (84 A+110 B+99 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a (3 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 \cos (c+d x))^{3/2} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{\left (4 a (336 A+374 B+429 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}{1155}\\ &=\frac{8 a^2 (336 A+374 B+429 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3465 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (336 A+374 B+429 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (84 A+110 B+99 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a (3 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 \cos (c+d x))^{3/2} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{\left (8 a (336 A+374 B+429 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}{3465}\\ &=\frac{16 a^2 (336 A+374 B+429 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3465 d \sqrt{a+a \cos (c+d x)}}+\frac{8 a^2 (336 A+374 B+429 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3465 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (336 A+374 B+429 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (84 A+110 B+99 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a (3 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 \cos (c+d x))^{3/2} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{11 d}\\ \end{align*}

Mathematica [A]  time = 1.14666, size = 187, normalized size = 0.66 \[ \frac{a \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{11}{2}}(c+d x) \sqrt{a (\cos (c+d x)+1)} ((12684 A+12386 B+12441 C) \cos (c+d x)+(4368 A+4862 B+4422 C) \cos (2 (c+d x))+4368 A \cos (3 (c+d x))+672 A \cos (4 (c+d x))+672 A \cos (5 (c+d x))+4956 A+4862 B \cos (3 (c+d x))+748 B \cos (4 (c+d x))+748 B \cos (5 (c+d x))+4114 B+5577 C \cos (3 (c+d x))+858 C \cos (4 (c+d x))+858 C \cos (5 (c+d x))+3564 C)}{6930 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(4956*A + 4114*B + 3564*C + (12684*A + 12386*B + 12441*C)*Cos[c + d*x] + (4368*A
+ 4862*B + 4422*C)*Cos[2*(c + d*x)] + 4368*A*Cos[3*(c + d*x)] + 4862*B*Cos[3*(c + d*x)] + 5577*C*Cos[3*(c + d
*x)] + 672*A*Cos[4*(c + d*x)] + 748*B*Cos[4*(c + d*x)] + 858*C*Cos[4*(c + d*x)] + 672*A*Cos[5*(c + d*x)] + 748
*B*Cos[5*(c + d*x)] + 858*C*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 = 0.221, size = 205, normalized size = 0.7 \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}+3432\,C \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}+1716\,C \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}+1287\,C \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}+495\,C \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+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*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+3432*C*cos(d*x+c)^5+1344*A*cos(d*x+c)^4+1
496*B*cos(d*x+c)^4+1716*C*cos(d*x+c)^4+1008*A*cos(d*x+c)^3+1122*B*cos(d*x+c)^3+1287*C*cos(d*x+c)^3+840*A*cos(d
*x+c)^2+935*B*cos(d*x+c)^2+495*C*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 = 1.94937, size = 1438, normalized size = 5.06 \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)+C*cos(d*x+c)^2)*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)) + 33*(105*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)^1
1 + 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/(c
os(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 = 1.55082, size = 443, normalized size = 1.56 \begin{align*} \frac{2 \,{\left (8 \,{\left (336 \, A + 374 \, B + 429 \, C\right )} a \cos \left (d x + c\right )^{5} + 4 \,{\left (336 \, A + 374 \, B + 429 \, C\right )} a \cos \left (d x + c\right )^{4} + 3 \,{\left (336 \, A + 374 \, B + 429 \, C\right )} a \cos \left (d x + c\right )^{3} + 5 \,{\left (168 \, A + 187 \, B + 99 \, C\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)+C*cos(d*x+c)^2)*sec(d*x+c)^(13/2),x, algorithm="fricas")

[Out]

2/3465*(8*(336*A + 374*B + 429*C)*a*cos(d*x + c)^5 + 4*(336*A + 374*B + 429*C)*a*cos(d*x + c)^4 + 3*(336*A + 3
74*B + 429*C)*a*cos(d*x + c)^3 + 5*(168*A + 187*B + 99*C)*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)+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} + 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)+C*cos(d*x+c)^2)*sec(d*x+c)^(13/2),x, algorithm="giac")

[Out]

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