3.1320 \(\int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx\)

Optimal. Leaf size=184 \[ \frac {2 a^2 (4 A+6 B+5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (104 A+126 B+175 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a (3 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{35 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d} \]

[Out]

2/7*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(7/2)*sin(d*x+c)/d+2/15*a^2*(4*A+6*B+5*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/
d/(a+a*cos(d*x+c))^(1/2)+2/35*a*(3*A+7*B)*sec(d*x+c)^(5/2)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d+2/105*a^2*(104*
A+126*B+175*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.71, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4221, 3043, 2975, 2980, 2771} \[ \frac {2 a^2 (4 A+6 B+5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (104 A+126 B+175 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a (3 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{35 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d} \]

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]^(9/2),x]

[Out]

(2*a^2*(104*A + 126*B + 175*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(4*A
+ 6*B + 5*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(3*A + 7*B)*Sqrt[a + a*C
os[c + d*x]]*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(35*d) + (2*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(7/2)*Sin[
c + d*x])/(7*d)

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]

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 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 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]

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 {9}{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 {9}{2}}(c+d x)} \, dx\\ &=\frac {2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 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+7 B)+\frac {1}{2} a (2 A+7 C) \cos (c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{7 a}\\ &=\frac {2 a (3 A+7 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 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 {7}{4} a^2 (4 A+6 B+5 C)+\frac {1}{4} a^2 (16 A+14 B+35 C) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{35 a}\\ &=\frac {2 a^2 (4 A+6 B+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (3 A+7 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{105} \left (a (104 A+126 B+175 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\\ &=\frac {2 a^2 (104 A+126 B+175 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (4 A+6 B+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (3 A+7 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 0.77, size = 122, normalized size = 0.66 \[ \frac {a \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a (\cos (c+d x)+1)} ((468 A+462 B+525 C) \cos (c+d x)+2 (52 A+63 B+35 C) \cos (2 (c+d x))+104 A \cos (3 (c+d x))+164 A+126 B \cos (3 (c+d x))+126 B+175 C \cos (3 (c+d x))+70 C)}{210 d} \]

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]^(9/2),x]

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(164*A + 126*B + 70*C + (468*A + 462*B + 525*C)*Cos[c + d*x] + 2*(52*A + 63*B +
35*C)*Cos[2*(c + d*x)] + 104*A*Cos[3*(c + d*x)] + 126*B*Cos[3*(c + d*x)] + 175*C*Cos[3*(c + d*x)])*Sec[c + d*x
]^(7/2)*Tan[(c + d*x)/2])/(210*d)

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fricas [A]  time = 0.46, size = 112, normalized size = 0.61 \[ \frac {2 \, {\left ({\left (104 \, A + 126 \, B + 175 \, C\right )} a \cos \left (d x + c\right )^{3} + {\left (52 \, A + 63 \, B + 35 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (13 \, A + 7 \, B\right )} a \cos \left (d x + c\right ) + 15 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )} \sqrt {\cos \left (d x + c\right )}} \]

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)^(9/2),x, algorithm="fricas")

[Out]

2/105*((104*A + 126*B + 175*C)*a*cos(d*x + c)^3 + (52*A + 63*B + 35*C)*a*cos(d*x + c)^2 + 3*(13*A + 7*B)*a*cos
(d*x + c) + 15*A*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/((d*cos(d*x + c)^4 + d*cos(d*x + c)^3)*sqrt(cos(d*x
+ c)))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)^(9/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.52, size = 139, normalized size = 0.76 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (104 A \left (\cos ^{3}\left (d x +c \right )\right )+126 B \left (\cos ^{3}\left (d x +c \right )\right )+175 C \left (\cos ^{3}\left (d x +c \right )\right )+52 A \left (\cos ^{2}\left (d x +c \right )\right )+63 B \left (\cos ^{2}\left (d x +c \right )\right )+35 C \left (\cos ^{2}\left (d x +c \right )\right )+39 A \cos \left (d x +c \right )+21 B \cos \left (d x +c \right )+15 A \right ) \cos \left (d x +c \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {9}{2}} a}{105 d \sin \left (d x +c \right )} \]

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)^(9/2),x)

[Out]

-2/105/d*(-1+cos(d*x+c))*(104*A*cos(d*x+c)^3+126*B*cos(d*x+c)^3+175*C*cos(d*x+c)^3+52*A*cos(d*x+c)^2+63*B*cos(
d*x+c)^2+35*C*cos(d*x+c)^2+39*A*cos(d*x+c)+21*B*cos(d*x+c)+15*A)*cos(d*x+c)*(a*(1+cos(d*x+c)))^(1/2)*(1/cos(d*
x+c))^(9/2)/sin(d*x+c)*a

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maxima [B]  time = 0.65, size = 788, normalized size = 4.28 \[ \text {result too large to display} \]

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)^(9/2),x, algorithm="maxima")

[Out]

4/105*((105*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 245*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 + 273*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 171*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*
x + c) + 1)^7 + 38*sqrt(2)*a^(3/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*A*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2
+ 1)^3/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(3*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1))
+ 21*(5*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 15*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1
)^3 + 17*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 9*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x + c)
+ 1)^7 + 2*sqrt(2)*a^(3/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*B*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/
((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(3*sin(d*x + c)^2/(c
os(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1)) + 35*(3
*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 11*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15
*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 9*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 +
2*sqrt(2)*a^(3/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*C*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/((sin(d*
x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(3*sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1)))/d

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mupad [B]  time = 7.08, size = 308, normalized size = 1.67 \[ -\frac {\sqrt {\frac {1}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {4\,C\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}{d}+\frac {4\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (4\,A+6\,B+11\,C\right )}{3\,d}-\frac {4\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (52\,A+48\,B+65\,C\right )}{15\,d}-\frac {4\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (104\,A+126\,B+175\,C\right )}{105\,d}\right )}{6\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+6\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )+2\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )+2\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(c + d*x))^(9/2)*(a + a*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)

[Out]

-((1/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((4*C*a*exp((c*7i)/2 + (d*x*7i)/2)*sin((5*c)/2 + (
5*d*x)/2)*(a + a*cos(c + d*x))^(1/2))/d + (4*a*exp((c*7i)/2 + (d*x*7i)/2)*sin(c/2 + (d*x)/2)*(a + a*cos(c + d*
x))^(1/2)*(4*A + 6*B + 11*C))/(3*d) - (4*a*exp((c*7i)/2 + (d*x*7i)/2)*sin((3*c)/2 + (3*d*x)/2)*(a + a*cos(c +
d*x))^(1/2)*(52*A + 48*B + 65*C))/(15*d) - (4*a*exp((c*7i)/2 + (d*x*7i)/2)*sin((7*c)/2 + (7*d*x)/2)*(a + a*cos
(c + d*x))^(1/2)*(104*A + 126*B + 175*C))/(105*d)))/(6*exp((c*7i)/2 + (d*x*7i)/2)*cos(c/2 + (d*x)/2) + 6*exp((
c*7i)/2 + (d*x*7i)/2)*cos((3*c)/2 + (3*d*x)/2) + 2*exp((c*7i)/2 + (d*x*7i)/2)*cos((5*c)/2 + (5*d*x)/2) + 2*exp
((c*7i)/2 + (d*x*7i)/2)*cos((7*c)/2 + (7*d*x)/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**(9/2),x)

[Out]

Timed out

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