\(\int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx\) [1311]

Optimal result

Integrand size = 45, antiderivative size = 178 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {4 a (24 A+28 B+35 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (24 A+28 B+35 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (A+7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \]

[Out]

2/105*a*(24*A+28*B+35*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/35*a*(A+7*B)*sec(d*x+c)^(5/2)*
sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/7*A*sec(d*x+c)^(7/2)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d+4/105*a*(24*A+2
8*B+35*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4306, 3122, 3059, 2851, 2850} \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {2 a (24 A+28 B+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {4 a (24 A+28 B+35 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a (A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{35 d \sqrt {a \cos (c+d x)+a}}+\frac {2 A \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d} \]

[In]

Int[Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(9/2),x]

[Out]

(4*a*(24*A + 28*B + 35*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(24*A + 28*
B + 35*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(A + 7*B)*Sec[c + d*x]^(5/2
)*Sin[c + d*x])/(35*d*Sqrt[a + a*Cos[c + d*x]]) + (2*A*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(7/2)*Sin[c + d*x
])/(7*d)

Rule 2850

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 2851

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 3059

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 3122

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 4306

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*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)} \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 \sqrt {a+a \cos (c+d x)} \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 {\sqrt {a+a \cos (c+d x)} \left (\frac {1}{2} a (A+7 B)+\frac {1}{2} a (4 A+7 C) \cos (c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{7 a} \\ & = \frac {2 a (A+7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{35} \left ((24 A+28 B+35 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 {2 a (24 A+28 B+35 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (A+7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{105} \left (2 (24 A+28 B+35 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 {4 a (24 A+28 B+35 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (24 A+28 B+35 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (A+7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.68 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} (54 A+28 B+35 C+3 (36 A+42 B+35 C) \cos (c+d x)+(24 A+28 B+35 C) \cos (2 (c+d x))+24 A \cos (3 (c+d x))+28 B \cos (3 (c+d x))+35 C \cos (3 (c+d x))) \sec ^{\frac {7}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{105 d} \]

[In]

Integrate[Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(9/2),x]

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*(54*A + 28*B + 35*C + 3*(36*A + 42*B + 35*C)*Cos[c + d*x] + (24*A + 28*B + 35*C)*C
os[2*(c + d*x)] + 24*A*Cos[3*(c + d*x)] + 28*B*Cos[3*(c + d*x)] + 35*C*Cos[3*(c + d*x)])*Sec[c + d*x]^(7/2)*Ta
n[(c + d*x)/2])/(105*d)

Maple [A] (verified)

Time = 1.73 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.68

method	result	size
default	\(-\frac {2 \left (\cos \left (d x +c \right )-1\right ) \left (\left (48 \left (\cos ^{3}\left (d x +c \right )\right )+24 \left (\cos ^{2}\left (d x +c \right )\right )+18 \cos \left (d x +c \right )+15\right ) A +\cos \left (d x +c \right ) \left (56 \left (\cos ^{2}\left (d x +c \right )\right )+28 \cos \left (d x +c \right )+21\right ) B +\left (\cos ^{2}\left (d x +c \right )\right ) \left (70 \cos \left (d x +c \right )+35\right ) C \right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\sec ^{\frac {9}{2}}\left (d x +c \right )\right ) \cot \left (d x +c \right )}{105 d}\)	\(121\)
parts	\(-\frac {2 A \left (\sec ^{\frac {9}{2}}\left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (16 \left (\cos ^{4}\left (d x +c \right )\right )-8 \left (\cos ^{3}\left (d x +c \right )\right )-2 \left (\cos ^{2}\left (d x +c \right )\right )-\cos \left (d x +c \right )-5\right ) \cot \left (d x +c \right )}{35 d}+\frac {2 B \sin \left (d x +c \right ) \left (8 \left (\cos ^{2}\left (d x +c \right )\right )+4 \cos \left (d x +c \right )+3\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\sec ^{\frac {9}{2}}\left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )}{15 d \left (1+\cos \left (d x +c \right )\right )}+\frac {2 C \sin \left (d x +c \right ) \left (2 \cos \left (d x +c \right )+1\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\sec ^{\frac {9}{2}}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{3 d \left (1+\cos \left (d x +c \right )\right )}\)	\(204\)

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2)*(a+a*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/105/d*(cos(d*x+c)-1)*((48*cos(d*x+c)^3+24*cos(d*x+c)^2+18*cos(d*x+c)+15)*A+cos(d*x+c)*(56*cos(d*x+c)^2+28*c
os(d*x+c)+21)*B+cos(d*x+c)^2*(70*cos(d*x+c)+35)*C)*((1+cos(d*x+c))*a)^(1/2)*sec(d*x+c)^(9/2)*cot(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.61 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {2 \, {\left (2 \, {\left (24 \, A + 28 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (24 \, A + 28 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (6 \, A + 7 \, B\right )} \cos \left (d x + c\right ) + 15 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )} \sqrt {\cos \left (d x + c\right )}} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/105*(2*(24*A + 28*B + 35*C)*cos(d*x + c)^3 + (24*A + 28*B + 35*C)*cos(d*x + c)^2 + 3*(6*A + 7*B)*cos(d*x + c
) + 15*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)))

Sympy [F(-1)]

Timed out. \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(9/2)*(a+a*cos(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 848 vs. \(2 (154) = 308\).

Time = 0.57 (sec) , antiderivative size = 848, normalized size of antiderivative = 4.76 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\text {Too large to display} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

2/105*(3*A*(35*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 70*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 + 84*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 58*sqrt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*
x + c) + 1)^7 + 9*sqrt(2)*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
1)^4/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(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/(cos(d*x + c) + 1)^6 + sin(
d*x + c)^8/(cos(d*x + c) + 1)^8 + 1)) + 7*B*(15*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 40*sqrt(2)*s
qrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 42*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 24*sqrt(
2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 7*sqrt(2)*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*(sin(d
*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^4/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c
) + 1) + 1)^(9/2)*(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/(cos(d*x + c) + 1)^6 + sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 1)) + 35*C*(3*sqrt(2)*sqrt(a)*sin(d*x + c)/
(cos(d*x + c) + 1) - 10*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 12*sqrt(2)*sqrt(a)*sin(d*x + c)^
5/(cos(d*x + c) + 1)^5 - 6*sqrt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + sqrt(2)*sqrt(a)*sin(d*x + c)^
9/(cos(d*x + c) + 1)^9)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^4/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/
2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(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/(cos(d*x + c) + 1)^6 + sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 1)))/d

Giac [F(-1)]

Timed out. \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 6.85 (sec) , antiderivative size = 465, normalized size of antiderivative = 2.61 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\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 {\sqrt {a+a\,\left (\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}\right )}\,\left (96\,A+112\,B+140\,C\right )\,1{}\mathrm {i}}{105\,d}-\frac {{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,\sqrt {a+a\,\left (\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}\right )}\,\left (280\,B+140\,C\right )\,1{}\mathrm {i}}{105\,d}+\frac {{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,\sqrt {a+a\,\left (\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}\right )}\,\left (280\,B+140\,C\right )\,1{}\mathrm {i}}{105\,d}-\frac {{\mathrm {e}}^{c\,7{}\mathrm {i}+d\,x\,7{}\mathrm {i}}\,\sqrt {a+a\,\left (\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}\right )}\,\left (96\,A+112\,B+140\,C\right )\,1{}\mathrm {i}}{105\,d}+\frac {{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,\sqrt {a+a\,\left (\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}\right )}\,\left (336\,A+392\,B+280\,C\right )\,1{}\mathrm {i}}{105\,d}-\frac {{\mathrm {e}}^{c\,5{}\mathrm {i}+d\,x\,5{}\mathrm {i}}\,\sqrt {a+a\,\left (\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}\right )}\,\left (336\,A+392\,B+280\,C\right )\,1{}\mathrm {i}}{105\,d}\right )}{{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+3\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+3\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}+3\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}+3\,{\mathrm {e}}^{c\,5{}\mathrm {i}+d\,x\,5{}\mathrm {i}}+{\mathrm {e}}^{c\,6{}\mathrm {i}+d\,x\,6{}\mathrm {i}}+{\mathrm {e}}^{c\,7{}\mathrm {i}+d\,x\,7{}\mathrm {i}}+1} \]

[In]

int((1/cos(c + d*x))^(9/2)*(a + a*cos(c + d*x))^(1/2)*(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)*(((a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1
i)/2))^(1/2)*(96*A + 112*B + 140*C)*1i)/(105*d) - (exp(c*3i + d*x*3i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1
i + d*x*1i)/2))^(1/2)*(280*B + 140*C)*1i)/(105*d) + (exp(c*4i + d*x*4i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c
*1i + d*x*1i)/2))^(1/2)*(280*B + 140*C)*1i)/(105*d) - (exp(c*7i + d*x*7i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp
(c*1i + d*x*1i)/2))^(1/2)*(96*A + 112*B + 140*C)*1i)/(105*d) + (exp(c*2i + d*x*2i)*(a + a*(exp(- c*1i - d*x*1i
)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(336*A + 392*B + 280*C)*1i)/(105*d) - (exp(c*5i + d*x*5i)*(a + a*(exp(- c*1
i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(336*A + 392*B + 280*C)*1i)/(105*d)))/(exp(c*1i + d*x*1i) + 3*exp
(c*2i + d*x*2i) + 3*exp(c*3i + d*x*3i) + 3*exp(c*4i + d*x*4i) + 3*exp(c*5i + d*x*5i) + exp(c*6i + d*x*6i) + ex
p(c*7i + d*x*7i) + 1)