\(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx\) [513]

Optimal result

Integrand size = 35, antiderivative size = 181 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {2 a^3 (230 A+301 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (10 A+11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (10 A+7 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \]

[Out]

2/7*a*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(7/2)*sin(d*x+c)/d+2/15*a^3*(10*A+11*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/
d/(a+a*cos(d*x+c))^(1/2)+2/35*a^2*(10*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^3*(2
30*A+301*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3040, 3054, 3059, 2850} \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {2 a^3 (10 A+11 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^3 (230 A+301 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (10 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{35 d}+\frac {2 a A \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d} \]

[In]

Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(9/2),x]

[Out]

(2*a^3*(230*A + 301*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(10*A + 11*B
)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(10*A + 7*B)*Sqrt[a + a*Cos[c + d*
x]]*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(35*d) + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*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 3040

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 3054

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
*Sin[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 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]

Rubi steps \begin{align*} \text {integral}& = \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 {9}{2}}(c+d x)} \, dx \\ & = \frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{7} \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 (10 A+7 B)+\frac {1}{2} a (2 A+7 B) \cos (c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (10 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+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{35} \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 (10 A+11 B)+\frac {1}{4} a^2 (30 A+49 B) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a^3 (10 A+11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (10 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+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^2 (230 A+301 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 \\ & = \frac {2 a^3 (230 A+301 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (10 A+11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (10 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+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.57 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} (290 A+196 B+(930 A+987 B) \cos (c+d x)+2 (115 A+98 B) \cos (2 (c+d x))+230 A \cos (3 (c+d x))+301 B \cos (3 (c+d x))) \sec ^{\frac {7}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{210 d} \]

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(9/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(290*A + 196*B + (930*A + 987*B)*Cos[c + d*x] + 2*(115*A + 98*B)*Cos[2*(c + d*
x)] + 230*A*Cos[3*(c + d*x)] + 301*B*Cos[3*(c + d*x)])*Sec[c + d*x]^(7/2)*Tan[(c + d*x)/2])/(210*d)

Maple [A] (verified)

Time = 2.09 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.57

\[-\frac {2 a^{2} \cot \left (d x +c \right ) \left (\cos \left (d x +c \right )-1\right ) \left (\left (230 \left (\cos ^{3}\left (d x +c \right )\right )+115 \left (\cos ^{2}\left (d x +c \right )\right )+60 \cos \left (d x +c \right )+15\right ) A +\cos \left (d x +c \right ) \left (301 \left (\cos ^{2}\left (d x +c \right )\right )+98 \cos \left (d x +c \right )+21\right ) B \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sec ^{\frac {9}{2}}\left (d x +c \right )\right )}{105 d}\]

[In]

int((a+cos(d*x+c)*a)^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(9/2),x)

[Out]

-2/105*a^2/d*cot(d*x+c)*(cos(d*x+c)-1)*((230*cos(d*x+c)^3+115*cos(d*x+c)^2+60*cos(d*x+c)+15)*A+cos(d*x+c)*(301
*cos(d*x+c)^2+98*cos(d*x+c)+21)*B)*(a*(1+cos(d*x+c)))^(1/2)*sec(d*x+c)^(9/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.63 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {2 \, {\left ({\left (230 \, A + 301 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (115 \, A + 98 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (20 \, A + 7 \, B\right )} a^{2} \cos \left (d x + c\right ) + 15 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )} \sqrt {\cos \left (d x + c\right )}} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

2/105*((230*A + 301*B)*a^2*cos(d*x + c)^3 + (115*A + 98*B)*a^2*cos(d*x + c)^2 + 3*(20*A + 7*B)*a^2*cos(d*x + c
) + 15*A*a^2)*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. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)**(9/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 488 vs. \(2 (157) = 314\).

Time = 0.36 (sec) , antiderivative size = 488, normalized size of antiderivative = 2.70 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {8 \, {\left (\frac {5 \, {\left (\frac {21 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {56 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {63 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {36 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {8 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}\right )} A {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}} {\left (\frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}} + \frac {7 \, {\left (\frac {15 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {50 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {63 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {36 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {8 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}\right )} B {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}} {\left (\frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}}\right )}}{105 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

8/105*(5*(21*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 56*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 + 63*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 36*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x
+ c) + 1)^7 + 8*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*A*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
1)^2/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(2*sin(d*x + c)
^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1)) + 7*(15*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(
d*x + c) + 1) - 50*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 - 36*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 8*sqrt(2)*a^(5/2)*sin(d*x + c)^9/
(cos(d*x + c) + 1)^9)*B*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/
2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d
*x + c) + 1)^4 + 1)))/d

Giac [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(9/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 5.25 (sec) , antiderivative size = 579, normalized size of antiderivative = 3.20 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \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 {a^2\,\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 (230\,A+301\,B\right )\,2{}\mathrm {i}}{105\,d}-\frac {B\,a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\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 )}\,2{}\mathrm {i}}{d}+\frac {B\,a^2\,{\mathrm {e}}^{c\,6{}\mathrm {i}+d\,x\,6{}\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 )}\,2{}\mathrm {i}}{d}-\frac {a^2\,{\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 (10\,A+17\,B\right )\,2{}\mathrm {i}}{3\,d}+\frac {a^2\,{\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 (10\,A+17\,B\right )\,2{}\mathrm {i}}{3\,d}+\frac {a^2\,{\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 (100\,A+113\,B\right )\,2{}\mathrm {i}}{15\,d}-\frac {a^2\,{\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 (100\,A+113\,B\right )\,2{}\mathrm {i}}{15\,d}-\frac {a^2\,{\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 (230\,A+301\,B\right )\,2{}\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((A + B*cos(c + d*x))*(1/cos(c + d*x))^(9/2)*(a + a*cos(c + d*x))^(5/2),x)

[Out]

((1/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((a^2*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d
*x*1i)/2))^(1/2)*(230*A + 301*B)*2i)/(105*d) - (B*a^2*exp(c*1i + d*x*1i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(
c*1i + d*x*1i)/2))^(1/2)*2i)/d + (B*a^2*exp(c*6i + d*x*6i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)
/2))^(1/2)*2i)/d - (a^2*exp(c*3i + d*x*3i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(10*A
+ 17*B)*2i)/(3*d) + (a^2*exp(c*4i + d*x*4i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(10
*A + 17*B)*2i)/(3*d) + (a^2*exp(c*2i + d*x*2i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(
100*A + 113*B)*2i)/(15*d) - (a^2*exp(c*5i + d*x*5i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1
/2)*(100*A + 113*B)*2i)/(15*d) - (a^2*exp(c*7i + d*x*7i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2
))^(1/2)*(230*A + 301*B)*2i)/(105*d)))/(exp(c*1i + d*x*1i) + 3*exp(c*2i + d*x*2i) + 3*exp(c*3i + d*x*3i) + 3*e
xp(c*4i + d*x*4i) + 3*exp(c*5i + d*x*5i) + exp(c*6i + d*x*6i) + exp(c*7i + d*x*7i) + 1)