3.637 \(\int (a+b \cos (c+d x))^{3/2} (A+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=436 \[ -\frac {b \left (3 A b^2-4 a^2 (13 A+20 C)\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{64 a^2 d}-\frac {b \left (A b^2-4 a^2 (19 A+28 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{64 a d \sqrt {a+b \cos (c+d x)}}+\frac {b \left (3 A b^2-4 a^2 (13 A+20 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{64 a^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (4 a^2 (3 A+4 C)+A b^2\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{32 a d}+\frac {\left (16 a^4 (3 A+4 C)+24 a^2 b^2 (A+2 C)+3 A b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{64 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^{3/2}}{4 d}+\frac {A b \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{8 d} \]

[Out]

1/64*b*(3*A*b^2-4*a^2*(13*A+20*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c
),2^(1/2)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/a^2/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-1/64*b*(A*b^2-4*a^2*(19
*A+28*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2)
)*((a+b*cos(d*x+c))/(a+b))^(1/2)/a/d/(a+b*cos(d*x+c))^(1/2)+1/64*(3*A*b^4+24*a^2*b^2*(A+2*C)+16*a^4*(3*A+4*C))
*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(b/(a+b))^(1/2))*((a+
b*cos(d*x+c))/(a+b))^(1/2)/a^2/d/(a+b*cos(d*x+c))^(1/2)+1/4*A*(a+b*cos(d*x+c))^(3/2)*sec(d*x+c)^3*tan(d*x+c)/d
-1/64*b*(3*A*b^2-4*a^2*(13*A+20*C))*(a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/a^2/d+1/32*(A*b^2+4*a^2*(3*A+4*C))*sec(d
*x+c)*(a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/a/d+1/8*A*b*sec(d*x+c)^2*(a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 1.81, antiderivative size = 436, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3048, 3047, 3055, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ -\frac {b \left (3 A b^2-4 a^2 (13 A+20 C)\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{64 a^2 d}-\frac {b \left (A b^2-4 a^2 (19 A+28 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{64 a d \sqrt {a+b \cos (c+d x)}}+\frac {b \left (3 A b^2-4 a^2 (13 A+20 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{64 a^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (24 a^2 b^2 (A+2 C)+16 a^4 (3 A+4 C)+3 A b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{64 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (4 a^2 (3 A+4 C)+A b^2\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{32 a d}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^{3/2}}{4 d}+\frac {A b \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(b*(3*A*b^2 - 4*a^2*(13*A + 20*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(64*a^2*d*S
qrt[(a + b*Cos[c + d*x])/(a + b)]) - (b*(A*b^2 - 4*a^2*(19*A + 28*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Ellip
ticF[(c + d*x)/2, (2*b)/(a + b)])/(64*a*d*Sqrt[a + b*Cos[c + d*x]]) + ((3*A*b^4 + 24*a^2*b^2*(A + 2*C) + 16*a^
4*(3*A + 4*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(64*a^2*d*Sqrt[a
+ b*Cos[c + d*x]]) - (b*(3*A*b^2 - 4*a^2*(13*A + 20*C))*Sqrt[a + b*Cos[c + d*x]]*Tan[c + d*x])/(64*a^2*d) + ((
A*b^2 + 4*a^2*(3*A + 4*C))*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(32*a*d) + (A*b*Sqrt[a + b*Cos[
c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x])/(8*d) + (A*(a + b*Cos[c + d*x])^(3/2)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d
)

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 2807

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 3002

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3047

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/(d*(n + 1)*(
c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c
*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)
))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2,
0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3048

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/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3055

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[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3059

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int \sqrt {a+b \cos (c+d x)} \left (\frac {3 A b}{2}+a (3 A+4 C) \cos (c+d x)+\frac {1}{2} b (3 A+8 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {A b \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{8 d}+\frac {A (a+b \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{12} \int \frac {\left (\frac {3}{4} \left (A b^2+4 a^2 (3 A+4 C)\right )+\frac {3}{2} a b (11 A+16 C) \cos (c+d x)+\frac {3}{4} b^2 (9 A+16 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {\left (A b^2+4 a^2 (3 A+4 C)\right ) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{32 a d}+\frac {A b \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{8 d}+\frac {A (a+b \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int \frac {\left (-\frac {3}{8} b \left (3 A b^2-a^2 (52 A+80 C)\right )+\frac {3}{4} a \left (4 a^2 (3 A+4 C)+b^2 (19 A+32 C)\right ) \cos (c+d x)+\frac {3}{8} b \left (A b^2+4 a^2 (3 A+4 C)\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{24 a}\\ &=-\frac {b \left (3 A b^2-4 a^2 (13 A+20 C)\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{64 a^2 d}+\frac {\left (A b^2+4 a^2 (3 A+4 C)\right ) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{32 a d}+\frac {A b \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{8 d}+\frac {A (a+b \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int \frac {\left (\frac {3}{16} \left (3 A b^4+24 a^2 b^2 (A+2 C)+16 a^4 (3 A+4 C)\right )+\frac {3}{8} a b \left (A b^2+4 a^2 (3 A+4 C)\right ) \cos (c+d x)+\frac {3}{16} b^2 \left (3 A b^2-4 a^2 (13 A+20 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{24 a^2}\\ &=-\frac {b \left (3 A b^2-4 a^2 (13 A+20 C)\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{64 a^2 d}+\frac {\left (A b^2+4 a^2 (3 A+4 C)\right ) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{32 a d}+\frac {A b \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{8 d}+\frac {A (a+b \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {\int \frac {\left (-\frac {3}{16} b \left (3 A b^4+24 a^2 b^2 (A+2 C)+16 a^4 (3 A+4 C)\right )+\frac {3}{16} a b^2 \left (A b^2-4 a^2 (19 A+28 C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{24 a^2 b}-\frac {1}{128} \left (b \left (A \left (52-\frac {3 b^2}{a^2}\right )+80 C\right )\right ) \int \sqrt {a+b \cos (c+d x)} \, dx\\ &=-\frac {b \left (3 A b^2-4 a^2 (13 A+20 C)\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{64 a^2 d}+\frac {\left (A b^2+4 a^2 (3 A+4 C)\right ) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{32 a d}+\frac {A b \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{8 d}+\frac {A (a+b \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\left (3 A b^4+24 a^2 b^2 (A+2 C)+16 a^4 (3 A+4 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{128 a^2}-\frac {\left (b \left (A b^2-4 a^2 (19 A+28 C)\right )\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{128 a}-\frac {\left (b \left (A \left (52-\frac {3 b^2}{a^2}\right )+80 C\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{128 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\\ &=-\frac {b \left (A \left (52-\frac {3 b^2}{a^2}\right )+80 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{64 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {b \left (3 A b^2-4 a^2 (13 A+20 C)\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{64 a^2 d}+\frac {\left (A b^2+4 a^2 (3 A+4 C)\right ) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{32 a d}+\frac {A b \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{8 d}+\frac {A (a+b \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\left (\left (3 A b^4+24 a^2 b^2 (A+2 C)+16 a^4 (3 A+4 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{128 a^2 \sqrt {a+b \cos (c+d x)}}-\frac {\left (b \left (A b^2-4 a^2 (19 A+28 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{128 a \sqrt {a+b \cos (c+d x)}}\\ &=-\frac {b \left (A \left (52-\frac {3 b^2}{a^2}\right )+80 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{64 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {b \left (A b^2-4 a^2 (19 A+28 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{64 a d \sqrt {a+b \cos (c+d x)}}+\frac {\left (3 A b^4+24 a^2 b^2 (A+2 C)+16 a^4 (3 A+4 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{64 a^2 d \sqrt {a+b \cos (c+d x)}}-\frac {b \left (3 A b^2-4 a^2 (13 A+20 C)\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{64 a^2 d}+\frac {\left (A b^2+4 a^2 (3 A+4 C)\right ) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{32 a d}+\frac {A b \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{8 d}+\frac {A (a+b \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [C]  time = 6.80, size = 696, normalized size = 1.60 \[ \frac {\sqrt {a+b \cos (c+d x)} \left (\frac {\sec (c+d x) \left (52 a^2 A b \sin (c+d x)+80 a^2 b C \sin (c+d x)-3 A b^3 \sin (c+d x)\right )}{64 a^2}+\frac {\sec ^2(c+d x) \left (12 a^2 A \sin (c+d x)+16 a^2 C \sin (c+d x)+A b^2 \sin (c+d x)\right )}{32 a}+\frac {1}{4} a A \tan (c+d x) \sec ^3(c+d x)+\frac {3}{8} A b \tan (c+d x) \sec ^2(c+d x)\right )}{d}+\frac {\frac {2 \left (48 a^3 A b+64 a^3 b C+4 a A b^3\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {2 i \left (-52 a^2 A b^2-80 a^2 b^2 C+3 A b^4\right ) \sin (c+d x) \cos (2 (c+d x)) \sqrt {\frac {b-b \cos (c+d x)}{a+b}} \sqrt {-\frac {b \cos (c+d x)+b}{a-b}} \left (2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (2 a F\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )-b \Pi \left (\frac {a+b}{a};i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )\right )}{a \sqrt {-\frac {1}{a+b}} \sqrt {1-\cos ^2(c+d x)} \sqrt {-\frac {a^2-2 a (a+b \cos (c+d x))+(a+b \cos (c+d x))^2-b^2}{b^2}} \left (2 a^2-4 a (a+b \cos (c+d x))+2 (a+b \cos (c+d x))^2-b^2\right )}+\frac {2 \left (96 a^4 A+128 a^4 C-4 a^2 A b^2+16 a^2 b^2 C+9 A b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}}{256 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

((2*(48*a^3*A*b + 4*a*A*b^3 + 64*a^3*b*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a +
b)])/Sqrt[a + b*Cos[c + d*x]] + (2*(96*a^4*A - 4*a^2*A*b^2 + 9*A*b^4 + 128*a^4*C + 16*a^2*b^2*C)*Sqrt[(a + b*
Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] - ((2*I)*(-52*a^2*A
*b^2 + 3*A*b^4 - 80*a^2*b^2*C)*Sqrt[(b - b*Cos[c + d*x])/(a + b)]*Sqrt[-((b + b*Cos[c + d*x])/(a - b))]*Cos[2*
(c + d*x)]*(2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] +
b*(2*a*EllipticF[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] - b*EllipticPi[(a +
b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)]))*Sin[c + d*x])/(a*Sqrt[-(a +
b)^(-1)]*Sqrt[1 - Cos[c + d*x]^2]*Sqrt[-((a^2 - b^2 - 2*a*(a + b*Cos[c + d*x]) + (a + b*Cos[c + d*x])^2)/b^2)
]*(2*a^2 - b^2 - 4*a*(a + b*Cos[c + d*x]) + 2*(a + b*Cos[c + d*x])^2)))/(256*a^2*d) + (Sqrt[a + b*Cos[c + d*x]
]*((Sec[c + d*x]^2*(12*a^2*A*Sin[c + d*x] + A*b^2*Sin[c + d*x] + 16*a^2*C*Sin[c + d*x]))/(32*a) + (Sec[c + d*x
]*(52*a^2*A*b*Sin[c + d*x] - 3*A*b^3*Sin[c + d*x] + 80*a^2*b*C*Sin[c + d*x]))/(64*a^2) + (3*A*b*Sec[c + d*x]^2
*Tan[c + d*x])/8 + (a*A*Sec[c + d*x]^3*Tan[c + d*x])/4))/d

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fricas [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+b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^(3/2)*sec(d*x + c)^5, x)

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maple [B]  time = 9.52, size = 3534, normalized size = 8.11 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*(A*b^2+C*a^2)*(-1/2/a*cos(1/2*d*x+1/2*c)*(-2
*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^2+3/4*b/a^2*cos(1/2*d*x+1
/2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)-1/8*b/a*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x
+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+3/8/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1
/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*EllipticE(c
os(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-3/8*b^2/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)
/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/
(a-b))^(1/2))-1/2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/
2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))-3/8/a^2*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*
x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*b^2)+4*A*a*b*(-1/3/a*cos(1/2*d*x+1/2*c)*
(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^3+5/12*b/a^2*cos(1/2*d
*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^2-1/24*(16*a
^2+15*b^2)/a^3*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+
1/2*c)^2-1)+5/48*b^2/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2
*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/3*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*
x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/
2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(
1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/3/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(
1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^
(1/2))-5/16*b^2/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+
1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+5/16/a^3*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*
x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+1/4/a*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((
2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipt
icPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))+5/16*b^3/a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)
^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2
*c),2,(-2*b/(a-b))^(1/2)))-2*b^2*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(
-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2)
)+4*C*a*b*(-1/a*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x
+1/2*c)^2-1)+1/2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2
*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/2*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)
^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/2/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+
1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*
d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/2/a*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/
2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(
1/2)))+2*a^2*A*(-1/4/a*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(
1/2*d*x+1/2*c)^2-1)^4+7/24*b/a^2*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/
2)/(2*cos(1/2*d*x+1/2*c)^2-1)^3-1/96*(36*a^2+35*b^2)/a^3*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*s
in(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^2+5/192*b*(20*a^2+21*b^2)/a^4*cos(1/2*d*x+1/2*c)*(-2*sin
(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)-7/96*b/a*(sin(1/2*d*x+1/2*c)^
2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(
1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-35/384*b^3/a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*
d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/
2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+25/96/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(
1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^
(1/2))-25/96*b^2/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x
+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+35/128/a^3*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2
*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3-35/128*b^4/a^4*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1
/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-3/8*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2
*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c
),2,(-2*b/(a-b))^(1/2))-3/16/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2
*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*
b^2-35/128/a^4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c
)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*b^4))/sin(1/2*d*x+
1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^(3/2)*sec(d*x + c)^5, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(3/2))/cos(c + d*x)^5,x)

[Out]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(3/2))/cos(c + d*x)^5, x)

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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+b*cos(d*x+c))**(3/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)

[Out]

Timed out

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