\(\int (a+b \cos (c+d x))^{3/2} (A+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\) [635]

Optimal result

Integrand size = 35, antiderivative size = 276 \[ \int (a+b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=-\frac {b (5 A-8 C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {a b (7 A+8 C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (3 A b^2+4 a^2 (A+2 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}+\frac {3 A b \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

-1/4*b*(5*A-8*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)/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+1/4*a*b*(7*A+8*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)/d/(a
+b*cos(d*x+c))^(1/2)+1/4*(3*A*b^2+4*a^2*(A+2*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(si
n(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)/d/(a+b*cos(d*x+c))^(1/2)+1/2*A*(a+b
*cos(d*x+c))^(3/2)*sec(d*x+c)*tan(d*x+c)/d+3/4*A*b*(a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 1.08 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3127, 3126, 3138, 2734, 2732, 3081, 2742, 2740, 2886, 2884} \[ \int (a+b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {\left (4 a^2 (A+2 C)+3 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}+\frac {a b (7 A+8 C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}-\frac {b (5 A-8 C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{4 d}+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d} \]

[In]

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

[Out]

-1/4*(b*(5*A - 8*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[(a + b*Cos[c + d*x
])/(a + b)]) + (a*b*(7*A + 8*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(4*d
*Sqrt[a + b*Cos[c + d*x]]) + ((3*A*b^2 + 4*a^2*(A + 2*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c
+ d*x)/2, (2*b)/(a + b)])/(4*d*Sqrt[a + b*Cos[c + d*x]]) + (3*A*b*Sqrt[a + b*Cos[c + d*x]]*Tan[c + d*x])/(4*d)
+ (A*(a + b*Cos[c + d*x])^(3/2)*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 2732

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

Rule 2734

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

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

Rule 2742

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(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 2886

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/
(c + d))*Sin[e + f*x]]), 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 3081

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 3126

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 3127

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*Si
n[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 3138

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*} \text {integral}& = \frac {A (a+b \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int \sqrt {a+b \cos (c+d x)} \left (\frac {3 A b}{2}+a (A+2 C) \cos (c+d x)-\frac {1}{2} b (A-4 C) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx \\ & = \frac {3 A b \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int \frac {\left (\frac {1}{4} \left (3 A b^2+4 a^2 (A+2 C)\right )+\frac {1}{2} a b (A+8 C) \cos (c+d x)-\frac {1}{4} b^2 (5 A-8 C) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx \\ & = \frac {3 A b \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{2 d}-\frac {\int \frac {\left (-\frac {1}{4} b \left (3 A b^2+4 a^2 (A+2 C)\right )-\frac {1}{4} a b^2 (7 A+8 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{2 b}-\frac {1}{8} (b (5 A-8 C)) \int \sqrt {a+b \cos (c+d x)} \, dx \\ & = \frac {3 A b \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{8} (a b (7 A+8 C)) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx-\frac {1}{8} \left (-3 A b^2-4 a^2 (A+2 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx-\frac {\left (b (5 A-8 C) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{8 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \\ & = -\frac {b (5 A-8 C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {3 A b \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\left (a b (7 A+8 C) \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}{8 \sqrt {a+b \cos (c+d x)}}-\frac {\left (\left (-3 A b^2-4 a^2 (A+2 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}{8 \sqrt {a+b \cos (c+d x)}} \\ & = -\frac {b (5 A-8 C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {a b (7 A+8 C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (3 A b^2+4 a^2 (A+2 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}+\frac {3 A b \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{2 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.70 (sec) , antiderivative size = 411, normalized size of antiderivative = 1.49 \[ \int (a+b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {\frac {8 a b (A+8 C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (8 a^2 (A+2 C)+b^2 (A+8 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {2 i (5 A-8 C) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} \csc (c+d x) \left (-2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (-2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )+b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right )}{a \sqrt {-\frac {1}{a+b}}}+4 A \sqrt {a+b \cos (c+d x)} (2 a+5 b \cos (c+d x)) \sec (c+d x) \tan (c+d x)}{16 d} \]

[In]

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

[Out]

((8*a*b*(A + 8*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*(8*a^2*(A + 2*C) + b^2*(A + 8*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)*(5*A - 8*C)*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[(b*(
1 + Cos[c + d*x]))/(-a + b)]*Csc[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)])))/(a*Sqrt[-(a + b)^(-1)]) + 4*A*Sqrt[a + b*Cos[c + d*x]]*(2*a + 5*b*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*
x])/(16*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1230\) vs. \(2(337)=674\).

Time = 18.07 (sec) , antiderivative size = 1231, normalized size of antiderivative = 4.46

method	result	size
parts	\(\text {Expression too large to display}\)	\(1231\)
default	\(\text {Expression too large to display}\)	\(1526\)

[In]

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

[Out]

-1/4*A*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-40*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
6*b^2+(28*a*b+40*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-4*a^2-14*a*b-10*b^2)*sin(1/2*d*x+1/2*c)^2*cos(
1/2*d*x+1/2*c)+4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(7*b*Ellipti
cF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-5*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a+5*b^2*Ellip
ticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-4*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a^2-3*Ellipt
icPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*b^2)*sin(1/2*d*x+1/2*c)^4-4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(
a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(7*b*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-5*b*Ellip
ticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a+5*b^2*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-4*Ellipti
cPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a^2-3*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*b^2)*si
n(1/2*d*x+1/2*c)^2+7*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(
a+b)/(a-b))^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a+5*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*
b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-4*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a
-b))^(1/2))*a^2-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticPi(
cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*b^2)/(2*cos(1/2*d*x+1/2*c)^2-1)^2/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*si
n(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-2*b*sin(1/2*d*x+1/2*c)^2+a+b)^(1/2)/d-2*C*((2*b*cos(1/2*d*x+1/2
*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/
2)*(b*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a+b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-
b^2*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a^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)/sin(1/2*d*x+1/2*c)/(-2*b*sin(1/2*d*x+1/2*c)^2+a+b
)^(1/2)/d

Fricas [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int (a+b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (a+b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \]

[In]

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

[Out]

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