\(\int (a+b \cos (c+d x))^{5/2} (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\) [832]

Optimal result

Integrand size = 42, antiderivative size = 315 \[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=-\frac {\left (9 a b B+4 a^2 C-8 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (11 a^2 b B+8 b^3 B+4 a^3 C+16 a b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}+\frac {a \left (4 a^2 B+15 b^2 B+20 a b C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}+\frac {a (7 b B+4 a C) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 d}+\frac {a B (a+b \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

-1/4*(9*B*a*b+4*C*a^2-8*C*b^2)*(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*(11*B*a^2*b+8*B*b^3+4*C*a^3
+16*C*a*b^2)*(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*a*(4*B*a^2+15*B*b^2+20*C*a*b)*(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)/d/(a+b*cos(d*x+c))^(1/2)+1/2*a*B*(a+b*cos(d*x+c))^(3/2)*sec(d*x+c)*tan(d*x+c)/d+1/4*a*(7*B*b+4*C*a)
*(a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 1.30 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.262, Rules used = {3108, 3068, 3126, 3138, 2734, 2732, 3081, 2742, 2740, 2886, 2884} \[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=-\frac {\left (4 a^2 C+9 a b B-8 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {a \left (4 a^2 B+20 a b C+15 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (4 a^3 C+11 a^2 b B+16 a b^2 C+8 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{4 d}+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d} \]

[In]

Int[(a + b*Cos[c + d*x])^(5/2)*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

-1/4*((9*a*b*B + 4*a^2*C - 8*b^2*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)]) + ((11*a^2*b*B + 8*b^3*B + 4*a^3*C + 16*a*b^2*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]]) + (a*(4*a^2*B + 15*b^2*B + 20*a*b*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]]) +
(a*(7*b*B + 4*a*C)*Sqrt[a + b*Cos[c + d*x]]*Tan[c + d*x])/(4*d) + (a*B*(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 3068

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

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 3108

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 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 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}& = \int (a+b \cos (c+d x))^{5/2} (B+C \cos (c+d x)) \sec ^3(c+d x) \, dx \\ & = \frac {a B (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 {1}{2} a (7 b B+4 a C)+\left (a^2 B+2 b^2 B+4 a b C\right ) \cos (c+d x)-\frac {1}{2} b (a B-4 b C) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx \\ & = \frac {a (7 b B+4 a C) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 d}+\frac {a B (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} a \left (4 a^2 B+15 b^2 B+20 a b C\right )+\frac {1}{2} b \left (a^2 B+4 b^2 B+12 a b C\right ) \cos (c+d x)-\frac {1}{4} b \left (9 a b B+4 a^2 C-8 b^2 C\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx \\ & = \frac {a (7 b B+4 a C) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 d}+\frac {a B (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} a b \left (4 a^2 B+15 b^2 B+20 a b C\right )-\frac {1}{4} b \left (11 a^2 b B+8 b^3 B+4 a^3 C+16 a b^2 C\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{2 b}+\frac {1}{8} \left (-9 a b B-4 a^2 C+8 b^2 C\right ) \int \sqrt {a+b \cos (c+d x)} \, dx \\ & = \frac {a (7 b B+4 a C) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 d}+\frac {a B (a+b \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{8} \left (a \left (4 a^2 B+15 b^2 B+20 a b C\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx-\frac {1}{8} \left (-11 a^2 b B-8 b^3 B-4 a^3 C-16 a b^2 C\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx+\frac {\left (\left (-9 a b B-4 a^2 C+8 b^2 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}{8 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \\ & = -\frac {\left (9 a b B+4 a^2 C-8 b^2 C\right ) \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 (7 b B+4 a C) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 d}+\frac {a B (a+b \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\left (a \left (4 a^2 B+15 b^2 B+20 a b 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 {\left (\left (-11 a^2 b B-8 b^3 B-4 a^3 C-16 a b^2 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}{8 \sqrt {a+b \cos (c+d x)}} \\ & = -\frac {\left (9 a b B+4 a^2 C-8 b^2 C\right ) \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 {\left (11 a^2 b B+8 b^3 B+4 a^3 C+16 a b^2 C\right ) \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 {a \left (4 a^2 B+15 b^2 B+20 a b 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 {a (7 b B+4 a C) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 d}+\frac {a B (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 = 6.38 (sec) , antiderivative size = 451, normalized size of antiderivative = 1.43 \[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {\frac {8 b \left (a^2 B+4 b^2 B+12 a b C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (8 a^3 B+21 a b^2 B+36 a^2 b C+8 b^3 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 i \left (-9 a b B-4 a^2 C+8 b^2 C\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} \csc (c+d x) \left (-2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (-2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )+b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right )}{a b \sqrt {-\frac {1}{a+b}}}+4 a \sqrt {a+b \cos (c+d x)} (2 a B+(9 b B+4 a C) \cos (c+d x)) \sec (c+d x) \tan (c+d x)}{16 d} \]

[In]

Integrate[(a + b*Cos[c + d*x])^(5/2)*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

((8*b*(a^2*B + 4*b^2*B + 12*a*b*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/S
qrt[a + b*Cos[c + d*x]] + (2*(8*a^3*B + 21*a*b^2*B + 36*a^2*b*C + 8*b^3*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)*(-9*a*b*B - 4*a^2*C + 8*b^2*C)*Sq
rt[-((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)*Ellipt
icE[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*(-2*a*EllipticF[I*ArcSinh[Sq
rt[-(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*b*Sqrt[-(a + b)^(-1)]) + 4*a*Sqrt[a + b*Cos[c + d*x]
]*(2*a*B + (9*b*B + 4*a*C)*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. \(1741\) vs. \(2(376)=752\).

Time = 64.69 (sec) , antiderivative size = 1742, normalized size of antiderivative = 5.53

method	result	size
default	\(\text {Expression too large to display}\)	\(1742\)
parts	\(\text {Expression too large to display}\)	\(2096\)

[In]

int((a+cos(d*x+c)*b)^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x,method=_RETURNVERBOSE)

[Out]

-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B*b^3*(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)/(-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))-2*C*b^3*(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)/(-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))+6*C*a*b^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)/(-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))-2*C*b^2*(a-b)*(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)/(-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))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b
))^(1/2)))+2*B*a^3*(-1/2*cos(1/2*d*x+1/2*c)/a*(-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*b*cos(1/2*d*x+1/2*c)^2+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*b*cos(1/2*d*x+1/2*c)^2+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))-3/8*b^2/a^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)/(-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*b*cos(1/2*
d*x+1/2*c)^2+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*b*cos(1/2*d*x+1/2*c)^2+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)+2*a^2*(3*B*b+C*a)*(-cos(1/2*d*x+1/2*c)/a*(-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*b*cos(1/2*d*x+1/2*c)^2+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*b*cos(1/2*d*x+1/2*c)^2+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*b*cos(1/2*d*x+1/2*c)^2+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*b*cos(1/2*d*x+1/2*
c)^2+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)))-6*a*b*(B*b+C*a)*(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)/(-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)))/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. \[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+b*cos(d*x+c))^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+b*cos(d*x+c))**(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

Timed out

Maxima [F]

\[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{4} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c) + a)^(5/2)*sec(d*x + c)^4, x)

Giac [F]

\[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{4} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c) + a)^(5/2)*sec(d*x + c)^4, x)

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^4} \,d x \]

[In]

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(5/2))/cos(c + d*x)^4,x)

[Out]

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(5/2))/cos(c + d*x)^4, x)