\(\int (a+b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\) [642]

Optimal result

Integrand size = 35, antiderivative size = 327 \[ \int (a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=-\frac {\left (a^2 (15 A-46 C)-6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {a \left (a^2 (15 A-16 C)+4 b^2 (15 A+4 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 d \sqrt {a+b \cos (c+d x)}}+\frac {5 a^2 A b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {a b (15 A-16 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^{5/2} \tan (c+d x)}{d} \]

[Out]

-1/5*b*(5*A-2*C)*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/d-1/15*a*b*(15*A-16*C)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d-
1/15*(a^2*(15*A-46*C)-6*b^2*(5*A+3*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)+1/15*a*(a^2*(15*A-16*C)
+4*b^2*(15*A+4*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)+5*a^2*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)+A*(a+b*cos(d*x+c))^(5/2)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 1.40 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3127, 3128, 3138, 2734, 2732, 3081, 2742, 2740, 2886, 2884} \[ \int (a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {a \left (a^2 (15 A-16 C)+4 b^2 (15 A+4 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 d \sqrt {a+b \cos (c+d x)}}-\frac {\left (a^2 (15 A-46 C)-6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {5 a^2 A b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}-\frac {a b (15 A-16 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{15 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d} \]

[In]

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

[Out]

-1/15*((a^2*(15*A - 46*C) - 6*b^2*(5*A + 3*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*(a^2*(15*A - 16*C) + 4*b^2*(15*A + 4*C))*Sqrt[(a + b*Cos[c + d*x]
)/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(15*d*Sqrt[a + b*Cos[c + d*x]]) + (5*a^2*A*b*Sqrt[(a + b*Cos
[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]) - (a*b*(15*A - 16*
C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*d) - (b*(5*A - 2*C)*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*
d) + (A*(a + b*Cos[c + d*x])^(5/2)*Tan[c + d*x])/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 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 3128

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] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e
+ f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*
x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n +
2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

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))^{5/2} \tan (c+d x)}{d}+\int (a+b \cos (c+d x))^{3/2} \left (\frac {5 A b}{2}+a C \cos (c+d x)-\frac {1}{2} b (5 A-2 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx \\ & = -\frac {b (5 A-2 C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^{5/2} \tan (c+d x)}{d}+\frac {2}{5} \int \sqrt {a+b \cos (c+d x)} \left (\frac {25 a A b}{4}+\frac {1}{2} \left (5 A b^2+5 a^2 C+3 b^2 C\right ) \cos (c+d x)-\frac {1}{4} a b (15 A-16 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx \\ & = -\frac {a b (15 A-16 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^{5/2} \tan (c+d x)}{d}+\frac {4}{15} \int \frac {\left (\frac {75}{8} a^2 A b+\frac {1}{4} a \left (45 A b^2+15 a^2 C+17 b^2 C\right ) \cos (c+d x)-\frac {1}{8} b \left (a^2 (15 A-46 C)-6 b^2 (5 A+3 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx \\ & = -\frac {a b (15 A-16 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^{5/2} \tan (c+d x)}{d}-\frac {4 \int \frac {\left (-\frac {75}{8} a^2 A b^2-\frac {1}{8} a b \left (a^2 (15 A-16 C)+4 b^2 (15 A+4 C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b}+\frac {1}{30} \left (-a^2 (15 A-46 C)+6 b^2 (5 A+3 C)\right ) \int \sqrt {a+b \cos (c+d x)} \, dx \\ & = -\frac {a b (15 A-16 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^{5/2} \tan (c+d x)}{d}+\frac {1}{2} \left (5 a^2 A b\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx+\frac {1}{30} \left (a \left (a^2 (15 A-16 C)+4 b^2 (15 A+4 C)\right )\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx+\frac {\left (\left (-a^2 (15 A-46 C)+6 b^2 (5 A+3 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}{30 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \\ & = -\frac {\left (a^2 (15 A-46 C)-6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {a b (15 A-16 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^{5/2} \tan (c+d x)}{d}+\frac {\left (5 a^2 A b \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}{2 \sqrt {a+b \cos (c+d x)}}+\frac {\left (a \left (a^2 (15 A-16 C)+4 b^2 (15 A+4 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}{30 \sqrt {a+b \cos (c+d x)}} \\ & = -\frac {\left (a^2 (15 A-46 C)-6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {a \left (a^2 (15 A-16 C)+4 b^2 (15 A+4 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 )}{15 d \sqrt {a+b \cos (c+d x)}}+\frac {5 a^2 A b \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 )}{d \sqrt {a+b \cos (c+d x)}}-\frac {a b (15 A-16 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^{5/2} \tan (c+d x)}{d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 17.46 (sec) , antiderivative size = 462, normalized size of antiderivative = 1.41 \[ \int (a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {\frac {8 a \left (45 A b^2+15 a^2 C+17 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (6 b^2 (5 A+3 C)+a^2 (135 A+46 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 i \left (6 b^2 (5 A+3 C)+a^2 (-15 A+46 C)\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {-\frac {b (1+\cos (c+d x))}{a-b}} \csc (c+d x) \left (-2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (-2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )+b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right )}{a b \sqrt {-\frac {1}{a+b}}}+4 \sqrt {a+b \cos (c+d x)} \left (22 a b C \sin (c+d x)+3 b^2 C \sin (2 (c+d x))+15 a^2 A \tan (c+d x)\right )}{60 d} \]

[In]

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

[Out]

((8*a*(45*A*b^2 + 15*a^2*C + 17*b^2*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*b*(6*b^2*(5*A + 3*C) + a^2*(135*A + 46*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)*(6*b^2*(5*A + 3*C) + a^2*(-15*A
+ 46*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*b*Sqrt[-(a + b)^(-1)]) + 4*Sqrt[a + b*C
os[c + d*x]]*(22*a*b*C*Sin[c + d*x] + 3*b^2*C*Sin[2*(c + d*x)] + 15*a^2*A*Tan[c + d*x]))/(60*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1623\) vs. \(2(388)=776\).

Time = 30.68 (sec) , antiderivative size = 1624, normalized size of antiderivative = 4.97

method	result	size
parts	\(\text {Expression too large to display}\)	\(1624\)
default	\(\text {Expression too large to display}\)	\(1714\)

[In]

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

[Out]

-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)*(4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a^2*
b+(-2*a^3-2*a^2*b)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-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)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+4*EllipticF(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))*a^3+EllipticE(cos(1/2*d*x+1/
2*c),(-2*b/(a-b))^(1/2))*a^2*b+2*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2-2*EllipticE(cos(1/2*d*
x+1/2*c),(-2*b/(a-b))^(1/2))*b^3-5*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a^2*b)*sin(1/2*d*x+1/2*
c)^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)*EllipticF(cos(1/2*d*x+1/
2*c),(-2*b/(a-b))^(1/2))*a^3+4*a*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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-(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))*a^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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*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))*a*b^2-2*(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)*El
lipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3-5*a^2*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)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2)))/(2*cos(1/2*d*x+1/2*c)^2-1)/
(-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-2/15*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)*(24*cos(1/2*d*x+1/2*c)^7*b^3+56*c
os(1/2*d*x+1/2*c)^5*a*b^2-48*cos(1/2*d*x+1/2*c)^5*b^3+22*cos(1/2*d*x+1/2*c)^3*a^2*b-84*cos(1/2*d*x+1/2*c)^3*a*
b^2+30*cos(1/2*d*x+1/2*c)^3*b^3-8*a^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
)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+8*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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+23*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3-23*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^
2*b+9*(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)*EllipticE(cos(1/2*d*x+1/2*c),(
-2*b/(a-b))^(1/2))*a*b^2-9*(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)*EllipticE
(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3-22*cos(1/2*d*x+1/2*c)*a^2*b+28*cos(1/2*d*x+1/2*c)*a*b^2-6*cos(1/2*
d*x+1/2*c)*b^3)/(-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]

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

[In]

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

[Out]

integral((C*b^2*cos(d*x + c)^4 + 2*C*a*b*cos(d*x + c)^3 + 2*A*a*b*cos(d*x + c) + A*a^2 + (C*a^2 + A*b^2)*cos(d
*x + c)^2)*sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^2, x)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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