3.643 \(\int (a+b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\)
Optimal. Leaf size=329 \[ \frac {b \left (a^2 (33 A+16 C)+8 b^2 (3 A+C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{12 d \sqrt {a+b \cos (c+d x)}}+\frac {a \left (4 a^2 (A+2 C)+15 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}-\frac {b^2 (21 A-8 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{12 d}-\frac {a b (27 A-56 C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{12 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {5 A b \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{4 d}+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{5/2}}{2 d} \]
[Out]
-1/12*b^2*(21*A-8*C)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d-1/12*a*b*(27*A-56*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)+1/12*b*(8*b^2*(3*A+C)+a^2*(33*A+16*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti
cF(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*(
15*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(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)+5/4*A*b*(a+b*cos(d*x+c))^(3/2)*tan(
d*x+c)/d+1/2*A*(a+b*cos(d*x+c))^(5/2)*sec(d*x+c)*tan(d*x+c)/d
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Rubi [A] time = 1.27, antiderivative size = 329, normalized size of antiderivative =
1.00, number of steps used = 11, number of rules used = 11, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) =
0.314, Rules used = {3048, 3047, 3049, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805}
\[ \frac {b \left (a^2 (33 A+16 C)+8 b^2 (3 A+C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{12 d \sqrt {a+b \cos (c+d x)}}+\frac {a \left (4 a^2 (A+2 C)+15 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}-\frac {b^2 (21 A-8 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{12 d}-\frac {a b (27 A-56 C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{12 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {5 A b \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{4 d}+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{5/2}}{2 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]
[Out]
-(a*b*(27*A - 56*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(12*d*Sqrt[(a + b*Cos[c +
d*x])/(a + b)]) + (b*(8*b^2*(3*A + C) + a^2*(33*A + 16*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d
*x)/2, (2*b)/(a + b)])/(12*d*Sqrt[a + b*Cos[c + d*x]]) + (a*(15*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]]) - (b^2*(21*A - 8*C)*Sq
rt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(12*d) + (5*A*b*(a + b*Cos[c + d*x])^(3/2)*Tan[c + d*x])/(4*d) + (A*(a +
b*Cos[c + d*x])^(5/2)*Sec[c + d*x]*Tan[c + d*x])/(2*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 3049
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 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))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (a+b \cos (c+d x))^{3/2} \left (\frac {5 A b}{2}+a (A+2 C) \cos (c+d x)-\frac {1}{2} b (3 A-4 C) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {5 A b (a+b \cos (c+d x))^{3/2} \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int \sqrt {a+b \cos (c+d x)} \left (\frac {1}{4} \left (15 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 (21 A-8 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b^2 (21 A-8 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{12 d}+\frac {5 A b (a+b \cos (c+d x))^{3/2} \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{3} \int \frac {\left (\frac {3}{8} a \left (15 A b^2+4 a^2 (A+2 C)\right )+\frac {1}{4} b \left (4 b^2 (3 A+C)+3 a^2 (A+12 C)\right ) \cos (c+d x)-\frac {1}{8} a b^2 (27 A-56 C) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\\ &=-\frac {b^2 (21 A-8 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{12 d}+\frac {5 A b (a+b \cos (c+d x))^{3/2} \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{2 d}-\frac {\int \frac {\left (-\frac {3}{8} a b \left (15 A b^2+4 a^2 (A+2 C)\right )-\frac {1}{8} b^2 \left (8 b^2 (3 A+C)+a^2 (33 A+16 C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 b}-\frac {1}{24} (a b (27 A-56 C)) \int \sqrt {a+b \cos (c+d x)} \, dx\\ &=-\frac {b^2 (21 A-8 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{12 d}+\frac {5 A b (a+b \cos (c+d x))^{3/2} \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{8} \left (a \left (15 A b^2+4 a^2 (A+2 C)\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx+\frac {1}{24} \left (b \left (8 b^2 (3 A+C)+a^2 (33 A+16 C)\right )\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx-\frac {\left (a b (27 A-56 C) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{24 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\\ &=-\frac {a b (27 A-56 C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{12 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {b^2 (21 A-8 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{12 d}+\frac {5 A b (a+b \cos (c+d x))^{3/2} \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\left (a \left (15 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 {\left (b \left (8 b^2 (3 A+C)+a^2 (33 A+16 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}{24 \sqrt {a+b \cos (c+d x)}}\\ &=-\frac {a b (27 A-56 C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{12 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {b \left (8 b^2 (3 A+C)+a^2 (33 A+16 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 )}{12 d \sqrt {a+b \cos (c+d x)}}+\frac {a \left (15 A b^2+4 a^2 (A+2 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 )}{4 d \sqrt {a+b \cos (c+d x)}}-\frac {b^2 (21 A-8 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{12 d}+\frac {5 A b (a+b \cos (c+d x))^{3/2} \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}
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Mathematica [C] time = 4.17, size = 445, normalized size = 1.35 \[ \frac {\frac {8 b \left (3 a^2 (A+12 C)+4 b^2 (3 A+C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 a \left (24 a^2 (A+2 C)+7 b^2 (9 A+8 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+4 \sec (c+d x) \sqrt {a+b \cos (c+d x)} \left (6 a^2 A \tan (c+d x)+27 a A b \sin (c+d x)+4 b^2 C \sin (2 (c+d x))\right )-\frac {2 i (27 A-56 C) \csc (c+d x) \sqrt {-\frac {b (\cos (c+d x)-1)}{a+b}} \sqrt {\frac {b (\cos (c+d x)+1)}{b-a}} \left (b \left (b \Pi \left (\frac {a+b}{a};i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )}{\sqrt {-\frac {1}{a+b}}}}{48 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]
[Out]
((8*b*(4*b^2*(3*A + C) + 3*a^2*(A + 12*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*a*(24*a^2*(A + 2*C) + 7*b^2*(9*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)*(27*A - 56*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)])))/Sqrt[-(a + b)^(-1)] + 4*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]*(27*a*A
*b*Sin[c + d*x] + 4*b^2*C*Sin[2*(c + d*x)] + 6*a^2*A*Tan[c + d*x]))/(48*d)
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fricas [F] time = 7.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} + 2 \, C a b \cos \left (d x + c\right )^{3} + 2 \, A a b \cos \left (d x + c\right ) + A a^{2} + {\left (C a^{2} + A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,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)^3, x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(5/2)*(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)^(5/2)*sec(d*x + c)^3, x)
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maple [B] time = 3.07, size = 1897, normalized size = 5.77 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x)
[Out]
-1/12*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(128*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^8+(-216*A*a*b^2-64*C*a*b^2-192*C*b^3)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(132*A*a^2*b+216*A*a*b^2+64
*C*a*b^2+96*C*b^3)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-12*A*a^3-66*A*a^2*b-54*A*a*b^2-16*C*a*b^2-16*C*b^
3)*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)*(27*A*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b-27*A*EllipticE(cos(1/2*d*x+1/2*c)
,(-2*b/(a-b))^(1/2))*a*b^2-33*A*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b-24*A*EllipticF(cos(1/2*
d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+12*A*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a^3+45*A*EllipticP
i(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a*b^2-56*C*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b+5
6*C*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2-16*C*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2
))*a^2*b-8*C*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+24*C*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(
a-b))^(1/2))*a^3)*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)*(27*A*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b-27*A*EllipticE(cos(1/2*d*x+1/2*c),(-2
*b/(a-b))^(1/2))*a*b^2-33*A*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b-24*A*EllipticF(cos(1/2*d*x+
1/2*c),(-2*b/(a-b))^(1/2))*b^3+12*A*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a^3+45*A*EllipticPi(co
s(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a*b^2-56*C*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b+56*C*
EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2-16*C*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a
^2*b-8*C*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+24*C*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b)
)^(1/2))*a^3)*sin(1/2*d*x+1/2*c)^2-27*A*(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+27*A*(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+33*A*(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^2*b+24*A*b^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)*El
lipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-12*A*(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^3-45*A*(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*b
^2+56*C*(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-56*C*(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+16*C*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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+8*b^3*C*(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))-24*C*(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^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)/(2
*cos(1/2*d*x+1/2*c)^2-1)^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(5/2)*(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)^(5/2)*sec(d*x + c)^3, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \]
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))^(5/2))/cos(c + d*x)^3,x)
[Out]
int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(5/2))/cos(c + d*x)^3, 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))**(5/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**3,x)
[Out]
Timed out
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