3.1032 \(\int (a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x) \, dx\)

Optimal. Leaf size=383 \[ \frac{2 \sin (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)}}{105 d}-\frac{2 \left (-10 a^2 b^2 (7 A-C)+56 a^3 b B+15 a^4 C-56 a b^3 B-5 b^4 (7 A+5 C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (161 a^2 b B+15 a^3 C+5 a b^2 (49 A+29 C)+63 b^3 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 a^3 A \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \cos (c+d x)}}+\frac{2 (5 a C+7 b B) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 d}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 d} \]

[Out]

(2*(161*a^2*b*B + 63*b^3*B + 15*a^3*C + 5*a*b^2*(49*A + 29*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2,
(2*b)/(a + b)])/(105*b*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(56*a^3*b*B - 56*a*b^3*B - 10*a^2*b^2*(7*A
- C) + 15*a^4*C - 5*b^4*(7*A + 5*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])
/(105*b*d*Sqrt[a + b*Cos[c + d*x]]) + (2*a^3*A*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]]) + (2*(35*A*b^2 + 56*a*b*B + 15*a^2*C + 25*b^2*C)*Sqrt[a + b*Cos[c
+ d*x]]*Sin[c + d*x])/(105*d) + (2*(7*b*B + 5*a*C)*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*d) + (2*C*(a +
b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*d)

________________________________________________________________________________________

Rubi [A]  time = 1.3781, antiderivative size = 383, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.22, Rules used = {3049, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ \frac{2 \sin (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)}}{105 d}-\frac{2 \left (-10 a^2 b^2 (7 A-C)+56 a^3 b B+15 a^4 C-56 a b^3 B-5 b^4 (7 A+5 C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (161 a^2 b B+15 a^3 C+5 a b^2 (49 A+29 C)+63 b^3 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 a^3 A \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \cos (c+d x)}}+\frac{2 (5 a C+7 b B) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 d}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(161*a^2*b*B + 63*b^3*B + 15*a^3*C + 5*a*b^2*(49*A + 29*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2,
(2*b)/(a + b)])/(105*b*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(56*a^3*b*B - 56*a*b^3*B - 10*a^2*b^2*(7*A
- C) + 15*a^4*C - 5*b^4*(7*A + 5*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])
/(105*b*d*Sqrt[a + b*Cos[c + d*x]]) + (2*a^3*A*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]]) + (2*(35*A*b^2 + 56*a*b*B + 15*a^2*C + 25*b^2*C)*Sqrt[a + b*Cos[c
+ d*x]]*Sin[c + d*x])/(105*d) + (2*(7*b*B + 5*a*C)*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*d) + (2*C*(a +
b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*d)

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]

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 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 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 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 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 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 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]

Rubi steps

\begin{align*} \int (a+b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx &=\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{2}{7} \int (a+b \cos (c+d x))^{3/2} \left (\frac{7 a A}{2}+\frac{1}{2} (7 A b+7 a B+5 b C) \cos (c+d x)+\frac{1}{2} (7 b B+5 a C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{2 (7 b B+5 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{4}{35} \int \sqrt{a+b \cos (c+d x)} \left (\frac{35 a^2 A}{4}+\frac{1}{4} \left (70 a A b+35 a^2 B+21 b^2 B+40 a b C\right ) \cos (c+d x)+\frac{1}{4} \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{2 \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 (7 b B+5 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{8}{105} \int \frac{\left (\frac{105 a^3 A}{8}+\frac{1}{8} \left (105 a^3 B+119 a b^2 B+45 a^2 b (7 A+3 C)+5 b^3 (7 A+5 C)\right ) \cos (c+d x)+\frac{1}{8} \left (161 a^2 b B+63 b^3 B+15 a^3 C+5 a b^2 (49 A+29 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx\\ &=\frac{2 \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 (7 b B+5 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}-\frac{8 \int \frac{\left (-\frac{105}{8} a^3 A b+\frac{1}{8} \left (56 a^3 b B-56 a b^3 B-10 a^2 b^2 (7 A-C)+15 a^4 C-5 b^4 (7 A+5 C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{105 b}+\frac{\left (161 a^2 b B+63 b^3 B+15 a^3 C+5 a b^2 (49 A+29 C)\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{105 b}\\ &=\frac{2 \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 (7 b B+5 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\left (a^3 A\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx-\frac{\left (56 a^3 b B-56 a b^3 B-10 a^2 b^2 (7 A-C)+15 a^4 C-5 b^4 (7 A+5 C)\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{105 b}+\frac{\left (\left (161 a^2 b B+63 b^3 B+15 a^3 C+5 a b^2 (49 A+29 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}{105 b \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}\\ &=\frac{2 \left (161 a^2 b B+63 b^3 B+15 a^3 C+5 a b^2 (49 A+29 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 (7 b B+5 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{\left (a^3 A \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}{\sqrt{a+b \cos (c+d x)}}-\frac{\left (\left (56 a^3 b B-56 a b^3 B-10 a^2 b^2 (7 A-C)+15 a^4 C-5 b^4 (7 A+5 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}{105 b \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 \left (161 a^2 b B+63 b^3 B+15 a^3 C+5 a b^2 (49 A+29 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 \left (56 a^3 b B-56 a b^3 B-10 a^2 b^2 (7 A-C)+15 a^4 C-5 b^4 (7 A+5 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 )}{105 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 a^3 A \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 )}{d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 (7 b B+5 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}\\ \end{align*}

Mathematica [C]  time = 4.03676, size = 526, normalized size = 1.37 \[ \frac{2 \sin (c+d x) \sqrt{a+b \cos (c+d x)} \left (90 a^2 C+6 b (15 a C+7 b B) \cos (c+d x)+154 a b B+70 A b^2+15 b^2 C \cos (2 (c+d x))+65 b^2 C\right )+\frac{4 \left (45 a^2 b (7 A+3 C)+105 a^3 B+119 a b^2 B+5 b^3 (7 A+5 C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{\sqrt{a+b \cos (c+d x)}}+\frac{2 \left (15 a^3 (14 A+C)+161 a^2 b B+5 a b^2 (49 A+29 C)+63 b^3 B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{\sqrt{a+b \cos (c+d x)}}+\frac{2 i \csc (c+d x) \left (161 a^2 b B+15 a^3 C+5 a b^2 (49 A+29 C)+63 b^3 B\right ) \sqrt{-\frac{b (\cos (c+d x)-1)}{a+b}} \sqrt{-\frac{b (\cos (c+d x)+1)}{a-b}} \left (b \left (b \Pi \left (\frac{a+b}{a};i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )}{a b^2 \sqrt{-\frac{1}{a+b}}}}{210 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*(105*a^3*B + 119*a*b^2*B + 45*a^2*b*(7*A + 3*C) + 5*b^3*(7*A + 5*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*El
lipticF[(c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + (2*(161*a^2*b*B + 63*b^3*B + 15*a^3*(14*A + C)
+ 5*a*b^2*(49*A + 29*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)*(161*a^2*b*B + 63*b^3*B + 15*a^3*C + 5*a*b^2*(49*A + 29*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^2*Sqrt[-(a + b)^(-1)]) + 2*Sqrt[a + b*Cos[c + d*x]]*(70*A*b^2 + 154*
a*b*B + 90*a^2*C + 65*b^2*C + 6*b*(7*b*B + 15*a*C)*Cos[c + d*x] + 15*b^2*C*Cos[2*(c + d*x)])*Sin[c + d*x])/(21
0*d)

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Maple [B]  time = 1.082, size = 1713, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*C*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^8+(-168*B*b^4-480*C*a*b^3-360*C*b^4)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(140*A*b^4+392*B*a*b^3+168*
B*b^4+360*C*a^2*b^2+480*C*a*b^3+280*C*b^4)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-70*A*a*b^3-70*A*b^4-154*B
*a^2*b^2-196*B*a*b^3-42*B*b^4-90*C*a^3*b-180*C*a^2*b^2-170*C*a*b^3-80*C*b^4)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+
1/2*c)+70*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^2+35*A*b^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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+245*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^2-245*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^3-105*A*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)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*b-56*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))*a^3*b+56*B*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))*b^3+161*B*EllipticE(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)*si
n(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*a^3*b-161*B*EllipticE(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)*a^2*b^2+63*B*EllipticE(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)*a*b^3-6
3*B*EllipticE(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)*b^4-15*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))*a^4-10*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))*a^2*b^2+25*C*b^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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b)
)^(1/2))+15*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^4-15*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^3*b+145*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^2-145*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^3)/b/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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