3.1034 \(\int (a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\)
Optimal. Leaf size=372 \[ \frac{\left (a^2 b (33 A+16 C)+12 a^3 B+48 a b^2 B+8 b^3 (3 A+C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{12 d \sqrt{a+b \cos (c+d x)}}-\frac{\left (12 a^2 B+a b (27 A-56 C)-24 b^2 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{12 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{a \left (4 a^2 (A+2 C)+20 a b B+15 A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{4 d \sqrt{a+b \cos (c+d x)}}-\frac{b \sin (c+d x) (12 a B+21 A b-8 b C) \sqrt{a+b \cos (c+d x)}}{12 d}+\frac{(4 a B+5 A b) \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{4 d}+\frac{A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{5/2}}{2 d} \]
[Out]
-((12*a^2*B - 24*b^2*B + a*b*(27*A - 56*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(1
2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + ((12*a^3*B + 48*a*b^2*B + 8*b^3*(3*A + C) + a^2*b*(33*A + 16*C))*Sqr
t[(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*(1
5*A*b^2 + 20*a*b*B + 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*(21*A*b + 12*a*B - 8*b*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(12
*d) + ((5*A*b + 4*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)
________________________________________________________________________________________
Rubi [A] time = 1.44759, antiderivative size = 372, normalized size of antiderivative
= 1., number of steps used = 11, number of rules used = 10, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\)
= 0.233, Rules used = {3047, 3049, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805}
\[ \frac{\left (a^2 b (33 A+16 C)+12 a^3 B+48 a b^2 B+8 b^3 (3 A+C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{12 d \sqrt{a+b \cos (c+d x)}}-\frac{\left (12 a^2 B+a b (27 A-56 C)-24 b^2 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{12 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{a \left (4 a^2 (A+2 C)+20 a b B+15 A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{4 d \sqrt{a+b \cos (c+d x)}}-\frac{b \sin (c+d x) (12 a B+21 A b-8 b C) \sqrt{a+b \cos (c+d x)}}{12 d}+\frac{(4 a B+5 A b) \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{4 d}+\frac{A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{5/2}}{2 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]
[Out]
-((12*a^2*B - 24*b^2*B + a*b*(27*A - 56*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(1
2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + ((12*a^3*B + 48*a*b^2*B + 8*b^3*(3*A + C) + a^2*b*(33*A + 16*C))*Sqr
t[(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*(1
5*A*b^2 + 20*a*b*B + 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*(21*A*b + 12*a*B - 8*b*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(12
*d) + ((5*A*b + 4*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 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 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 ^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{1}{2} (5 A b+4 a B)+(2 b B+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+4 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+20 a b B+4 a^2 (A+2 C)\right )-\frac{1}{2} b (a A-4 b B-8 a C) \cos (c+d x)-\frac{1}{4} b (21 A b+12 a B-8 b C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b (21 A b+12 a B-8 b C) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{12 d}+\frac{(5 A b+4 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+20 a b B+4 a^2 (A+2 C)\right )+\frac{1}{4} b \left (36 a b B+4 b^2 (3 A+C)+3 a^2 (A+12 C)\right ) \cos (c+d x)-\frac{1}{8} b \left (27 a A b+12 a^2 B-24 b^2 B-56 a b C\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx\\ &=-\frac{b (21 A b+12 a B-8 b C) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{12 d}+\frac{(5 A b+4 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+20 a b B+4 a^2 (A+2 C)\right )-\frac{1}{8} b \left (12 a^3 B+48 a b^2 B+8 b^3 (3 A+C)+a^2 b (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} \left (-27 a A b-12 a^2 B+24 b^2 B+56 a b C\right ) \int \sqrt{a+b \cos (c+d x)} \, dx\\ &=-\frac{b (21 A b+12 a B-8 b C) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{12 d}+\frac{(5 A b+4 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+20 a b B+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 (-12 a^3 B-48 a b^2 B-8 b^3 (3 A+C)-a^2 b (33 A+16 C)\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx+\frac{\left (\left (-27 a A b-12 a^2 B+24 b^2 B+56 a b 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}{24 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}\\ &=-\frac{\left (27 a A b+12 a^2 B-24 b^2 B-56 a b C\right ) \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 (21 A b+12 a B-8 b C) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{12 d}+\frac{(5 A b+4 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+20 a b B+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 (\left (-12 a^3 B-48 a b^2 B-8 b^3 (3 A+C)-a^2 b (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{\left (27 a A b+12 a^2 B-24 b^2 B-56 a b C\right ) \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{\left (12 a^3 B+48 a b^2 B+8 b^3 (3 A+C)+a^2 b (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+20 a b B+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 (21 A b+12 a B-8 b C) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{12 d}+\frac{(5 A b+4 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*}
Mathematica [C] time = 5.0116, size = 492, normalized size = 1.32 \[ \frac{\frac{8 b \left (3 a^2 (A+12 C)+36 a b B+4 b^2 (3 A+C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{\sqrt{a+b \cos (c+d x)}}+\frac{2 \left (24 a^3 (A+2 C)+108 a^2 b B+7 a b^2 (9 A+8 C)+24 b^3 B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{\sqrt{a+b \cos (c+d x)}}+4 \sec (c+d x) \sqrt{a+b \cos (c+d x)} \left (6 a^2 A \tan (c+d x)+3 a (4 a B+9 A b) \sin (c+d x)+4 b^2 C \sin (2 (c+d x))\right )+\frac{2 i \csc (c+d x) \left (-12 a^2 B-27 a A b+56 a b C+24 b^2 B\right ) \sqrt{-\frac{b (\cos (c+d x)-1)}{a+b}} \sqrt{\frac{b (\cos (c+d x)+1)}{b-a}} \left (b \left (b \Pi \left (\frac{a+b}{a};i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )}{a b \sqrt{-\frac{1}{a+b}}}}{48 d} \]
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]^3,x]
[Out]
((8*b*(36*a*b*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*(108*a^2*b*B + 24*b^3*B + 24*a^3*(A + 2*C) + 7*a*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*A*b - 12*a^2*B + 24*b^2*B + 56*a*b*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*Cos[c + d*x]]*Sec[c + d*x]*(3*a*(9*A*b + 4*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)
________________________________________________________________________________________
Maple [B] time = 2.752, size = 2375, normalized size = 6.4 \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)^3,x)
[Out]
-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(8*C*b^3*(-1/6/b*cos(1/2*d*x+1/2*c)*(-2*b*sin(1
/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+1/6*(a-b)/b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2
*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(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))-1/12/b^2*(-2*a+6*b)*(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-
b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(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*b^3+6*C*a*b^2-4*C*b^3)*(a-b)*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x
+1/2*c)^2)^(1/2)/b*(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*A*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1
/2*c)^4+(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*a*b^2*B*(sin(1/2*
d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(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*b^3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(
1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(co
s(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+6*a^2*b*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a
-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(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*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d
*x+1/2*c)^4+(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*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(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*a^2*(3*A*b+B*a)*(-1/a*cos(1/2*d*x+1/2*c)*
(-2*b*sin(1/2*d*x+1/2*c)^4+(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*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(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*cos(1/2*d*x+1/2*c
)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(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*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*
sin(1/2*d*x+1/2*c)^4+(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*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(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)))-2*a*(3*A*b^2+3*B*a*b+C*a^2)*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(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))+2*A*a^3*(-1/2/a*cos(1/2*d*x+1/2*c
)*(-2*b*sin(1/2*d*x+1/2*c)^4+(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*b*sin(1/2*d*x+1/2*c)^4+(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*(si
n(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(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*
cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*Ellipt
icE(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*cos(1/2*d*x+1/2*c)^2*b
+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(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*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2
*d*x+1/2*c)^4+(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*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(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))/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(-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)^3,x, algorithm="maxima")
[Out]
Timed out
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} +{\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} + A a^{2} +{\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sqrt{b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3}, x\right ) \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)^3,x, algorithm="fricas")
[Out]
integral((C*b^2*cos(d*x + c)^4 + (2*C*a*b + B*b^2)*cos(d*x + c)^3 + A*a^2 + (C*a^2 + 2*B*a*b + A*b^2)*cos(d*x
+ c)^2 + (B*a^2 + 2*A*a*b)*cos(d*x + c))*sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^3, x)
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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)**3,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 )^{3}\,{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)^3,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)^3, x)