3.641 \(\int (a+b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x)) \sec (c+d x) \, dx\)
Optimal. Leaf size=342 \[ \frac{2 \left (3 a^2 C+b^2 (7 A+5 C)\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{21 d}+\frac{2 \left (2 a^2 b^2 (7 A-C)-3 a^4 C+b^4 (7 A+5 C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{21 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 a \left (3 a^2 C+49 A b^2+29 b^2 C\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{21 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 a^3 A \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \cos (c+d x)}}+\frac{2 a C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{7 d}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 d} \]
[Out]
(2*a*(49*A*b^2 + 3*a^2*C + 29*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(21*b*d*S
qrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(2*a^2*b^2*(7*A - C) - 3*a^4*C + b^4*(7*A + 5*C))*Sqrt[(a + b*Cos[c +
d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(21*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*(3*a^2*C
+ b^2*(7*A + 5*C))*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2*a*C*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d
*x])/(7*d) + (2*C*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*d)
________________________________________________________________________________________
Rubi [A] time = 1.22306, antiderivative size = 342, normalized size of antiderivative
= 1., number of steps used = 11, number of rules used = 10, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\)
= 0.303, Rules used = {3050, 3049, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805}
\[ \frac{2 \left (3 a^2 C+b^2 (7 A+5 C)\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{21 d}+\frac{2 \left (2 a^2 b^2 (7 A-C)-3 a^4 C+b^4 (7 A+5 C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{21 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 a \left (3 a^2 C+49 A b^2+29 b^2 C\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{21 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 a^3 A \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \cos (c+d x)}}+\frac{2 a C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{7 d}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x],x]
[Out]
(2*a*(49*A*b^2 + 3*a^2*C + 29*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(21*b*d*S
qrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(2*a^2*b^2*(7*A - C) - 3*a^4*C + b^4*(7*A + 5*C))*Sqrt[(a + b*Cos[c +
d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(21*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*(3*a^2*C
+ b^2*(7*A + 5*C))*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2*a*C*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d
*x])/(7*d) + (2*C*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*d)
Rule 3050
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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 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+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} b (7 A+5 C) \cos (c+d x)+\frac{5}{2} a C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{2 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 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{5}{2} a b (7 A+4 C) \cos (c+d x)+\frac{5}{4} \left (3 a^2 C+b^2 (7 A+5 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{2 \left (3 a^2 C+b^2 (7 A+5 C)\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 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{5}{8} b \left (9 a^2 (7 A+3 C)+b^2 (7 A+5 C)\right ) \cos (c+d x)+\frac{5}{8} a \left (49 A b^2+3 a^2 C+29 b^2 C\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx\\ &=\frac{2 \left (3 a^2 C+b^2 (7 A+5 C)\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 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{5}{8} \left (2 a^2 b^2 (7 A-C)-3 a^4 C+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 (a \left (49 A b^2+3 a^2 C+29 b^2 C\right )\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{21 b}\\ &=\frac{2 \left (3 a^2 C+b^2 (7 A+5 C)\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 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 (2 a^2 b^2 (7 A-C)-3 a^4 C+b^4 (7 A+5 C)\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{21 b}+\frac{\left (a \left (49 A b^2+3 a^2 C+29 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}{21 b \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}\\ &=\frac{2 a \left (49 A b^2+3 a^2 C+29 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 )}{21 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 \left (3 a^2 C+b^2 (7 A+5 C)\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 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 (2 a^2 b^2 (7 A-C)-3 a^4 C+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}{21 b \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 a \left (49 A b^2+3 a^2 C+29 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 )}{21 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 \left (2 a^2 b^2 (7 A-C)-3 a^4 C+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 )}{21 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 (3 a^2 C+b^2 (7 A+5 C)\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 d}+\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}\\ \end{align*}
Mathematica [C] time = 3.08505, size = 468, normalized size = 1.37 \[ \frac{2 \sin (c+d x) \sqrt{a+b \cos (c+d x)} \left (18 a^2 C+18 a b C \cos (c+d x)+14 A b^2+3 b^2 C \cos (2 (c+d x))+13 b^2 C\right )+\frac{4 b \left (9 a^2 (7 A+3 C)+b^2 (7 A+5 C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{\sqrt{a+b \cos (c+d x)}}+\frac{2 a \left (3 a^2 (14 A+C)+b^2 (49 A+29 C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{\sqrt{a+b \cos (c+d x)}}+\frac{2 i \left (3 a^2 C+49 A b^2+29 b^2 C\right ) \csc (c+d x) \sqrt{-\frac{b (\cos (c+d x)-1)}{a+b}} \sqrt{\frac{b (\cos (c+d x)+1)}{b-a}} \left (b \left (b \Pi \left (\frac{a+b}{a};i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )}{b^2 \sqrt{-\frac{1}{a+b}}}}{42 d} \]
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],x]
[Out]
((4*b*(9*a^2*(7*A + 3*C) + b^2*(7*A + 5*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*(3*a^2*(14*A + C) + 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)*(49*A*b^2 + 3*a^2*C + 29*b^2*
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)*E
llipticE[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*(-2*a*EllipticF[I*ArcSi
nh[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)])))/(b^2*Sqrt[-(a + b)^(-1)]) + 2*Sqrt[a + b*Cos[c + d
*x]]*(14*A*b^2 + 18*a^2*C + 13*b^2*C + 18*a*b*C*Cos[c + d*x] + 3*b^2*C*Cos[2*(c + d*x)])*Sin[c + d*x])/(42*d)
________________________________________________________________________________________
Maple [B] time = 0.456, size = 1209, normalized size = 3.5 \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+C*cos(d*x+c)^2)*sec(d*x+c),x)
[Out]
-2/21*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(48*C*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^8+(-96*C*a*b^3-72*C*b^4)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(28*A*b^4+72*C*a^2*b^2+96*C*a*b^3+56*C*b^
4)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-14*A*a*b^3-14*A*b^4-18*C*a^3*b-36*C*a^2*b^2-34*C*a*b^3-16*C*b^4)*
sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+49*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^2-49*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^3-21*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+14*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+7*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))+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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2
))*a^4-3*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)*EllipticE(cos(1/2*
d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b+29*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^2-29*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^3-3*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))*a^4-2*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)*Ellipt
icF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^2+5*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)))/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} + 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+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^(5/2)*sec(d*x + c), x)
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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} + 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 ), 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+C*cos(d*x+c)^2)*sec(d*x+c),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), 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+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} + 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+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^(5/2)*sec(d*x + c), x)