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 \[ \frac {2 a^3 A \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (3 a^2 C+b^2 (7 A+5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{21 d}+\frac {2 a \left (3 a^2 C+49 A b^2+29 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{21 b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (-3 a^4 C+2 a^2 b^2 (7 A-C)+b^4 (7 A+5 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{21 b d \sqrt {a+b \cos (c+d x)}}+\frac {2 a C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{7 d}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 d} \]

[Out]

2/7*a*C*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/7*C*(a+b*cos(d*x+c))^(5/2)*sin(d*x+c)/d+2/21*(3*a^2*C+b^2*(7*A+5
*C))*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d+2/21*a*(49*A*b^2+3*C*a^2+29*C*b^2)*(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)/b/d/((a+b*cos(d*x+c
))/(a+b))^(1/2)+2/21*(2*a^2*b^2*(7*A-C)-3*a^4*C+b^4*(7*A+5*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)/b/d/(a+b*cos(d*x+c))^(1/
2)+2*a^3*A*(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)

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Rubi [A]  time = 1.22, antiderivative size = 342, normalized size of antiderivative = 1.00, 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 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 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 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 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 (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*}

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Mathematica [C]  time = 3.08, 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)

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fricas [F]  time = 1.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} + 2 \, C a b \cos \left (d x + c\right )^{3} + 2 \, A a b \cos \left (d x + c\right ) + A a^{2} + {\left (C a^{2} + A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right ), x\right ) \]

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),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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )\,{d x} \]

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),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)

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maple [B]  time = 2.63, size = 1209, normalized size = 3.54 \[ \text {result too large to display} \]

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),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)+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))+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-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)*EllipticF(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))+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)/b/(-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*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )\,{d x} \]

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),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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{\cos \left (c+d\,x\right )} \,d x \]

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),x)

[Out]

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

[Out]

Timed out

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