3.1033 \(\int (a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\)

Optimal. Leaf size=357 \[ \frac {\left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {a^2 (2 a B+5 A b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {\left (a^3 (15 A-16 C)+20 a^2 b B+4 a b^2 (15 A+4 C)+10 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {a+b \cos (c+d x)}}-\frac {b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{15 d}-\frac {b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d} \]

[Out]

-1/5*b*(5*A-2*C)*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/d-1/15*b*(15*A*a-10*B*b-16*C*a)*sin(d*x+c)*(a+b*cos(d*x+c))
^(1/2)/d+1/15*(70*a*b*B-a^2*(15*A-46*C)+6*b^2*(5*A+3*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip
ticE(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)+1/15*
(20*a^2*b*B+10*b^3*B+a^3*(15*A-16*C)+4*a*b^2*(15*A+4*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip
ticF(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)+a^2*(
5*A*b+2*B*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)+A*(a+b*cos(d*x+c))^(5/2)*tan(d*x+c)/d

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Rubi [A]  time = 1.40, antiderivative size = 357, normalized size of antiderivative = 1.00, 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} \[ \frac {\left (a^3 (15 A-16 C)+20 a^2 b B+4 a b^2 (15 A+4 C)+10 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (a^2 (-(15 A-46 C))+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {a^2 (2 a B+5 A b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {b \sin (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \cos (c+d x)}}{15 d}-\frac {b (5 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{5/2}}{d} \]

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]^2,x]

[Out]

((70*a*b*B - a^2*(15*A - 46*C) + 6*b^2*(5*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a +
b)])/(15*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + ((20*a^2*b*B + 10*b^3*B + a^3*(15*A - 16*C) + 4*a*b^2*(15*A
+ 4*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(15*d*Sqrt[a + b*Cos[c + d*x
]]) + (a^2*(5*A*b + 2*a*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(d*Sq
rt[a + b*Cos[c + d*x]]) - (b*(15*a*A - 10*b*B - 16*a*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*d) - (b*(5*
A - 2*C)*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + (A*(a + b*Cos[c + d*x])^(5/2)*Tan[c + d*x])/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 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+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^{5/2} \tan (c+d x)}{d}+\int (a+b \cos (c+d x))^{3/2} \left (\frac {1}{2} (5 A b+2 a B)+(b B+a C) \cos (c+d x)-\frac {1}{2} b (5 A-2 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b (5 A-2 C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^{5/2} \tan (c+d x)}{d}+\frac {2}{5} \int \sqrt {a+b \cos (c+d x)} \left (\frac {5}{4} a (5 A b+2 a B)+\frac {1}{2} \left (5 A b^2+10 a b B+5 a^2 C+3 b^2 C\right ) \cos (c+d x)-\frac {1}{4} b (15 a A-10 b B-16 a C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b (15 a A-10 b B-16 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^{5/2} \tan (c+d x)}{d}+\frac {4}{15} \int \frac {\left (\frac {15}{8} a^2 (5 A b+2 a B)+\frac {1}{4} \left (45 a^2 b B+5 b^3 B+15 a^3 C+a b^2 (45 A+17 C)\right ) \cos (c+d x)+\frac {1}{8} b \left (70 a b B-a^2 (15 A-46 C)+6 b^2 (5 A+3 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\\ &=-\frac {b (15 a A-10 b B-16 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^{5/2} \tan (c+d x)}{d}-\frac {4 \int \frac {\left (-\frac {15}{8} a^2 b (5 A b+2 a B)-\frac {1}{8} b \left (20 a^2 b B+10 b^3 B+a^3 (15 A-16 C)+4 a b^2 (15 A+4 C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b}+\frac {1}{30} \left (70 a b B-a^2 (15 A-46 C)+6 b^2 (5 A+3 C)\right ) \int \sqrt {a+b \cos (c+d x)} \, dx\\ &=-\frac {b (15 a A-10 b B-16 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^{5/2} \tan (c+d x)}{d}+\frac {1}{2} \left (a^2 (5 A b+2 a B)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx-\frac {1}{30} \left (-20 a^2 b B-10 b^3 B-a^3 (15 A-16 C)-4 a b^2 (15 A+4 C)\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx+\frac {\left (\left (70 a b B-a^2 (15 A-46 C)+6 b^2 (5 A+3 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}{30 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\\ &=\frac {\left (70 a b B-a^2 (15 A-46 C)+6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {b (15 a A-10 b B-16 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^{5/2} \tan (c+d x)}{d}+\frac {\left (a^2 (5 A b+2 a B) \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}{2 \sqrt {a+b \cos (c+d x)}}-\frac {\left (\left (-20 a^2 b B-10 b^3 B-a^3 (15 A-16 C)-4 a b^2 (15 A+4 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}{30 \sqrt {a+b \cos (c+d x)}}\\ &=\frac {\left (70 a b B-a^2 (15 A-46 C)+6 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (20 a^2 b B+10 b^3 B+a^3 (15 A-16 C)+4 a b^2 (15 A+4 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 )}{15 d \sqrt {a+b \cos (c+d x)}}+\frac {a^2 (5 A b+2 a B) \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 {b (15 a A-10 b B-16 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^{5/2} \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [C]  time = 4.25, size = 502, normalized size = 1.41 \[ \frac {60 \sqrt {a+b \cos (c+d x)} \left (a^2 A \tan (c+d x)+\frac {2}{15} b (11 a C+5 b B) \sin (c+d x)+\frac {1}{5} b^2 C \sin (2 (c+d x))\right )+\frac {2 i \csc (c+d x) \left (a^2 (46 C-15 A)+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {-\frac {b (\cos (c+d x)-1)}{a+b}} \sqrt {-\frac {b (\cos (c+d x)+1)}{a-b}} \left (b \left (b \Pi \left (\frac {a+b}{a};i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )}{a b \sqrt {-\frac {1}{a+b}}}+\frac {8 \left (15 a^3 C+45 a^2 b B+a b^2 (45 A+17 C)+5 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (60 a^3 B+a^2 b (135 A+46 C)+70 a b^2 B+6 b^3 (5 A+3 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}}{60 d} \]

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]^2,x]

[Out]

((8*(45*a^2*b*B + 5*b^3*B + 15*a^3*C + a*b^2*(45*A + 17*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*(60*a^3*B + 70*a*b^2*B + 6*b^3*(5*A + 3*C) + a^2*b*(135*
A + 46*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)*(70*a*b*B + 6*b^2*(5*A + 3*C) + a^2*(-15*A + 46*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)]) + 60*Sqrt[a + b*Cos[c + d*x]]*((2*b*(5*b*B + 11*a*C)*Sin[c + d*x])/15
+ (b^2*C*Sin[2*(c + d*x)])/5 + a^2*A*Tan[c + d*x]))/(60*d)

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fricas [F]  time = 8.28, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} + {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} + A a^{2} + {\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{2}, x\right ) \]

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)^2,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)^2, x)

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

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)^2,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)^2, x)

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maple [B]  time = 4.02, size = 2274, normalized size = 6.37 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(96*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^8+(-80*B*b^3-224*C*a*b^2-144*C*b^3)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(60*A*a^2*b+40*B*a*b^2+80*B*b^
3+88*C*a^2*b+224*C*a*b^2+72*C*b^3)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-30*A*a^3-30*A*a^2*b-20*B*a*b^2-20
*B*b^3-44*C*a^2*b-56*C*a*b^2-12*C*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-2*(-2*b/(a-b)*sin(1/2*d*x+1/2*c
)^2+(a+b)/(a-b))^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(15*A*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3
+60*A*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2-15*A*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1
/2))*a^3+15*A*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b+30*A*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(
a-b))^(1/2))*a*b^2-30*A*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3-75*A*EllipticPi(cos(1/2*d*x+1/2*c
),2,(-2*b/(a-b))^(1/2))*a^2*b+20*B*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b+10*B*EllipticF(cos(1
/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+70*B*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b-70*B*Ellipti
cE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2-30*B*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a^3-1
6*C*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+16*C*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))
*a*b^2+46*C*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3-46*C*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b)
)^(1/2))*a^2*b+18*C*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2-18*C*EllipticE(cos(1/2*d*x+1/2*c),(
-2*b/(a-b))^(1/2))*b^3)*sin(1/2*d*x+1/2*c)^2+15*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^3+60*A*a*b^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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-15*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^3+15*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+30*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^2-30*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))
*b^3-75*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^2*b+20*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^2*b+10*B*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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+70*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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/
(a-b))^(1/2))*a^2*b-70*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)*Elli
pticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2-30*a^3*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)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))-16*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^
3+16*a*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))*b^2+46*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^3-46*C*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b+18*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^2-18*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))*b^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*co
s(1/2*d*x+1/2*c)^2-1)/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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2}\,{d x} \]

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)^2,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)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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