3.317 \(\int (a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx\)
Optimal. Leaf size=376 \[ \frac {\left (16 a^2 A+54 a b B+33 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{24 d}-\frac {\left (16 a^2 A+54 a b B+33 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (16 a^3 A+66 a^2 b B+59 a A b^2+48 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (8 a^3 B+20 a^2 A b+30 a b^2 B+5 A b^3\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{8 d \sqrt {a+b \cos (c+d x)}}+\frac {a (2 a B+3 A b) \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{4 d}+\frac {a A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{3/2}}{3 d} \]
[Out]
-1/24*(16*A*a^2+33*A*b^2+54*B*a*b)*(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)/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+1/24*(16*A*a^3+59*A*a*b^2+6
6*B*a^2*b+48*B*b^3)*(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)/d/(a+b*cos(d*x+c))^(1/2)+1/8*(20*A*a^2*b+5*A*b^3+8*B*a^3+30*B*a*b^2
)*(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)+1/3*a*A*(a+b*cos(d*x+c))^(3/2)*sec(d*x+c)^2*tan(d*x+c)/d+
1/24*(16*A*a^2+33*A*b^2+54*B*a*b)*(a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/d+1/4*a*(3*A*b+2*B*a)*sec(d*x+c)*(a+b*cos(
d*x+c))^(1/2)*tan(d*x+c)/d
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Rubi [A] time = 1.43, antiderivative size = 376, normalized size of antiderivative =
1.00, number of steps used = 11, number of rules used = 11, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) =
0.333, Rules used = {2989, 3047, 3055, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805}
\[ \frac {\left (16 a^2 A+54 a b B+33 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{24 d}+\frac {\left (16 a^3 A+66 a^2 b B+59 a A b^2+48 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 d \sqrt {a+b \cos (c+d x)}}-\frac {\left (16 a^2 A+54 a b B+33 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (20 a^2 A b+8 a^3 B+30 a b^2 B+5 A b^3\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{8 d \sqrt {a+b \cos (c+d x)}}+\frac {a (2 a B+3 A b) \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{4 d}+\frac {a A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{3/2}}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^4,x]
[Out]
-((16*a^2*A + 33*A*b^2 + 54*a*b*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(24*d*Sqrt[
(a + b*Cos[c + d*x])/(a + b)]) + ((16*a^3*A + 59*a*A*b^2 + 66*a^2*b*B + 48*b^3*B)*Sqrt[(a + b*Cos[c + d*x])/(a
+ b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(24*d*Sqrt[a + b*Cos[c + d*x]]) + ((20*a^2*A*b + 5*A*b^3 + 8*a^3
*B + 30*a*b^2*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(8*d*Sqrt[a + b
*Cos[c + d*x]]) + ((16*a^2*A + 33*A*b^2 + 54*a*b*B)*Sqrt[a + b*Cos[c + d*x]]*Tan[c + d*x])/(24*d) + (a*(3*A*b
+ 2*a*B)*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(4*d) + (a*A*(a + b*Cos[c + d*x])^(3/2)*Sec[c + d
*x]^2*Tan[c + d*x])/(3*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 2989
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((b*c - a*d)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)
*(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 - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (
A*b + a*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)
*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] && GtQ[m, 1] && LtQ[n, -1]
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 3055
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[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 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} (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx &=\frac {a A (a+b \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int \sqrt {a+b \cos (c+d x)} \left (\frac {3}{2} a (3 A b+2 a B)+\left (2 a^2 A+3 A b^2+6 a b B\right ) \cos (c+d x)+\frac {1}{2} b (a A+6 b B) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a (3 A b+2 a B) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a A (a+b \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{6} \int \frac {\left (\frac {1}{4} a \left (16 a^2 A+33 A b^2+54 a b B\right )+\frac {1}{2} \left (19 a^2 A b+12 A b^3+6 a^3 B+36 a b^2 B\right ) \cos (c+d x)+\frac {1}{4} b \left (13 a A b+6 a^2 B+24 b^2 B\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {\left (16 a^2 A+33 A b^2+54 a b B\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 d}+\frac {a (3 A b+2 a B) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a A (a+b \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int \frac {\left (\frac {3}{8} a \left (20 a^2 A b+5 A b^3+8 a^3 B+30 a b^2 B\right )+\frac {1}{4} a b \left (13 a A b+6 a^2 B+24 b^2 B\right ) \cos (c+d x)-\frac {1}{8} a b \left (16 a^2 A+33 A b^2+54 a b B\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{6 a}\\ &=\frac {\left (16 a^2 A+33 A b^2+54 a b B\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 d}+\frac {a (3 A b+2 a B) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a A (a+b \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {\int \frac {\left (-\frac {3}{8} a b \left (20 a^2 A b+5 A b^3+8 a^3 B+30 a b^2 B\right )-\frac {1}{8} a b \left (16 a^3 A+59 a A b^2+66 a^2 b B+48 b^3 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{6 a b}+\frac {1}{48} \left (-16 a^2 A-33 A b^2-54 a b B\right ) \int \sqrt {a+b \cos (c+d x)} \, dx\\ &=\frac {\left (16 a^2 A+33 A b^2+54 a b B\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 d}+\frac {a (3 A b+2 a B) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a A (a+b \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{16} \left (-20 a^2 A b-5 A b^3-8 a^3 B-30 a b^2 B\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx-\frac {1}{48} \left (-16 a^3 A-59 a A b^2-66 a^2 b B-48 b^3 B\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx+\frac {\left (\left (-16 a^2 A-33 A b^2-54 a b B\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{48 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\\ &=-\frac {\left (16 a^2 A+33 A b^2+54 a b B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (16 a^2 A+33 A b^2+54 a b B\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 d}+\frac {a (3 A b+2 a B) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a A (a+b \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {\left (\left (-20 a^2 A b-5 A b^3-8 a^3 B-30 a b^2 B\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}{16 \sqrt {a+b \cos (c+d x)}}-\frac {\left (\left (-16 a^3 A-59 a A b^2-66 a^2 b B-48 b^3 B\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}{48 \sqrt {a+b \cos (c+d x)}}\\ &=-\frac {\left (16 a^2 A+33 A b^2+54 a b B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (16 a^3 A+59 a A b^2+66 a^2 b B+48 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 )}{24 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (20 a^2 A b+5 A b^3+8 a^3 B+30 a b^2 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 )}{8 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (16 a^2 A+33 A b^2+54 a b B\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 d}+\frac {a (3 A b+2 a B) \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a A (a+b \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}
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Mathematica [C] time = 6.05, size = 486, normalized size = 1.29 \[ \frac {\frac {8 b \left (6 a^2 B+13 a A b+24 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+4 \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)} \left (\left (8 a^2 A+27 a b B+\frac {33 A b^2}{2}\right ) \sin (2 (c+d x))+8 a^2 A \tan (c+d x)+2 a (6 a B+13 A b) \sin (c+d x)\right )-\frac {2 i \left (16 a^2 A+54 a b B+33 A b^2\right ) \csc (c+d x) \sqrt {-\frac {b (\cos (c+d x)-1)}{a+b}} \sqrt {\frac {b (\cos (c+d x)+1)}{b-a}} \left (b \left (b \Pi \left (\frac {a+b}{a};i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )}{a b \sqrt {-\frac {1}{a+b}}}+\frac {2 \left (48 a^3 B+104 a^2 A b+126 a b^2 B-3 A b^3\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}}{96 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^4,x]
[Out]
((8*b*(13*a*A*b + 6*a^2*B + 24*b^2*B)*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*(104*a^2*A*b - 3*A*b^3 + 48*a^3*B + 126*a*b^2*B)*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)*(16*a^2*A + 33*A*b^2 + 54*a
*b*B)*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*Ar
cSinh[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]^2*(2*a*(13*A*b + 6*a*B)*Sin[c + d*x] + (8*a^2*A + (33*A*b^2)/2 + 27*a*b*B)*Sin[2*(c + d*x
)] + 8*a^2*A*Tan[c + d*x]))/(96*d)
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fricas [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))*sec(d*x+c)^4,x, algorithm="fricas")
[Out]
Timed out
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{4}\,{d x} \]
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))*sec(d*x+c)^4,x, algorithm="giac")
[Out]
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(5/2)*sec(d*x + c)^4, x)
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maple [B] time = 4.61, size = 2438, normalized size = 6.48 \[ \text {Expression too large to display} \]
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))*sec(d*x+c)^4,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)*(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*sin(1/2*d*x+1/2*c)^4*b+(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^2*(A*b+3*B*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*sin(1/2*d*x+1/2*c)^4*b+(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^2*(3*A*b+B*a)*(-1/2/a*cos(1/2*d*x+1/2*c)*(-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*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*sin(1/2*d*x+1/2*c)^4*b+(a+b)*si
n(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*(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*sin(1/2*d*x+1/2*c)^4*b+(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*s
in(1/2*d*x+1/2*c)^4*b+(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))-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*sin(1/2*d*x+1/2*c)^4*b+
(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*sin(1/2*d*x+1/2*c)^4*b+(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*(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*sin(1/2*d*x+1/2*c)^4*b+(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)+2*A*a^3*(-1/3/a*cos(1/2*d*x+1/2*c)*(-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*cos(1/2*d*x+1/2*c)^2-1)^3+5/12*b/a^2*cos(1/2*d*x+1/2*c)*(-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*cos(1/2*d*x+1/2*c)^2-1)^2-1/24*(16*a^2+15*b^2)/a^3*cos(1/2*d*x+1/2*c)*(-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*cos(1/2*d*x+1/2*c)^2-1)+5/48*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*sin(1/2*d*x+1/2*c)^4*b+(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))+1/3*(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*sin(1/2*d*x+1/2*c)^4*b+(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))-1/3*(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*si
n(1/2*d*x+1/2*c)^4*b+(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))+1/3/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*sin(1/2*d*x+1/2*c)^4*b+(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))-5/16*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*sin(1/2*d*x+1/2*c)^4*b+(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))+5/16/a^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*sin(1/2*d*x+1/2*c)^4*b+(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))*b^3+1/4/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*sin(1/2*d*x+1/2*c)^4*b+(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))+5/16*b^3/a^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*sin(1/2*d*x+1/
2*c)^4*b+(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)))+6*a*b*(A*b+B*a
)*(-1/a*cos(1/2*d*x+1/2*c)*(-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*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*sin(1/2*d*x+1/2*c)^4*b+
(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*sin(1/2*d*x+1/2*c)^4*b+(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*sin(1/2*d*x+1/2*c)^4*b+(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*s
in(1/2*d*x+1/2*c)^4*b+(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))))/
sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{4}\,{d x} \]
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))*sec(d*x+c)^4,x, algorithm="maxima")
[Out]
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(5/2)*sec(d*x + c)^4, x)
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^4} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^(5/2))/cos(c + d*x)^4,x)
[Out]
int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^(5/2))/cos(c + d*x)^4, 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))*sec(d*x+c)**4,x)
[Out]
Timed out
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