\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\) [621]

Optimal result

Integrand size = 35, antiderivative size = 423 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \left (12 a^2 A b+48 A b^3-5 a^3 B-40 a b^2 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{15 a^4 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (9 a^4 A+24 a^2 A b^2-48 A b^4-25 a^3 b B+40 a b^3 B\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{15 a^4 \left (a^2-b^2\right ) d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (9 a^2 A b-24 A b^3-5 a^3 B+20 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d}+\frac {2 \left (a^2 A-6 A b^2+5 a b B\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d} \]

[Out]

2*b*(A*b-B*a)*cos(d*x+c)^(3/2)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)-2/15*(12*A*a^2*b+48*A*b^3-5*B*a
^3-40*B*a*b^2)*(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)*(a/(a+b))^
(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)/a^4/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+2/5*(A*a^2-6*A*b^2+5*B*a*b
)*cos(d*x+c)^(3/2)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a^2/(a^2-b^2)/d-2/15*(9*A*a^2*b-24*A*b^3-5*B*a^3+20*B*a*b
^2)*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/a^3/(a^2-b^2)/d+2/15*(9*A*a^4+24*A*a^2*b^2-48*A*b^4-25*
B*a^3*b+40*B*a*b^3)*(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)*(a/(a
+b))^(1/2))*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/a^4/(a^2-b^2)/d/((b+a*cos(d*x+c))/(a+b))^(1/2)

Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3034, 4115, 4189, 4120, 3941, 2734, 2732, 3943, 2742, 2740} \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \left (a^2 A+5 a b B-6 A b^2\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{5 a^2 d \left (a^2-b^2\right )}+\frac {2 b (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (-5 a^3 B+9 a^2 A b+20 a b^2 B-24 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}{15 a^3 d \left (a^2-b^2\right )}-\frac {2 \left (-5 a^3 B+12 a^2 A b-40 a b^2 B+48 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{15 a^4 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (9 a^4 A-25 a^3 b B+24 a^2 A b^2+40 a b^3 B-48 A b^4\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a^4 d \left (a^2-b^2\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \]

[In]

Int[(Cos[c + d*x]^(5/2)*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(-2*(12*a^2*A*b + 48*A*b^3 - 5*a^3*B - 40*a*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (
2*a)/(a + b)])/(15*a^4*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(9*a^4*A + 24*a^2*A*b^2 - 48*A*b^4
- 25*a^3*b*B + 40*a*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/
(15*a^4*(a^2 - b^2)*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (2*b*(A*b - a*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/
(a*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) - (2*(9*a^2*A*b - 24*A*b^3 - 5*a^3*B + 20*a*b^2*B)*Sqrt[Cos[c + d*x
]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(15*a^3*(a^2 - b^2)*d) + (2*(a^2*A - 6*A*b^2 + 5*a*b*B)*Cos[c + d*x]
^(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(5*a^2*(a^2 - b^2)*d)

Rule 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2742

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 3034

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[(a + b*Csc[e + f*x])^m*((
c + d*Csc[e + f*x])^n/(g*Csc[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 3941

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3943

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[Sqrt[d*C
sc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x]]/Sqrt[a + b*Csc[e + f*x]]), Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4115

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[b*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(
a*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*
x])^n*Simp[A*(a^2*(m + 1) - b^2*(m + n + 1)) + a*b*B*n - a*(A*b - a*B)*(m + 1)*Csc[e + f*x] + b*(A*b - a*B)*(m
+ n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b
^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] && ILtQ[n, 0])

Rule 4120

Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]

Rule 4189

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1
)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}} \, dx \\ & = \frac {2 b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} \left (-a^2 A+6 A b^2-5 a b B\right )+\frac {1}{2} a (A b-a B) \sec (c+d x)-2 b (A b-a B) \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx}{a \left (a^2-b^2\right )} \\ & = \frac {2 b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (a^2 A-6 A b^2+5 a b B\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} \left (-9 a^2 A b+24 A b^3+5 a^3 B-20 a b^2 B\right )+\frac {1}{4} a \left (3 a^2 A+2 A b^2-5 a b B\right ) \sec (c+d x)+\frac {1}{2} b \left (a^2 A-6 A b^2+5 a b B\right ) \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx}{5 a^2 \left (a^2-b^2\right )} \\ & = \frac {2 b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (9 a^2 A b-24 A b^3-5 a^3 B+20 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d}+\frac {2 \left (a^2 A-6 A b^2+5 a b B\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d}-\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{8} \left (-9 a^4 A-24 a^2 A b^2+48 A b^4+25 a^3 b B-40 a b^3 B\right )+\frac {1}{8} a \left (3 a^2 A b+12 A b^3-5 a^3 B-10 a b^2 B\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )} \\ & = \frac {2 b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (9 a^2 A b-24 A b^3-5 a^3 B+20 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d}+\frac {2 \left (a^2 A-6 A b^2+5 a b B\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d}-\frac {\left (\left (12 a^2 A b+48 A b^3-5 a^3 B-40 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 a^4}-\frac {\left (\left (-9 a^4 A-24 a^2 A b^2+48 A b^4+25 a^3 b B-40 a b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{15 a^4 \left (a^2-b^2\right )} \\ & = \frac {2 b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (9 a^2 A b-24 A b^3-5 a^3 B+20 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d}+\frac {2 \left (a^2 A-6 A b^2+5 a b B\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d}-\frac {\left (\left (12 a^2 A b+48 A b^3-5 a^3 B-40 a b^2 B\right ) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{15 a^4 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (\left (-9 a^4 A-24 a^2 A b^2+48 A b^4+25 a^3 b B-40 a b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{15 a^4 \left (a^2-b^2\right ) \sqrt {b+a \cos (c+d x)}} \\ & = \frac {2 b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (9 a^2 A b-24 A b^3-5 a^3 B+20 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d}+\frac {2 \left (a^2 A-6 A b^2+5 a b B\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d}-\frac {\left (\left (12 a^2 A b+48 A b^3-5 a^3 B-40 a b^2 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{15 a^4 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (\left (-9 a^4 A-24 a^2 A b^2+48 A b^4+25 a^3 b B-40 a b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{15 a^4 \left (a^2-b^2\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}}} \\ & = -\frac {2 \left (12 a^2 A b+48 A b^3-5 a^3 B-40 a b^2 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{15 a^4 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (9 a^4 A+24 a^2 A b^2-48 A b^4-25 a^3 b B+40 a b^3 B\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{15 a^4 \left (a^2-b^2\right ) d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (9 a^2 A b-24 A b^3-5 a^3 B+20 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d}+\frac {2 \left (a^2 A-6 A b^2+5 a b B\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 27.47 (sec) , antiderivative size = 533, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {(b+a \cos (c+d x))^2 \left (\frac {2 (-9 A b+5 a B) \sin (c+d x)}{15 a^3}+\frac {2 \left (A b^4 \sin (c+d x)-a b^3 B \sin (c+d x)\right )}{a^3 \left (a^2-b^2\right ) (b+a \cos (c+d x))}+\frac {A \sin (2 (c+d x))}{5 a^2}\right )}{d \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}-\frac {2 \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (-i (a+b) \left (9 a^4 A+24 a^2 A b^2-48 A b^4-25 a^3 b B+40 a b^3 B\right ) E\left (i \text {arcsinh}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+i a (a+b) \left (-48 A b^3-6 a^2 b (2 A+5 B)+a^3 (9 A+5 B)+4 a b^2 (9 A+10 B)\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-\left (9 a^4 A+24 a^2 A b^2-48 A b^4-25 a^3 b B+40 a b^3 B\right ) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{15 a^4 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}} \]

[In]

Integrate[(Cos[c + d*x]^(5/2)*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

((b + a*Cos[c + d*x])^2*((2*(-9*A*b + 5*a*B)*Sin[c + d*x])/(15*a^3) + (2*(A*b^4*Sin[c + d*x] - a*b^3*B*Sin[c +
d*x]))/(a^3*(a^2 - b^2)*(b + a*Cos[c + d*x])) + (A*Sin[2*(c + d*x)])/(5*a^2)))/(d*Cos[c + d*x]^(3/2)*(a + b*S
ec[c + d*x])^(3/2)) - (2*Cos[c + d*x]^(3/2)*(b + a*Cos[c + d*x])*Sec[c + d*x]^(3/2)*(Cos[(c + d*x)/2]^2*Sec[c
+ d*x])^(3/2)*((-I)*(a + b)*(9*a^4*A + 24*a^2*A*b^2 - 48*A*b^4 - 25*a^3*b*B + 40*a*b^3*B)*EllipticE[I*ArcSinh[
Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)
] + I*a*(a + b)*(-48*A*b^3 - 6*a^2*b*(2*A + 5*B) + a^3*(9*A + 5*B) + 4*a*b^2*(9*A + 10*B))*EllipticF[I*ArcSinh
[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b
)] - (9*a^4*A + 24*a^2*A*b^2 - 48*A*b^4 - 25*a^3*b*B + 40*a*b^3*B)*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(
3/2)*Tan[(c + d*x)/2]))/(15*a^4*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2913\) vs. \(2(449)=898\).

Time = 3.96 (sec) , antiderivative size = 2914, normalized size of antiderivative = 6.89

method	result	size
default	\(\text {Expression too large to display}\)	\(2914\)

[In]

int(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/15/d*(9*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+
c)),(-(a+b)/(a-b))^(1/2))*a^4-9*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2
)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^4+48*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Ellip
ticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*b^4-5*B*(1/(a+b)*(b+a*cos(d*x+c))/(1+co
s(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^4+9*A*(1/(a+b)*
(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1
/2))*a^4*cos(d*x+c)-9*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x
+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^4*cos(d*x+c)+48*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Elli
pticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*b^4*cos(d*x+c)-5*B*(1/(a+b)*(b+a*cos(d
*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^4*c
os(d*x+c)+48*A*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*b^4*sin(d*x+c)+48*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+c
os(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a*b^3-24*A*(1/(a
+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b)
)^(1/2))*a^2*b^2-30*B*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c
)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^3*b-40*B*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b
)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b^2+25*B*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x
+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^3*b-40*B*(1/(a+b)*(b
+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2
))*a*b^3+12*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*
x+c)),(-(a+b)/(a-b))^(1/2))*a^3*b+36*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))
^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b^2-6*A*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*
a^2*b^2*cos(d*x+c)^2*sin(d*x+c)+5*B*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*a^3*b*cos(d*x+c)^2*sin(d*x+c)
+3*A*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*a^3*b*cos(d*x+c)*sin(d*x+c)+18*A*(1/(1+cos(d*x+c)))^(1/2)*((
a-b)/(a+b))^(1/2)*a^2*b^2*cos(d*x+c)*sin(d*x+c)+9*A*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*a^3*b*sin(d*x
+c)+24*A*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*a*b^3*sin(d*x+c)+5*B*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+
b))^(1/2)*a^3*b*sin(d*x+c)-20*B*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*a^2*b^2*sin(d*x+c)-40*B*(1/(1+cos
(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*a*b^3*sin(d*x+c)+12*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Elli
pticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^3*b*cos(d*x+c)+36*A*(1/(a+b)*(b+a*co
s(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^
2*b^2*cos(d*x+c)+48*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c
)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a*b^3*cos(d*x+c)-24*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Elli
pticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b^2*cos(d*x+c)-30*B*(1/(a+b)*(b+a*
cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*
a^3*b*cos(d*x+c)-40*B*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c
)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b^2*cos(d*x+c)+25*B*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*El
lipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^3*b*cos(d*x+c)-40*B*(1/(a+b)*(b+a*
cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*
a*b^3*cos(d*x+c)+3*A*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*a^4*cos(d*x+c)^3*sin(d*x+c)+3*A*(1/(1+cos(d*
x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*a^4*cos(d*x+c)^2*sin(d*x+c)+5*B*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*
a^4*cos(d*x+c)^2*sin(d*x+c)+9*A*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*a^4*cos(d*x+c)*sin(d*x+c)+24*A*(1
/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*a*b^3*cos(d*x+c)*sin(d*x+c)-15*B*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a
+b))^(1/2)*a^3*b*cos(d*x+c)*sin(d*x+c)-20*B*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*a^2*b^2*cos(d*x+c)*si
n(d*x+c)+3*A*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*a^3*b*cos(d*x+c)^3*sin(d*x+c)-3*A*(1/(1+cos(d*x+c)))
^(1/2)*((a-b)/(a+b))^(1/2)*a^3*b*cos(d*x+c)^2*sin(d*x+c)+5*B*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*a^4*
cos(d*x+c)*sin(d*x+c))*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/(1/(1+cos(d*x+c)))^(1/2)/(b+a*cos(d*x+c))/((a-b
)/(a+b))^(1/2)/(a+b)/a^4/(1+cos(d*x+c))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.19 (sec) , antiderivative size = 928, normalized size of antiderivative = 2.19 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/45*(6*(5*B*a^5*b - 9*A*a^4*b^2 - 20*B*a^3*b^3 + 24*A*a^2*b^4 + 3*(A*a^6 - A*a^4*b^2)*cos(d*x + c)^2 + (5*B*a
^6 - 6*A*a^5*b - 5*B*a^4*b^2 + 6*A*a^3*b^3)*cos(d*x + c))*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*sqrt(cos(d*x
+ c))*sin(d*x + c) - (sqrt(2)*(15*I*B*a^6 - 27*I*A*a^5*b + 80*I*B*a^4*b^2 - 84*I*A*a^3*b^3 - 80*I*B*a^2*b^4 +
96*I*A*a*b^5)*cos(d*x + c) + sqrt(2)*(15*I*B*a^5*b - 27*I*A*a^4*b^2 + 80*I*B*a^3*b^3 - 84*I*A*a^2*b^4 - 80*I*
B*a*b^5 + 96*I*A*b^6))*sqrt(a)*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(
3*a*cos(d*x + c) + 3*I*a*sin(d*x + c) + 2*b)/a) - (sqrt(2)*(-15*I*B*a^6 + 27*I*A*a^5*b - 80*I*B*a^4*b^2 + 84*I
*A*a^3*b^3 + 80*I*B*a^2*b^4 - 96*I*A*a*b^5)*cos(d*x + c) + sqrt(2)*(-15*I*B*a^5*b + 27*I*A*a^4*b^2 - 80*I*B*a^
3*b^3 + 84*I*A*a^2*b^4 + 80*I*B*a*b^5 - 96*I*A*b^6))*sqrt(a)*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/2
7*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a) + 3*(sqrt(2)*(9*I*A*a^6 - 25*I*B
*a^5*b + 24*I*A*a^4*b^2 + 40*I*B*a^3*b^3 - 48*I*A*a^2*b^4)*cos(d*x + c) + sqrt(2)*(9*I*A*a^5*b - 25*I*B*a^4*b^
2 + 24*I*A*a^3*b^3 + 40*I*B*a^2*b^4 - 48*I*A*a*b^5))*sqrt(a)*weierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9
*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*
x + c) + 3*I*a*sin(d*x + c) + 2*b)/a)) + 3*(sqrt(2)*(-9*I*A*a^6 + 25*I*B*a^5*b - 24*I*A*a^4*b^2 - 40*I*B*a^3*b
^3 + 48*I*A*a^2*b^4)*cos(d*x + c) + sqrt(2)*(-9*I*A*a^5*b + 25*I*B*a^4*b^2 - 24*I*A*a^3*b^3 - 40*I*B*a^2*b^4 +
48*I*A*a*b^5))*sqrt(a)*weierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierstrassPInve
rse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a)
))/((a^8 - a^6*b^2)*d*cos(d*x + c) + (a^7*b - a^5*b^3)*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(5/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*cos(d*x + c)^(5/2)/(b*sec(d*x + c) + a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int((cos(c + d*x)^(5/2)*(A + B/cos(c + d*x)))/(a + b/cos(c + d*x))^(3/2),x)

[Out]

int((cos(c + d*x)^(5/2)*(A + B/cos(c + d*x)))/(a + b/cos(c + d*x))^(3/2), x)