\(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [509]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 33, antiderivative size = 221 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {7 (17 A-7 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(33 A-13 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(33 A-13 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 a^3 d}-\frac {(A-B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(2 A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \cos (c+d x))^2}-\frac {7 (17 A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

-7/10*(17*A-7*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))/a^3/d+1
/6*(33*A-13*B)*(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^3/d-1/5
*(A-B)*cos(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3-1/3*(2*A-B)*cos(d*x+c)^(5/2)*sin(d*x+c)/a/d/(a+a*cos(d
*x+c))^2-7/30*(17*A-7*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))+1/6*(33*A-13*B)*sin(d*x+c)*cos(d*x
+c)^(1/2)/a^3/d

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3033, 3056, 2827, 2719, 2715, 2720} \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {(33 A-13 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {7 (17 A-7 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {7 (17 A-7 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{30 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {(33 A-13 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 a^3 d}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {(2 A-B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 a d (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

(-7*(17*A - 7*B)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + ((33*A - 13*B)*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) +
 ((33*A - 13*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(6*a^3*d) - ((A - B)*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(5*d*(a
 + a*Cos[c + d*x])^3) - ((2*A - B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(3*a*d*(a + a*Cos[c + d*x])^2) - (7*(17*A
- 7*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(30*d*(a^3 + a^3*Cos[c + d*x]))

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3033

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^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3056

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^n/(a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cos ^{\frac {7}{2}}(c+d x) (B+A \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx \\ & = -\frac {(A-B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (-\frac {7}{2} a (A-B)+\frac {1}{2} a (13 A-3 B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A-B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(2 A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (-\frac {25}{2} a^2 (2 A-B)+\frac {3}{2} a^2 (23 A-8 B) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4} \\ & = -\frac {(A-B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(2 A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \cos (c+d x))^2}-\frac {7 (17 A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \sqrt {\cos (c+d x)} \left (-\frac {21}{4} a^3 (17 A-7 B)+\frac {15}{4} a^3 (33 A-13 B) \cos (c+d x)\right ) \, dx}{15 a^6} \\ & = -\frac {(A-B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(2 A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \cos (c+d x))^2}-\frac {7 (17 A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(33 A-13 B) \int \cos ^{\frac {3}{2}}(c+d x) \, dx}{4 a^3}-\frac {(7 (17 A-7 B)) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3} \\ & = -\frac {7 (17 A-7 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(33 A-13 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 a^3 d}-\frac {(A-B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(2 A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \cos (c+d x))^2}-\frac {7 (17 A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(33 A-13 B) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3} \\ & = -\frac {7 (17 A-7 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(33 A-13 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(33 A-13 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 a^3 d}-\frac {(A-B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(2 A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \cos (c+d x))^2}-\frac {7 (17 A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 6.05 (sec) , antiderivative size = 463, normalized size of antiderivative = 2.10 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {\sqrt {\cos (c+d x)} \csc ^5(c+d x) \left (-396 A+396 B-680 A \cos (c+d x)+680 B \cos (c+d x)+792 A \cos ^2(c+d x)-792 B \cos ^2(c+d x)+680 A \cos ^3(c+d x)-680 B \cos ^3(c+d x)-440 A \cos ^4(c+d x)+440 B \cos ^4(c+d x)-180 A \cos ^5(c+d x)+180 B \cos ^5(c+d x)+20 A \cos ^6(c+d x)-20 B \cos ^6(c+d x)+66 A \sin ^2(c+d x)-266 B \sin ^2(c+d x)-280 B \cos (c+d x) \sin ^2(c+d x)-20 B \cos ^4(c+d x) \sin ^2(c+d x)+165 A \sin ^4(c+d x)-65 B \sin ^4(c+d x)+5 (33 A-13 B) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cos ^2(c+d x)\right ) \sin ^4(c+d x) \sqrt {\sin ^2(c+d x)}+280 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},\cos ^2(c+d x)\right ) \sin ^4(c+d x) \sqrt {\sin ^2(c+d x)}+680 A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{2},\frac {7}{4},\cos ^2(c+d x)\right ) \sin ^4(c+d x) \sqrt {\sin ^2(c+d x)}-680 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{2},\frac {7}{4},\cos ^2(c+d x)\right ) \sin ^4(c+d x) \sqrt {\sin ^2(c+d x)}+60 B \sin ^2(2 (c+d x))+15 B \csc (c+d x) \sin ^3(2 (c+d x))\right )}{30 a^3 d} \]

[In]

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

[Out]

-1/30*(Sqrt[Cos[c + d*x]]*Csc[c + d*x]^5*(-396*A + 396*B - 680*A*Cos[c + d*x] + 680*B*Cos[c + d*x] + 792*A*Cos
[c + d*x]^2 - 792*B*Cos[c + d*x]^2 + 680*A*Cos[c + d*x]^3 - 680*B*Cos[c + d*x]^3 - 440*A*Cos[c + d*x]^4 + 440*
B*Cos[c + d*x]^4 - 180*A*Cos[c + d*x]^5 + 180*B*Cos[c + d*x]^5 + 20*A*Cos[c + d*x]^6 - 20*B*Cos[c + d*x]^6 + 6
6*A*Sin[c + d*x]^2 - 266*B*Sin[c + d*x]^2 - 280*B*Cos[c + d*x]*Sin[c + d*x]^2 - 20*B*Cos[c + d*x]^4*Sin[c + d*
x]^2 + 165*A*Sin[c + d*x]^4 - 65*B*Sin[c + d*x]^4 + 5*(33*A - 13*B)*Hypergeometric2F1[1/4, 1/2, 5/4, Cos[c + d
*x]^2]*Sin[c + d*x]^4*Sqrt[Sin[c + d*x]^2] + 280*B*Cos[c + d*x]*Hypergeometric2F1[3/4, 5/2, 7/4, Cos[c + d*x]^
2]*Sin[c + d*x]^4*Sqrt[Sin[c + d*x]^2] + 680*A*Cos[c + d*x]*Hypergeometric2F1[3/4, 7/2, 7/4, Cos[c + d*x]^2]*S
in[c + d*x]^4*Sqrt[Sin[c + d*x]^2] - 680*B*Cos[c + d*x]*Hypergeometric2F1[3/4, 7/2, 7/4, Cos[c + d*x]^2]*Sin[c
 + d*x]^4*Sqrt[Sin[c + d*x]^2] + 60*B*Sin[2*(c + d*x)]^2 + 15*B*Csc[c + d*x]*Sin[2*(c + d*x)]^3))/(a^3*d)

Maple [A] (verified)

Time = 10.62 (sec) , antiderivative size = 465, normalized size of antiderivative = 2.10

method result size
default \(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (160 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+468 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+330 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+714 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-348 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-130 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-294 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-1058 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+578 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+474 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-264 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-47 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+37 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 A -3 B \right )}{60 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(465\)

[In]

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

[Out]

-1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(160*A*cos(1/2*d*x+1/2*c)^10+468*A*cos(1/2*d*x+1
/2*c)^8+330*A*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(co
s(1/2*d*x+1/2*c),2^(1/2))+714*A*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^
(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-348*B*cos(1/2*d*x+1/2*c)^8-130*B*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-294*B*cos(1/2*d*x+1/2
*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-105
8*A*cos(1/2*d*x+1/2*c)^6+578*B*cos(1/2*d*x+1/2*c)^6+474*A*cos(1/2*d*x+1/2*c)^4-264*B*cos(1/2*d*x+1/2*c)^4-47*A
*cos(1/2*d*x+1/2*c)^2+37*B*cos(1/2*d*x+1/2*c)^2+3*A-3*B)/a^3/cos(1/2*d*x+1/2*c)^5/(-2*sin(1/2*d*x+1/2*c)^4+sin
(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

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

Time = 0.12 (sec) , antiderivative size = 478, normalized size of antiderivative = 2.16 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {2 \, {\left (20 \, A \cos \left (d x + c\right )^{3} + 3 \, {\left (79 \, A - 29 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (188 \, A - 73 \, B\right )} \cos \left (d x + c\right ) + 165 \, A - 65 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 \, {\left (\sqrt {2} {\left (33 i \, A - 13 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (33 i \, A - 13 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (33 i \, A - 13 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (33 i \, A - 13 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 \, {\left (\sqrt {2} {\left (-33 i \, A + 13 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-33 i \, A + 13 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-33 i \, A + 13 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-33 i \, A + 13 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 21 \, {\left (\sqrt {2} {\left (17 i \, A - 7 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (17 i \, A - 7 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (17 i \, A - 7 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (17 i \, A - 7 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 \, {\left (\sqrt {2} {\left (-17 i \, A + 7 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-17 i \, A + 7 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-17 i \, A + 7 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-17 i \, A + 7 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

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

[Out]

1/60*(2*(20*A*cos(d*x + c)^3 + 3*(79*A - 29*B)*cos(d*x + c)^2 + 2*(188*A - 73*B)*cos(d*x + c) + 165*A - 65*B)*
sqrt(cos(d*x + c))*sin(d*x + c) - 5*(sqrt(2)*(33*I*A - 13*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(33*I*A - 13*I*B)*co
s(d*x + c)^2 + 3*sqrt(2)*(33*I*A - 13*I*B)*cos(d*x + c) + sqrt(2)*(33*I*A - 13*I*B))*weierstrassPInverse(-4, 0
, cos(d*x + c) + I*sin(d*x + c)) - 5*(sqrt(2)*(-33*I*A + 13*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-33*I*A + 13*I*B)
*cos(d*x + c)^2 + 3*sqrt(2)*(-33*I*A + 13*I*B)*cos(d*x + c) + sqrt(2)*(-33*I*A + 13*I*B))*weierstrassPInverse(
-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 21*(sqrt(2)*(17*I*A - 7*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(17*I*A - 7*I*
B)*cos(d*x + c)^2 + 3*sqrt(2)*(17*I*A - 7*I*B)*cos(d*x + c) + sqrt(2)*(17*I*A - 7*I*B))*weierstrassZeta(-4, 0,
 weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*(sqrt(2)*(-17*I*A + 7*I*B)*cos(d*x + c)^3 + 3
*sqrt(2)*(-17*I*A + 7*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-17*I*A + 7*I*B)*cos(d*x + c) + sqrt(2)*(-17*I*A + 7*I*
B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^3*d*cos(d*x + c)^3 +
 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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