\(\int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [510]

Optimal result

Integrand size = 33, antiderivative size = 188 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {(49 A-9 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(13 A-3 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(13 A-3 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

1/10*(49*A-9*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*(13*A-3*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)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3-1/15*(8*A-3*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*cos(
d*x+c))^2-1/6*(13*A-3*B)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a^3+a^3*cos(d*x+c))

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3033, 3056, 2827, 2720, 2719} \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {(13 A-3 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(49 A-9 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(13 A-3 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {(8 A-3 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

((49*A - 9*B)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) - ((13*A - 3*B)*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) - ((A
- B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((8*A - 3*B)*Cos[c + d*x]^(3/2)*Sin[c +
d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((13*A - 3*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(6*d*(a^3 + a^3*Cos[c +
d*x]))

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 {5}{2}}(c+d x) (B+A \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx \\ & = -\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (-\frac {5}{2} a (A-B)+\frac {1}{2} a (11 A-B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\sqrt {\cos (c+d x)} \left (-\frac {3}{2} a^2 (8 A-3 B)+\frac {1}{2} a^2 (41 A-6 B) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4} \\ & = -\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(13 A-3 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \frac {-\frac {5}{4} a^3 (13 A-3 B)+\frac {3}{4} a^3 (49 A-9 B) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{15 a^6} \\ & = -\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(13 A-3 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(49 A-9 B) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}-\frac {(13 A-3 B) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3} \\ & = \frac {(49 A-9 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(13 A-3 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(13 A-3 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 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 = 5.44 (sec) , antiderivative size = 422, normalized size of antiderivative = 2.24 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {\sqrt {\cos (c+d x)} \csc ^5(c+d x) \left (156 A-156 B+280 A \cos (c+d x)-280 B \cos (c+d x)-312 A \cos ^2(c+d x)+312 B \cos ^2(c+d x)-280 A \cos ^3(c+d x)+280 B \cos ^3(c+d x)+180 A \cos ^4(c+d x)-180 B \cos ^4(c+d x)+60 A \cos ^5(c+d x)-60 B \cos ^5(c+d x)-26 A \sin ^2(c+d x)+126 B \sin ^2(c+d x)-60 B \cos ^3(c+d x) \sin ^2(c+d x)-65 A \sin ^4(c+d x)+15 B \sin ^4(c+d x)-5 (13 A-3 B) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cos ^2(c+d x)\right ) \sin ^4(c+d x) \sqrt {\sin ^2(c+d x)}-160 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},\cos ^2(c+d x)\right ) \sin ^4(c+d x) \sqrt {\sin ^2(c+d x)}-280 A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{2},\frac {7}{4},\cos ^2(c+d x)\right ) \sin ^4(c+d x) \sqrt {\sin ^2(c+d x)}+280 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{2},\frac {7}{4},\cos ^2(c+d x)\right ) \sin ^4(c+d x) \sqrt {\sin ^2(c+d x)}+80 B \sin (c+d x) \sin (2 (c+d x))-30 B \sin ^2(2 (c+d x))\right )}{30 a^3 d} \]

[In]

Integrate[(Sqrt[Cos[c + d*x]]*(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*(156*A - 156*B + 280*A*Cos[c + d*x] - 280*B*Cos[c + d*x] - 312*A*Cos[
c + d*x]^2 + 312*B*Cos[c + d*x]^2 - 280*A*Cos[c + d*x]^3 + 280*B*Cos[c + d*x]^3 + 180*A*Cos[c + d*x]^4 - 180*B
*Cos[c + d*x]^4 + 60*A*Cos[c + d*x]^5 - 60*B*Cos[c + d*x]^5 - 26*A*Sin[c + d*x]^2 + 126*B*Sin[c + d*x]^2 - 60*
B*Cos[c + d*x]^3*Sin[c + d*x]^2 - 65*A*Sin[c + d*x]^4 + 15*B*Sin[c + d*x]^4 - 5*(13*A - 3*B)*Hypergeometric2F1
[1/4, 1/2, 5/4, Cos[c + d*x]^2]*Sin[c + d*x]^4*Sqrt[Sin[c + d*x]^2] - 160*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] - 280*A*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] + 280*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] + 80*B*Sin[c + d*x]*Sin[2*(c + d*x)] - 30*B*Sin[2*(c +
d*x)]^2))/(a^3*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(450\) vs. \(2(224)=448\).

Time = 10.58 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.40

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 (348 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+130 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 )+294 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 )-108 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-30 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 )-54 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 )-578 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+198 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+264 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-114 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-37 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+27 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}\)	\(451\)

[In]

int((A+B*sec(d*x+c))*cos(d*x+c)^(1/2)/(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)*(348*A*cos(1/2*d*x+1/2*c)^8+130*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(cos(1/2*d*x+1/2*c),2^(1/2))+294
*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))-108*B*cos(1/2*d*x+1/2*c)^8-30*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))-54*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))-578*A*cos(1/2*d*x+1/2*c)^6+198*B*c
os(1/2*d*x+1/2*c)^6+264*A*cos(1/2*d*x+1/2*c)^4-114*B*cos(1/2*d*x+1/2*c)^4-37*A*cos(1/2*d*x+1/2*c)^2+27*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 = 467, normalized size of antiderivative = 2.48 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {2 \, {\left (3 \, {\left (29 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (73 \, A - 18 \, B\right )} \cos \left (d x + c\right ) + 65 \, A - 15 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 5 \, {\left (\sqrt {2} {\left (-13 i \, A + 3 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-13 i \, A + 3 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-13 i \, A + 3 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-13 i \, A + 3 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 (13 i \, A - 3 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (13 i \, A - 3 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (13 i \, A - 3 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (13 i \, A - 3 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (\sqrt {2} {\left (-49 i \, A + 9 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-49 i \, A + 9 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-49 i \, A + 9 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-49 i \, A + 9 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 ) + 3 \, {\left (\sqrt {2} {\left (49 i \, A - 9 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (49 i \, A - 9 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (49 i \, A - 9 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (49 i \, A - 9 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((A+B*sec(d*x+c))*cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(2*(3*(29*A - 9*B)*cos(d*x + c)^2 + 2*(73*A - 18*B)*cos(d*x + c) + 65*A - 15*B)*sqrt(cos(d*x + c))*sin(d
*x + c) + 5*(sqrt(2)*(-13*I*A + 3*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-13*I*A + 3*I*B)*cos(d*x + c)^2 + 3*sqrt(2)
*(-13*I*A + 3*I*B)*cos(d*x + c) + sqrt(2)*(-13*I*A + 3*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d
*x + c)) + 5*(sqrt(2)*(13*I*A - 3*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(13*I*A - 3*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*
(13*I*A - 3*I*B)*cos(d*x + c) + sqrt(2)*(13*I*A - 3*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x
+ c)) + 3*(sqrt(2)*(-49*I*A + 9*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-49*I*A + 9*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(
-49*I*A + 9*I*B)*cos(d*x + c) + sqrt(2)*(-49*I*A + 9*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, c
os(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(49*I*A - 9*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(49*I*A - 9*I*B)*cos(d
*x + c)^2 + 3*sqrt(2)*(49*I*A - 9*I*B)*cos(d*x + c) + sqrt(2)*(49*I*A - 9*I*B))*weierstrassZeta(-4, 0, weierst
rassPInverse(-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]

\[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \sqrt {\cos {\left (c + d x \right )}}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sqrt {\cos {\left (c + d x \right )}} \sec {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

[In]

integrate((A+B*sec(d*x+c))*cos(d*x+c)**(1/2)/(a+a*sec(d*x+c))**3,x)

[Out]

(Integral(A*sqrt(cos(c + d*x))/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*sqr
t(cos(c + d*x))*sec(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3

Maxima [F]

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

[In]

integrate((A+B*sec(d*x+c))*cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((A+B*sec(d*x+c))*cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int((cos(c + d*x)^(1/2)*(A + B/cos(c + d*x)))/(a + a/cos(c + d*x))^3,x)

[Out]

int((cos(c + d*x)^(1/2)*(A + B/cos(c + d*x)))/(a + a/cos(c + d*x))^3, x)