\(\int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^3} \, dx\) [511]

Optimal result

Integrand size = 33, antiderivative size = 182 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^3} \, dx=-\frac {(9 A+B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(3 A+B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(A-B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(6 A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(9 A+B) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3033, 3056, 3057, 2827, 2720, 2719} \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^3} \, dx=\frac {(3 A+B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(9 A+B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(9 A+B) \sin (c+d x) \sqrt {\cos (c+d x)}}{10 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {(6 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{15 a d (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

-1/10*((9*A + B)*EllipticE[(c + d*x)/2, 2])/(a^3*d) + ((3*A + B)*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) - ((A -
B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((6*A - B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])
/(15*a*d*(a + a*Cos[c + d*x])^2) + ((9*A + B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(10*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])

Rule 3057

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 {3}{2}}(c+d x) (B+A \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx \\ & = -\frac {(A-B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {\sqrt {\cos (c+d x)} \left (-\frac {3}{2} a (A-B)+\frac {1}{2} a (9 A+B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A-B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(6 A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {-\frac {1}{2} a^2 (6 A-B)+\frac {1}{2} a^2 (21 A+4 B) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))} \, dx}{15 a^4} \\ & = -\frac {(A-B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(6 A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(9 A+B) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \frac {\frac {5}{4} a^3 (3 A+B)-\frac {3}{4} a^3 (9 A+B) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{15 a^6} \\ & = -\frac {(A-B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(6 A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(9 A+B) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(3 A+B) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3}-\frac {(9 A+B) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3} \\ & = -\frac {(9 A+B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(3 A+B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(A-B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(6 A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(9 A+B) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 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 = 4.37 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.08 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^3} \, dx=-\frac {\sqrt {\cos (c+d x)} \csc ^5(c+d x) \left (-252 A+252 B-360 A \cos (c+d x)+360 B \cos (c+d x)+504 A \cos ^2(c+d x)-504 B \cos ^2(c+d x)+420 A \cos ^3(c+d x)-420 B \cos ^3(c+d x)-420 A \cos ^4(c+d x)+420 B \cos ^4(c+d x)+42 A \sin ^2(c+d x)-322 B \sin ^2(c+d x)-280 B \cos (c+d x) \sin ^2(c+d x)+105 A \sin ^4(c+d x)+35 B \sin ^4(c+d x)+35 (3 A+B) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cos ^2(c+d x)\right ) \sin ^4(c+d x) \sqrt {\sin ^2(c+d x)}+280 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},\cos ^2(c+d x)\right ) \sin ^4(c+d x) \sqrt {\sin ^2(c+d x)}+360 A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{2},\frac {7}{4},\cos ^2(c+d x)\right ) \sin ^4(c+d x) \sqrt {\sin ^2(c+d x)}-360 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{2},\frac {7}{4},\cos ^2(c+d x)\right ) \sin ^4(c+d x) \sqrt {\sin ^2(c+d x)}+105 B \sin ^2(2 (c+d x))\right )}{210 a^3 d} \]

[In]

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

[Out]

-1/210*(Sqrt[Cos[c + d*x]]*Csc[c + d*x]^5*(-252*A + 252*B - 360*A*Cos[c + d*x] + 360*B*Cos[c + d*x] + 504*A*Co
s[c + d*x]^2 - 504*B*Cos[c + d*x]^2 + 420*A*Cos[c + d*x]^3 - 420*B*Cos[c + d*x]^3 - 420*A*Cos[c + d*x]^4 + 420
*B*Cos[c + d*x]^4 + 42*A*Sin[c + d*x]^2 - 322*B*Sin[c + d*x]^2 - 280*B*Cos[c + d*x]*Sin[c + d*x]^2 + 105*A*Sin
[c + d*x]^4 + 35*B*Sin[c + d*x]^4 + 35*(3*A + 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*S
qrt[Sin[c + d*x]^2] + 360*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] - 360*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] + 105*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. \(2(218)=436\).

Time = 9.94 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.48

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 (108 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+30 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 )+54 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 )+12 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+10 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 )+6 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 )-198 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+114 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-27 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+17 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))/(a+a*sec(d*x+c))^3/cos(d*x+c)^(1/2),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)*(108*A*cos(1/2*d*x+1/2*c)^8+30*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))+54*
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))+12*B*cos(1/2*d*x+1/2*c)^8+10*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))+6*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))-198*A*cos(1/2*d*x+1/2*c)^6-2*B*cos(1/
2*d*x+1/2*c)^6+114*A*cos(1/2*d*x+1/2*c)^4-24*B*cos(1/2*d*x+1/2*c)^4-27*A*cos(1/2*d*x+1/2*c)^2+17*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 = 0.10 (sec) , antiderivative size = 465, normalized size of antiderivative = 2.55 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^3} \, dx=\frac {2 \, {\left (3 \, {\left (9 \, A + B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (18 \, A + 7 \, B\right )} \cos \left (d x + c\right ) + 15 \, A + 5 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 \, {\left (\sqrt {2} {\left (3 i \, A + i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (3 i \, A + i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (3 i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A + 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 (-3 i \, A - i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-3 i \, A - i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-3 i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A - 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 (9 i \, A + i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (9 i \, A + i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (9 i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (9 i \, A + 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 (-9 i \, A - i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-9 i \, A - i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-9 i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-9 i \, A - 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))/(a+a*sec(d*x+c))^3/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

1/60*(2*(3*(9*A + B)*cos(d*x + c)^2 + 2*(18*A + 7*B)*cos(d*x + c) + 15*A + 5*B)*sqrt(cos(d*x + c))*sin(d*x + c
) - 5*(sqrt(2)*(3*I*A + I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(3*I*A + I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(3*I*A + I*B)
*cos(d*x + c) + sqrt(2)*(3*I*A + I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*(sqrt(2)*
(-3*I*A - I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-3*I*A - I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-3*I*A - I*B)*cos(d*x + c
) + sqrt(2)*(-3*I*A - I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*(sqrt(2)*(9*I*A + I*
B)*cos(d*x + c)^3 + 3*sqrt(2)*(9*I*A + I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(9*I*A + I*B)*cos(d*x + c) + sqrt(2)*(9
*I*A + I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(-
9*I*A - I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-9*I*A - I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-9*I*A - I*B)*cos(d*x + c)
+ sqrt(2)*(-9*I*A - 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. \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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