\(\int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx\) [221]

Optimal result

Integrand size = 33, antiderivative size = 228 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx=\frac {(49 A-9 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {(13 A-3 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}-\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(8 A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(13 A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4105, 3872, 3856, 2719, 2720} \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx=-\frac {(13 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{6 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {(13 A-3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(49 A-9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(8 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3} \]

[In]

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

[Out]

((49*A - 9*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) - ((13*A - 3*B)*Sqrt
[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) - ((A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d
*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((8*A - 3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])
^2) - ((13*A - 3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(6*d*(a^3 + a^3*Sec[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 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4105

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\frac {1}{2} a (11 A-B)-\frac {5}{2} a (A-B) \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(8 A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {\frac {1}{2} a^2 (41 A-6 B)-\frac {3}{2} a^2 (8 A-3 B) \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))} \, dx}{15 a^4} \\ & = -\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(8 A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(13 A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\int \frac {\frac {3}{4} a^3 (49 A-9 B)-\frac {5}{4} a^3 (13 A-3 B) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{15 a^6} \\ & = -\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(8 A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(13 A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(49 A-9 B) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{20 a^3}-\frac {(13 A-3 B) \int \sqrt {\sec (c+d x)} \, dx}{12 a^3} \\ & = -\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(8 A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(13 A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\left ((49 A-9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}-\frac {\left ((13 A-3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3} \\ & = \frac {(49 A-9 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {(13 A-3 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}-\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(8 A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(13 A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (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 = 6.60 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.60 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx=-\frac {e^{-i d x} \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x)) (\cos (d x)+i \sin (d x)) \left (160 (13 A-3 B) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+i (49 A-9 B) e^{-\frac {3}{2} i (c+d x)} \left (1+e^{i (c+d x)}\right )^5 \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+2 \cos (c+d x) \left (-30 i (49 A-9 B) \cos \left (\frac {1}{2} (c+d x)\right )-15 i (49 A-9 B) \cos \left (\frac {3}{2} (c+d x)\right )-147 i A \cos \left (\frac {5}{2} (c+d x)\right )+27 i B \cos \left (\frac {5}{2} (c+d x)\right )+142 A \sin \left (\frac {1}{2} (c+d x)\right )-42 B \sin \left (\frac {1}{2} (c+d x)\right )+205 A \sin \left (\frac {3}{2} (c+d x)\right )-45 B \sin \left (\frac {3}{2} (c+d x)\right )+87 A \sin \left (\frac {5}{2} (c+d x)\right )-27 B \sin \left (\frac {5}{2} (c+d x)\right )\right )\right )}{120 a^3 d (B+A \cos (c+d x)) (1+\sec (c+d x))^3} \]

[In]

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

[Out]

-1/120*(Cos[(c + d*x)/2]*Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x])*(Cos[d*x] + I*Sin[d*x])*(160*(13*A - 3*B)*Cos
[(c + d*x)/2]^5*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (I*(49*A - 9*B)*(1 + E^(I*(c + d*x)))^5*Sqrt[1
+ E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(((3*I)/2)*(c + d*x)) + 2*Cos
[c + d*x]*((-30*I)*(49*A - 9*B)*Cos[(c + d*x)/2] - (15*I)*(49*A - 9*B)*Cos[(3*(c + d*x))/2] - (147*I)*A*Cos[(5
*(c + d*x))/2] + (27*I)*B*Cos[(5*(c + d*x))/2] + 142*A*Sin[(c + d*x)/2] - 42*B*Sin[(c + d*x)/2] + 205*A*Sin[(3
*(c + d*x))/2] - 45*B*Sin[(3*(c + d*x))/2] + 87*A*Sin[(5*(c + d*x))/2] - 27*B*Sin[(5*(c + d*x))/2])))/(a^3*d*E
^(I*d*x)*(B + A*Cos[c + d*x])*(1 + Sec[c + d*x])^3)

Maple [A] (verified)

Time = 11.36 (sec) , antiderivative size = 451, normalized size of antiderivative = 1.98

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))/(a+a*sec(d*x+c))^3/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/60/a^3*((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*cos(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*c
os(1/2*d*x+1/2*c)^2+3*A-3*B)/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 = 2.10 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx=-\frac {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 ) + \frac {2 \, {\left (3 \, {\left (29 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (73 \, A - 18 \, B\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (13 \, A - 3 \, B\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\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/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/60*(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, 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, weierstras
sPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*(29*A - 9*B)*cos(d*x + c)^3 + 2*(73*A - 18*B)*cos(d*x
+ c)^2 + 5*(13*A - 3*B)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(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]

\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + 3 \sec ^{\frac {5}{2}}{\left (c + d x \right )} + 3 \sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + 3 \sec ^{\frac {5}{2}}{\left (c + d x \right )} + 3 \sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sqrt {\sec {\left (c + d x \right )}}}\, dx}{a^{3}} \]

[In]

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

[Out]

(Integral(A/(sec(c + d*x)**(7/2) + 3*sec(c + d*x)**(5/2) + 3*sec(c + d*x)**(3/2) + sqrt(sec(c + d*x))), x) + I
ntegral(B*sec(c + d*x)/(sec(c + d*x)**(7/2) + 3*sec(c + d*x)**(5/2) + 3*sec(c + d*x)**(3/2) + sqrt(sec(c + d*x
))), x))/a**3

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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