\(\int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\) [214]

Optimal result

Integrand size = 33, antiderivative size = 211 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {(7 A-4 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {5 (2 A-B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {5 (2 A-B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(7 A-4 B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)} (1+\sec (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \]

[Out]

5/3*(2*A-B)*sin(d*x+c)/a^2/d/sec(d*x+c)^(1/2)-1/3*(7*A-4*B)*sin(d*x+c)/a^2/d/(1+sec(d*x+c))/sec(d*x+c)^(1/2)-1
/3*(A-B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^2/sec(d*x+c)^(1/2)-(7*A-4*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^2/d+5/3*(2*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/
d

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 211, 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 = {4105, 3872, 3854, 3856, 2720, 2719} \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {5 (2 A-B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(7 A-4 B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}+\frac {5 (2 A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}-\frac {(7 A-4 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {(A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2} \]

[In]

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

[Out]

-(((7*A - 4*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d)) + (5*(2*A - B)*Sqrt[C
os[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + (5*(2*A - B)*Sin[c + d*x])/(3*a^2*d*Sqr
t[Sec[c + d*x]]) - ((7*A - 4*B)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]*(1 + Sec[c + d*x])) - ((A - B)*Sin[c
+ d*x])/(3*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2)

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 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

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) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac {\int \frac {\frac {3}{2} a (3 A-B)-\frac {5}{2} a (A-B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{3 a^2} \\ & = -\frac {(7 A-4 B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)} (1+\sec (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac {\int \frac {\frac {15}{2} a^2 (2 A-B)-\frac {3}{2} a^2 (7 A-4 B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{3 a^4} \\ & = -\frac {(7 A-4 B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)} (1+\sec (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2}-\frac {(7 A-4 B) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a^2}+\frac {(5 (2 A-B)) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{2 a^2} \\ & = \frac {5 (2 A-B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(7 A-4 B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)} (1+\sec (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac {(5 (2 A-B)) \int \sqrt {\sec (c+d x)} \, dx}{6 a^2}-\frac {\left ((7 A-4 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^2} \\ & = -\frac {(7 A-4 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {5 (2 A-B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(7 A-4 B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)} (1+\sec (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac {\left (5 (2 A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^2} \\ & = -\frac {(7 A-4 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {5 (2 A-B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {5 (2 A-B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(7 A-4 B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)} (1+\sec (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 8.65 (sec) , antiderivative size = 899, normalized size of antiderivative = 4.26 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {7 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sec \left (\frac {c}{2}\right ) \sec (c+d x) (A+B \sec (c+d x))}{3 d (B+A \cos (c+d x)) (a+a \sec (c+d x))^2}-\frac {4 \sqrt {2} B e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sec \left (\frac {c}{2}\right ) \sec (c+d x) (A+B \sec (c+d x))}{3 d (B+A \cos (c+d x)) (a+a \sec (c+d x))^2}+\frac {20 A \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec \left (\frac {c}{2}\right ) \sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x)) \sin (c)}{3 d (B+A \cos (c+d x)) (a+a \sec (c+d x))^2}-\frac {10 B \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec \left (\frac {c}{2}\right ) \sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x)) \sin (c)}{3 d (B+A \cos (c+d x)) (a+a \sec (c+d x))^2}+\frac {\cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x)) \left (-\frac {2 (-5 A+3 B-2 A \cos (2 c)+B \cos (2 c)) \cos (d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )}{d}+\frac {4 A \cos (2 d x) \sin (2 c)}{3 d}-\frac {2 \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-A \sin \left (\frac {d x}{2}\right )+B \sin \left (\frac {d x}{2}\right )\right )}{3 d}+\frac {4 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-10 A \sin \left (\frac {d x}{2}\right )+7 B \sin \left (\frac {d x}{2}\right )\right )}{3 d}+\frac {8 (-2 A+B) \cos (c) \sin (d x)}{d}+\frac {4 A \cos (2 c) \sin (2 d x)}{3 d}+\frac {4 (-10 A+7 B) \tan \left (\frac {c}{2}\right )}{3 d}-\frac {2 (-A+B) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{3 d}\right )}{(B+A \cos (c+d x)) (a+a \sec (c+d x))^2} \]

[In]

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

[Out]

(7*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^
4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/
4, -E^((2*I)*(c + d*x))])*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x]))/(3*d*E^(I*d*x)*(B + A*Cos[c + d*x])*(a +
a*Sec[c + d*x])^2) - (4*Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x)
)]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeo
metric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x]))/(3*d*E^(I*d*x)*(B
+ A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2) + (20*A*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(
c + d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x])*Sin[c])/(3*d*(B + A*Cos[c + d*x])*(a + a*Sec[c
+ d*x])^2) - (10*B*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sec[c
+ d*x]^(3/2)*(A + B*Sec[c + d*x])*Sin[c])/(3*d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2) + (Cos[c/2 + (d*x)
/2]^4*Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x])*((-2*(-5*A + 3*B - 2*A*Cos[2*c] + B*Cos[2*c])*Cos[d*x]*Csc[c/2]*
Sec[c/2])/d + (4*A*Cos[2*d*x]*Sin[2*c])/(3*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(-(A*Sin[(d*x)/2]) + B*Sin[(d
*x)/2]))/(3*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(-10*A*Sin[(d*x)/2] + 7*B*Sin[(d*x)/2]))/(3*d) + (8*(-2*A + B)
*Cos[c]*Sin[d*x])/d + (4*A*Cos[2*c]*Sin[2*d*x])/(3*d) + (4*(-10*A + 7*B)*Tan[c/2])/(3*d) - (2*(-A + B)*Sec[c/2
+ (d*x)/2]^2*Tan[c/2])/(3*d)))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2)

Maple [A] (verified)

Time = 9.87 (sec) , antiderivative size = 435, normalized size of antiderivative = 2.06

method	result	size
default	\(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (16 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+12 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+20 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+42 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-10 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-48 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+38 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+21 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-15 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-A +B \right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(435\)

[In]

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

[Out]

-1/6/a^2*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*A*cos(1/2*d*x+1/2*c)^8+12*A*cos(1/2*d*x+1
/2*c)^6+20*A*cos(1/2*d*x+1/2*c)^3*(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))+42*A*cos(1/2*d*x+1/2*c)^3*(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))-24*B*cos(1/2*d*x+1/2*c)^6-10*B*cos(1/2*d*x+1/2*c)^3*(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))-24*B*cos(1/2*d*x+1/2*c)^3
*(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))-48*A*cos
(1/2*d*x+1/2*c)^4+38*B*cos(1/2*d*x+1/2*c)^4+21*A*cos(1/2*d*x+1/2*c)^2-15*B*cos(1/2*d*x+1/2*c)^2-A+B)/cos(1/2*d
*x+1/2*c)^3/(-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 = 377, normalized size of antiderivative = 1.79 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {5 \, {\left (\sqrt {2} {\left (2 i \, A - i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (2 i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (2 i \, A - i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (-2 i \, A + i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-2 i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-2 i \, A + i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (\sqrt {2} {\left (7 i \, A - 4 i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (7 i \, A - 4 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (7 i \, A - 4 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (\sqrt {2} {\left (-7 i \, A + 4 i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-7 i \, A + 4 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-7 i \, A + 4 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (2 \, A \cos \left (d x + c\right )^{3} + {\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (2 \, A - B\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

-1/6*(5*(sqrt(2)*(2*I*A - I*B)*cos(d*x + c)^2 + 2*sqrt(2)*(2*I*A - I*B)*cos(d*x + c) + sqrt(2)*(2*I*A - I*B))*
weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-2*I*A + I*B)*cos(d*x + c)^2 + 2*sqrt(
2)*(-2*I*A + I*B)*cos(d*x + c) + sqrt(2)*(-2*I*A + I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x +
c)) + 3*(sqrt(2)*(7*I*A - 4*I*B)*cos(d*x + c)^2 + 2*sqrt(2)*(7*I*A - 4*I*B)*cos(d*x + c) + sqrt(2)*(7*I*A - 4
*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(-7*I*A
+ 4*I*B)*cos(d*x + c)^2 + 2*sqrt(2)*(-7*I*A + 4*I*B)*cos(d*x + c) + sqrt(2)*(-7*I*A + 4*I*B))*weierstrassZeta(
-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(2*A*cos(d*x + c)^3 + (13*A - 6*B)*cos(d
*x + c)^2 + 5*(2*A - B)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x
+ c) + a^2*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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