\(\int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\) [242]

Optimal result

Integrand size = 35, antiderivative size = 236 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {4 (14 A+5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a^2 d}-\frac {5 (3 A+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {4 (14 A+5 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (3 A+C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(3 A+C) \sin (c+d x)}{a^2 d \sec ^{\frac {3}{2}}(c+d x) (1+\sec (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \]

[Out]

4/15*(14*A+5*C)*sin(d*x+c)/a^2/d/sec(d*x+c)^(3/2)-(3*A+C)*sin(d*x+c)/a^2/d/sec(d*x+c)^(3/2)/(1+sec(d*x+c))-1/3
*(A+C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2-5/3*(3*A+C)*sin(d*x+c)/a^2/d/sec(d*x+c)^(1/2)+4/5*(14*
A+5*C)*(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*(3*A+C)*(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 = 0.44 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4170, 4105, 3872, 3854, 3856, 2719, 2720} \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {(3 A+C) \sin (c+d x)}{a^2 d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}+\frac {4 (14 A+5 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (3 A+C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {5 (3 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {4 (14 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac {(A+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2} \]

[In]

Int[(A + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

(4*(14*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a^2*d) - (5*(3*A + C)*Sqrt
[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + (4*(14*A + 5*C)*Sin[c + d*x])/(15*a^2
*d*Sec[c + d*x]^(3/2)) - (5*(3*A + C)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]) - ((3*A + C)*Sin[c + d*x])/(a
^2*d*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])) - ((A + C)*Sin[c + d*x])/(3*d*Sec[c + d*x]^(3/2)*(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]

Rule 4170

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(A+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {\int \frac {-\frac {1}{2} a (11 A+5 C)+\frac {1}{2} a (7 A+C) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{3 a^2} \\ & = -\frac {(3 A+C) \sin (c+d x)}{a^2 d \sec ^{\frac {3}{2}}(c+d x) (1+\sec (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {\int \frac {-2 a^2 (14 A+5 C)+\frac {15}{2} a^2 (3 A+C) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{3 a^4} \\ & = -\frac {(3 A+C) \sin (c+d x)}{a^2 d \sec ^{\frac {3}{2}}(c+d x) (1+\sec (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {(5 (3 A+C)) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{2 a^2}+\frac {(2 (14 A+5 C)) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{3 a^2} \\ & = \frac {4 (14 A+5 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (3 A+C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(3 A+C) \sin (c+d x)}{a^2 d \sec ^{\frac {3}{2}}(c+d x) (1+\sec (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {(5 (3 A+C)) \int \sqrt {\sec (c+d x)} \, dx}{6 a^2}+\frac {(2 (14 A+5 C)) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{5 a^2} \\ & = \frac {4 (14 A+5 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (3 A+C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(3 A+C) \sin (c+d x)}{a^2 d \sec ^{\frac {3}{2}}(c+d x) (1+\sec (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {\left (5 (3 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^2}+\frac {\left (2 (14 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 a^2} \\ & = \frac {4 (14 A+5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a^2 d}-\frac {5 (3 A+C) \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 {4 (14 A+5 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (3 A+C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(3 A+C) \sin (c+d x)}{a^2 d \sec ^{\frac {3}{2}}(c+d x) (1+\sec (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(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 = 6.68 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.28 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {e^{-i d x} \cos \left (\frac {1}{2} (c+d x)\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c) (\cos (d x)+i \sin (d x)) \left (400 (3 A+C) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+8 i (14 A+5 C) e^{-\frac {1}{2} i (c+d x)} \left (1+e^{i (c+d x)}\right )^3 \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+2 \cos (c+d x) \left (-72 i (14 A+5 C) \cos \left (\frac {1}{2} (c+d x)\right )-24 i (14 A+5 C) \cos \left (\frac {3}{2} (c+d x)\right )+2 (158 A+50 C+(179 A+60 C) \cos (c+d x)+8 A \cos (2 (c+d x))-3 A \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{120 a^2 d (1+\sec (c+d x))^2} \]

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

-1/120*(Cos[(c + d*x)/2]*Csc[c/2]*Sec[c/2]*Sec[c + d*x]^(5/2)*Sin[c]*(Cos[d*x] + I*Sin[d*x])*(400*(3*A + C)*Co
s[(c + d*x)/2]^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + ((8*I)*(14*A + 5*C)*(1 + E^(I*(c + d*x)))^3*Sq
rt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^((I/2)*(c + d*x)) + 2*Co
s[c + d*x]*((-72*I)*(14*A + 5*C)*Cos[(c + d*x)/2] - (24*I)*(14*A + 5*C)*Cos[(3*(c + d*x))/2] + 2*(158*A + 50*C
+ (179*A + 60*C)*Cos[c + d*x] + 8*A*Cos[2*(c + d*x)] - 3*A*Cos[3*(c + d*x)])*Sin[(c + d*x)/2])))/(a^2*d*E^(I*
d*x)*(1 + Sec[c + d*x])^2)

Maple [A] (verified)

Time = 3.58 (sec) , antiderivative size = 451, normalized size of antiderivative = 1.91

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 (96 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-352 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+120 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-150 A \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 ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-336 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 )-120 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-50 C \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 )-120 C \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 )+266 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+190 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-135 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-75 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+5 A +5 C \right )}{30 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}\)	\(451\)

[In]

int((A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/30*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(96*A*cos(1/2*d*x+1/2*c)^10-352*A*cos(1/2*d*x+1/
2*c)^8+120*A*cos(1/2*d*x+1/2*c)^6-150*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*Ellipti
cF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3-336*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))-120*C*cos(1/2*d*x+1/2*c)^6-50*C*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))-120*C*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))+266*A*cos(1/2*d*x+1/2*c)^4+190*C*cos(1/2*d*x+1/2*c)^4-135*A*cos(1/2*d*x+1/2*c)^2-75*C*co
s(1/2*d*x+1/2*c)^2+5*A+5*C)/a^2/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 = 387, normalized size of antiderivative = 1.64 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {25 \, {\left (\sqrt {2} {\left (-3 i \, A - i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-3 i \, A - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A - i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 25 \, {\left (\sqrt {2} {\left (3 i \, A + i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (3 i \, A + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A + i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 12 \, {\left (\sqrt {2} {\left (-14 i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-14 i \, A - 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-14 i \, A - 5 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 12 \, {\left (\sqrt {2} {\left (14 i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (14 i \, A + 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (14 i \, A + 5 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (6 \, A \cos \left (d x + c\right )^{4} - 8 \, A \cos \left (d x + c\right )^{3} - 2 \, {\left (47 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{2} - 25 \, {\left (3 \, A + C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

integrate((A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/30*(25*(sqrt(2)*(-3*I*A - I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(-3*I*A - I*C)*cos(d*x + c) + sqrt(2)*(-3*I*A - I
*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 25*(sqrt(2)*(3*I*A + I*C)*cos(d*x + c)^2 + 2*
sqrt(2)*(3*I*A + I*C)*cos(d*x + c) + sqrt(2)*(3*I*A + I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*
x + c)) + 12*(sqrt(2)*(-14*I*A - 5*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(-14*I*A - 5*I*C)*cos(d*x + c) + sqrt(2)*(-
14*I*A - 5*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 12*(sqrt(
2)*(14*I*A + 5*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(14*I*A + 5*I*C)*cos(d*x + c) + sqrt(2)*(14*I*A + 5*I*C))*weier
strassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(6*A*cos(d*x + c)^4 - 8*A*cos
(d*x + c)^3 - 2*(47*A + 15*C)*cos(d*x + c)^2 - 25*(3*A + C)*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]

\[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A}{\sec ^{\frac {9}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {7}{2}}{\left (c + d x \right )} + \sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec ^{\frac {9}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {7}{2}}{\left (c + d x \right )} + \sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx}{a^{2}} \]

[In]

integrate((A+C*sec(d*x+c)**2)/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c))**2,x)

[Out]

(Integral(A/(sec(c + d*x)**(9/2) + 2*sec(c + d*x)**(7/2) + sec(c + d*x)**(5/2)), x) + Integral(C*sec(c + d*x)*
*2/(sec(c + d*x)**(9/2) + 2*sec(c + d*x)**(7/2) + sec(c + d*x)**(5/2)), x))/a**2

Maxima [F(-1)]

Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate((A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

Giac [F]

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

[In]

integrate((A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)/((a*sec(d*x + c) + a)^2*sec(d*x + c)^(5/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int((A + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))^2*(1/cos(c + d*x))^(5/2)),x)

[Out]

int((A + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))^2*(1/cos(c + d*x))^(5/2)), x)