\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx\) [1328]

Optimal result

Integrand size = 43, antiderivative size = 387 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=-\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 C)-5 a^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{(a-b) b^3 (a+b)^2 d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))} \]

[Out]

-(3*B*a^2*b-2*B*b^3-a*b^2*(A-4*C)-5*a^3*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))/b^3/(a^2-b^2)/d+1/3*(3*A*b^2-3*B*a*b+5*C*a^2-2*C*b^2)*(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))/b^2/(a^2-b^2)/d-(3*A*b^4+3*B*a^3*b-5*B*a*b^3-a^2*b^2*(A-7*C)-
5*a^4*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/(a-b
)/b^3/(a+b)^2/d+1/3*(3*A*b^2-3*B*a*b+5*C*a^2-2*C*b^2)*sin(d*x+c)/b^2/(a^2-b^2)/d/cos(d*x+c)^(3/2)-(A*b^2-a*(B*
b-C*a))*sin(d*x+c)/b/(a^2-b^2)/d/cos(d*x+c)^(3/2)/(b+a*cos(d*x+c))+(3*B*a^2*b-2*B*b^3-a*b^2*(A-4*C)-5*a^3*C)*s
in(d*x+c)/b^3/(a^2-b^2)/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 1.65 (sec) , antiderivative size = 387, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4197, 3134, 3138, 2719, 3081, 2720, 2884} \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )}-\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b)}+\frac {\sin (c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x)}-\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right )}+\frac {\sin (c+d x) \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)}}-\frac {\left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{b^3 d (a-b) (a+b)^2} \]

[In]

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

[Out]

-(((3*a^2*b*B - 2*b^3*B - a*b^2*(A - 4*C) - 5*a^3*C)*EllipticE[(c + d*x)/2, 2])/(b^3*(a^2 - b^2)*d)) + ((3*A*b
^2 - 3*a*b*B + 5*a^2*C - 2*b^2*C)*EllipticF[(c + d*x)/2, 2])/(3*b^2*(a^2 - b^2)*d) - ((3*A*b^4 + 3*a^3*b*B - 5
*a*b^3*B - a^2*b^2*(A - 7*C) - 5*a^4*C)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/((a - b)*b^3*(a + b)^2*d) +
((3*A*b^2 - 3*a*b*B + 5*a^2*C - 2*b^2*C)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)) + ((3*a^2*b*B
- 2*b^3*B - a*b^2*(A - 4*C) - 5*a^3*C)*Sin[c + d*x])/(b^3*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]) - ((A*b^2 - a*(b*
B - a*C))*Sin[c + d*x])/(b*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*(b + a*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 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 3081

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3134

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*
(b*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(
b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3138

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 4197

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] &&  !IntegerQ[n] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \frac {C+B \cos (c+d x)+A \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b+a \cos (c+d x))^2} \, dx \\ & = -\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac {\int \frac {\frac {1}{2} \left (-3 A b^2+3 a b B-5 a^2 C+2 b^2 C\right )+b (b B-a (A+C)) \cos (c+d x)+\frac {3}{2} \left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{b \left (a^2-b^2\right )} \\ & = \frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac {2 \int \frac {-\frac {3}{4} \left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right )+\frac {1}{2} b \left (3 A b^2-3 a b B+2 a^2 C+b^2 C\right ) \cos (c+d x)-\frac {1}{4} a \left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{3 b^2 \left (a^2-b^2\right )} \\ & = \frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac {4 \int \frac {\frac {1}{8} \left (9 a^3 b B-12 a b^3 B-a^2 b^2 (3 A-16 C)-15 a^4 C+2 b^4 (3 A+C)\right )+\frac {1}{4} b \left (6 a^2 b B-3 b^3 B-a b^2 (3 A-7 C)-10 a^3 C\right ) \cos (c+d x)+\frac {3}{8} a \left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 b^3 \left (a^2-b^2\right )} \\ & = \frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}+\frac {4 \int \frac {-\frac {1}{8} a \left (9 a^3 b B-12 a b^3 B-a^2 b^2 (3 A-16 C)-15 a^4 C+2 b^4 (3 A+C)\right )+\frac {1}{8} a^2 b \left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 a b^3 \left (a^2-b^2\right )}-\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )} \\ & = -\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac {\left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 C)-5 a^4 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 b^3 \left (a^2-b^2\right )}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 b^2 \left (a^2-b^2\right )} \\ & = -\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 C)-5 a^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{(a-b) b^3 (a+b)^2 d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 9.16 (sec) , antiderivative size = 472, normalized size of antiderivative = 1.22 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\frac {\frac {2 \left (9 a^2 A b^2-12 A b^4-27 a^3 b B+30 a b^3 B+45 a^4 C-44 a^2 b^2 C-4 b^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {\left (12 a A b^3-24 a^2 b^2 B+12 b^4 B+40 a^3 b C-28 a b^3 C\right ) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{a}+\frac {2 \left (3 a^2 A b^2-9 a^3 b B+6 a b^3 B+15 a^4 C-12 a^2 b^2 C\right ) \cos (2 (c+d x)) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {1-\cos ^2(c+d x)} \left (-1+2 \cos ^2(c+d x)\right )}}{12 (a-b) b^3 (a+b) d}+\frac {\sqrt {\cos (c+d x)} \left (\frac {2 \sec (c+d x) (b B \sin (c+d x)-2 a C \sin (c+d x))}{b^3}+\frac {a^2 A b^2 \sin (c+d x)-a^3 b B \sin (c+d x)+a^4 C \sin (c+d x)}{b^3 \left (-a^2+b^2\right ) (b+a \cos (c+d x))}+\frac {2 C \sec (c+d x) \tan (c+d x)}{3 b^2}\right )}{d} \]

[In]

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

[Out]

((2*(9*a^2*A*b^2 - 12*A*b^4 - 27*a^3*b*B + 30*a*b^3*B + 45*a^4*C - 44*a^2*b^2*C - 4*b^4*C)*EllipticPi[(2*a)/(a
+ b), (c + d*x)/2, 2])/(a + b) + ((12*a*A*b^3 - 24*a^2*b^2*B + 12*b^4*B + 40*a^3*b*C - 28*a*b^3*C)*(2*Ellipti
cF[(c + d*x)/2, 2] - (2*b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)))/a + (2*(3*a^2*A*b^2 - 9*a^3*b*B
+ 6*a*b^3*B + 15*a^4*C - 12*a^2*b^2*C)*Cos[2*(c + d*x)]*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2
*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt[Cos[c + d*
x]]], -1])*Sin[c + d*x])/(a^2*b*Sqrt[1 - Cos[c + d*x]^2]*(-1 + 2*Cos[c + d*x]^2)))/(12*(a - b)*b^3*(a + b)*d)
+ (Sqrt[Cos[c + d*x]]*((2*Sec[c + d*x]*(b*B*Sin[c + d*x] - 2*a*C*Sin[c + d*x]))/b^3 + (a^2*A*b^2*Sin[c + d*x]
- a^3*b*B*Sin[c + d*x] + a^4*C*Sin[c + d*x])/(b^3*(-a^2 + b^2)*(b + a*Cos[c + d*x])) + (2*C*Sec[c + d*x]*Tan[c
+ d*x])/(3*b^2)))/d

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1003\) vs. \(2(455)=910\).

Time = 4.78 (sec) , antiderivative size = 1004, normalized size of antiderivative = 2.59

method	result	size
default	\(\text {Expression too large to display}\)	\(1004\)

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C/b^2*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1
/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*
d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
)))+2*(B*b-2*C*a)/b^3/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2
*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1
/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2))+2*a^2*(B*b-2*C*a)/b^3/(a^2-a*b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-
2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/
2*c),2*a/(a-b),2^(1/2))+2*(A*b^2-B*a*b+C*a^2)/b^2*(a^2/b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4
+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*a-a+b)-1/2/(a+b)/b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1
/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(
1/2))+1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^
4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*
d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b)*a^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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/2*b
/(a^2-b^2)/(a^2-a*b)*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)/(-2*sin(1/2*d*x+1/2*c)^4
+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),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 [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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