3.1328 \(\int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=387 \[ \frac{\text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )}-\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b B-5 a^3 C-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right )}-\frac{\left (-a^2 b^2 (A-7 C)+3 a^3 b B-5 a^4 C-5 a b^3 B+3 A b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2}-\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+b)}+\frac{\sin (c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x)}+\frac{\sin (c+d x) \left (3 a^2 b B-5 a^3 C-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)}} \]

[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]))

________________________________________________________________________________________

Rubi [A]  time = 1.4961, antiderivative size = 387, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {4112, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )}-\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b B-5 a^3 C-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right )}-\frac{\left (-a^2 b^2 (A-7 C)+3 a^3 b B-5 a^4 C-5 a b^3 B+3 A b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2}-\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+b)}+\frac{\sin (c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x)}+\frac{\sin (c+d x) \left (3 a^2 b B-5 a^3 C-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[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 4112

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]

Rule 3055

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] + Dis
t[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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3059

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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 3002

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 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[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]

Rubi steps

\begin{align*} \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+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 ) F\left (\left .\frac{1}{2} (c+d x)\right |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 ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |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]  time = 7.17638, size = 474, normalized size = 1.22 \[ \frac{\frac{\left (-24 a^2 b^2 B+40 a^3 b C+12 a A b^3-28 a b^3 C+12 b^4 B\right ) \left (2 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-\frac{2 b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{a}+\frac{\sin (c+d x) \cos (2 (c+d x)) \left (3 a^2 A b^2-12 a^2 b^2 C-9 a^3 b B+15 a^4 C+6 a b^3 B\right ) \left (4 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )-2 \left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-4 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt{1-\cos ^2(c+d x)} \left (2 \cos ^2(c+d x)-1\right )}+\frac{2 \left (9 a^2 A b^2-44 a^2 b^2 C-27 a^3 b B+45 a^4 C+30 a b^3 B-12 A b^4-4 b^4 C\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}}{12 b^3 d (a-b) (a+b)}+\frac{\sqrt{\cos (c+d x)} \left (\frac{a^2 A b^2 \sin (c+d x)-a^3 b B \sin (c+d x)+a^4 C \sin (c+d x)}{b^3 \left (b^2-a^2\right ) (a \cos (c+d x)+b)}+\frac{2 \sec (c+d x) (b B \sin (c+d x)-2 a C \sin (c+d x))}{b^3}+\frac{2 C \tan (c+d x) \sec (c+d x)}{3 b^2}\right )}{d} \]

Antiderivative was successfully verified.

[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 + ((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)]*(-4*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 4*b
*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] - 2*(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

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Maple [B]  time = 12.576, size = 1031, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

[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*(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)))+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(co
s(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+2*(B*b-2*C*a)/b^3*(-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(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))+2*(-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)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*s
in(1/2*d*x+1/2*c)^2-1))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)