3.1331 \(\int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=423 \[ \frac{\text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (-a^2 b^2 (5 A-3 C)+a^4 (8 A+3 C)-7 a^3 b B+a b^3 B+3 A b^4\right )}{4 a^3 d \left (a^2-b^2\right )^2}-\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-a^2 b^2 (9 A+5 C)+5 a^3 b B+a^4 (-C)+a b^3 B+3 A b^4\right )}{4 a^2 b d \left (a^2-b^2\right )^2}-\frac{\left (5 a^4 b^2 (3 A+2 C)-3 a^2 b^4 (2 A-C)-10 a^3 b^3 B-3 a^5 b B+a^6 (-C)+a b^5 B+3 A b^6\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 b d (a-b)^2 (a+b)^3}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)} \left (-a^2 b^2 (9 A+5 C)+5 a^3 b B+a^4 (-C)+a b^3 B+3 A b^4\right )}{4 a b d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2} \]

[Out]

-((3*A*b^4 + 5*a^3*b*B + a*b^3*B - a^4*C - a^2*b^2*(9*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(4*a^2*b*(a^2 - b^2
)^2*d) + ((3*A*b^4 - 7*a^3*b*B + a*b^3*B - a^2*b^2*(5*A - 3*C) + a^4*(8*A + 3*C))*EllipticF[(c + d*x)/2, 2])/(
4*a^3*(a^2 - b^2)^2*d) - ((3*A*b^6 - 3*a^5*b*B - 10*a^3*b^3*B + a*b^5*B - 3*a^2*b^4*(2*A - C) - a^6*C + 5*a^4*
b^2*(3*A + 2*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(4*a^3*(a - b)^2*b*(a + b)^3*d) + ((A*b^2 - a*(b*B
- a*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(b + a*Cos[c + d*x])^2) + ((3*A*b^4 + 5*a^3*b*B +
a*b^3*B - a^4*C - a^2*b^2*(9*A + 5*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*a*b*(a^2 - b^2)^2*d*(b + a*Cos[c +
d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.35867, antiderivative size = 423, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used = {4112, 3047, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-a^2 b^2 (5 A-3 C)+a^4 (8 A+3 C)-7 a^3 b B+a b^3 B+3 A b^4\right )}{4 a^3 d \left (a^2-b^2\right )^2}-\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-a^2 b^2 (9 A+5 C)+5 a^3 b B+a^4 (-C)+a b^3 B+3 A b^4\right )}{4 a^2 b d \left (a^2-b^2\right )^2}-\frac{\left (5 a^4 b^2 (3 A+2 C)-3 a^2 b^4 (2 A-C)-10 a^3 b^3 B-3 a^5 b B+a^6 (-C)+a b^5 B+3 A b^6\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 b d (a-b)^2 (a+b)^3}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)} \left (-a^2 b^2 (9 A+5 C)+5 a^3 b B+a^4 (-C)+a b^3 B+3 A b^4\right )}{4 a b d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*A*b^4 + 5*a^3*b*B + a*b^3*B - a^4*C - a^2*b^2*(9*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(4*a^2*b*(a^2 - b^2
)^2*d) + ((3*A*b^4 - 7*a^3*b*B + a*b^3*B - a^2*b^2*(5*A - 3*C) + a^4*(8*A + 3*C))*EllipticF[(c + d*x)/2, 2])/(
4*a^3*(a^2 - b^2)^2*d) - ((3*A*b^6 - 3*a^5*b*B - 10*a^3*b^3*B + a*b^5*B - 3*a^2*b^4*(2*A - C) - a^6*C + 5*a^4*
b^2*(3*A + 2*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(4*a^3*(a - b)^2*b*(a + b)^3*d) + ((A*b^2 - a*(b*B
- a*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(b + a*Cos[c + d*x])^2) + ((3*A*b^4 + 5*a^3*b*B +
a*b^3*B - a^4*C - a^2*b^2*(9*A + 5*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*a*b*(a^2 - b^2)^2*d*(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 3047

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[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(
c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c
*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)
))*Sin[e + f*x]^2, 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] && GtQ[m, 0] && LtQ[n, -1]

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)}{\sqrt{\cos (c+d x)} (a+b \sec (c+d x))^3} \, dx &=\int \frac{\sqrt{\cos (c+d x)} \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{(b+a \cos (c+d x))^3} \, dx\\ &=\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac{\int \frac{\frac{1}{2} \left (A b^2-a (b B-a C)\right )-2 a (A b-a B+b C) \cos (c+d x)-\frac{1}{2} \left (3 A b^2+a b B-a^2 (4 A+C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac{\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac{\int \frac{\frac{1}{4} \left (-A b^4-3 a^3 b B-3 a b^3 B-a^4 C+7 a^2 b^2 (A+C)\right )+a b \left (3 a b B-a^2 (2 A+C)-b^2 (A+2 C)\right ) \cos (c+d x)+\frac{1}{4} \left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a b \left (a^2-b^2\right )^2}\\ &=\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac{\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac{\int \frac{\frac{1}{4} a \left (A b^4+3 a^3 b B+3 a b^3 B+a^4 C-7 a^2 b^2 (A+C)\right )+\frac{1}{4} b \left (3 A b^4-7 a^3 b B+a b^3 B-a^2 b^2 (5 A-3 C)+a^4 (8 A+3 C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^2 b \left (a^2-b^2\right )^2}-\frac{\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) \int \sqrt{\cos (c+d x)} \, dx}{8 a^2 b \left (a^2-b^2\right )^2}\\ &=-\frac{\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^2 b \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac{\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac{\left (3 A b^6-3 a^5 b B-10 a^3 b^3 B+a b^5 B-3 a^2 b^4 (2 A-C)-a^6 C+5 a^4 b^2 (3 A+2 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 a^3 b \left (a^2-b^2\right )^2}+\frac{\left (3 A b^4-7 a^3 b B+a b^3 B-a^2 b^2 (5 A-3 C)+a^4 (8 A+3 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{8 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^2 b \left (a^2-b^2\right )^2 d}+\frac{\left (3 A b^4-7 a^3 b B+a b^3 B-a^2 b^2 (5 A-3 C)+a^4 (8 A+3 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac{\left (3 A b^6-3 a^5 b B-10 a^3 b^3 B+a b^5 B-3 a^2 b^4 (2 A-C)-a^6 C+5 a^4 b^2 (3 A+2 C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 (a-b)^2 b (a+b)^3 d}+\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac{\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 5.813, size = 428, normalized size = 1.01 \[ \frac{\frac{\frac{16 b \left (a^2 (2 A+C)-3 a b B+b^2 (A+2 C)\right ) \left ((a+b) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{a+b}-\frac{2 \sin (c+d x) \left (a^2 b^2 (9 A+5 C)-5 a^3 b B+a^4 C-a b^3 B-3 A b^4\right ) \left (-2 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt{\sin ^2(c+d x)}}+\frac{2 \left (-a^2 b^2 (5 A+9 C)+a^3 b B+3 a^4 C+5 a b^3 B-A b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}}{(a-b)^2 (a+b)^2}-\frac{4 \sin (c+d x) \sqrt{\cos (c+d x)} \left (a \cos (c+d x) \left (a^2 b^2 (9 A+5 C)-5 a^3 b B+a^4 C-a b^3 B-3 A b^4\right )-b \left (-7 a^2 b^2 (A+C)+3 a^3 b B+a^4 C+3 a b^3 B+A b^4\right )\right )}{\left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}}{16 a b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*Sqrt[Cos[c + d*x]]*(-(b*(A*b^4 + 3*a^3*b*B + 3*a*b^3*B + a^4*C - 7*a^2*b^2*(A + C))) + a*(-3*A*b^4 - 5*a^
3*b*B - a*b^3*B + a^4*C + a^2*b^2*(9*A + 5*C))*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(b + a*Cos[c + d*x])
^2) + ((2*(-(A*b^4) + a^3*b*B + 5*a*b^3*B + 3*a^4*C - a^2*b^2*(5*A + 9*C))*EllipticPi[(2*a)/(a + b), (c + d*x)
/2, 2])/(a + b) + (16*b*(-3*a*b*B + a^2*(2*A + C) + b^2*(A + 2*C))*((a + b)*EllipticF[(c + d*x)/2, 2] - b*Elli
pticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a + b) - (2*(-3*A*b^4 - 5*a^3*b*B - a*b^3*B + a^4*C + a^2*b^2*(9*A +
5*C))*(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[Sin[c + d*x]^2
]))/((a - b)^2*(a + b)^2))/(16*a*b*d)

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Maple [B]  time = 11.946, size = 1972, normalized size = 4.7 \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)/(a+b*sec(d*x+c))^3/cos(d*x+c)^(1/2),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*A/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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)+2/a^3*(3*A*b^2-2*B*a*b+C*a^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*(-3*A*b+B*a)/a^2/(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*b*(A*b^2-B*a*b+C*a^2)/a^3*(1/2*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)^2+3/4*a^2*(a^2-3*b^2)/b^2/(a^2-b^
2)^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)-3/
8/(a+b)/(a^2-b^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))*a^2-1/4/(a+b)/(a^2-b^2)/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(co
s(1/2*d*x+1/2*c),2^(1/2))*a+7/8/(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))+3/8*a^3/b^2/(a^2-b
^2)^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))-9/8*a/(a^2-b^2)^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)
)-3/8*a^3/b^2/(a^2-b^2)^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))+9/8*a/(a^2-b^2)^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))-3/8/(a-b)/(a+b)/(a^2-b^2)/b^2/(a^2-a*b)*a^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)/(-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))+3/4/(a-b)/(a+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))-15/8
/(a-b)/(a+b)/(a^2-b^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

________________________________________________________________________________________

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)/(a+b*sec(d*x+c))^3/cos(d*x+c)^(1/2),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)/(a+b*sec(d*x+c))^3/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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)/(a+b*sec(d*x+c))**3/cos(d*x+c)**(1/2),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 )}^{3} \sqrt{\cos \left (d x + c\right )}}\,{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)/(a+b*sec(d*x+c))^3/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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