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

Optimal. Leaf size=346 \[ \frac{\left (-5 a^2 A b^2+8 a^4 A-7 a^3 b B+a b^3 B+3 A b^4\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{4 a^3 d \left (a^2-b^2\right )^2}+\frac{\left (9 a^2 A b-5 a^3 B-a b^2 B-3 A b^3\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^2 d \left (a^2-b^2\right )^2}-\frac{\left (-6 a^2 A b^3+15 a^4 A b-10 a^3 b^2 B-3 a^5 B+a b^4 B+3 A b^5\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 d (a-b)^2 (a+b)^3}-\frac{\left (9 a^2 A b-5 a^3 B-a b^2 B-3 A b^3\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{4 a d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac{b (A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2} \]

[Out]

((9*a^2*A*b - 3*A*b^3 - 5*a^3*B - a*b^2*B)*EllipticE[(c + d*x)/2, 2])/(4*a^2*(a^2 - b^2)^2*d) + ((8*a^4*A - 5*
a^2*A*b^2 + 3*A*b^4 - 7*a^3*b*B + a*b^3*B)*EllipticF[(c + d*x)/2, 2])/(4*a^3*(a^2 - b^2)^2*d) - ((15*a^4*A*b -
6*a^2*A*b^3 + 3*A*b^5 - 3*a^5*B - 10*a^3*b^2*B + a*b^4*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(4*a^3*(
a - b)^2*(a + b)^3*d) + (b*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(b + a*Cos[c + d*x]
)^2) - ((9*a^2*A*b - 3*A*b^3 - 5*a^3*B - a*b^2*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*a*(a^2 - b^2)^2*d*(b + a
*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.10186, antiderivative size = 346, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {2954, 2989, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac{\left (-5 a^2 A b^2+8 a^4 A-7 a^3 b B+a b^3 B+3 A b^4\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 d \left (a^2-b^2\right )^2}+\frac{\left (9 a^2 A b-5 a^3 B-a b^2 B-3 A b^3\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^2 d \left (a^2-b^2\right )^2}-\frac{\left (-6 a^2 A b^3+15 a^4 A b-10 a^3 b^2 B-3 a^5 B+a b^4 B+3 A b^5\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 d (a-b)^2 (a+b)^3}-\frac{\left (9 a^2 A b-5 a^3 B-a b^2 B-3 A b^3\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{4 a d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac{b (A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((9*a^2*A*b - 3*A*b^3 - 5*a^3*B - a*b^2*B)*EllipticE[(c + d*x)/2, 2])/(4*a^2*(a^2 - b^2)^2*d) + ((8*a^4*A - 5*
a^2*A*b^2 + 3*A*b^4 - 7*a^3*b*B + a*b^3*B)*EllipticF[(c + d*x)/2, 2])/(4*a^3*(a^2 - b^2)^2*d) - ((15*a^4*A*b -
6*a^2*A*b^3 + 3*A*b^5 - 3*a^5*B - 10*a^3*b^2*B + a*b^4*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(4*a^3*(
a - b)^2*(a + b)^3*d) + (b*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(b + a*Cos[c + d*x]
)^2) - ((9*a^2*A*b - 3*A*b^3 - 5*a^3*B - a*b^2*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*a*(a^2 - b^2)^2*d*(b + a
*Cos[c + d*x]))

Rule 2954

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 2989

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((b*c - a*d)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)
*(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 - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (
A*b + a*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)
*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] && GtQ[m, 1] && 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)}{\sqrt{\cos (c+d x)} (a+b \sec (c+d x))^3} \, dx &=\int \frac{\cos ^{\frac{3}{2}}(c+d x) (B+A \cos (c+d x))}{(b+a \cos (c+d x))^3} \, dx\\ &=\frac{b (A b-a B) \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} b (A b-a B)-2 a (A b-a B) \cos (c+d x)+\frac{1}{2} \left (4 a^2 A-3 A b^2-a b B\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{b (A b-a B) \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 (9 a^2 A b-3 A b^3-5 a^3 B-a b^2 B\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac{\int \frac{\frac{1}{4} b \left (7 a^2 A b-A b^3-3 a^3 B-3 a b^2 B\right )-a b \left (2 a^2 A+A b^2-3 a b B\right ) \cos (c+d x)-\frac{1}{4} b \left (9 a^2 A b-3 A b^3-5 a^3 B-a b^2 B\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{b (A b-a B) \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 (9 a^2 A b-3 A b^3-5 a^3 B-a b^2 B\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac{\int \frac{-\frac{1}{4} a b \left (7 a^2 A b-A b^3-3 a^3 B-3 a b^2 B\right )+\frac{1}{4} b \left (8 a^4 A-5 a^2 A b^2+3 A b^4-7 a^3 b B+a b^3 B\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 (9 a^2 A b-3 A b^3-5 a^3 B-a b^2 B\right ) \int \sqrt{\cos (c+d x)} \, dx}{8 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (9 a^2 A b-3 A b^3-5 a^3 B-a b^2 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^2 \left (a^2-b^2\right )^2 d}+\frac{b (A b-a B) \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 (9 a^2 A b-3 A b^3-5 a^3 B-a b^2 B\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac{\left (8 a^4 A-5 a^2 A b^2+3 A b^4-7 a^3 b B+a b^3 B\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{8 a^3 \left (a^2-b^2\right )^2}-\frac{\left (15 a^4 A b-6 a^2 A b^3+3 A b^5-3 a^5 B-10 a^3 b^2 B+a b^4 B\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (9 a^2 A b-3 A b^3-5 a^3 B-a b^2 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^2 \left (a^2-b^2\right )^2 d}+\frac{\left (8 a^4 A-5 a^2 A b^2+3 A b^4-7 a^3 b B+a b^3 B\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 (15 a^4 A b-6 a^2 A b^3+3 A b^5-3 a^5 B-10 a^3 b^2 B+a b^4 B\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 (a-b)^2 (a+b)^3 d}+\frac{b (A b-a B) \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 (9 a^2 A b-3 A b^3-5 a^3 B-a b^2 B\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 4.49601, size = 361, normalized size = 1.04 \[ \frac{\frac{\frac{8 \left (2 a^2 A-3 a b B+A b^2\right ) \left ((a+b) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{a+b}+\frac{\left (-9 a^2 A b+5 a^3 B+a b^2 B+3 A b^3\right ) \sin (c+d x) \left (-2 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt{\sin ^2(c+d x)}}+\frac{\left (-5 a^2 A b+a^3 B+5 a b^2 B-A b^3\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}}{(a-b)^2 (a+b)^2}+\frac{2 \sin (c+d x) \sqrt{\cos (c+d x)} \left (a \left (-9 a^2 A b+5 a^3 B+a b^2 B+3 A b^3\right ) \cos (c+d x)+b \left (-7 a^2 A b+3 a^3 B+3 a b^2 B+A b^3\right )\right )}{\left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}}{8 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 8.999, size = 1959, normalized size = 5.7 \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))/(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*b*(3*A*b-2*B*a)*(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)*(s
in(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*(si
n(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^2*(A*b-B*a)/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)*El
lipticF(cos(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)*Ellipt
icE(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*c
os(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)))-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)))/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))/(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))/(a+b*sec(d*x+c))^3/cos(d*x+c)^(1/2),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))/(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{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))/(a+b*sec(d*x+c))^3/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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