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

Optimal. Leaf size=367 \[ -\frac{\left (-33 a^2 A b^3+24 a^4 A b+5 a^3 b^2 B-8 a^5 B-3 a b^4 B+15 A b^5\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{4 a^4 d \left (a^2-b^2\right )^2}+\frac{\left (-29 a^2 A b^2+8 a^4 A+9 a^3 b B-3 a b^3 B+15 A b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 d \left (a^2-b^2\right )^2}+\frac{b \left (-38 a^2 A b^3+35 a^4 A b+6 a^3 b^2 B-15 a^5 B-3 a b^4 B+15 A b^5\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 d (a-b)^2 (a+b)^3}+\frac{b (A b-a B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac{b \left (11 a^2 A b-7 a^3 B+a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{4 a^2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)} \]

[Out]

((8*a^4*A - 29*a^2*A*b^2 + 15*A*b^4 + 9*a^3*b*B - 3*a*b^3*B)*EllipticE[(c + d*x)/2, 2])/(4*a^3*(a^2 - b^2)^2*d
) - ((24*a^4*A*b - 33*a^2*A*b^3 + 15*A*b^5 - 8*a^5*B + 5*a^3*b^2*B - 3*a*b^4*B)*EllipticF[(c + d*x)/2, 2])/(4*
a^4*(a^2 - b^2)^2*d) + (b*(35*a^4*A*b - 38*a^2*A*b^3 + 15*A*b^5 - 15*a^5*B + 6*a^3*b^2*B - 3*a*b^4*B)*Elliptic
Pi[(2*a)/(a + b), (c + d*x)/2, 2])/(4*a^4*(a - b)^2*(a + b)^3*d) + (b*(A*b - a*B)*Cos[c + d*x]^(3/2)*Sin[c + d
*x])/(2*a*(a^2 - b^2)*d*(b + a*Cos[c + d*x])^2) + (b*(11*a^2*A*b - 5*A*b^3 - 7*a^3*B + a*b^2*B)*Sqrt[Cos[c + d
*x]]*Sin[c + d*x])/(4*a^2*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.10805, antiderivative size = 367, 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, 3047, 3059, 2639, 3002, 2641, 2805} \[ -\frac{\left (-33 a^2 A b^3+24 a^4 A b+5 a^3 b^2 B-8 a^5 B-3 a b^4 B+15 A b^5\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 d \left (a^2-b^2\right )^2}+\frac{\left (-29 a^2 A b^2+8 a^4 A+9 a^3 b B-3 a b^3 B+15 A b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 d \left (a^2-b^2\right )^2}+\frac{b \left (-38 a^2 A b^3+35 a^4 A b+6 a^3 b^2 B-15 a^5 B-3 a b^4 B+15 A b^5\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 d (a-b)^2 (a+b)^3}+\frac{b (A b-a B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac{b \left (11 a^2 A b-7 a^3 B+a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{4 a^2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^4*A - 29*a^2*A*b^2 + 15*A*b^4 + 9*a^3*b*B - 3*a*b^3*B)*EllipticE[(c + d*x)/2, 2])/(4*a^3*(a^2 - b^2)^2*d
) - ((24*a^4*A*b - 33*a^2*A*b^3 + 15*A*b^5 - 8*a^5*B + 5*a^3*b^2*B - 3*a*b^4*B)*EllipticF[(c + d*x)/2, 2])/(4*
a^4*(a^2 - b^2)^2*d) + (b*(35*a^4*A*b - 38*a^2*A*b^3 + 15*A*b^5 - 15*a^5*B + 6*a^3*b^2*B - 3*a*b^4*B)*Elliptic
Pi[(2*a)/(a + b), (c + d*x)/2, 2])/(4*a^4*(a - b)^2*(a + b)^3*d) + (b*(A*b - a*B)*Cos[c + d*x]^(3/2)*Sin[c + d
*x])/(2*a*(a^2 - b^2)*d*(b + a*Cos[c + d*x])^2) + (b*(11*a^2*A*b - 5*A*b^3 - 7*a^3*B + a*b^2*B)*Sqrt[Cos[c + d
*x]]*Sin[c + d*x])/(4*a^2*(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 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 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{\sqrt{\cos (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx &=\int \frac{\cos ^{\frac{5}{2}}(c+d x) (B+A \cos (c+d x))}{(b+a \cos (c+d x))^3} \, dx\\ &=\frac{b (A b-a B) \cos ^{\frac{3}{2}}(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{\sqrt{\cos (c+d x)} \left (\frac{3}{2} b (A b-a B)-2 a (A b-a B) \cos (c+d x)+\frac{1}{2} \left (4 a^2 A-5 A b^2+a b B\right ) \cos ^2(c+d x)\right )}{(b+a \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{b (A b-a B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac{b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac{\int \frac{\frac{1}{4} b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right )-a \left (4 a^2 A b-A b^3-2 a^3 B-a b^2 B\right ) \cos (c+d x)+\frac{1}{4} \left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{b (A b-a B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac{b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac{\int \frac{-\frac{1}{4} a b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right )+\frac{1}{4} \left (24 a^4 A b-33 a^2 A b^3+15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^3 \left (a^2-b^2\right )^2}+\frac{\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \int \sqrt{\cos (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac{b (A b-a B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac{b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac{\left (24 a^4 A b-33 a^2 A b^3+15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{8 a^4 \left (a^2-b^2\right )^2}+\frac{\left (b \left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 a b^4 B\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 a^4 \left (a^2-b^2\right )^2}\\ &=\frac{\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac{\left (24 a^4 A b-33 a^2 A b^3+15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac{b \left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 a b^4 B\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 (a-b)^2 (a+b)^3 d}+\frac{b (A b-a B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac{b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 4.70144, size = 394, normalized size = 1.07 \[ \frac{\frac{\frac{16 \left (-4 a^2 A b+2 a^3 B+a b^2 B+A b^3\right ) \left ((a+b) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{a+b}-\frac{2 \left (-29 a^2 A b^2+8 a^4 A+9 a^3 b B-3 a b^3 B+15 A b^4\right ) \sin (c+d x) \left (-2 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt{\sin ^2(c+d x)}}+\frac{2 \left (-7 a^2 A b^2+8 a^4 A-5 a^3 b B-a b^3 B+5 A b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}}{(a-b)^2 (a+b)^2}-\frac{4 b \sin (c+d x) \sqrt{\cos (c+d x)} \left (a \left (-13 a^2 A b+9 a^3 B-3 a b^2 B+7 A b^3\right ) \cos (c+d x)+b \left (-11 a^2 A b+7 a^3 B-a b^2 B+5 A b^3\right )\right )}{\left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}}{16 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*b*Sqrt[Cos[c + d*x]]*(b*(-11*a^2*A*b + 5*A*b^3 + 7*a^3*B - a*b^2*B) + a*(-13*a^2*A*b + 7*A*b^3 + 9*a^3*B
- 3*a*b^2*B)*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(b + a*Cos[c + d*x])^2) + ((2*(8*a^4*A - 7*a^2*A*b^2 +
5*A*b^4 - 5*a^3*b*B - a*b^3*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + (16*(-4*a^2*A*b + A*b^3 +
2*a^3*B + a*b^2*B)*((a + b)*EllipticF[(c + d*x)/2, 2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a + b)
- (2*(8*a^4*A - 29*a^2*A*b^2 + 15*A*b^4 + 9*a^3*b*B - 3*a*b^3*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[C
os[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^2*d)

________________________________________________________________________________________

Maple [B]  time = 10.316, size = 2000, normalized size = 5.5 \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))*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,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^4/(-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)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1
/2))*b+A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a-B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a)-2/a^4*b^2*(4*A*b-3
*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)*(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*s
in(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*b^3*(A*b-B*a)/a^4*(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(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*(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))-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)))-6/a^3*b*(2*A*b-B*a)/(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))*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt{\cos \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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