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

Optimal. Leaf size=365 \[ \frac{\left (16 a^2 A b^2+2 a^4 A-12 a^3 b B+9 a b^3 B-15 A b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a^4 d \left (a^2-b^2\right )}+\frac{b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))}+\frac{\left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)}}-\frac{\left (4 a^2 A b-2 a^3 B+3 a b^2 B-5 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}-\frac{b^2 \left (7 a^2 A b-5 a^3 B+3 a b^2 B-5 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 d (a-b) (a+b)^2} \]

[Out]

-(((4*a^2*A*b - 5*A*b^3 - 2*a^3*B + 3*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]]
)/(a^3*(a^2 - b^2)*d)) + ((2*a^4*A + 16*a^2*A*b^2 - 15*A*b^4 - 12*a^3*b*B + 9*a*b^3*B)*Sqrt[Cos[c + d*x]]*Elli
pticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^4*(a^2 - b^2)*d) - (b^2*(7*a^2*A*b - 5*A*b^3 - 5*a^3*B + 3*a*b^
2*B)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^4*(a - b)*(a + b)^2*d
) + ((2*a^2*A - 5*A*b^2 + 3*a*b*B)*Sin[c + d*x])/(3*a^2*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]) + (b*(A*b - a*B)*Sin
[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.859338, antiderivative size = 365, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4030, 4104, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac{b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))}+\frac{\left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)}}+\frac{\left (16 a^2 A b^2+2 a^4 A-12 a^3 b B+9 a b^3 B-15 A b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^4 d \left (a^2-b^2\right )}-\frac{\left (4 a^2 A b-2 a^3 B+3 a b^2 B-5 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}-\frac{b^2 \left (7 a^2 A b-5 a^3 B+3 a b^2 B-5 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 d (a-b) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((4*a^2*A*b - 5*A*b^3 - 2*a^3*B + 3*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]]
)/(a^3*(a^2 - b^2)*d)) + ((2*a^4*A + 16*a^2*A*b^2 - 15*A*b^4 - 12*a^3*b*B + 9*a*b^3*B)*Sqrt[Cos[c + d*x]]*Elli
pticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^4*(a^2 - b^2)*d) - (b^2*(7*a^2*A*b - 5*A*b^3 - 5*a^3*B + 3*a*b^
2*B)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^4*(a - b)*(a + b)^2*d
) + ((2*a^2*A - 5*A*b^2 + 3*a*b*B)*Sin[c + d*x])/(3*a^2*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]) + (b*(A*b - a*B)*Sin
[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x]))

Rule 4030

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(b*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/
(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*
x])^n*Simp[A*(a^2*(m + 1) - b^2*(m + n + 1)) + a*b*B*n - a*(A*b - a*B)*(m + 1)*Csc[e + f*x] + b*(A*b - a*B)*(m
+ n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b
^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] && ILtQ[n, 0])

Rule 4104

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 4106

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2), Int[(d*Csc[
e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Dist[1/a^2, Int[(a*A - (A*b - a*B)*Csc[e + f*x])/Sqrt[d*Csc[e +
f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rule 3849

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[d*Sqrt[d*S
in[e + f*x]]*Sqrt[d*Csc[e + f*x]], Int[1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d
, e, f}, x] && NeQ[a^2 - b^2, 0]

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]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

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 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]

Rubi steps

\begin{align*} \int \frac{A+B \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx &=\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))}-\frac{\int \frac{\frac{1}{2} \left (-2 a^2 A+5 A b^2-3 a b B\right )+a (A b-a B) \sec (c+d x)-\frac{3}{2} b (A b-a B) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{\left (2 a^2 A-5 A b^2+3 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}+\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))}+\frac{2 \int \frac{-\frac{3}{4} \left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right )+\frac{1}{2} a \left (a^2 A+2 A b^2-3 a b B\right ) \sec (c+d x)+\frac{1}{4} b \left (2 a^2 A-5 A b^2+3 a b B\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=\frac{\left (2 a^2 A-5 A b^2+3 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}+\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))}+\frac{2 \int \frac{-\frac{3}{4} a \left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right )-\left (-\frac{1}{2} a^2 \left (a^2 A+2 A b^2-3 a b B\right )-\frac{3}{4} b \left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{3 a^4 \left (a^2-b^2\right )}-\frac{\left (b^2 \left (7 a^2 A b-5 A b^3-5 a^3 B+3 a b^2 B\right )\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^4 \left (a^2-b^2\right )}\\ &=\frac{\left (2 a^2 A-5 A b^2+3 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}+\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))}-\frac{\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 a^3 \left (a^2-b^2\right )}+\frac{\left (2 a^4 A+16 a^2 A b^2-15 A b^4-12 a^3 b B+9 a b^3 B\right ) \int \sqrt{\sec (c+d x)} \, dx}{6 a^4 \left (a^2-b^2\right )}-\frac{\left (b^2 \left (7 a^2 A b-5 A b^3-5 a^3 B+3 a b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^4 \left (a^2-b^2\right )}\\ &=-\frac{b^2 \left (7 a^2 A b-5 A b^3-5 a^3 B+3 a b^2 B\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^4 (a-b) (a+b)^2 d}+\frac{\left (2 a^2 A-5 A b^2+3 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}+\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))}-\frac{\left (\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}+\frac{\left (\left (2 a^4 A+16 a^2 A b^2-15 A b^4-12 a^3 b B+9 a b^3 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^4 \left (a^2-b^2\right )}\\ &=-\frac{\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (2 a^4 A+16 a^2 A b^2-15 A b^4-12 a^3 b B+9 a b^3 B\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^4 \left (a^2-b^2\right ) d}-\frac{b^2 \left (7 a^2 A b-5 A b^3-5 a^3 B+3 a b^2 B\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^4 (a-b) (a+b)^2 d}+\frac{\left (2 a^2 A-5 A b^2+3 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}+\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 7.14556, size = 704, normalized size = 1.93 \[ \frac{\frac{2 \left (-8 a^2 A b+6 a^3 B-3 a b^2 B+5 A b^3\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} (a+b \sec (c+d x)) \left (\text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right ),-1\right )+\Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right )}{b \left (1-\cos ^2(c+d x)\right ) (a \cos (c+d x)+b)}-\frac{2 \left (-12 a^2 A b+6 a^3 B-9 a b^2 B+15 A b^3\right ) \sin (c+d x) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (a (a-2 b) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right ),-1\right )+a^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-2 b^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-2 a b \sec ^2(c+d x)+2 a b \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+2 a b\right )}{a^2 b \left (1-\cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} \left (2-\sec ^2(c+d x)\right ) (a \cos (c+d x)+b)}-\frac{2 \left (4 a^3 A-12 a^2 b B+8 a A b^2\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} (a+b \sec (c+d x)) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )}{a \left (1-\cos ^2(c+d x)\right ) (a \cos (c+d x)+b)}}{12 a^2 d (a-b) (a+b)}+\frac{\sqrt{\sec (c+d x)} \left (\frac{b^2 (A b-a B) \sin (c+d x)}{a^3 \left (b^2-a^2\right )}-\frac{a b^3 B \sin (c+d x)-A b^4 \sin (c+d x)}{a^3 \left (a^2-b^2\right ) (a \cos (c+d x)+b)}+\frac{A \sin (2 (c+d x))}{3 a^2}\right )}{d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-2*(4*a^3*A + 8*a*A*b^2 - 12*a^2*b*B)*Cos[c + d*x]^2*EllipticPi[-(b/a), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*(a
+ b*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(a*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(-
8*a^2*A*b + 5*A*b^3 + 6*a^3*B - 3*a*b^2*B)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] + Ellipti
cPi[-(b/a), -ArcSin[Sqrt[Sec[c + d*x]]], -1])*(a + b*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(b*(
b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) - (2*(-12*a^2*A*b + 15*A*b^3 + 6*a^3*B - 9*a*b^2*B)*Cos[2*(c + d*x)]
*(a + b*Sec[c + d*x])*(2*a*b - 2*a*b*Sec[c + d*x]^2 + 2*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec
[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + a*(a - 2*b)*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]
*Sqrt[1 - Sec[c + d*x]^2] + a^2*EllipticPi[-(b/a), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1
- Sec[c + d*x]^2] - 2*b^2*EllipticPi[-(b/a), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[
c + d*x]^2])*Sin[c + d*x])/(a^2*b*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*
x]^2)))/(12*a^2*(a - b)*(a + b)*d) + (Sqrt[Sec[c + d*x]]*((b^2*(A*b - a*B)*Sin[c + d*x])/(a^3*(-a^2 + b^2)) -
(-(A*b^4*Sin[c + d*x]) + a*b^3*B*Sin[c + d*x])/(a^3*(a^2 - b^2)*(b + a*Cos[c + d*x])) + (A*Sin[2*(c + d*x)])/(
3*a^2)))/d

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Maple [B]  time = 7.161, size = 1059, normalized size = 2.9 \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))/sec(d*x+c)^(3/2)/(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)*(2/3/a^4*(4*A*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^4+a^2*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
))+9*A*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))
+6*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b-2
*A*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-6*B*a*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)
^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)
*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*b^3*(A*b-B*
a)/a^4*(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^3*b^2*(4*A*b-3*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

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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))/sec(d*x+c)^(3/2)/(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))/sec(d*x+c)^(3/2)/(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))/sec(d*x+c)**(3/2)/(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{B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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