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

Optimal. Leaf size=552 \[ -\frac{\cot (c+d x) \left (-2 a^2 (A+2 B-4 C)+a b (5 A-12 B)+15 A b^2\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{4 a^3 d \sqrt{a+b}}+\frac{b \tan (c+d x) \left (-a^2 (7 A b-8 b C)+4 a^3 B-12 a b^2 B+15 A b^3\right )}{4 a^3 d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{\cot (c+d x) \left (-a^2 (7 A b-8 b C)+4 a^3 B-12 a b^2 B+15 A b^3\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{4 a^3 b d \sqrt{a+b}}-\frac{\sqrt{a+b} \cot (c+d x) \left (4 a^2 (A+2 C)-12 a b B+15 A b^2\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{4 a^4 d}-\frac{(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt{a+b \sec (c+d x)}}+\frac{A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt{a+b \sec (c+d x)}} \]

[Out]

((15*A*b^3 + 4*a^3*B - 12*a*b^2*B - a^2*(7*A*b - 8*b*C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]
]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])
/(4*a^3*b*Sqrt[a + b]*d) - ((15*A*b^2 + a*b*(5*A - 12*B) - 2*a^2*(A + 2*B - 4*C))*Cot[c + d*x]*EllipticF[ArcSi
n[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 +
Sec[c + d*x]))/(a - b))])/(4*a^3*Sqrt[a + b]*d) - (Sqrt[a + b]*(15*A*b^2 - 12*a*b*B + 4*a^2*(A + 2*C))*Cot[c +
d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c
+ d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(4*a^4*d) - ((5*A*b - 4*a*B)*Sin[c + d*x])/(4*a^2*d
*Sqrt[a + b*Sec[c + d*x]]) + (A*Cos[c + d*x]*Sin[c + d*x])/(2*a*d*Sqrt[a + b*Sec[c + d*x]]) + (b*(15*A*b^3 + 4
*a^3*B - 12*a*b^2*B - a^2*(7*A*b - 8*b*C))*Tan[c + d*x])/(4*a^3*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.24644, antiderivative size = 552, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {4104, 4060, 4058, 3921, 3784, 3832, 4004} \[ \frac{b \tan (c+d x) \left (-a^2 (7 A b-8 b C)+4 a^3 B-12 a b^2 B+15 A b^3\right )}{4 a^3 d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}-\frac{\cot (c+d x) \left (-2 a^2 (A+2 B-4 C)+a b (5 A-12 B)+15 A b^2\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{4 a^3 d \sqrt{a+b}}+\frac{\cot (c+d x) \left (-a^2 (7 A b-8 b C)+4 a^3 B-12 a b^2 B+15 A b^3\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{4 a^3 b d \sqrt{a+b}}-\frac{\sqrt{a+b} \cot (c+d x) \left (4 a^2 (A+2 C)-12 a b B+15 A b^2\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{4 a^4 d}-\frac{(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt{a+b \sec (c+d x)}}+\frac{A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt{a+b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((15*A*b^3 + 4*a^3*B - 12*a*b^2*B - a^2*(7*A*b - 8*b*C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]
]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])
/(4*a^3*b*Sqrt[a + b]*d) - ((15*A*b^2 + a*b*(5*A - 12*B) - 2*a^2*(A + 2*B - 4*C))*Cot[c + d*x]*EllipticF[ArcSi
n[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 +
Sec[c + d*x]))/(a - b))])/(4*a^3*Sqrt[a + b]*d) - (Sqrt[a + b]*(15*A*b^2 - 12*a*b*B + 4*a^2*(A + 2*C))*Cot[c +
d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c
+ d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(4*a^4*d) - ((5*A*b - 4*a*B)*Sin[c + d*x])/(4*a^2*d
*Sqrt[a + b*Sec[c + d*x]]) + (A*Cos[c + d*x]*Sin[c + d*x])/(2*a*d*Sqrt[a + b*Sec[c + d*x]]) + (b*(15*A*b^3 + 4
*a^3*B - 12*a*b^2*B - a^2*(7*A*b - 8*b*C))*Tan[c + d*x])/(4*a^3*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]])

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 4060

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

Rule 4058

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

Rule 3921

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

Rule 3784

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(2*Rt[a + b, 2]*Sqrt[(b*(1 - Csc[c + d*x])
)/(a + b)]*Sqrt[-((b*(1 + Csc[c + d*x]))/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[c + d*x]]/Rt[a
+ b, 2]], (a + b)/(a - b)])/(a*d*Cot[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx &=\frac{A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt{a+b \sec (c+d x)}}-\frac{\int \frac{\cos (c+d x) \left (\frac{1}{2} (5 A b-4 a B)-a (A+2 C) \sec (c+d x)-\frac{3}{2} A b \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{2 a}\\ &=-\frac{(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt{a+b \sec (c+d x)}}+\frac{A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt{a+b \sec (c+d x)}}+\frac{\int \frac{\frac{1}{4} \left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right )+\frac{3}{2} a A b \sec (c+d x)-\frac{1}{4} b (5 A b-4 a B) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{2 a^2}\\ &=-\frac{(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt{a+b \sec (c+d x)}}+\frac{A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt{a+b \sec (c+d x)}}+\frac{b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \tan (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{\int \frac{-\frac{1}{8} \left (a^2-b^2\right ) \left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right )+\frac{1}{4} a b \left (5 A b^2-4 a b B-a^2 (A-4 C)\right ) \sec (c+d x)+\frac{1}{8} b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \sec ^2(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{a^3 \left (a^2-b^2\right )}\\ &=-\frac{(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt{a+b \sec (c+d x)}}+\frac{A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt{a+b \sec (c+d x)}}+\frac{b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \tan (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{\int \frac{-\frac{1}{8} \left (a^2-b^2\right ) \left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right )+\left (\frac{1}{4} a b \left (5 A b^2-4 a b B-a^2 (A-4 C)\right )-\frac{1}{8} b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right )\right ) \sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{a^3 \left (a^2-b^2\right )}-\frac{\left (b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right )\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{8 a^3 \left (a^2-b^2\right )}\\ &=\frac{\left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{4 a^3 b \sqrt{a+b} d}-\frac{(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt{a+b \sec (c+d x)}}+\frac{A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt{a+b \sec (c+d x)}}+\frac{b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \tan (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{\left (b \left (15 A b^2+a b (5 A-12 B)-2 a^2 (A+2 B-4 C)\right )\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{8 a^3 (a+b)}+\frac{\left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right ) \int \frac{1}{\sqrt{a+b \sec (c+d x)}} \, dx}{8 a^3}\\ &=\frac{\left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{4 a^3 b \sqrt{a+b} d}-\frac{\left (15 A b^2+a b (5 A-12 B)-2 a^2 (A+2 B-4 C)\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{4 a^3 \sqrt{a+b} d}-\frac{\sqrt{a+b} \left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right ) \cot (c+d x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{4 a^4 d}-\frac{(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt{a+b \sec (c+d x)}}+\frac{A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt{a+b \sec (c+d x)}}+\frac{b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \tan (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 17.0449, size = 490, normalized size = 0.89 \[ \frac{\sqrt{a+b \sec (c+d x)} \left (\frac{a \sin (2 (c+d x)) \left (b \left (a^2 (A-4 C)+4 a b B-5 A b^2\right )+a A \left (a^2-b^2\right ) \cos (c+d x)\right )}{a \cos (c+d x)+b}+\cos (c+d x) \left (\tan \left (\frac{1}{2} (c+d x)\right ) \left (a^2 (8 b C-7 A b)+4 a^3 B-12 a b^2 B+15 A b^3\right )+\frac{i (a-b) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} \left (2 \left (a^2 b (A-8 B+8 C)+2 a^3 (A+2 C)+2 a b^2 (5 A-6 B)+15 A b^3\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{\frac{b-a}{a+b}} \tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a+b}{a-b}\right )+\left (a^2 b (7 A-8 C)-4 a^3 B+12 a b^2 B-15 A b^3\right ) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b-a}{a+b}} \tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a+b}{a-b}\right )-2 (a+b) \left (4 a^2 (A+2 C)-12 a b B+15 A b^2\right ) \Pi \left (-\frac{a+b}{a-b};i \sinh ^{-1}\left (\sqrt{\frac{b-a}{a+b}} \tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a+b}{a-b}\right )\right )}{\sqrt{\frac{b-a}{a+b}} \sqrt{\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )} (a \cos (c+d x)+b)}\right )\right )}{4 a^3 d \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*Sec[c + d*x]]*((a*(b*(-5*A*b^2 + 4*a*b*B + a^2*(A - 4*C)) + a*A*(a^2 - b^2)*Cos[c + d*x])*Sin[2*(c
+ d*x)])/(b + a*Cos[c + d*x]) + Cos[c + d*x]*((I*(a - b)*((-15*A*b^3 - 4*a^3*B + 12*a*b^2*B + a^2*b*(7*A - 8*
C))*EllipticE[I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)] + 2*(15*A*b^3 + 2*a*b^2*(5*
A - 6*B) + 2*a^3*(A + 2*C) + a^2*b*(A - 8*B + 8*C))*EllipticF[I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2
]], (a + b)/(a - b)] - 2*(a + b)*(15*A*b^2 - 12*a*b*B + 4*a^2*(A + 2*C))*EllipticPi[-((a + b)/(a - b)), I*ArcS
inh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)])*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)
/(a + b)])/(Sqrt[(-a + b)/(a + b)]*(b + a*Cos[c + d*x])*Sqrt[Cos[c + d*x]*Sec[(c + d*x)/2]^2]) + (15*A*b^3 + 4
*a^3*B - 12*a*b^2*B + a^2*(-7*A*b + 8*b*C))*Tan[(c + d*x)/2])))/(4*a^3*(a^2 - b^2)*d)

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Maple [B]  time = 0.56, size = 5176, normalized size = 9.4 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + B \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + A \cos \left (d x + c\right )^{2}\right )} \sqrt{b \sec \left (d x + c\right ) + a}}{b^{2} \sec \left (d x + c\right )^{2} + 2 \, a b \sec \left (d x + c\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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