3.971 \(\int \frac{\sec ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=449 \[ \frac{2 \cot (c+d x) \left (2 a^2 b (B-3 C)-8 a^3 C+a b^2 (A+3 B+9 C)-3 b^3 (A+B-C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{3 b^3 d \sqrt{a+b} \left (a^2-b^2\right )}+\frac{2 \tan (c+d x) \left (a^2 b^2 (A+9 C)+2 a^3 b B-5 a^4 C-6 a b^3 B+3 A b^4\right )}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}+\frac{2 a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac{2 \cot (c+d x) \left (a^2 b^2 (A+15 C)+2 a^3 b B-8 a^4 C-6 a b^3 B+3 b^4 (A-C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^4 d \sqrt{a+b} \left (a^2-b^2\right )} \]

[Out]

(2*(2*a^3*b*B - 6*a*b^3*B + 3*b^4*(A - C) - 8*a^4*C + a^2*b^2*(A + 15*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))])/(3*b^4*Sqrt[a + b]*(a^2 - b^2)*d) + (2*(2*a^2*b*(B - 3*C) - 3*b^3*(A + B - C) - 8*a^3*C + a*
b^2*(A + 3*B + 9*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqr
t[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*b^3*Sqrt[a + b]*(a^2 - b^2)*d) +
(2*a*(A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) + (2*(3*A*b^4 + 2
*a^3*b*B - 6*a*b^3*B - 5*a^4*C + a^2*b^2*(A + 9*C))*Tan[c + d*x])/(3*b^2*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*
x]])

________________________________________________________________________________________

Rubi [A]  time = 1.04373, antiderivative size = 449, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {4090, 4080, 4005, 3832, 4004} \[ \frac{2 \tan (c+d x) \left (a^2 b^2 (A+9 C)+2 a^3 b B-5 a^4 C-6 a b^3 B+3 A b^4\right )}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}+\frac{2 a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac{2 \cot (c+d x) \left (2 a^2 b (B-3 C)-8 a^3 C+a b^2 (A+3 B+9 C)-3 b^3 (A+B-C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^3 d \sqrt{a+b} \left (a^2-b^2\right )}+\frac{2 \cot (c+d x) \left (a^2 b^2 (A+15 C)+2 a^3 b B-8 a^4 C-6 a b^3 B+3 b^4 (A-C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^4 d \sqrt{a+b} \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^(5/2),x]

[Out]

(2*(2*a^3*b*B - 6*a*b^3*B + 3*b^4*(A - C) - 8*a^4*C + a^2*b^2*(A + 15*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))])/(3*b^4*Sqrt[a + b]*(a^2 - b^2)*d) + (2*(2*a^2*b*(B - 3*C) - 3*b^3*(A + B - C) - 8*a^3*C + a*
b^2*(A + 3*B + 9*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqr
t[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*b^3*Sqrt[a + b]*(a^2 - b^2)*d) +
(2*a*(A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) + (2*(3*A*b^4 + 2
*a^3*b*B - 6*a*b^3*B - 5*a^4*C + a^2*b^2*(A + 9*C))*Tan[c + d*x])/(3*b^2*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*
x]])

Rule 4090

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e
+ f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*C
sc[e + f*x])^(m + 1)*Simp[b*(m + 1)*(-(a*(b*B - a*C)) + A*b^2) + (b*B*(a^2 + b^2*(m + 1)) - a*(A*b^2*(m + 2) +
C*(a^2 + b^2*(m + 1))))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^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 4080

Int[csc[(e_.) + (f_.)*(x_)]*((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))/(b*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e +
f*x])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 4005

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, 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}, x] && NeQ[a^2 - b^2, 0] && 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{\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx &=\frac{2 a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 \int \frac{\sec (c+d x) \left (-\frac{3}{2} b \left (A b^2-a (b B-a C)\right )+\frac{1}{2} \left (2 a^2 b B-3 b^3 B-2 a^3 C+a b^2 (A+3 C)\right ) \sec (c+d x)+\frac{3}{2} b \left (a^2-b^2\right ) C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=\frac{2 a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (3 A b^4+2 a^3 b B-6 a b^3 B-5 a^4 C+a^2 b^2 (A+9 C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}-\frac{4 \int \frac{\sec (c+d x) \left (-\frac{1}{4} b^2 \left (a^2 b B+3 b^3 B+2 a^3 C-2 a b^2 (2 A+3 C)\right )+\frac{1}{4} b \left (2 a^3 b B-6 a b^3 B+3 b^4 (A-C)-8 a^4 C+a^2 b^2 (A+15 C)\right ) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{2 a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (3 A b^4+2 a^3 b B-6 a b^3 B-5 a^4 C+a^2 b^2 (A+9 C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}+\frac{\left (2 a^2 b (B-3 C)-3 b^3 (A+B-C)-8 a^3 C+a b^2 (A+3 B+9 C)\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 (a-b) b^2 (a+b)^2}-\frac{\left (2 a^3 b B-6 a b^3 B+3 b^4 (A-C)-8 a^4 C+a^2 b^2 (A+15 C)\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 b^2 \left (a^2-b^2\right )^2}\\ &=\frac{2 \left (2 a^3 b B-6 a b^3 B+3 b^4 (A-C)-8 a^4 C+a^2 b^2 (A+15 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}}}{3 (a-b) b^4 (a+b)^{3/2} d}+\frac{2 \left (2 a^2 b (B-3 C)-3 b^3 (A+B-C)-8 a^3 C+a b^2 (A+3 B+9 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}}}{3 (a-b) b^3 (a+b)^{3/2} d}+\frac{2 a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (3 A b^4+2 a^3 b B-6 a b^3 B-5 a^4 C+a^2 b^2 (A+9 C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 27.6009, size = 4504, normalized size = 10.03 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^(5/2),x]

[Out]

((b + a*Cos[c + d*x])^3*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-4*(a^2*A*b^2 + 3*A*b^4 + 2*a^3
*b*B - 6*a*b^3*B - 8*a^4*C + 15*a^2*b^2*C - 3*b^4*C)*Sin[c + d*x])/(3*b^3*(-a^2 + b^2)^2) + (4*(A*b^2*Sin[c +
d*x] - a*b*B*Sin[c + d*x] + a^2*C*Sin[c + d*x]))/(3*b*(-a^2 + b^2)*(b + a*Cos[c + d*x])^2) + (4*(2*a^2*A*b^2*S
in[c + d*x] + 2*A*b^4*Sin[c + d*x] + a^3*b*B*Sin[c + d*x] - 5*a*b^3*B*Sin[c + d*x] - 4*a^4*C*Sin[c + d*x] + 8*
a^2*b^2*C*Sin[c + d*x]))/(3*b^2*(-a^2 + b^2)^2*(b + a*Cos[c + d*x]))))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[
2*c + 2*d*x])*(a + b*Sec[c + d*x])^(5/2)) - (4*(b + a*Cos[c + d*x])^2*((2*a^2*A)/(3*(-a^2 + b^2)^2*Sqrt[b + a*
Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*A*b^2)/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (
4*a^3*B)/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (4*a*b*B)/((-a^2 + b^2)^2*Sqrt[b +
a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (10*a^2*C)/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]])
- (16*a^4*C)/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*b^2*C)/((-a^2 + b^2)^2*S
qrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*a^3*A*Sqrt[Sec[c + d*x]])/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[
c + d*x]]) - (2*a*A*b*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (10*a^2*B*Sqrt[Sec[c +
d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (4*a^4*B*Sqrt[Sec[c + d*x]])/(3*b^2*(-a^2 + b^2)^2*Sqrt[
b + a*Cos[c + d*x]]) + (2*b^2*B*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (16*a^5*C*Sqrt
[Sec[c + d*x]])/(3*b^3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (34*a^3*C*Sqrt[Sec[c + d*x]])/(3*b*(-a^2 + b
^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (6*a*b*C*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (2*
a^3*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (2*a*A*b*Cos[2*(c +
d*x)]*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (4*a^2*B*Cos[2*(c + d*x)]*Sqrt[Sec[c +
d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (4*a^4*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*b^2*(-a^2
+ b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (16*a^5*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*b^3*(-a^2 + b^2)^2*Sqrt
[b + a*Cos[c + d*x]]) + (10*a^3*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*
x]]) - (2*a*b*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]))*Sqrt[Sec[c + d
*x]]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(2*(a + b)*(-2*a^3*b*B + 6*
a*b^3*B + 8*a^4*C + 3*b^4*(-A + C) - a^2*b^2*(A + 15*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos
[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(3
*b^3*(A - B - C) - 8*a^3*C + 2*a^2*b*(B + 3*C) + a*b^2*(A - 3*B + 9*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*
Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] +
(-2*a^3*b*B + 6*a*b^3*B + 8*a^4*C + 3*b^4*(-A + C) - a^2*b^2*(A + 15*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Se
c[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b^3*(a^2 - b^2)^2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*S
qrt[Sec[(c + d*x)/2]^2]*(a + b*Sec[c + d*x])^(5/2)*((-2*a*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sin[c + d*x]*(
2*(a + b)*(-2*a^3*b*B + 6*a*b^3*B + 8*a^4*C + 3*b^4*(-A + C) - a^2*b^2*(A + 15*C))*Sqrt[Cos[c + d*x]/(1 + Cos[
c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)
/(a + b)] + 2*b*(a + b)*(3*b^3*(A - B - C) - 8*a^3*C + 2*a^2*b*(B + 3*C) + a*b^2*(A - 3*B + 9*C))*Sqrt[Cos[c +
d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*
x)/2]], (a - b)/(a + b)] + (-2*a^3*b*B + 6*a*b^3*B + 8*a^4*C + 3*b^4*(-A + C) - a^2*b^2*(A + 15*C))*Cos[c + d*
x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b^3*(a^2 - b^2)^2*(b + a*Cos[c + d*x])^(3/2)*
Sqrt[Sec[(c + d*x)/2]^2]) + (2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Tan[(c + d*x)/2]*(2*(a + b)*(-2*a^3*b*B +
6*a*b^3*B + 8*a^4*C + 3*b^4*(-A + C) - a^2*b^2*(A + 15*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*
Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)
*(3*b^3*(A - B - C) - 8*a^3*C + 2*a^2*b*(B + 3*C) + a*b^2*(A - 3*B + 9*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x]
)]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)
] + (-2*a^3*b*B + 6*a*b^3*B + 8*a^4*C + 3*b^4*(-A + C) - a^2*b^2*(A + 15*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])
*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b^3*(a^2 - b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2])
- (4*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(((-2*a^3*b*B + 6*a*b^3*B + 8*a^4*C + 3*b^4*(-A + C) - a^2*b^2*(A
+ 15*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^4)/2 + ((a + b)*(-2*a^3*b*B + 6*a*b^3*B + 8*a^4*C
+ 3*b^4*(-A + C) - a^2*b^2*(A + 15*C))*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSi
n[Tan[(c + d*x)/2]], (a - b)/(a + b)]*((Cos[c + d*x]*Sin[c + d*x])/(1 + Cos[c + d*x])^2 - Sin[c + d*x]/(1 + Co
s[c + d*x])))/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])] + (b*(a + b)*(3*b^3*(A - B - C) - 8*a^3*C + 2*a^2*b*(B + 3
*C) + a*b^2*(A - 3*B + 9*C))*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c +
d*x)/2]], (a - b)/(a + b)]*((Cos[c + d*x]*Sin[c + d*x])/(1 + Cos[c + d*x])^2 - Sin[c + d*x]/(1 + Cos[c + d*x]
)))/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])] + ((a + b)*(-2*a^3*b*B + 6*a*b^3*B + 8*a^4*C + 3*b^4*(-A + C) - a^2*
b^2*(A + 15*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*(-(
(a*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x]))) + ((b + a*Cos[c + d*x])*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x
])^2)))/Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))] + (b*(a + b)*(3*b^3*(A - B - C) - 8*a^3*C + 2*
a^2*b*(B + 3*C) + a*b^2*(A - 3*B + 9*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*EllipticF[ArcSin[Tan[(c + d*x)/
2]], (a - b)/(a + b)]*(-((a*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x]))) + ((b + a*Cos[c + d*x])*Sin[c + d*x])/
((a + b)*(1 + Cos[c + d*x])^2)))/Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))] - a*(-2*a^3*b*B + 6*a
*b^3*B + 8*a^4*C + 3*b^4*(-A + C) - a^2*b^2*(A + 15*C))*Cos[c + d*x]*Sec[(c + d*x)/2]^2*Sin[c + d*x]*Tan[(c +
d*x)/2] - (-2*a^3*b*B + 6*a*b^3*B + 8*a^4*C + 3*b^4*(-A + C) - a^2*b^2*(A + 15*C))*(b + a*Cos[c + d*x])*Sec[(c
+ d*x)/2]^2*Sin[c + d*x]*Tan[(c + d*x)/2] + (-2*a^3*b*B + 6*a*b^3*B + 8*a^4*C + 3*b^4*(-A + C) - a^2*b^2*(A +
15*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]^2 + (b*(a + b)*(3*b^3*(A - B - C
) - 8*a^3*C + 2*a^2*b*(B + 3*C) + a*b^2*(A - 3*B + 9*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos
[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*Sec[(c + d*x)/2]^2)/(Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[1 - ((a - b)*T
an[(c + d*x)/2]^2)/(a + b)]) + ((a + b)*(-2*a^3*b*B + 6*a*b^3*B + 8*a^4*C + 3*b^4*(-A + C) - a^2*b^2*(A + 15*C
))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*Sec[(c + d*x)
/2]^2*Sqrt[1 - ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])/Sqrt[1 - Tan[(c + d*x)/2]^2]))/(3*b^3*(a^2 - b^2)^2*Sqrt
[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) - (2*(2*(a + b)*(-2*a^3*b*B + 6*a*b^3*B + 8*a^4*C + 3*b^4*(-A +
C) - a^2*b^2*(A + 15*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c
+ d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(3*b^3*(A - B - C) - 8*a^3*C + 2*
a^2*b*(B + 3*C) + a*b^2*(A - 3*B + 9*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a +
b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (-2*a^3*b*B + 6*a*b^3*B + 8*a^
4*C + 3*b^4*(-A + C) - a^2*b^2*(A + 15*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/
2])*(-(Cos[(c + d*x)/2]*Sec[c + d*x]*Sin[(c + d*x)/2]) + Cos[(c + d*x)/2]^2*Sec[c + d*x]*Tan[c + d*x]))/(3*b^3
*(a^2 - b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]])))

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Maple [B]  time = 0.845, size = 8858, normalized size = 19.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x)

[Out]

result too large to display

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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(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \sec \left (d x + c\right )^{4} + B \sec \left (d x + c\right )^{3} + A \sec \left (d x + c\right )^{2}\right )} \sqrt{b \sec \left (d x + c\right ) + a}}{b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^4 + B*sec(d*x + c)^3 + A*sec(d*x + c)^2)*sqrt(b*sec(d*x + c) + a)/(b^3*sec(d*x + c)^3
+ 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*sec(d*x + c) + a^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(5/2),x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**2/(a + b*sec(c + d*x))**(5/2), x)

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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 )} \sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^2/(b*sec(d*x + c) + a)^(5/2), x)