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

Optimal. Leaf size=510 \[ \frac{2 \cot (c+d x) \left (a^2 b (40 B-36 C)-48 a^3 C-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{15 b^4 d \sqrt{a+b}}-\frac{2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt{a+b \sec (c+d x)}}{5 b^2 d \left (a^2-b^2\right )}+\frac{2 \tan (c+d x) \left (20 a^2 b B-24 a^3 C-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt{a+b \sec (c+d x)}}{15 b^3 d \left (a^2-b^2\right )}+\frac{2 \cot (c+d x) \left (-6 a^2 b^2 (5 A-4 C)+40 a^3 b B-48 a^4 C-25 a b^3 B+3 b^4 (5 A+3 C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^5 d \sqrt{a+b}} \]

[Out]

(2*(40*a^3*b*B - 25*a*b^3*B - 6*a^2*b^2*(5*A - 4*C) - 48*a^4*C + 3*b^4*(5*A + 3*C))*Cot[c + d*x]*EllipticE[Arc
Sin[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))])/(15*b^5*Sqrt[a + b]*d) + (2*(a^2*b*(40*B - 36*C) - 48*a^3*C - 6*a*b^2*(5*A - 5*B +
2*C) - b^3*(15*A - 5*B + 9*C))*Cot[c + d*x]*EllipticF[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))])/(15*b^4*Sqrt[a + b]*d) -
(2*(A*b^2 - a*(b*B - a*C))*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(20*a
^2*b*B - 5*b^3*B - 3*a*b^2*(5*A - 3*C) - 24*a^3*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(15*b^3*(a^2 - b^2)*
d) + (2*(5*A*b^2 - 5*a*b*B + 6*a^2*C - b^2*C)*Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(5*b^2*(a^2
- b^2)*d)

________________________________________________________________________________________

Rubi [A]  time = 1.34519, antiderivative size = 510, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {4098, 4092, 4082, 4005, 3832, 4004} \[ -\frac{2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt{a+b \sec (c+d x)}}{5 b^2 d \left (a^2-b^2\right )}+\frac{2 \tan (c+d x) \left (20 a^2 b B-24 a^3 C-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt{a+b \sec (c+d x)}}{15 b^3 d \left (a^2-b^2\right )}+\frac{2 \cot (c+d x) \left (a^2 b (40 B-36 C)-48 a^3 C-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^4 d \sqrt{a+b}}+\frac{2 \cot (c+d x) \left (-6 a^2 b^2 (5 A-4 C)+40 a^3 b B-48 a^4 C-25 a b^3 B+3 b^4 (5 A+3 C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^5 d \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(40*a^3*b*B - 25*a*b^3*B - 6*a^2*b^2*(5*A - 4*C) - 48*a^4*C + 3*b^4*(5*A + 3*C))*Cot[c + d*x]*EllipticE[Arc
Sin[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))])/(15*b^5*Sqrt[a + b]*d) + (2*(a^2*b*(40*B - 36*C) - 48*a^3*C - 6*a*b^2*(5*A - 5*B +
2*C) - b^3*(15*A - 5*B + 9*C))*Cot[c + d*x]*EllipticF[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))])/(15*b^4*Sqrt[a + b]*d) -
(2*(A*b^2 - a*(b*B - a*C))*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(20*a
^2*b*B - 5*b^3*B - 3*a*b^2*(5*A - 3*C) - 24*a^3*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(15*b^3*(a^2 - b^2)*
d) + (2*(5*A*b^2 - 5*a*b*B + 6*a^2*C - b^2*C)*Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(5*b^2*(a^2
- b^2)*d)

Rule 4098

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[(d*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(
a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1))/(b*f*(a^2 - b^2)*(m + 1)), x] + Dist[d/(b*(a^2 - b^2)*(m
+ 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A*b^2*(n - 1) - a*(b*B - a*C)*(n - 1) +
b*(a*A - b*B + a*C)*(m + 1)*Csc[e + f*x] - (b*(A*b - a*B)*(m + n + 1) + C*(a^2*n + b^2*(m + 1)))*Csc[e + f*x]
^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 4092

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[(C*Csc[e + f*x]*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2)
+ A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B*(m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m
}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rule 4082

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[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m
+ 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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 ^3(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{2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{2 \int \frac{\sec ^2(c+d x) \left (2 \left (A b^2-a (b B-a C)\right )+\frac{1}{2} b (b B-a (A+C)) \sec (c+d x)-\frac{1}{2} \left (5 A b^2-5 a b B+6 a^2 C-b^2 C\right ) \sec ^2(c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (5 A b^2-5 a b B+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac{4 \int \frac{\sec (c+d x) \left (-\frac{1}{2} a \left (5 A b^2-5 a b B+6 a^2 C-b^2 C\right )+\frac{1}{4} b \left (5 A b^2-5 a b B+2 a^2 C+3 b^2 C\right ) \sec (c+d x)-\frac{1}{4} \left (20 a^2 b B-5 b^3 B-3 a b^2 (5 A-3 C)-24 a^3 C\right ) \sec ^2(c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{5 b^2 \left (a^2-b^2\right )}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (20 a^2 b B-5 b^3 B-3 a b^2 (5 A-3 C)-24 a^3 C\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 b^3 \left (a^2-b^2\right ) d}+\frac{2 \left (5 A b^2-5 a b B+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac{8 \int \frac{\sec (c+d x) \left (\frac{1}{8} b \left (10 a^2 b B+5 b^3 B-12 a^3 C-3 a b^2 (5 A+C)\right )+\frac{1}{8} \left (40 a^3 b B-25 a b^3 B-6 a^2 b^2 (5 A-4 C)-48 a^4 C+3 b^4 (5 A+3 C)\right ) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 b^3 \left (a^2-b^2\right )}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (20 a^2 b B-5 b^3 B-3 a b^2 (5 A-3 C)-24 a^3 C\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 b^3 \left (a^2-b^2\right ) d}+\frac{2 \left (5 A b^2-5 a b B+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac{\left (40 a^3 b B-25 a b^3 B-6 a^2 b^2 (5 A-4 C)-48 a^4 C+3 b^4 (5 A+3 C)\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 b^3 \left (a^2-b^2\right )}+\frac{\left (a^2 b (40 B-36 C)-48 a^3 C-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 b^3 (a+b)}\\ &=\frac{2 \left (40 a^3 b B-25 a b^3 B-6 a^2 b^2 (5 A-4 C)-48 a^4 C+3 b^4 (5 A+3 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}}}{15 b^5 \sqrt{a+b} d}+\frac{2 \left (a^2 b (40 B-36 C)-48 a^3 C-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 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}}}{15 b^4 \sqrt{a+b} d}-\frac{2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (20 a^2 b B-5 b^3 B-3 a b^2 (5 A-3 C)-24 a^3 C\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 b^3 \left (a^2-b^2\right ) d}+\frac{2 \left (5 A b^2-5 a b B+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 20.8391, size = 874, normalized size = 1.71 \[ \frac{\left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (\frac{4 \left (-48 C a^4+40 b B a^3-30 A b^2 a^2+24 b^2 C a^2-25 b^3 B a+15 A b^4+9 b^4 C\right ) \sin (c+d x)}{15 b^4 \left (b^2-a^2\right )}+\frac{4 \sec (c+d x) (5 b B \sin (c+d x)-9 a C \sin (c+d x))}{15 b^3}+\frac{4 \left (C \sin (c+d x) a^4-b B \sin (c+d x) a^3+A b^2 \sin (c+d x) a^2\right )}{b^3 \left (b^2-a^2\right ) (b+a \cos (c+d x))}+\frac{4 C \sec (c+d x) \tan (c+d x)}{5 b^2}\right ) (b+a \cos (c+d x))^2}{d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (a+b \sec (c+d x))^{3/2}}+\frac{4 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \sqrt{\frac{1}{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )}} \left ((a+b) \left (48 C a^4-40 b B a^3+6 b^2 (5 A-4 C) a^2+25 b^3 B a-3 b^4 (5 A+3 C)\right ) E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right ) \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b}{a+b}} \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )+b (a+b) \left (-48 C a^3+4 b (10 B+9 C) a^2-6 b^2 (5 A+5 B+2 C) a+b^3 (15 A+5 B+9 C)\right ) \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right ) \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b}{a+b}} \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )+\left (48 C a^4-40 b B a^3+6 b^2 (5 A-4 C) a^2+25 b^3 B a-3 b^4 (5 A+3 C)\right ) \tan \left (\frac{1}{2} (c+d x)\right ) \left (-b \tan ^4\left (\frac{1}{2} (c+d x)\right )+a \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )-1\right )^2+b\right )\right ) (b+a \cos (c+d x))^{3/2}}{15 b^4 \left (b^2-a^2\right ) d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )^{3/2} \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*(b + a*Cos[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*((a
+ b)*(-40*a^3*b*B + 25*a*b^3*B + 6*a^2*b^2*(5*A - 4*C) + 48*a^4*C - 3*b^4*(5*A + 3*C))*EllipticE[ArcSin[Tan[(c
+ d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d
*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + b*(a + b)*(-48*a^3*C - 6*a*b^2*(5*A + 5*B + 2*C) + b^3*(15*A + 5*B
+ 9*C) + 4*a^2*b*(10*B + 9*C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]
^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (-40*a^3*b*
B + 25*a*b^3*B + 6*a^2*b^2*(5*A - 4*C) + 48*a^4*C - 3*b^4*(5*A + 3*C))*Tan[(c + d*x)/2]*(b - b*Tan[(c + d*x)/2
]^4 + a*(-1 + Tan[(c + d*x)/2]^2)^2)))/(15*b^4*(-a^2 + b^2)*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]
)*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(1 + Tan[(c + d*x)/2]^2)^(3/2)*Sqrt[(a + b - a*Tan[(c + d*x)/2
]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]) + ((b + a*Cos[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c
+ d*x]^2)*((4*(-30*a^2*A*b^2 + 15*A*b^4 + 40*a^3*b*B - 25*a*b^3*B - 48*a^4*C + 24*a^2*b^2*C + 9*b^4*C)*Sin[c
+ d*x])/(15*b^4*(-a^2 + b^2)) + (4*Sec[c + d*x]*(5*b*B*Sin[c + d*x] - 9*a*C*Sin[c + d*x]))/(15*b^3) + (4*(a^2*
A*b^2*Sin[c + d*x] - a^3*b*B*Sin[c + d*x] + a^4*C*Sin[c + d*x]))/(b^3*(-a^2 + b^2)*(b + a*Cos[c + d*x])) + (4*
C*Sec[c + d*x]*Tan[c + d*x])/(5*b^2)))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x
])^(3/2))

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Maple [B]  time = 1.534, size = 5857, normalized size = 11.5 \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)^3*(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(-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)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),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 )^{5} + B \sec \left (d x + c\right )^{4} + A \sec \left (d x + c\right )^{3}\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(sec(d*x+c)^3*(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*sec(d*x + c)^5 + B*sec(d*x + c)^4 + A*sec(d*x + c)^3)*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]  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 ^{3}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**3/(a + b*sec(c + d*x))**(3/2), 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 )^{3}}{{\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(sec(d*x+c)^3*(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)*sec(d*x + c)^3/(b*sec(d*x + c) + a)^(3/2), x)