3.970 \(\int \frac{\sec ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=549 \[ -\frac{2 \cot (c+d x) \left (-2 a^2 b^2 (A-3 B-8 C)+a^3 b (8 B-12 C)-16 a^4 C-3 a b^3 (A+3 B-3 C)+b^4 (3 A-3 B+C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{3 b^4 d \sqrt{a+b} \left (a^2-b^2\right )}-\frac{2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac{2 \tan (c+d x) \left (2 a^2 C-a b B+A b^2-b^2 C\right ) \sqrt{a+b \sec (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}-\frac{2 a \tan (c+d x) \left (a \left (3 a^2 b B-6 a^3 C+10 a b^2 C-7 b^3 B\right )+4 A b^4\right )}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}-\frac{2 \cot (c+d x) \left (-2 a^3 b^2 (A-14 C)-15 a^2 b^3 B+8 a^4 b B-16 a^5 C+2 a b^4 (3 A-4 C)+3 b^5 B\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^5 d \sqrt{a+b} \left (a^2-b^2\right )} \]

[Out]

(-2*(8*a^4*b*B - 15*a^2*b^3*B + 3*b^5*B - 2*a^3*b^2*(A - 14*C) + 2*a*b^4*(3*A - 4*C) - 16*a^5*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^5*Sqrt[a + b]*(a^2 - b^2)*d) - (2*(a^3*b*(8*B - 12*C) - 2*a^2*b^
2*(A - 3*B - 8*C) - 3*a*b^3*(A + 3*B - 3*C) - 16*a^4*C + b^4*(3*A - 3*B + C))*Cot[c + d*x]*EllipticF[ArcSin[Sq
rt[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*(A*b^2 - a*(b*B - a*C))*Sec[c + d*x]^2*Tan[c + d*
x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) - (2*a*(4*A*b^4 + a*(3*a^2*b*B - 7*b^3*B - 6*a^3*C + 10*a*b
^2*C))*Tan[c + d*x])/(3*b^3*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(A*b^2 - a*b*B + 2*a^2*C - b^2*C)*S
qrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*b^3*(a^2 - b^2)*d)

________________________________________________________________________________________

Rubi [A]  time = 1.85034, antiderivative size = 549, 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, 4090, 4082, 4005, 3832, 4004} \[ -\frac{2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac{2 \tan (c+d x) \left (2 a^2 C-a b B+A b^2-b^2 C\right ) \sqrt{a+b \sec (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}-\frac{2 a \tan (c+d x) \left (a \left (3 a^2 b B-6 a^3 C+10 a b^2 C-7 b^3 B\right )+4 A b^4\right )}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}-\frac{2 \cot (c+d x) \left (-2 a^2 b^2 (A-3 B-8 C)+a^3 b (8 B-12 C)-16 a^4 C-3 a b^3 (A+3 B-3 C)+b^4 (3 A-3 B+C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^4 d \sqrt{a+b} \left (a^2-b^2\right )}-\frac{2 \cot (c+d x) \left (-2 a^3 b^2 (A-14 C)-15 a^2 b^3 B+8 a^4 b B-16 a^5 C+2 a b^4 (3 A-4 C)+3 b^5 B\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^5 d \sqrt{a+b} \left (a^2-b^2\right )} \]

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])^(5/2),x]

[Out]

(-2*(8*a^4*b*B - 15*a^2*b^3*B + 3*b^5*B - 2*a^3*b^2*(A - 14*C) + 2*a*b^4*(3*A - 4*C) - 16*a^5*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^5*Sqrt[a + b]*(a^2 - b^2)*d) - (2*(a^3*b*(8*B - 12*C) - 2*a^2*b^
2*(A - 3*B - 8*C) - 3*a*b^3*(A + 3*B - 3*C) - 16*a^4*C + b^4*(3*A - 3*B + C))*Cot[c + d*x]*EllipticF[ArcSin[Sq
rt[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*(A*b^2 - a*(b*B - a*C))*Sec[c + d*x]^2*Tan[c + d*
x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) - (2*a*(4*A*b^4 + a*(3*a^2*b*B - 7*b^3*B - 6*a^3*C + 10*a*b
^2*C))*Tan[c + d*x])/(3*b^3*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(A*b^2 - a*b*B + 2*a^2*C - b^2*C)*S
qrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*b^3*(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 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 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))^{5/2}} \, dx &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac{2 \int \frac{\sec ^2(c+d x) \left (2 \left (A b^2-a (b B-a C)\right )+\frac{3}{2} b (b B-a (A+C)) \sec (c+d x)-\frac{3}{2} \left (A b^2-a b B+2 a^2 C-b^2 C\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 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)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac{2 a \left (4 A b^4+a \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right )\right ) \tan (c+d x)}{3 b^3 \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 \left (4 A b^4+a \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right )\right )-\frac{1}{4} \left (6 a^4 b B-13 a^2 b^3 B+3 b^5 B+2 a b^4 (2 A-3 C)-12 a^5 C+22 a^3 b^2 C\right ) \sec (c+d x)-\frac{3}{4} b \left (a^2-b^2\right ) \left (A b^2-a b B+2 a^2 C-b^2 C\right ) \sec ^2(c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac{2 a \left (4 A b^4+a \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right )\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (A b^2-a b B+2 a^2 C-b^2 C\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}-\frac{8 \int \frac{\sec (c+d x) \left (-\frac{3}{8} b^2 \left (2 a^3 b B-6 a b^3 B-4 a^4 C+b^4 (3 A+C)+a^2 b^2 (A+7 C)\right )-\frac{3}{8} b \left (8 a^4 b B-15 a^2 b^3 B+3 b^5 B-2 a^3 b^2 (A-14 C)+2 a b^4 (3 A-4 C)-16 a^5 C\right ) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{9 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac{2 a \left (4 A b^4+a \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right )\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (A b^2-a b B+2 a^2 C-b^2 C\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac{\left (8 a^4 b B-15 a^2 b^3 B+3 b^5 B-2 a^3 b^2 (A-14 C)+2 a b^4 (3 A-4 C)-16 a^5 C\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}-\frac{\left (8 \left (\frac{3}{8} b \left (8 a^4 b B-15 a^2 b^3 B+3 b^5 B-2 a^3 b^2 (A-14 C)+2 a b^4 (3 A-4 C)-16 a^5 C\right )-\frac{3}{8} b^2 \left (2 a^3 b B-6 a b^3 B-4 a^4 C+b^4 (3 A+C)+a^2 b^2 (A+7 C)\right )\right )\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{9 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{2 \left (8 a^4 b B-15 a^2 b^3 B+3 b^5 B-2 a^3 b^2 (A-14 C)+2 a b^4 (3 A-4 C)-16 a^5 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^5 (a+b)^{3/2} d}-\frac{2 \left (a^3 b (8 B-12 C)-2 a^2 b^2 (A-3 B-8 C)-3 a b^3 (A+3 B-3 C)-16 a^4 C+b^4 (3 A-3 B+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^4 (a+b)^{3/2} d}-\frac{2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac{2 a \left (4 A b^4+a \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right )\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (A b^2-a b B+2 a^2 C-b^2 C\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 21.58, size = 989, normalized size = 1.8 \[ \frac{\sec (c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (-\frac{4 \left (16 C a^5-8 b B a^4+2 A b^2 a^3-28 b^2 C a^3+15 b^3 B a^2-6 A b^4 a+8 b^4 C a-3 b^5 B\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right )^2}-\frac{4 \left (C \sin (c+d x) a^3-b B \sin (c+d x) a^2+A b^2 \sin (c+d x) a\right )}{3 b^2 \left (b^2-a^2\right ) (b+a \cos (c+d x))^2}-\frac{4 \left (-7 C \sin (c+d x) a^5+4 b B \sin (c+d x) a^4-A b^2 \sin (c+d x) a^3+11 b^2 C \sin (c+d x) a^3-8 b^3 B \sin (c+d x) a^2+5 A b^4 \sin (c+d x) a\right )}{3 b^3 \left (b^2-a^2\right )^2 (b+a \cos (c+d x))}+\frac{4 C \tan (c+d x)}{3 b^3}\right ) (b+a \cos (c+d x))^3}{d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (a+b \sec (c+d x))^{5/2}}+\frac{4 \sqrt{\sec (c+d x)} \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 (16 C a^5-8 b B a^4+2 b^2 (A-14 C) a^3+15 b^3 B a^2+2 b^4 (4 C-3 A) a-3 b^5 B\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 (-16 C a^4+4 b (2 B+3 C) a^3-2 b^2 (A+3 B-8 C) a^2+3 b^3 (A-3 (B+C)) a+b^4 (3 A+3 B+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 (16 C a^5-8 b B a^4+2 b^2 (A-14 C) a^3+15 b^3 B a^2+2 b^4 (4 C-3 A) a-3 b^5 B\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))^{5/2}}{3 b^4 \left (a^2-b^2\right )^2 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (a+b \sec (c+d x))^{5/2} \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])^(5/2),x]

[Out]

(4*(b + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt[(1 - Tan[(c + d*
x)/2]^2)^(-1)]*((a + b)*(-8*a^4*b*B + 15*a^2*b^3*B - 3*b^5*B + 2*a^3*b^2*(A - 14*C) + 16*a^5*C + 2*a*b^4*(-3*A
+ 4*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)*(-2*a^2*b^2*(A + 3*B - 8
*C) - 16*a^4*C + b^4*(3*A + 3*B + C) + 4*a^3*b*(2*B + 3*C) + 3*a*b^3*(A - 3*(B + 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)] + (-8*a^4*b*B + 15*a^2*b^3*B - 3*b^5*B + 2*a^3*b^2*(A - 14*C) + 16*a^
5*C + 2*a*b^4*(-3*A + 4*C))*Tan[(c + d*x)/2]*(b - b*Tan[(c + d*x)/2]^4 + a*(-1 + Tan[(c + d*x)/2]^2)^2)))/(3*b
^4*(a^2 - b^2)^2*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)*(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])^3*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-4*(2*a^3*A*b^2 - 6*a*A*b^4 - 8*a^4*
b*B + 15*a^2*b^3*B - 3*b^5*B + 16*a^5*C - 28*a^3*b^2*C + 8*a*b^4*C)*Sin[c + d*x])/(3*b^4*(a^2 - b^2)^2) - (4*(
a*A*b^2*Sin[c + d*x] - a^2*b*B*Sin[c + d*x] + a^3*C*Sin[c + d*x]))/(3*b^2*(-a^2 + b^2)*(b + a*Cos[c + d*x])^2)
- (4*(-(a^3*A*b^2*Sin[c + d*x]) + 5*a*A*b^4*Sin[c + d*x] + 4*a^4*b*B*Sin[c + d*x] - 8*a^2*b^3*B*Sin[c + d*x]
- 7*a^5*C*Sin[c + d*x] + 11*a^3*b^2*C*Sin[c + d*x]))/(3*b^3*(-a^2 + b^2)^2*(b + a*Cos[c + d*x])) + (4*C*Tan[c
+ d*x])/(3*b^3)))/(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))

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Maple [B]  time = 1.746, size = 10856, normalized size = 19.8 \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))^(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)^3*(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 )^{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^{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)^3*(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)^5 + B*sec(d*x + c)^4 + A*sec(d*x + c)^3)*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 ^{3}{\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)**3*(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)**3/(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 )^{3}}{{\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)^3*(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)^3/(b*sec(d*x + c) + a)^(5/2), x)