3.1027 \(\int \frac{\sec ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=469 \[ \frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (-7 a^2 b^2 (A+C)+3 a^3 b B+a^4 C+3 a b^3 B+A b^4\right )}{4 a^2 b d \left (a^2-b^2\right )^2}-\frac{\sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{\sin (c+d x) \sqrt{\sec (c+d x)} \left (a^2 b^2 (5 A+9 C)-a^3 b B-3 a^4 C-5 a b^3 B+A b^4\right )}{4 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 b^2 (5 A+9 C)-a^3 b B-3 a^4 C-5 a b^3 B+A b^4\right )}{4 a b^2 d \left (a^2-b^2\right )^2}-\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (-3 a^4 b^2 (A-2 C)-5 a^2 b^4 (2 A+3 C)+10 a^3 b^3 B-a^5 b B-3 a^6 C+3 a b^5 B+A b^6\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^2 b^2 d (a-b)^2 (a+b)^3} \]

[Out]

-((A*b^4 - a^3*b*B - 5*a*b^3*B - 3*a^4*C + a^2*b^2*(5*A + 9*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*S
qrt[Sec[c + d*x]])/(4*a*b^2*(a^2 - b^2)^2*d) + ((A*b^4 + 3*a^3*b*B + 3*a*b^3*B + a^4*C - 7*a^2*b^2*(A + C))*Sq
rt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a^2*b*(a^2 - b^2)^2*d) - ((A*b^6 - a^5*b*B +
10*a^3*b^3*B + 3*a*b^5*B - 3*a^4*b^2*(A - 2*C) - 3*a^6*C - 5*a^2*b^4*(2*A + 3*C))*Sqrt[Cos[c + d*x]]*Elliptic
Pi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a^2*(a - b)^2*b^2*(a + b)^3*d) - ((A*b^2 - a*(b*B - a
*C))*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + ((A*b^4 - a^3*b*B - 5*a*b^3
*B - 3*a^4*C + a^2*b^2*(5*A + 9*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(4*b^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x
]))

________________________________________________________________________________________

Rubi [A]  time = 1.17975, antiderivative size = 469, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used = {4098, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ -\frac{\sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{\sin (c+d x) \sqrt{\sec (c+d x)} \left (a^2 b^2 (5 A+9 C)-a^3 b B-3 a^4 C-5 a b^3 B+A b^4\right )}{4 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-7 a^2 b^2 (A+C)+3 a^3 b B+a^4 C+3 a b^3 B+A b^4\right )}{4 a^2 b d \left (a^2-b^2\right )^2}-\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 b^2 (5 A+9 C)-a^3 b B-3 a^4 C-5 a b^3 B+A b^4\right )}{4 a b^2 d \left (a^2-b^2\right )^2}-\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (-3 a^4 b^2 (A-2 C)-5 a^2 b^4 (2 A+3 C)+10 a^3 b^3 B-a^5 b B-3 a^6 C+3 a b^5 B+A b^6\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^2 b^2 d (a-b)^2 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A*b^4 - a^3*b*B - 5*a*b^3*B - 3*a^4*C + a^2*b^2*(5*A + 9*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*S
qrt[Sec[c + d*x]])/(4*a*b^2*(a^2 - b^2)^2*d) + ((A*b^4 + 3*a^3*b*B + 3*a*b^3*B + a^4*C - 7*a^2*b^2*(A + C))*Sq
rt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a^2*b*(a^2 - b^2)^2*d) - ((A*b^6 - a^5*b*B +
10*a^3*b^3*B + 3*a*b^5*B - 3*a^4*b^2*(A - 2*C) - 3*a^6*C - 5*a^2*b^4*(2*A + 3*C))*Sqrt[Cos[c + d*x]]*Elliptic
Pi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*a^2*(a - b)^2*b^2*(a + b)^3*d) - ((A*b^2 - a*(b*B - a
*C))*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + ((A*b^4 - a^3*b*B - 5*a*b^3
*B - 3*a^4*C + a^2*b^2*(5*A + 9*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(4*b^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x
]))

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 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{\sec ^{\frac{3}{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} \, dx &=-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\int \frac{\sqrt{\sec (c+d x)} \left (\frac{1}{2} \left (A b^2-a (b B-a C)\right )+2 b (b B-a (A+C)) \sec (c+d x)+\frac{1}{2} \left (A b^2-a b B-3 a^2 C+4 b^2 C\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (A b^4-a^3 b B-5 a b^3 B-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\int \frac{\frac{1}{4} \left (-A b^4+a^3 b B+5 a b^3 B+3 a^4 C-a^2 b^2 (5 A+9 C)\right )+b \left (a^2 b B+2 b^3 B+a^3 C-a b^2 (3 A+4 C)\right ) \sec (c+d x)+\frac{1}{4} \left (a^3 b B-7 a b^3 B+a^2 b^2 (3 A-5 C)+3 a^4 C+b^4 (3 A+8 C)\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (A b^4-a^3 b B-5 a b^3 B-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\int \frac{\frac{1}{4} a \left (-A b^4+a^3 b B+5 a b^3 B+3 a^4 C-a^2 b^2 (5 A+9 C)\right )-\left (-a b \left (a^2 b B+2 b^3 B+a^3 C-a b^2 (3 A+4 C)\right )+\frac{1}{4} b \left (-A b^4+a^3 b B+5 a b^3 B+3 a^4 C-a^2 b^2 (5 A+9 C)\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{2 a^2 b^2 \left (a^2-b^2\right )^2}-\frac{\left (A b^6-a^5 b B+10 a^3 b^3 B+3 a b^5 B-3 a^4 b^2 (A-2 C)-3 a^6 C-5 a^2 b^4 (2 A+3 C)\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{8 a^2 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (A b^4-a^3 b B-5 a b^3 B-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (A b^4+3 a^3 b B+3 a b^3 B+a^4 C-7 a^2 b^2 (A+C)\right ) \int \sqrt{\sec (c+d x)} \, dx}{8 a^2 b \left (a^2-b^2\right )^2}-\frac{\left (A b^4-a^3 b B-5 a b^3 B-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{8 a b^2 \left (a^2-b^2\right )^2}-\frac{\left (\left (A b^6-a^5 b B+10 a^3 b^3 B+3 a b^5 B-3 a^4 b^2 (A-2 C)-3 a^6 C-5 a^2 b^4 (2 A+3 C)\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}{8 a^2 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (A b^6-a^5 b B+10 a^3 b^3 B+3 a b^5 B-3 a^4 b^2 (A-2 C)-3 a^6 C-5 a^2 b^4 (2 A+3 C)\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)}}{4 a^2 (a-b)^2 b^2 (a+b)^3 d}-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (A b^4-a^3 b B-5 a b^3 B-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (\left (A b^4+3 a^3 b B+3 a b^3 B+a^4 C-7 a^2 b^2 (A+C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{8 a^2 b \left (a^2-b^2\right )^2}-\frac{\left (\left (A b^4-a^3 b B-5 a b^3 B-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{8 a b^2 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (A b^4-a^3 b B-5 a b^3 B-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{4 a b^2 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^4+3 a^3 b B+3 a b^3 B+a^4 C-7 a^2 b^2 (A+C)\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{4 a^2 b \left (a^2-b^2\right )^2 d}-\frac{\left (A b^6-a^5 b B+10 a^3 b^3 B+3 a b^5 B-3 a^4 b^2 (A-2 C)-3 a^6 C-5 a^2 b^4 (2 A+3 C)\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)}}{4 a^2 (a-b)^2 b^2 (a+b)^3 d}-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{\left (A b^4-a^3 b B-5 a b^3 B-3 a^4 C+a^2 b^2 (5 A+9 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [B]  time = 7.23851, size = 1051, normalized size = 2.24 \[ \frac{\sec (c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (-\frac{2 \left (16 B b^4-24 a A b^3-32 a C b^3+8 a^2 B b^2+8 a^3 C b\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) (a+b \sec (c+d x)) \sqrt{1-\sec ^2(c+d x)} \sin (c+d x) \cos ^2(c+d x)}{a (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac{2 \left (9 C a^4+3 b B a^3+A b^2 a^2-19 b^2 C a^2-9 b^3 B a+5 A b^4+16 b^4 C\right ) \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 ) (a+b \sec (c+d x)) \sqrt{1-\sec ^2(c+d x)} \sin (c+d x) \cos ^2(c+d x)}{b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}-\frac{2 \left (3 C a^4+b B a^3-5 A b^2 a^2-9 b^2 C a^2+5 b^3 B a-A b^4\right ) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (\Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} a^2-2 b \sec ^2(c+d x) a+2 b a+2 b E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} a+(a-2 b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right ),-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} a-2 b^2 \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a^2 b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}\right ) (b+a \cos (c+d x))^3}{8 (a-b)^2 b^2 (a+b)^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}+\frac{\sec ^{\frac{3}{2}}(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (\frac{\left (-3 C a^4-b B a^3+5 A b^2 a^2+9 b^2 C a^2-5 b^3 B a+A b^4\right ) \sin (c+d x)}{2 a b^2 \left (b^2-a^2\right )^2}+\frac{C \sin (c+d x) a^2-b B \sin (c+d x) a+A b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}+\frac{C \sin (c+d x) a^4+3 b B \sin (c+d x) a^3-7 A b^2 \sin (c+d x) a^2-7 b^2 C \sin (c+d x) a^2+3 b^3 B \sin (c+d x) a+A b^4 \sin (c+d x)}{2 a b \left (b^2-a^2\right )^2 (b+a \cos (c+d x))}\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))^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^3,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)*((-2*(-24*a*A*b^3 + 8*a^2*b^2*B +
16*b^4*B + 8*a^3*b*C - 32*a*b^3*C)*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*(a^2*A
*b^2 + 5*A*b^4 + 3*a^3*b*B - 9*a*b^3*B + 9*a^4*C - 19*a^2*b^2*C + 16*b^4*C)*Cos[c + d*x]^2*(EllipticF[ArcSin[S
qrt[Sec[c + d*x]]], -1] + EllipticPi[-(b/a), -ArcSin[Sqrt[Sec[c + d*x]]], -1])*(a + b*Sec[c + d*x])*Sqrt[1 - S
ec[c + d*x]^2]*Sin[c + d*x])/(b*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) - (2*(-5*a^2*A*b^2 - A*b^4 + a^3*b*
B + 5*a*b^3*B + 3*a^4*C - 9*a^2*b^2*C)*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)*Ellip
ticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + a^2*EllipticPi[-(b/a), -Arc
Sin[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[Sq
rt[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))))/(8*(a - b)^2*b^2*(a + b)^2*d*(A + 2*C + 2*B*Co
s[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^3) + ((b + a*Cos[c + d*x])^3*Sec[c + d*x]^(3/2)*(A + B*S
ec[c + d*x] + C*Sec[c + d*x]^2)*(((5*a^2*A*b^2 + A*b^4 - a^3*b*B - 5*a*b^3*B - 3*a^4*C + 9*a^2*b^2*C)*Sin[c +
d*x])/(2*a*b^2*(-a^2 + b^2)^2) + (A*b^2*Sin[c + d*x] - a*b*B*Sin[c + d*x] + a^2*C*Sin[c + d*x])/(a*(a^2 - b^2)
*(b + a*Cos[c + d*x])^2) + (-7*a^2*A*b^2*Sin[c + d*x] + A*b^4*Sin[c + d*x] + 3*a^3*b*B*Sin[c + d*x] + 3*a*b^3*
B*Sin[c + d*x] + a^4*C*Sin[c + d*x] - 7*a^2*b^2*C*Sin[c + d*x])/(2*a*b*(-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])^3)

________________________________________________________________________________________

Maple [B]  time = 10.419, size = 1879, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,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*(-2*A*b+B*a)/a^2*(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)*Ell
ipticF(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/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))+2*(A*b^2-B*a*b+C*a^2)/a^2*(1/2*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)^2
+3/4*a^2*(a^2-3*b^2)/b^2/(a^2-b^2)^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)-3/8/(a+b)/(a^2-b^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))*a^2-1/4/(a+b
)/(a^2-b^2)/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))*a+7/8/(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))+3/8*a^3/b^2/(a^2-b^2)^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))-9/8*a/(a^2-b^2)^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)*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))-3/8*a^3/b^2/(a^2-b^2)^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))+9/8*a/(a
^2-b^2)^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))-3/8/(a-b)/(a+b)/(a^2-b^2)/b^2/(a^2-a*b)*a^5*(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)*Ellipt
icPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/4/(a-b)/(a+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))-15/8/(a-b)/(a+b)/(a^2-b^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))))/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(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,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(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,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(sec(d*x+c)**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**3,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 )} \sec \left (d x + c\right )^{\frac{3}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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