3.621 \(\int \frac{\sec ^{\frac{9}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=388 \[ -\frac{a \left (5 a^2-11 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{4 b^2 d \left (a^2-b^2\right )^2}-\frac{a^2 \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac{a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{4 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{\left (-29 a^2 b^2+15 a^4+8 b^4\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{4 b^3 d \left (a^2-b^2\right )^2}-\frac{\left (-29 a^2 b^2+15 a^4+8 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}-\frac{a \left (-38 a^2 b^2+15 a^4+35 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^3 d (a-b)^2 (a+b)^3} \]

[Out]

-((15*a^4 - 29*a^2*b^2 + 8*b^4)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*b^3*(a^2 -
b^2)^2*d) - (a*(5*a^2 - 11*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*b^2*(a^2
- b^2)^2*d) - (a*(15*a^4 - 38*a^2*b^2 + 35*b^4)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*S
qrt[Sec[c + d*x]])/(4*(a - b)^2*b^3*(a + b)^3*d) + ((15*a^4 - 29*a^2*b^2 + 8*b^4)*Sqrt[Sec[c + d*x]]*Sin[c + d
*x])/(4*b^3*(a^2 - b^2)^2*d) - (a^2*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2
) - (a^2*(5*a^2 - 11*b^2)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(4*b^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.961193, antiderivative size = 388, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {3845, 4098, 4102, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ -\frac{a^2 \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac{a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{4 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{\left (-29 a^2 b^2+15 a^4+8 b^4\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{4 b^3 d \left (a^2-b^2\right )^2}-\frac{a \left (5 a^2-11 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^2 d \left (a^2-b^2\right )^2}-\frac{\left (-29 a^2 b^2+15 a^4+8 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}-\frac{a \left (-38 a^2 b^2+15 a^4+35 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^3 d (a-b)^2 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(9/2)/(a + b*Sec[c + d*x])^3,x]

[Out]

-((15*a^4 - 29*a^2*b^2 + 8*b^4)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*b^3*(a^2 -
b^2)^2*d) - (a*(5*a^2 - 11*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*b^2*(a^2
- b^2)^2*d) - (a*(15*a^4 - 38*a^2*b^2 + 35*b^4)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*S
qrt[Sec[c + d*x]])/(4*(a - b)^2*b^3*(a + b)^3*d) + ((15*a^4 - 29*a^2*b^2 + 8*b^4)*Sqrt[Sec[c + d*x]]*Sin[c + d
*x])/(4*b^3*(a^2 - b^2)^2*d) - (a^2*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2
) - (a^2*(5*a^2 - 11*b^2)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(4*b^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

Rule 3845

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(a^2*
d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3))/(b*f*(m + 1)*(a^2 - b^2)), x] + Dist[d
^3/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3)*Simp[a^2*(n - 3) + a*b*(
m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*(m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && N
eQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n, 2]))

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 4102

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[(C*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(d*Csc[e + f*x])^(n - 1))/(b*f*(m + n + 1)), x] + Dist[d/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + (A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a*C
*n)*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] && 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{9}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx &=-\frac{a^2 \sec ^{\frac{5}{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{\sec ^{\frac{3}{2}}(c+d x) \left (\frac{3 a^2}{2}-2 a b \sec (c+d x)-\frac{1}{2} \left (5 a^2-4 b^2\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{a^2 \sec ^{\frac{5}{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{a^2 \left (5 a^2-11 b^2\right ) \sec ^{\frac{3}{2}}(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{\sqrt{\sec (c+d x)} \left (-\frac{1}{4} a^2 \left (5 a^2-11 b^2\right )+a b \left (a^2-4 b^2\right ) \sec (c+d x)+\frac{1}{4} \left (15 a^4-29 a^2 b^2+8 b^4\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac{a^2 \sec ^{\frac{5}{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{a^2 \left (5 a^2-11 b^2\right ) \sec ^{\frac{3}{2}}(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}{8} a \left (15 a^4-29 a^2 b^2+8 b^4\right )-\frac{1}{2} b \left (5 a^4-10 a^2 b^2+2 b^4\right ) \sec (c+d x)-\frac{3}{8} a \left (5 a^4-11 a^2 b^2+8 b^4\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{b^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac{a^2 \sec ^{\frac{5}{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{a^2 \left (5 a^2-11 b^2\right ) \sec ^{\frac{3}{2}}(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}{8} a^2 \left (15 a^4-29 a^2 b^2+8 b^4\right )-\left (\frac{1}{2} a b \left (5 a^4-10 a^2 b^2+2 b^4\right )-\frac{1}{8} a b \left (15 a^4-29 a^2 b^2+8 b^4\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{a^2 b^3 \left (a^2-b^2\right )^2}-\frac{\left (a \left (15 a^4-38 a^2 b^2+35 b^4\right )\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac{a^2 \sec ^{\frac{5}{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{a^2 \left (5 a^2-11 b^2\right ) \sec ^{\frac{3}{2}}(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 \left (5 a^2-11 b^2\right )\right ) \int \sqrt{\sec (c+d x)} \, dx}{8 b^2 \left (a^2-b^2\right )^2}-\frac{\left (15 a^4-29 a^2 b^2+8 b^4\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{8 b^3 \left (a^2-b^2\right )^2}-\frac{\left (a \left (15 a^4-38 a^2 b^2+35 b^4\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 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{a \left (15 a^4-38 a^2 b^2+35 b^4\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-b)^2 b^3 (a+b)^3 d}+\frac{\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac{a^2 \sec ^{\frac{5}{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{a^2 \left (5 a^2-11 b^2\right ) \sec ^{\frac{3}{2}}(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 \left (5 a^2-11 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{8 b^2 \left (a^2-b^2\right )^2}-\frac{\left (\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac{a \left (5 a^2-11 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{4 b^2 \left (a^2-b^2\right )^2 d}-\frac{a \left (15 a^4-38 a^2 b^2+35 b^4\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-b)^2 b^3 (a+b)^3 d}+\frac{\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac{a^2 \sec ^{\frac{5}{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{a^2 \left (5 a^2-11 b^2\right ) \sec ^{\frac{3}{2}}(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 [A]  time = 6.7834, size = 726, normalized size = 1.87 \[ \frac{\sqrt{\sec (c+d x)} \left (\frac{\left (-29 a^2 b^2+15 a^4+8 b^4\right ) \sin (c+d x)}{4 b^3 \left (b^2-a^2\right )^2}+\frac{a^2 \sin (c+d x)}{2 b \left (b^2-a^2\right ) (a \cos (c+d x)+b)^2}+\frac{11 a^2 b^2 \sin (c+d x)-5 a^4 \sin (c+d x)}{4 b^2 \left (b^2-a^2\right )^2 (a \cos (c+d x)+b)}\right )}{d}-\frac{\frac{2 \left (-95 a^3 b^2+45 a^5+56 a b^4\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} (a+b \sec (c+d x)) \left (\text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right ),-1\right )+\Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right )}{b \left (1-\cos ^2(c+d x)\right ) (a \cos (c+d x)+b)}-\frac{2 \left (-29 a^3 b^2+15 a^5+8 a b^4\right ) \sin (c+d x) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (a (a-2 b) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right ),-1\right )+a^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-2 b^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-2 a b \sec ^2(c+d x)+2 a b \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+2 a b\right )}{a^2 b \left (1-\cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} \left (2-\sec ^2(c+d x)\right ) (a \cos (c+d x)+b)}-\frac{2 \left (-80 a^2 b^3+40 a^4 b+16 b^5\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} (a+b \sec (c+d x)) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )}{a \left (1-\cos ^2(c+d x)\right ) (a \cos (c+d x)+b)}}{16 b^3 d (a-b)^2 (a+b)^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^(9/2)/(a + b*Sec[c + d*x])^3,x]

[Out]

-((-2*(40*a^4*b - 80*a^2*b^3 + 16*b^5)*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*(45
*a^5 - 95*a^3*b^2 + 56*a*b^4)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] + 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])/(b*(b + a*Cos[c +
d*x])*(1 - Cos[c + d*x]^2)) - (2*(15*a^5 - 29*a^3*b^2 + 8*a*b^4)*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)*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] +
a^2*EllipticPi[-(b/a), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 2*b^2*El
lipticPi[-(b/a), -ArcSin[Sqrt[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)))/(16*(a - b)^2*b^3*(a
+ b)^2*d) + (Sqrt[Sec[c + d*x]]*(((15*a^4 - 29*a^2*b^2 + 8*b^4)*Sin[c + d*x])/(4*b^3*(-a^2 + b^2)^2) + (a^2*S
in[c + d*x])/(2*b*(-a^2 + b^2)*(b + a*Cos[c + d*x])^2) + (-5*a^4*Sin[c + d*x] + 11*a^2*b^2*Sin[c + d*x])/(4*b^
2*(-a^2 + b^2)^2*(b + a*Cos[c + d*x]))))/d

________________________________________________________________________________________

Maple [B]  time = 7.598, size = 2014, normalized size = 5.2 \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)^(9/2)/(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*a/b*(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)*Elliptic
F(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)*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)*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(co
s(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)*EllipticPi(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)))+2*a^2/b^3/(
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*(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)*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*a/b/(a^2-b^2)*(s
in(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/b^3*(-(
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1
/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1
/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x
+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

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)^(9/2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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)^(9/2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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)**(9/2)/(a+b*sec(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{\frac{9}{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)^(9/2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^(9/2)/(b*sec(d*x + c) + a)^3, x)