\(\int \frac {1}{\cos ^{\frac {9}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx\) [836]

Optimal result

Integrand size = 23, antiderivative size = 328 \[ \int \frac {1}{\cos ^{\frac {9}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=-\frac {\left (15 a^4-29 a^2 b^2+8 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {a \left (5 a^2-11 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 b^2 \left (a^2-b^2\right )^2 d}-\frac {a \left (15 a^4-38 a^2 b^2+35 b^4\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 (a-b)^2 b^3 (a+b)^3 d}+\frac {\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}-\frac {a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac {a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))} \]

[Out]

-1/4*(15*a^4-29*a^2*b^2+8*b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^
(1/2))/b^3/(a^2-b^2)^2/d-1/4*a*(5*a^2-11*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/
2*d*x+1/2*c),2^(1/2))/b^2/(a^2-b^2)^2/d-1/4*a*(15*a^4-38*a^2*b^2+35*b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*
d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/(a-b)^2/b^3/(a+b)^3/d-1/2*a^2*sin(d*x+c)/b/(a^2-b^
2)/d/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^2-1/4*a^2*(5*a^2-11*b^2)*sin(d*x+c)/b^2/(a^2-b^2)^2/d/cos(d*x+c)^(3/2)/
(a+b*sec(d*x+c))+1/4*(15*a^4-29*a^2*b^2+8*b^4)*sin(d*x+c)/b^3/(a^2-b^2)^2/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 1.16 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {4349, 3930, 4183, 4187, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \[ \int \frac {1}{\cos ^{\frac {9}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=-\frac {a \left (5 a^2-11 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 b^2 d \left (a^2-b^2\right )^2}-\frac {a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x)}{4 b^2 d \left (a^2-b^2\right )^2 \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {a^2 \sin (c+d x)}{2 b d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac {\left (15 a^4-29 a^2 b^2+8 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}-\frac {a \left (15 a^4-38 a^2 b^2+35 b^4\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 b^3 d (a-b)^2 (a+b)^3}+\frac {\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sin (c+d x)}{4 b^3 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)}} \]

[In]

Int[1/(Cos[c + d*x]^(9/2)*(a + b*Sec[c + d*x])^3),x]

[Out]

-1/4*((15*a^4 - 29*a^2*b^2 + 8*b^4)*EllipticE[(c + d*x)/2, 2])/(b^3*(a^2 - b^2)^2*d) - (a*(5*a^2 - 11*b^2)*Ell
ipticF[(c + d*x)/2, 2])/(4*b^2*(a^2 - b^2)^2*d) - (a*(15*a^4 - 38*a^2*b^2 + 35*b^4)*EllipticPi[(2*a)/(a + b),
(c + d*x)/2, 2])/(4*(a - b)^2*b^3*(a + b)^3*d) + ((15*a^4 - 29*a^2*b^2 + 8*b^4)*Sin[c + d*x])/(4*b^3*(a^2 - b^
2)^2*d*Sqrt[Cos[c + d*x]]) - (a^2*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^2)
- (a^2*(5*a^2 - 11*b^2)*Sin[c + d*x])/(4*b^2*(a^2 - b^2)^2*d*Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x]))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(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 3856

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 3872

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 3930

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] &&
NeQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n, 2]))

Rule 3934

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 4183

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 4187

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 4191

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 4349

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx \\ & = -\frac {a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac {a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}-\frac {a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac {a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}-\frac {a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac {a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\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 ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}-\frac {a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac {a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\left (a \left (15 a^4-38 a^2 b^2+35 b^4\right )\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 {\left (a \left (5 a^2-11 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (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 \frac {1}{\sqrt {\sec (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 ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 (a-b)^2 b^3 (a+b)^3 d}+\frac {\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}-\frac {a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac {a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\left (a \left (5 a^2-11 b^2\right )\right ) \int \frac {1}{\sqrt {\cos (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 \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 ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {a \left (5 a^2-11 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{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 ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 (a-b)^2 b^3 (a+b)^3 d}+\frac {\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}-\frac {a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac {a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.03 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\cos ^{\frac {9}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=\frac {-\frac {\frac {2 \left (45 a^5-95 a^3 b^2+56 a b^4\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 b \left (5 a^4-10 a^2 b^2+2 b^4\right ) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{a}+\frac {2 \left (15 a^4-29 a^2 b^2+8 b^4\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}+4 \sqrt {\cos (c+d x)} \left (\frac {a^3 \left (9 a^2 b-15 b^3+a \left (7 a^2-13 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (b+a \cos (c+d x))^2}+8 \tan (c+d x)\right )}{16 b^3 d} \]

[In]

Integrate[1/(Cos[c + d*x]^(9/2)*(a + b*Sec[c + d*x])^3),x]

[Out]

(-(((2*(45*a^5 - 95*a^3*b^2 + 56*a*b^4)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*b*(5*a^4 - 10*
a^2*b^2 + 2*b^4)*(2*EllipticF[(c + d*x)/2, 2] - (2*b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)))/a +
(2*(15*a^4 - 29*a^2*b^2 + 8*b^4)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[Arc
Sin[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])
/(a*b*Sqrt[Sin[c + d*x]^2]))/((a - b)^2*(a + b)^2)) + 4*Sqrt[Cos[c + d*x]]*((a^3*(9*a^2*b - 15*b^3 + a*(7*a^2
- 13*b^2)*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(b + a*Cos[c + d*x])^2) + 8*Tan[c + d*x]))/(16*b^3*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1986\) vs. \(2(388)=776\).

Time = 32.90 (sec) , antiderivative size = 1987, normalized size of antiderivative = 6.06

method	result	size
default	\(\text {Expression too large to display}\)	\(1987\)

[In]

int(1/cos(d*x+c)^(9/2)/(a+b*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/b^3/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2
-1)*(-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)*sin(1/2*d*x+1/2*c)^2-(sin(1/2*d
*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(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*a*cos(1/2*d*x+1/2*c)^2-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*a*cos(1/2*d*x+1/2*c)^2-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)*El
lipticF(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)*Elli
pticPi(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/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*a*cos(1/2*d*x+1/2*c)^2-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)*(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^2/b^3/(a^2-a*b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*c
os(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

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\cos ^{\frac {9}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(1/cos(d*x+c)^(9/2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\cos ^{\frac {9}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(1/cos(d*x+c)**(9/2)/(a+b*sec(d*x+c))**3,x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {1}{\cos ^{\frac {9}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(1/cos(d*x+c)^(9/2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {1}{\cos ^{\frac {9}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate(1/cos(d*x+c)^(9/2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(1/((b*sec(d*x + c) + a)^3*cos(d*x + c)^(9/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\cos ^{\frac {9}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^{9/2}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \]

[In]

int(1/(cos(c + d*x)^(9/2)*(a + b/cos(c + d*x))^3),x)

[Out]

int(1/(cos(c + d*x)^(9/2)*(a + b/cos(c + d*x))^3), x)