\(\int \frac {1}{\cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\) [388]

Optimal result

Integrand size = 23, antiderivative size = 162 \[ \int \frac {1}{\cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {7 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {10 \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {7 \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}-\frac {7 \sin (c+d x)}{3 a^2 d \cos ^{\frac {5}{2}}(c+d x) (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^2} \]

[Out]

7*(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))/a^2/d+10/3*(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))/a^2/d+10/3*sin(d*x+c)/a^2/d/cos(d*x
+c)^(3/2)-7/3*sin(d*x+c)/a^2/d/cos(d*x+c)^(5/2)/(1+sec(d*x+c))-1/3*sin(d*x+c)/d/cos(d*x+c)^(7/2)/(a+a*sec(d*x+
c))^2-7*sin(d*x+c)/a^2/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {4349, 3901, 4104, 3872, 3853, 3856, 2719, 2720} \[ \int \frac {1}{\cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {7 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {10 \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {7 \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}-\frac {7 \sin (c+d x)}{3 a^2 d \cos ^{\frac {5}{2}}(c+d x) (\sec (c+d x)+1)}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^2} \]

[In]

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

[Out]

(7*EllipticE[(c + d*x)/2, 2])/(a^2*d) + (10*EllipticF[(c + d*x)/2, 2])/(3*a^2*d) + (10*Sin[c + d*x])/(3*a^2*d*
Cos[c + d*x]^(3/2)) - (7*Sin[c + d*x])/(a^2*d*Sqrt[Cos[c + d*x]]) - (7*Sin[c + d*x])/(3*a^2*d*Cos[c + d*x]^(5/
2)*(1 + Sec[c + d*x])) - Sin[c + d*x]/(3*d*Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^2)

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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

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 3901

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

Rule 4104

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(
a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A
*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,
d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 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+a \sec (c+d x))^2} \, dx \\ & = -\frac {\sin (c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (\frac {5 a}{2}-\frac {9}{2} a \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{3 a^2} \\ & = -\frac {7 \sin (c+d x)}{3 a^2 d \cos ^{\frac {5}{2}}(c+d x) (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \left (\frac {21 a^2}{2}-15 a^2 \sec (c+d x)\right ) \, dx}{3 a^4} \\ & = -\frac {7 \sin (c+d x)}{3 a^2 d \cos ^{\frac {5}{2}}(c+d x) (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {\left (7 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx}{2 a^2}+\frac {\left (5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {5}{2}}(c+d x) \, dx}{a^2} \\ & = \frac {10 \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {7 \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}-\frac {7 \sin (c+d x)}{3 a^2 d \cos ^{\frac {5}{2}}(c+d x) (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac {\left (5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx}{3 a^2}+\frac {\left (7 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a^2} \\ & = \frac {10 \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {7 \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}-\frac {7 \sin (c+d x)}{3 a^2 d \cos ^{\frac {5}{2}}(c+d x) (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac {5 \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 a^2}+\frac {7 \int \sqrt {\cos (c+d x)} \, dx}{2 a^2} \\ & = \frac {7 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {10 \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {7 \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}-\frac {7 \sin (c+d x)}{3 a^2 d \cos ^{\frac {5}{2}}(c+d x) (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^2} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 2.46 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.30 \[ \int \frac {1}{\cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {\left (82 \cos \left (\frac {1}{2} (c-d x)\right )+65 \cos \left (\frac {1}{2} (3 c+d x)\right )+68 \cos \left (\frac {1}{2} (c+3 d x)\right )+37 \cos \left (\frac {1}{2} (5 c+3 d x)\right )+53 \cos \left (\frac {1}{2} (3 c+5 d x)\right )+10 \cos \left (\frac {1}{2} (7 c+5 d x)\right )+21 \cos \left (\frac {1}{2} (5 c+7 d x)\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right )}{8 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {4 i \sqrt {2} e^{-i (c+d x)} \left (21 \left (1+e^{2 i (c+d x)}\right )+21 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )-10 e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right ) \sec ^2(c+d x)}{d \left (-1+e^{2 i c}\right ) \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}\right )}{3 a^2 (1+\sec (c+d x))^2} \]

[In]

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

[Out]

(Cos[(c + d*x)/2]^4*(-1/8*((82*Cos[(c - d*x)/2] + 65*Cos[(3*c + d*x)/2] + 68*Cos[(c + 3*d*x)/2] + 37*Cos[(5*c
+ 3*d*x)/2] + 53*Cos[(3*c + 5*d*x)/2] + 10*Cos[(7*c + 5*d*x)/2] + 21*Cos[(5*c + 7*d*x)/2])*Csc[c/2]*Sec[c/2]*S
ec[(c + d*x)/2]^3)/(d*Cos[c + d*x]^(7/2)) + ((4*I)*Sqrt[2]*(21*(1 + E^((2*I)*(c + d*x))) + 21*(-1 + E^((2*I)*c
))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] - 10*E^(I*(c + d*x))*
(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))])*Sec[c
+ d*x]^2)/(d*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))])))/(3*a^2*(1
+ Sec[c + d*x])^2)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(412\) vs. \(2(198)=396\).

Time = 11.27 (sec) , antiderivative size = 413, normalized size of antiderivative = 2.55

method	result	size
default	\(-\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\frac {\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {6 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {22 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {14 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {16 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}\right )}{2 a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(413\)

[In]

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

[Out]

-1/2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^2*(1/3*(-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)^3+6*(-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)-
22/3*(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+14*(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))-EllipticE(cos
(1/2*d*x+1/2*c),2^(1/2)))+16*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x
+1/2*c)^2)^(1/2)-2/3*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)/(cos(1/2*d*x+1/2*
c)^2-1/2)^2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 338, normalized size of antiderivative = 2.09 \[ \int \frac {1}{\cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {2 \, {\left (21 \, \cos \left (d x + c\right )^{3} + 32 \, \cos \left (d x + c\right )^{2} + 8 \, \cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 10 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{4} + 2 i \, \sqrt {2} \cos \left (d x + c\right )^{3} + i \, \sqrt {2} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 10 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{4} - 2 i \, \sqrt {2} \cos \left (d x + c\right )^{3} - i \, \sqrt {2} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{4} - 2 i \, \sqrt {2} \cos \left (d x + c\right )^{3} - i \, \sqrt {2} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{4} + 2 i \, \sqrt {2} \cos \left (d x + c\right )^{3} + i \, \sqrt {2} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

-1/6*(2*(21*cos(d*x + c)^3 + 32*cos(d*x + c)^2 + 8*cos(d*x + c) - 2)*sqrt(cos(d*x + c))*sin(d*x + c) + 10*(I*s
qrt(2)*cos(d*x + c)^4 + 2*I*sqrt(2)*cos(d*x + c)^3 + I*sqrt(2)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(
d*x + c) + I*sin(d*x + c)) + 10*(-I*sqrt(2)*cos(d*x + c)^4 - 2*I*sqrt(2)*cos(d*x + c)^3 - I*sqrt(2)*cos(d*x +
c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*(-I*sqrt(2)*cos(d*x + c)^4 - 2*I*sqrt(2)*
cos(d*x + c)^3 - I*sqrt(2)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*
sin(d*x + c))) + 21*(I*sqrt(2)*cos(d*x + c)^4 + 2*I*sqrt(2)*cos(d*x + c)^3 + I*sqrt(2)*cos(d*x + c)^2)*weierst
rassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*co
s(d*x + c)^3 + a^2*d*cos(d*x + c)^2)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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