3.13.31 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\) [1231]

Optimal. Leaf size=250 \[ -\frac {(20 A-35 B+56 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac {5 (A-2 B+3 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac {(20 A-35 B+56 C) \sin (c+d x)}{15 a^2 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {5 (A-2 B+3 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {(20 A-35 B+56 C) \sin (c+d x)}{5 a^2 d \sqrt {\cos (c+d x)}}-\frac {(A-2 B+3 C) \sin (c+d x)}{a^2 d \cos ^{\frac {5}{2}}(c+d x) (1+\cos (c+d x))}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2} \]

[Out]

-1/5*(20*A-35*B+56*C)*(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-5/3*(A-2*B+3*C)*(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+1/15*(20*A-35*B+56*C)*sin(d*x+c)/a^2/d/cos(d*x+c)^(5/2)-5/3*(A-2*B+3*C)*sin(d*x+c)/a^2/d/cos(d*x+c)^(3/2)-(A
-2*B+3*C)*sin(d*x+c)/a^2/d/cos(d*x+c)^(5/2)/(1+cos(d*x+c))-1/3*(A-B+C)*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*cos(
d*x+c))^2+1/5*(20*A-35*B+56*C)*sin(d*x+c)/a^2/d/cos(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.34, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4197, 3120, 3057, 2827, 2716, 2719, 2720} \begin {gather*} -\frac {5 (A-2 B+3 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac {(20 A-35 B+56 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac {(A-2 B+3 C) \sin (c+d x)}{a^2 d \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x)+1)}-\frac {5 (A-2 B+3 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {(20 A-35 B+56 C) \sin (c+d x)}{15 a^2 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {(20 A-35 B+56 C) \sin (c+d x)}{5 a^2 d \sqrt {\cos (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

-1/5*((20*A - 35*B + 56*C)*EllipticE[(c + d*x)/2, 2])/(a^2*d) - (5*(A - 2*B + 3*C)*EllipticF[(c + d*x)/2, 2])/
(3*a^2*d) + ((20*A - 35*B + 56*C)*Sin[c + d*x])/(15*a^2*d*Cos[c + d*x]^(5/2)) - (5*(A - 2*B + 3*C)*Sin[c + d*x
])/(3*a^2*d*Cos[c + d*x]^(3/2)) + ((20*A - 35*B + 56*C)*Sin[c + d*x])/(5*a^2*d*Sqrt[Cos[c + d*x]]) - ((A - 2*B
+ 3*C)*Sin[c + d*x])/(a^2*d*Cos[c + d*x]^(5/2)*(1 + Cos[c + d*x])) - ((A - B + C)*Sin[c + d*x])/(3*d*Cos[c +
d*x]^(5/2)*(a + a*Cos[c + d*x])^2)

Rule 2716

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

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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3057

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 3120

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + a*C)*Cos[e + f*x]*(a
+ b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Dist[1/(b*(b*c - a*d)*(2*m
+ 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c*(2*m + 1) - a*d*(m - n -
1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^
2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

Rule 4197

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] &&  !IntegerQ[n] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx &=\int \frac {C+B \cos (c+d x)+A \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx\\ &=-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2}+\frac {\int \frac {\frac {1}{2} a (5 A-5 B+11 C)-\frac {1}{2} a (A-7 B+7 C) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \cos (c+d x))} \, dx}{3 a^2}\\ &=-\frac {(A-2 B+3 C) \sin (c+d x)}{a^2 d \cos ^{\frac {5}{2}}(c+d x) (1+\cos (c+d x))}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2}+\frac {\int \frac {\frac {1}{2} a^2 (20 A-35 B+56 C)-\frac {15}{2} a^2 (A-2 B+3 C) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{3 a^4}\\ &=-\frac {(A-2 B+3 C) \sin (c+d x)}{a^2 d \cos ^{\frac {5}{2}}(c+d x) (1+\cos (c+d x))}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac {(5 (A-2 B+3 C)) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{2 a^2}+\frac {(20 A-35 B+56 C) \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{6 a^2}\\ &=\frac {(20 A-35 B+56 C) \sin (c+d x)}{15 a^2 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {5 (A-2 B+3 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(A-2 B+3 C) \sin (c+d x)}{a^2 d \cos ^{\frac {5}{2}}(c+d x) (1+\cos (c+d x))}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac {(5 (A-2 B+3 C)) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^2}+\frac {(20 A-35 B+56 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{10 a^2}\\ &=-\frac {5 (A-2 B+3 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac {(20 A-35 B+56 C) \sin (c+d x)}{15 a^2 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {5 (A-2 B+3 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {(20 A-35 B+56 C) \sin (c+d x)}{5 a^2 d \sqrt {\cos (c+d x)}}-\frac {(A-2 B+3 C) \sin (c+d x)}{a^2 d \cos ^{\frac {5}{2}}(c+d x) (1+\cos (c+d x))}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac {(20 A-35 B+56 C) \int \sqrt {\cos (c+d x)} \, dx}{10 a^2}\\ &=-\frac {(20 A-35 B+56 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac {5 (A-2 B+3 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac {(20 A-35 B+56 C) \sin (c+d x)}{15 a^2 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {5 (A-2 B+3 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {(20 A-35 B+56 C) \sin (c+d x)}{5 a^2 d \sqrt {\cos (c+d x)}}-\frac {(A-2 B+3 C) \sin (c+d x)}{a^2 d \cos ^{\frac {5}{2}}(c+d x) (1+\cos (c+d x))}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 8.41, size = 2164, normalized size = 8.66 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

((-4*I)*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*E^((2*I)*d*x)*Hyp
ergeometric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*
I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)
*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((
2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I
*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1
+ E^((2*I)*d*x))*Sin[c])))/((A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) + ((7*I
)*B*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*E^((2*I)*d*x)*Hypergeom
etric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1
+ E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1
+ E^((2*I)*d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d
*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]
*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^(
(2*I)*d*x))*Sin[c])))/((A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) - (((56*I)/5)
*C*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*E^((2*I)*d*x)*Hypergeome
tric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1
+ E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 +
E^((2*I)*d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*
x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*
Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((
2*I)*d*x))*Sin[c])))/((A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) + (20*A*Cos[c/
2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + B*Sec[
c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot
[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(A + 2*C + 2*B*Cos[c + d*x
] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^2) - (40*B*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Hyper
geometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*
Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTa
n[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[
1 + Cot[c]^2]*(a + a*Sec[c + d*x])^2) + (20*C*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4
}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqr
t[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x
- ArcTan[Cot[c]]]])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*
x])^2) + (Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((4*(10*A - 20*B + 3
6*C + 10*A*Cos[c] - 15*B*Cos[c] + 20*C*Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2
]^3*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(3*d) + (8*Sec[c/2]*Sec[c/2 + (d*x)/2]*(2*A*Sin[(d*x)/
2] - 3*B*Sin[(d*x)/2] + 4*C*Sin[(d*x)/2]))/d + (16*C*Sec[c]*Sec[c + d*x]^3*Sin[d*x])/(5*d) - (16*Sec[c]*Sec[c
+ d*x]*(-5*B*Sin[c] + 10*C*Sin[c] - 15*A*Sin[d*x] + 30*B*Sin[d*x] - 54*C*Sin[d*x]))/(15*d) + (16*Sec[c]*Sec[c
+ d*x]^2*(3*C*Sin[c] + 5*B*Sin[d*x] - 10*C*Sin[d*x]))/(15*d) + (4*(A - B + C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(
3*d)))/((A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2)

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1044\) vs. \(2(280)=560\).
time = 0.42, size = 1045, normalized size = 4.18

method	result	size
default	\(\text {Expression too large to display}\)	\(1045\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(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*((-2*A+4*B-6*C)*(cos(1/2*d*x+1/2*c)*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(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)))-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^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)+(-8*C+4*B)*(-1/6*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+1/3*(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*s
in(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)))+1/3*(-A+B-C)*(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)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(co
s(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-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)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2
*d*x+1/2*c)-12*sin(1/2*d*x+1/2*c)^6+20*sin(1/2*d*x+1/2*c)^4-7*sin(1/2*d*x+1/2*c)^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)/(sin(1/2*d*x+1/2*c)^2-1)+4/5*C/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/
2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(
2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x
+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3
*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(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)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(4*A-8*B+12*C)/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*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(2*sin(1/2*d*x+1
/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*co
s(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.04, size = 498, normalized size = 1.99 \begin {gather*} \frac {2 \, {\left (3 \, {\left (20 \, A - 35 \, B + 56 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (95 \, A - 160 \, B + 261 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (15 \, A - 20 \, B + 37 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, B - 4 \, C\right )} \cos \left (d x + c\right ) + 6 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 25 \, {\left (\sqrt {2} {\left (-i \, A + 2 i \, B - 3 i \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, \sqrt {2} {\left (-i \, A + 2 i \, B - 3 i \, C\right )} \cos \left (d x + c\right )^{4} + \sqrt {2} {\left (-i \, A + 2 i \, B - 3 i \, C\right )} \cos \left (d x + c\right )^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 25 \, {\left (\sqrt {2} {\left (i \, A - 2 i \, B + 3 i \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, \sqrt {2} {\left (i \, A - 2 i \, B + 3 i \, C\right )} \cos \left (d x + c\right )^{4} + \sqrt {2} {\left (i \, A - 2 i \, B + 3 i \, C\right )} \cos \left (d x + c\right )^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (\sqrt {2} {\left (20 i \, A - 35 i \, B + 56 i \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, \sqrt {2} {\left (20 i \, A - 35 i \, B + 56 i \, C\right )} \cos \left (d x + c\right )^{4} + \sqrt {2} {\left (20 i \, A - 35 i \, B + 56 i \, C\right )} \cos \left (d x + c\right )^{3}\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 ) - 3 \, {\left (\sqrt {2} {\left (-20 i \, A + 35 i \, B - 56 i \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, \sqrt {2} {\left (-20 i \, A + 35 i \, B - 56 i \, C\right )} \cos \left (d x + c\right )^{4} + \sqrt {2} {\left (-20 i \, A + 35 i \, B - 56 i \, C\right )} \cos \left (d x + c\right )^{3}\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 )}{30 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/30*(2*(3*(20*A - 35*B + 56*C)*cos(d*x + c)^4 + (95*A - 160*B + 261*C)*cos(d*x + c)^3 + 2*(15*A - 20*B + 37*C
)*cos(d*x + c)^2 + 2*(5*B - 4*C)*cos(d*x + c) + 6*C)*sqrt(cos(d*x + c))*sin(d*x + c) - 25*(sqrt(2)*(-I*A + 2*I
*B - 3*I*C)*cos(d*x + c)^5 + 2*sqrt(2)*(-I*A + 2*I*B - 3*I*C)*cos(d*x + c)^4 + sqrt(2)*(-I*A + 2*I*B - 3*I*C)*
cos(d*x + c)^3)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 25*(sqrt(2)*(I*A - 2*I*B + 3*I*C)*
cos(d*x + c)^5 + 2*sqrt(2)*(I*A - 2*I*B + 3*I*C)*cos(d*x + c)^4 + sqrt(2)*(I*A - 2*I*B + 3*I*C)*cos(d*x + c)^3
)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*(sqrt(2)*(20*I*A - 35*I*B + 56*I*C)*cos(d*x +
c)^5 + 2*sqrt(2)*(20*I*A - 35*I*B + 56*I*C)*cos(d*x + c)^4 + sqrt(2)*(20*I*A - 35*I*B + 56*I*C)*cos(d*x + c)^3
)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(-20*I*A + 35
*I*B - 56*I*C)*cos(d*x + c)^5 + 2*sqrt(2)*(-20*I*A + 35*I*B - 56*I*C)*cos(d*x + c)^4 + sqrt(2)*(-20*I*A + 35*I
*B - 56*I*C)*cos(d*x + c)^3)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)))
)/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)**(7/2)/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^2*cos(d*x + c)^(7/2)), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{7/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(7/2)*(a + a/cos(c + d*x))^2),x)

[Out]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(7/2)*(a + a/cos(c + d*x))^2), x)

________________________________________________________________________________________