3.1097 \(\int \frac {(a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=230 \[ \frac {4 a^2 (7 A+5 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {16 a^2 (3 A+2 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^2 (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a^2 (21 A+19 C) \sin (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {16 a^2 (3 A+2 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {8 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{63 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)} \]
[Out]
-16/15*a^2*(3*A+2*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))/d+4
/21*a^2*(7*A+5*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))/d+2/10
5*a^2*(21*A+19*C)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+4/21*a^2*(7*A+5*C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/9*C*(a+a*co
s(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(9/2)+8/63*C*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(7/2)+16/15*a^2*
(3*A+2*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)
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Rubi [A] time = 0.54, antiderivative size = 230, normalized size of antiderivative = 1.00,
number of steps used = 10, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used
= {4114, 3044, 2975, 2968, 3021, 2748, 2636, 2641, 2639} \[ \frac {4 a^2 (7 A+5 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {16 a^2 (3 A+2 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^2 (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a^2 (21 A+19 C) \sin (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {16 a^2 (3 A+2 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {8 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{63 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac {9}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[((a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2))/Cos[c + d*x]^(3/2),x]
[Out]
(-16*a^2*(3*A + 2*C)*EllipticE[(c + d*x)/2, 2])/(15*d) + (4*a^2*(7*A + 5*C)*EllipticF[(c + d*x)/2, 2])/(21*d)
+ (2*a^2*(21*A + 19*C)*Sin[c + d*x])/(105*d*Cos[c + d*x]^(5/2)) + (4*a^2*(7*A + 5*C)*Sin[c + d*x])/(21*d*Cos[c
+ d*x]^(3/2)) + (16*a^2*(3*A + 2*C)*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]) + (2*C*(a + a*Cos[c + d*x])^2*Sin
[c + d*x])/(9*d*Cos[c + d*x]^(9/2)) + (8*C*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(63*d*Cos[c + d*x]^(7/2))
Rule 2636
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 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]
Rule 2748
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 2968
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Rule 2975
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^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])
Rule 3021
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Rule 3044
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n +
2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b
*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0
])
Rule 4114
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (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 + A
*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && !IntegerQ[n] && IntegerQ[m]
Rubi steps
\begin {align*} \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx &=\int \frac {(a+a \cos (c+d x))^2 \left (C+A \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\\ &=\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \cos (c+d x))^2 \left (2 a C+\frac {3}{2} a (3 A+C) \cos (c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx}{9 a}\\ &=\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {4 \int \frac {(a+a \cos (c+d x)) \left (\frac {3}{4} a^2 (21 A+19 C)+\frac {3}{4} a^2 (21 A+11 C) \cos (c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{63 a}\\ &=\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {4 \int \frac {\frac {3}{4} a^3 (21 A+19 C)+\left (\frac {3}{4} a^3 (21 A+11 C)+\frac {3}{4} a^3 (21 A+19 C)\right ) \cos (c+d x)+\frac {3}{4} a^3 (21 A+11 C) \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{63 a}\\ &=\frac {2 a^2 (21 A+19 C) \sin (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {8 \int \frac {\frac {45}{4} a^3 (7 A+5 C)+21 a^3 (3 A+2 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{315 a}\\ &=\frac {2 a^2 (21 A+19 C) \sin (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {1}{15} \left (8 a^2 (3 A+2 C)\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx+\frac {1}{7} \left (2 a^2 (7 A+5 C)\right ) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 (21 A+19 C) \sin (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {16 a^2 (3 A+2 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {1}{15} \left (8 a^2 (3 A+2 C)\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} \left (2 a^2 (7 A+5 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {16 a^2 (3 A+2 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^2 (7 A+5 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 a^2 (21 A+19 C) \sin (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {16 a^2 (3 A+2 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \cos ^{\frac {7}{2}}(c+d x)}\\ \end {align*}
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Mathematica [C] time = 6.77, size = 1137, normalized size = 4.94 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[((a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2))/Cos[c + d*x]^(3/2),x]
[Out]
(Cos[c + d*x]^(9/2)*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*((8*(3*A + 2*C)*Csc[c]*
Sec[c])/(15*d) + (C*Sec[c]*Sec[c + d*x]^5*Sin[d*x])/(9*d) + (Sec[c]*Sec[c + d*x]^4*(7*C*Sin[c] + 18*C*Sin[d*x]
))/(63*d) + (2*Sec[c]*Sec[c + d*x]*(35*A*Sin[c] + 25*C*Sin[c] + 84*A*Sin[d*x] + 56*C*Sin[d*x]))/(105*d) + (Sec
[c]*Sec[c + d*x]^3*(90*C*Sin[c] + 63*A*Sin[d*x] + 112*C*Sin[d*x]))/(315*d) + (Sec[c]*Sec[c + d*x]^2*(63*A*Sin[
c] + 112*C*Sin[c] + 210*A*Sin[d*x] + 150*C*Sin[d*x]))/(315*d)))/(A + 2*C + A*Cos[2*c + 2*d*x]) - (2*A*Cos[c +
d*x]^4*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Se
c[c + d*x])^2*(A + 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 + A*Cos[2
*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (10*C*Cos[c + d*x]^4*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x -
ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + 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 + Si
n[d*x - ArcTan[Cot[c]]]])/(21*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) + (4*A*Cos[c + d*x]^4*Csc[c
]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4},
Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + C
os[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Si
n[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])
/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(5*d*(A + 2*C + A*Cos[2*c
+ 2*d*x])) + (8*C*Cos[c + d*x]^4*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*((H
ypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 -
Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 +
Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x
+ ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + T
an[c]^2]]))/(15*d*(A + 2*C + A*Cos[2*c + 2*d*x]))
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fricas [F] time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C a^{2} \sec \left (d x + c\right )^{4} + 2 \, C a^{2} \sec \left (d x + c\right )^{3} + {\left (A + C\right )} a^{2} \sec \left (d x + c\right )^{2} + 2 \, A a^{2} \sec \left (d x + c\right ) + A a^{2}}{\cos \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="fricas")
[Out]
integral((C*a^2*sec(d*x + c)^4 + 2*C*a^2*sec(d*x + c)^3 + (A + C)*a^2*sec(d*x + c)^2 + 2*A*a^2*sec(d*x + c) +
A*a^2)/cos(d*x + c)^(3/2), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^2/cos(d*x + c)^(3/2), x)
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maple [B] time = 18.49, size = 1168, normalized size = 5.08 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x)
[Out]
-8*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(-1/5*(1/4*A+1/4*C)/(8*sin(1/2*d*x+1/2*c)^6-1
2*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*(12*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)*(sin(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)^6-12*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)*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s
in(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-8*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)+1/4*A*(-(-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)+1/2*C*(-1/56*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)/(-1/2+cos(1/2*d*x+1/2*c)^2)^4-5/42*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)/
(-1/2+cos(1/2*d*x+1/2*c)^2)^2+5/21*(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*(-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)/(-1/2+cos(1/2*d*x+1/2*c)^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*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)))+1/4*C*(-1/144*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)/(
-1/2+cos(1/2*d*x+1/2*c)^2)^5-7/180*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)/(-1
/2+cos(1/2*d*x+1/2*c)^2)^3-14/15*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)+7/15*(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))-7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co
s(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)))))/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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="maxima")
[Out]
Timed out
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mupad [B] time = 6.76, size = 482, normalized size = 2.10 \[ \frac {4\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )\,\left (\frac {7\,A\,a^2\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {4\,C\,a^2\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {3\,C\,a^2\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}\right )}{21\,d}-\frac {8\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {7}{4};\ {\cos \left (c+d\,x\right )}^2\right )\,\left (\frac {9\,A\,a^2\,\sin \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {16\,C\,a^2\,\sin \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {5\,C\,a^2\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}\right )}{135\,d}+\frac {2\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )\,\left (\frac {81\,A\,a^2\,\sin \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {9\,A\,a^2\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {64\,C\,a^2\,\sin \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {21\,C\,a^2\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {5\,C\,a^2\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{9/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}\right )}{45\,d}+\frac {16\,C\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {5}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{21\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^2)/cos(c + d*x)^(3/2),x)
[Out]
(4*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2)*((7*A*a^2*sin(c + d*x))/(cos(c + d*x)^(3/2)*(1 - cos(c + d*x)^2
)^(1/2)) + (4*C*a^2*sin(c + d*x))/(cos(c + d*x)^(3/2)*(1 - cos(c + d*x)^2)^(1/2)) + (3*C*a^2*sin(c + d*x))/(co
s(c + d*x)^(7/2)*(1 - cos(c + d*x)^2)^(1/2))))/(21*d) - (8*hypergeom([-1/4, 1/2], 7/4, cos(c + d*x)^2)*((9*A*a
^2*sin(c + d*x))/(cos(c + d*x)^(1/2)*(1 - cos(c + d*x)^2)^(1/2)) + (16*C*a^2*sin(c + d*x))/(cos(c + d*x)^(1/2)
*(1 - cos(c + d*x)^2)^(1/2)) + (5*C*a^2*sin(c + d*x))/(cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2))))/(135*d
) + (2*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2)*((81*A*a^2*sin(c + d*x))/(cos(c + d*x)^(1/2)*(1 - cos(c + d
*x)^2)^(1/2)) + (9*A*a^2*sin(c + d*x))/(cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2)) + (64*C*a^2*sin(c + d*x
))/(cos(c + d*x)^(1/2)*(1 - cos(c + d*x)^2)^(1/2)) + (21*C*a^2*sin(c + d*x))/(cos(c + d*x)^(5/2)*(1 - cos(c +
d*x)^2)^(1/2)) + (5*C*a^2*sin(c + d*x))/(cos(c + d*x)^(9/2)*(1 - cos(c + d*x)^2)^(1/2))))/(45*d) + (16*C*a^2*s
in(c + d*x)*hypergeom([-3/4, 1/2], 5/4, cos(c + d*x)^2))/(21*d*cos(c + d*x)^(3/2)*(1 - cos(c + d*x)^2)^(1/2))
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int \frac {A}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {2 A \sec {\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {A \sec ^{2}{\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {2 C \sec ^{3}{\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {C \sec ^{4}{\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))**2*(A+C*sec(d*x+c)**2)/cos(d*x+c)**(3/2),x)
[Out]
a**2*(Integral(A/cos(c + d*x)**(3/2), x) + Integral(2*A*sec(c + d*x)/cos(c + d*x)**(3/2), x) + Integral(A*sec(
c + d*x)**2/cos(c + d*x)**(3/2), x) + Integral(C*sec(c + d*x)**2/cos(c + d*x)**(3/2), x) + Integral(2*C*sec(c
+ d*x)**3/cos(c + d*x)**(3/2), x) + Integral(C*sec(c + d*x)**4/cos(c + d*x)**(3/2), x))
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