3.13.19 \(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\) [1219]
Optimal. Leaf size=134 \[ -\frac {(3 A-3 B+C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {(5 A-3 B+3 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}+\frac {(5 A-3 B+3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}-\frac {(A-B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))} \]
[Out]
-(3*A-3*B+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/d+1/3*(5*
A-3*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/d-(A-B+C)*c
os(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))+1/3*(5*A-3*B+3*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d
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Rubi [A]
time = 0.19, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {4197, 3120,
2827, 2719, 2715, 2720} \begin {gather*} \frac {(5 A-3 B+3 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}-\frac {(3 A-3 B+C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}+\frac {(5 A-3 B+3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]),x]
[Out]
-(((3*A - 3*B + C)*EllipticE[(c + d*x)/2, 2])/(a*d)) + ((5*A - 3*B + 3*C)*EllipticF[(c + d*x)/2, 2])/(3*a*d) +
((5*A - 3*B + 3*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a*d) - ((A - B + C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(
d*(a + a*Cos[c + d*x]))
Rule 2715
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[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 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 {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx &=\int \frac {\sqrt {\cos (c+d x)} \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx\\ &=-\frac {(A-B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac {\int \sqrt {\cos (c+d x)} \left (-\frac {1}{2} a (3 A-3 B+C)+\frac {1}{2} a (5 A-3 B+3 C) \cos (c+d x)\right ) \, dx}{a^2}\\ &=-\frac {(A-B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}-\frac {(3 A-3 B+C) \int \sqrt {\cos (c+d x)} \, dx}{2 a}+\frac {(5 A-3 B+3 C) \int \cos ^{\frac {3}{2}}(c+d x) \, dx}{2 a}\\ &=-\frac {(3 A-3 B+C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {(5 A-3 B+3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}-\frac {(A-B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac {(5 A-3 B+3 C) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a}\\ &=-\frac {(3 A-3 B+C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {(5 A-3 B+3 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}+\frac {(5 A-3 B+3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}-\frac {(A-B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in
optimal.
time = 6.49, size = 2008, normalized size = 14.99 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]),x]
[Out]
(((-3*I)/2)*A*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*
E^((2*I)*d*x)*Hypergeometric2F1[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])) + (((3*I)/2)*B*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x
]^2)*((2*E^((2*I)*d*x)*Hypergeometric2F1[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*Hypergeometric
2F1[-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])) - ((I/2)*C*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c
+ d*x]^2)*((2*E^((2*I)*d*x)*Hypergeometric2F1[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*Hypergeom
etric2F1[-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])) + (Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((4*(
A - B + C + 2*A*Cos[c] - 2*B*Cos[c])*Csc[c])/d + (8*A*Cos[d*x]*Sin[c])/(3*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*
(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/d + (8*A*Cos[c]*Sin[d*x])/(3*d)))/((A + 2*C + 2*B*Cos[c +
d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) - (10*A*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*Hypergeome
tricPFQ[{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 + C
ot[c]^2]*(a + a*Sec[c + d*x])) + (2*B*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*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 + Si
n[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*C*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - Arc
Tan[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]
]]])/(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]))
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Maple [A]
time = 0.16, size = 300, normalized size = 2.24
| | |
| method |
result |
size |
| | |
| default |
\(-\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (5 A \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+9 A \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-9 B \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-8 A \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (18 A -6 B +6 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-7 A +3 B -3 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{3 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) |
\(300\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)
[Out]
-1/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(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)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*A*EllipticE(cos(1/2*d*x+1/2*c),2
^(1/2))-3*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticF(cos(
1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-8*A*sin(1/2*d*x+1/2*c)^6+(18*A-6*B+6*C)*sin
(1/2*d*x+1/2*c)^4+(-7*A+3*B-3*C)*sin(1/2*d*x+1/2*c)^2)/a/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*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="maxima")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(a*sec(d*x + c) + a), x)
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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 0.51, size = 277, normalized size = 2.07 \begin {gather*} \frac {2 \, {\left (2 \, A \cos \left (d x + c\right ) + 5 \, A - 3 \, B + 3 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (\sqrt {2} {\left (-5 i \, A + 3 i \, B - 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A + 3 i \, B - 3 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (5 i \, A - 3 i \, B + 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A - 3 i \, B + 3 i \, C\right )}\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 (3 i \, A - 3 i \, B + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A - 3 i \, B + i \, C\right )}\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 (-3 i \, A + 3 i \, B - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A + 3 i \, B - i \, C\right )}\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 d \cos \left (d x + c\right ) + a d\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="fricas")
[Out]
1/6*(2*(2*A*cos(d*x + c) + 5*A - 3*B + 3*C)*sqrt(cos(d*x + c))*sin(d*x + c) + (sqrt(2)*(-5*I*A + 3*I*B - 3*I*C
)*cos(d*x + c) + sqrt(2)*(-5*I*A + 3*I*B - 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) +
(sqrt(2)*(5*I*A - 3*I*B + 3*I*C)*cos(d*x + c) + sqrt(2)*(5*I*A - 3*I*B + 3*I*C))*weierstrassPInverse(-4, 0, c
os(d*x + c) - I*sin(d*x + c)) - 3*(sqrt(2)*(3*I*A - 3*I*B + I*C)*cos(d*x + c) + sqrt(2)*(3*I*A - 3*I*B + I*C))
*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(-3*I*A + 3*I*
B - I*C)*cos(d*x + c) + sqrt(2)*(-3*I*A + 3*I*B - I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(
d*x + c) - I*sin(d*x + c))))/(a*d*cos(d*x + c) + a*d)
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)),x)
[Out]
Exception raised: SystemError >> excessive stack use: stack is 3006 deep
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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(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(a*sec(d*x + c) + a), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((cos(c + d*x)^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x)),x)
[Out]
int((cos(c + d*x)^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x)), x)
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