3.1234 \(\int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^3} \, dx\)
Optimal. Leaf size=201 \[ -\frac {(13 A-3 B-C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac {(49 A-9 B-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(13 A-3 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {(8 A-3 B-2 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]
[Out]
1/10*(49*A-9*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^3/d-
1/6*(13*A-3*B-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^3/d-1
/5*(A-B+C)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3-1/15*(8*A-3*B-2*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d
/(a+a*cos(d*x+c))^2-1/6*(13*A-3*B-C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a^3+a^3*cos(d*x+c))
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Rubi [A] time = 0.61, antiderivative size = 201, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used =
{4112, 3041, 2977, 2748, 2641, 2639} \[ -\frac {(13 A-3 B-C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac {(49 A-9 B-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(13 A-3 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {(8 A-3 B-2 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
[In]
Int[(Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]
[Out]
((49*A - 9*B - C)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) - ((13*A - 3*B - C)*EllipticF[(c + d*x)/2, 2])/(6*a^3*
d) - ((A - B + C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((8*A - 3*B - 2*C)*Cos[c + d
*x]^(3/2)*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((13*A - 3*B - C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(
6*d*(a^3 + a^3*Cos[c + d*x]))
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 2977
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + 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] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Rule 3041
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 4112
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 {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx\\ &=-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (-\frac {5}{2} a (A-B-C)+\frac {1}{2} a (11 A-B+C) \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B-2 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\sqrt {\cos (c+d x)} \left (-\frac {3}{2} a^2 (8 A-3 B-2 C)+\frac {1}{2} a^2 (41 A-6 B+C) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B-2 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(13 A-3 B-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \frac {-\frac {5}{4} a^3 (13 A-3 B-C)+\frac {3}{4} a^3 (49 A-9 B-C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{15 a^6}\\ &=-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B-2 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(13 A-3 B-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(49 A-9 B-C) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}-\frac {(13 A-3 B-C) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3}\\ &=\frac {(49 A-9 B-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(13 A-3 B-C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B-2 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(13 A-3 B-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}
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Mathematica [C] time = 7.06, size = 2175, normalized size = 10.82 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]
[Out]
(((49*I)/5)*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]*(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) - (((9*I)/5)*B*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]*(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*Hypergeometr
ic2F1[-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/5)*C*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Se
c[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*Hyper
geometric2F1[-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) + (52*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x
- ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(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])^3) - (4*B*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]
]^2]*Sec[c/2]*Sec[c + d*x]*(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])^3) - (4*C*
Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c
+ d*x]*(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])^3) + (Cos[c/2 + (d*x)/2]^6*
(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-8*(29*A - 9*B - C + 20*A*Cos[c])*Csc[c])/(5*d) - (8*Sec[c/2]*Sec[c/
2 + (d*x)/2]*(29*A*Sin[(d*x)/2] - 9*B*Sin[(d*x)/2] - C*Sin[(d*x)/2]))/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^5
*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(5*d) + (8*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(14*A*Sin[(d*x)/
2] - 9*B*Sin[(d*x)/2] + 4*C*Sin[(d*x)/2]))/(15*d) + (8*(14*A - 9*B + 4*C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d
) - (4*(A - B + C)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d)))/(Sqrt[Cos[c + d*x]]*(A + 2*C + 2*B*Cos[c + d*x] + A*
Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3)
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{a^{3} \sec \left (d x + c\right )^{3} + 3 \, a^{3} \sec \left (d x + c\right )^{2} + 3 \, a^{3} \sec \left (d x + c\right ) + a^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")
[Out]
integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(cos(d*x + c))/(a^3*sec(d*x + c)^3 + 3*a^3*sec(d*x + c)^2
+ 3*a^3*sec(d*x + c) + a^3), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(cos(d*x + c))/(a*sec(d*x + c) + a)^3, x)
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maple [B] time = 6.09, size = 624, normalized size = 3.10 \[ \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 (348 A \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+130 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+294 A \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-108 B \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-30 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-54 B \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 C \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 C \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-6 C \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-578 A \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+198 B \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 C \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+264 A \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-114 B \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 C \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-37 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+27 B \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-17 C \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 A -3 B +3 C \right )}{60 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x)
[Out]
1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(348*A*cos(1/2*d*x+1/2*c)^8+130*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5+294
*A*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1
/2*c),2^(1/2))-108*B*cos(1/2*d*x+1/2*c)^8-30*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)*
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5-54*B*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-12*C*cos(1/2*d*x+1/2*c)^8-10*C*cos
(1/2*d*x+1/2*c)^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)*EllipticF(cos(1/2*d*x+1/2*c),
2^(1/2))-6*C*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos
(1/2*d*x+1/2*c),2^(1/2))-578*A*cos(1/2*d*x+1/2*c)^6+198*B*cos(1/2*d*x+1/2*c)^6+2*C*cos(1/2*d*x+1/2*c)^6+264*A*
cos(1/2*d*x+1/2*c)^4-114*B*cos(1/2*d*x+1/2*c)^4+24*C*cos(1/2*d*x+1/2*c)^4-37*A*cos(1/2*d*x+1/2*c)^2+27*B*cos(1
/2*d*x+1/2*c)^2-17*C*cos(1/2*d*x+1/2*c)^2+3*A-3*B+3*C)/a^3/cos(1/2*d*x+1/2*c)^5/(-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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(cos(d*x + c))/(a*sec(d*x + c) + a)^3, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((cos(c + d*x)^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^3,x)
[Out]
int((cos(c + d*x)^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^3, x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sqrt {\cos {\left (c + d x \right )}}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sqrt {\cos {\left (c + d x \right )}} \sec {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sqrt {\cos {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**3,x)
[Out]
(Integral(A*sqrt(cos(c + d*x))/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*sqr
t(cos(c + d*x))*sec(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(C*sqrt(
cos(c + d*x))*sec(c + d*x)**2/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3
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