3.509 \(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=221 \[ \frac {(33 A-13 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac {7 (17 A-7 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {7 (17 A-7 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{30 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {(33 A-13 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 a^3 d}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {(2 A-B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 a d (a \cos (c+d x)+a)^2} \]

[Out]

-7/10*(17*A-7*B)*(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*(33*A-13*B)*(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)*cos(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3-1/3*(2*A-B)*cos(d*x+c)^(5/2)*sin(d*x+c)/a/d/(a+a*cos(d
*x+c))^2-7/30*(17*A-7*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))+1/6*(33*A-13*B)*sin(d*x+c)*cos(d*x
+c)^(1/2)/a^3/d

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Rubi [A]  time = 0.58, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2954, 2977, 2748, 2639, 2635, 2641} \[ \frac {(33 A-13 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac {7 (17 A-7 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {7 (17 A-7 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{30 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {(33 A-13 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 a^3 d}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {(2 A-B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 a d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

(-7*(17*A - 7*B)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + ((33*A - 13*B)*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) +
((33*A - 13*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(6*a^3*d) - ((A - B)*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(5*d*(a
+ a*Cos[c + d*x])^3) - ((2*A - B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(3*a*d*(a + a*Cos[c + d*x])^2) - (7*(17*A
- 7*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(30*d*(a^3 + a^3*Cos[c + d*x]))

Rule 2635

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] && Integer
Q[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 2954

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

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])

Rubi steps

\begin {align*} \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx &=\int \frac {\cos ^{\frac {7}{2}}(c+d x) (B+A \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx\\ &=-\frac {(A-B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (-\frac {7}{2} a (A-B)+\frac {1}{2} a (13 A-3 B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(2 A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (-\frac {25}{2} a^2 (2 A-B)+\frac {3}{2} a^2 (23 A-8 B) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A-B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(2 A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \cos (c+d x))^2}-\frac {7 (17 A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \sqrt {\cos (c+d x)} \left (-\frac {21}{4} a^3 (17 A-7 B)+\frac {15}{4} a^3 (33 A-13 B) \cos (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {(A-B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(2 A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \cos (c+d x))^2}-\frac {7 (17 A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(33 A-13 B) \int \cos ^{\frac {3}{2}}(c+d x) \, dx}{4 a^3}-\frac {(7 (17 A-7 B)) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}\\ &=-\frac {7 (17 A-7 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(33 A-13 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 a^3 d}-\frac {(A-B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(2 A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \cos (c+d x))^2}-\frac {7 (17 A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(33 A-13 B) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3}\\ &=-\frac {7 (17 A-7 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(33 A-13 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac {(33 A-13 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 a^3 d}-\frac {(A-B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(2 A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \cos (c+d x))^2}-\frac {7 (17 A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [C]  time = 7.08, size = 1448, normalized size = 6.55 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

(((-119*I)/10)*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*((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])))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) + (((49*I)/10)*B*Cos[c/2 + (d*x)/
2]^6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*((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])))/((
B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) - (22*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]^2*(A + B*Sec[c + d*x])*Sec[d*x - ArcTan[Cot[c]]]*S
qrt[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*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^3) + (26*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]^2*(A
+ B*Sec[c + d*x])*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*(B + A*Cos[c + 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])*((-4*(-59*A + 29*B - 60*A*Cos[c] + 20
*B*Cos[c])*Csc[c])/(5*d) + (16*A*Cos[d*x]*Sin[c])/(3*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(-(A*Sin[(d*x)/2])
+ B*Sin[(d*x)/2]))/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(-19*A*Sin[(d*x)/2] + 14*B*Sin[(d*x)/2]))/(15*d) -
(4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(-59*A*Sin[(d*x)/2] + 29*B*Sin[(d*x)/2]))/(5*d) + (16*A*Cos[c]*Sin[d*x])/(3*d)
+ (4*(-19*A + 14*B)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) - (2*(-A + B)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d))
)/(Cos[c + d*x]^(3/2)*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3)

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \cos \left (d x + c\right ) \sec \left (d x + c\right ) + A \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}}{a^{3} \sec \left (d x + c\right )^{3} + 3 \, a^{3} \sec \left (d x + c\right )^{2} + 3 \, a^{3} \sec \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

integral((B*cos(d*x + c)*sec(d*x + c) + A*cos(d*x + c))*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 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(a*sec(d*x + c) + a)^3, x)

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maple [A]  time = 5.70, size = 465, normalized size = 2.10 \[ -\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 (160 A \left (\cos ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+468 A \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+330 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 )+714 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 )-348 B \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-130 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 )-294 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 )-1058 A \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+578 B \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+474 A \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-264 B \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-47 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+37 B \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 A -3 B \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)^(3/2)*(A+B*sec(d*x+c))/(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)*(160*A*cos(1/2*d*x+1/2*c)^10+468*A*cos(1/2*d*x+1
/2*c)^8+330*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+714*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))-348*B*cos(1/2*d*x+1/2*c)^8-130*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)^5-294*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))-105
8*A*cos(1/2*d*x+1/2*c)^6+578*B*cos(1/2*d*x+1/2*c)^6+474*A*cos(1/2*d*x+1/2*c)^4-264*B*cos(1/2*d*x+1/2*c)^4-47*A
*cos(1/2*d*x+1/2*c)^2+37*B*cos(1/2*d*x+1/2*c)^2+3*A-3*B)/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 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(a*sec(d*x + c) + a)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^(3/2)*(A + B/cos(c + d*x)))/(a + a/cos(c + d*x))^3,x)

[Out]

int((cos(c + d*x)^(3/2)*(A + B/cos(c + d*x)))/(a + a/cos(c + d*x))^3, x)

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sympy [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(cos(d*x+c)**(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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