3.3.45 \(\int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [245]

Optimal. Leaf size=220 \[ -\frac {(A-9 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}+\frac {(A+3 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {2 (2 A-3 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(A-9 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

-1/5*(A+C)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^3+2/15*(2*A-3*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+
a*sec(d*x+c))^2+1/10*(A-9*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a^3+a^3*sec(d*x+c))-1/10*(A-9*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^3/d+1
/6*(A+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))*cos(d*x+c)^(1
/2)*sec(d*x+c)^(1/2)/a^3/d

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Rubi [A]
time = 0.34, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4170, 4104, 3872, 3856, 2719, 2720} \begin {gather*} \frac {(A-9 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{10 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {(A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac {(A-9 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {2 (2 A-3 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/10*((A - 9*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^3*d) + ((A + 3*C)*Sqrt[Co
s[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) - ((A + C)*Sec[c + d*x]^(5/2)*Sin[c + d*x]
)/(5*d*(a + a*Sec[c + d*x])^3) + (2*(2*A - 3*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^
2) + ((A - 9*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(10*d*(a^3 + a^3*Sec[c + d*x]))

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 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4104

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

Rule 4170

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(-a)*(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*
(2*m + 1))), x] + Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*C*n + A*b
*(2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x
] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (-\frac {1}{2} a (7 A-3 C)+\frac {1}{2} a (A-9 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {2 (2 A-3 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {\int \frac {\sqrt {\sec (c+d x)} \left (-a^2 (2 A-3 C)-\frac {1}{2} a^2 (A+21 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {2 (2 A-3 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(A-9 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\int \frac {\frac {3}{4} a^3 (A-9 C)-\frac {5}{4} a^3 (A+3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{15 a^6}\\ &=-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {2 (2 A-3 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(A-9 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(A-9 C) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{20 a^3}+\frac {(A+3 C) \int \sqrt {\sec (c+d x)} \, dx}{12 a^3}\\ &=-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {2 (2 A-3 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(A-9 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\left ((A-9 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}+\frac {\left ((A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3}\\ &=-\frac {(A-9 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}+\frac {(A+3 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {2 (2 A-3 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(A-9 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 7.04, size = 952, normalized size = 4.33 \begin {gather*} \frac {2 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )\right ) \sec \left (\frac {c}{2}\right ) \sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{15 d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^3}-\frac {6 \sqrt {2} C e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )\right ) \sec \left (\frac {c}{2}\right ) \sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{5 d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^3}+\frac {4 A \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sec \left (\frac {c}{2}\right ) \sec ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \sin (c)}{3 d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^3}+\frac {4 C \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sec \left (\frac {c}{2}\right ) \sec ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \sin (c)}{d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^3}+\frac {\cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (\frac {4 (A-9 C) \cos (d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )}{5 d}-\frac {8 \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (7 A \sin \left (\frac {d x}{2}\right )-3 C \sin \left (\frac {d x}{2}\right )\right )}{15 d}+\frac {4 \sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )+C \sin \left (\frac {d x}{2}\right )\right )}{5 d}+\frac {8 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )+3 C \sin \left (\frac {d x}{2}\right )\right )}{3 d}+\frac {8 (A+3 C) \tan \left (\frac {c}{2}\right )}{3 d}-\frac {8 (7 A-3 C) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{15 d}+\frac {4 (A+C) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{5 d}\right )}{(A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^
6*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/
4, -E^((2*I)*(c + d*x))])*Sec[c/2]*Sec[c + d*x]*(A + C*Sec[c + d*x]^2))/(15*d*E^(I*d*x)*(A + 2*C + A*Cos[2*c +
2*d*x])*(a + a*Sec[c + d*x])^3) - (6*Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2
*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)
*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*Sec[c + d*x]*(A + C*Sec[c + d*x]^2))/(5*
d*E^(I*d*x)*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3) + (4*A*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*
x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2)*Sin[c])/(3*d*(A + 2*
C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3) + (4*C*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]*Ellipt
icF[(c + d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2)*Sin[c])/(d*(A + 2*C + A*Cos[2*c + 2*d*x
])*(a + a*Sec[c + d*x])^3) + (Cos[c/2 + (d*x)/2]^6*Sec[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2)*((4*(A - 9*C)*Cos
[d*x]*Csc[c/2]*Sec[c/2])/(5*d) - (8*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(7*A*Sin[(d*x)/2] - 3*C*Sin[(d*x)/2]))/(15*d
) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(5*d) + (8*Sec[c/2]*Sec[c/2 + (d*x)/2]
*(A*Sin[(d*x)/2] + 3*C*Sin[(d*x)/2]))/(3*d) + (8*(A + 3*C)*Tan[c/2])/(3*d) - (8*(7*A - 3*C)*Sec[c/2 + (d*x)/2]
^2*Tan[c/2])/(15*d) + (4*(A + C)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d)))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a + a
*Sec[c + d*x])^3)

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Maple [A]
time = 3.69, size = 451, normalized size = 2.05

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 (12 A \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 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}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+6 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 C \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+30 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 )-54 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 )-2 A \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+138 C \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 A \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 )+17 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 C \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 A -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}\)	\(451\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[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)*(12*A*cos(1/2*d*x+1/2*c)^8+10*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+6*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*C*cos(1/2*d*x+1/2*c)^8+30*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))-54*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))-2*A*cos(1/2*d*x+1/2*c)^6+138*C*cos(1/
2*d*x+1/2*c)^6-24*A*cos(1/2*d*x+1/2*c)^4-24*C*cos(1/2*d*x+1/2*c)^4+17*A*cos(1/2*d*x+1/2*c)^2-3*C*cos(1/2*d*x+1
/2*c)^2-3*A-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(-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(sec(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.61, size = 474, normalized size = 2.15 \begin {gather*} -\frac {5 \, {\left (\sqrt {2} {\left (i \, A + 3 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (i \, A + 3 i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (i \, A + 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 3 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (-i \, A - 3 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-i \, A - 3 i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-i \, A - 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 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 (i \, A - 9 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (i \, A - 9 i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (i \, A - 9 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - 9 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 (-i \, A + 9 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-i \, A + 9 i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-i \, A + 9 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + 9 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 ) - \frac {2 \, {\left (3 \, {\left (A - 9 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (7 \, A - 33 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (A - 9 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(5*(sqrt(2)*(I*A + 3*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A + 3*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(I*A + 3*I
*C)*cos(d*x + c) + sqrt(2)*(I*A + 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(
2)*(-I*A - 3*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A - 3*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A - 3*I*C)*cos(d*x
+ c) + sqrt(2)*(-I*A - 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(I*A - 9
*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A - 9*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(I*A - 9*I*C)*cos(d*x + c) + sqrt(2)
*(I*A - 9*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)
*(-I*A + 9*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A + 9*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A + 9*I*C)*cos(d*x +
c) + sqrt(2)*(-I*A + 9*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)))
- 2*(3*(A - 9*C)*cos(d*x + c)^3 + 2*(7*A - 33*C)*cos(d*x + c)^2 + 5*(A - 9*C)*cos(d*x + c))*sin(d*x + c)/sqrt
(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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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(sec(d*x+c)**(3/2)*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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

[Out]

integrate((C*sec(d*x + c)^2 + A)*sec(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 \begin {gather*} \int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C/cos(c + d*x)^2)*(1/cos(c + d*x))^(3/2))/(a + a/cos(c + d*x))^3,x)

[Out]

int(((A + C/cos(c + d*x)^2)*(1/cos(c + d*x))^(3/2))/(a + a/cos(c + d*x))^3, x)

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