3.3.15 \(\int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\) [215]

Optimal. Leaf size=244 \[ \frac {7 (8 A-5 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a^2 d}-\frac {5 (3 A-2 B) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {7 (8 A-5 B) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (3 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(3 A-2 B) \sin (c+d x)}{a^2 d \sec ^{\frac {3}{2}}(c+d x) (1+\sec (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \]

[Out]

7/15*(8*A-5*B)*sin(d*x+c)/a^2/d/sec(d*x+c)^(3/2)-(3*A-2*B)*sin(d*x+c)/a^2/d/sec(d*x+c)^(3/2)/(1+sec(d*x+c))-1/
3*(A-B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2-5/3*(3*A-2*B)*sin(d*x+c)/a^2/d/sec(d*x+c)^(1/2)+7/5*(
8*A-5*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))*cos(d*x+c)^(1/2
)*sec(d*x+c)^(1/2)/a^2/d-5/3*(3*A-2*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/d

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Rubi [A]
time = 0.27, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4105, 3872, 3854, 3856, 2719, 2720} \begin {gather*} -\frac {(3 A-2 B) \sin (c+d x)}{a^2 d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}+\frac {7 (8 A-5 B) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (3 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {5 (3 A-2 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac {7 (8 A-5 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

(7*(8*A - 5*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a^2*d) - (5*(3*A - 2*B)*Sqr
t[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + (7*(8*A - 5*B)*Sin[c + d*x])/(15*a^2
*d*Sec[c + d*x]^(3/2)) - (5*(3*A - 2*B)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]) - ((3*A - 2*B)*Sin[c + d*x]
)/(a^2*d*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])) - ((A - B)*Sin[c + d*x])/(3*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c
+ d*x])^2)

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 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

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 4105

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

Rubi steps

\begin {align*} \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx &=-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac {\int \frac {\frac {1}{2} a (11 A-5 B)-\frac {7}{2} a (A-B) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{3 a^2}\\ &=-\frac {(3 A-2 B) \sin (c+d x)}{a^2 d \sec ^{\frac {3}{2}}(c+d x) (1+\sec (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac {\int \frac {\frac {7}{2} a^2 (8 A-5 B)-\frac {15}{2} a^2 (3 A-2 B) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{3 a^4}\\ &=-\frac {(3 A-2 B) \sin (c+d x)}{a^2 d \sec ^{\frac {3}{2}}(c+d x) (1+\sec (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac {(7 (8 A-5 B)) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{6 a^2}-\frac {(5 (3 A-2 B)) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{2 a^2}\\ &=\frac {7 (8 A-5 B) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (3 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(3 A-2 B) \sin (c+d x)}{a^2 d \sec ^{\frac {3}{2}}(c+d x) (1+\sec (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac {(7 (8 A-5 B)) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{10 a^2}-\frac {(5 (3 A-2 B)) \int \sqrt {\sec (c+d x)} \, dx}{6 a^2}\\ &=\frac {7 (8 A-5 B) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (3 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(3 A-2 B) \sin (c+d x)}{a^2 d \sec ^{\frac {3}{2}}(c+d x) (1+\sec (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac {\left (7 (8 A-5 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{10 a^2}-\frac {\left (5 (3 A-2 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^2}\\ &=\frac {7 (8 A-5 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a^2 d}-\frac {5 (3 A-2 B) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {7 (8 A-5 B) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (3 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(3 A-2 B) \sin (c+d x)}{a^2 d \sec ^{\frac {3}{2}}(c+d x) (1+\sec (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 7.01, size = 946, normalized size = 3.88 \begin {gather*} -\frac {56 \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 ^4\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) (A+B \sec (c+d x))}{15 d (B+A \cos (c+d x)) (a+a \sec (c+d x))^2}+\frac {7 \sqrt {2} B 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 ^4\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) (A+B \sec (c+d x))}{3 d (B+A \cos (c+d x)) (a+a \sec (c+d x))^2}-\frac {10 A \cos ^4\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) (A+B \sec (c+d x)) \sin (c)}{d (B+A \cos (c+d x)) (a+a \sec (c+d x))^2}+\frac {20 B \cos ^4\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) (A+B \sec (c+d x)) \sin (c)}{3 d (B+A \cos (c+d x)) (a+a \sec (c+d x))^2}+\frac {\cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x)) \left (\frac {(-151 A+100 B-73 A \cos (2 c)+40 B \cos (2 c)) \cos (d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )}{10 d}+\frac {4 (-2 A+B) \cos (2 d x) \sin (2 c)}{3 d}+\frac {2 A \cos (3 d x) \sin (3 c)}{5 d}+\frac {2 \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-A \sin \left (\frac {d x}{2}\right )+B \sin \left (\frac {d x}{2}\right )\right )}{3 d}-\frac {4 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-13 A \sin \left (\frac {d x}{2}\right )+10 B \sin \left (\frac {d x}{2}\right )\right )}{3 d}-\frac {2 (-73 A+40 B) \cos (c) \sin (d x)}{5 d}+\frac {4 (-2 A+B) \cos (2 c) \sin (2 d x)}{3 d}+\frac {2 A \cos (3 c) \sin (3 d x)}{5 d}-\frac {4 (-13 A+10 B) \tan \left (\frac {c}{2}\right )}{3 d}+\frac {2 (-A+B) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{3 d}\right )}{(B+A \cos (c+d x)) (a+a \sec (c+d x))^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

(-56*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
]^4*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 + B*Sec[c + d*x]))/(15*d*E^(I*d*x)*(B + A*Cos[c + d*x])*(
a + a*Sec[c + d*x])^2) + (7*Sqrt[2]*B*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]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hyper
geometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x]))/(3*d*E^(I*d*x)*
(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2) - (10*A*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*Elliptic
F[(c + d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x])*Sin[c])/(d*(B + A*Cos[c + d*x])*(a + a*Sec[
c + d*x])^2) + (20*B*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sec[c
+ d*x]^(3/2)*(A + B*Sec[c + d*x])*Sin[c])/(3*d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2) + (Cos[c/2 + (d*x
)/2]^4*Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x])*(((-151*A + 100*B - 73*A*Cos[2*c] + 40*B*Cos[2*c])*Cos[d*x]*Csc
[c/2]*Sec[c/2])/(10*d) + (4*(-2*A + B)*Cos[2*d*x]*Sin[2*c])/(3*d) + (2*A*Cos[3*d*x]*Sin[3*c])/(5*d) + (2*Sec[c
/2]*Sec[c/2 + (d*x)/2]^3*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/(3*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(-13*A*S
in[(d*x)/2] + 10*B*Sin[(d*x)/2]))/(3*d) - (2*(-73*A + 40*B)*Cos[c]*Sin[d*x])/(5*d) + (4*(-2*A + B)*Cos[2*c]*Si
n[2*d*x])/(3*d) + (2*A*Cos[3*c]*Sin[3*d*x])/(5*d) - (4*(-13*A + 10*B)*Tan[c/2])/(3*d) + (2*(-A + B)*Sec[c/2 +
(d*x)/2]^2*Tan[c/2])/(3*d)))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2)

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Maple [A]
time = 1.61, size = 465, normalized size = 1.91

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 (96 A \left (\cos ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-352 A \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+80 B \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+120 A \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-150 A \left (\cos ^{3}\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 )-336 A \left (\cos ^{3}\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 )+60 B \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+100 B \left (\cos ^{3}\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 )+210 B \left (\cos ^{3}\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 )+266 A \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-240 B \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-135 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+105 B \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 A -5 B \right )}{30 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \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}\)	\(465\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/30/a^2*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(96*A*cos(1/2*d*x+1/2*c)^10-352*A*cos(1/2*d*
x+1/2*c)^8+80*B*cos(1/2*d*x+1/2*c)^8+120*A*cos(1/2*d*x+1/2*c)^6-150*A*cos(1/2*d*x+1/2*c)^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-336*A*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+60*B*cos(1
/2*d*x+1/2*c)^6+100*B*cos(1/2*d*x+1/2*c)^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)*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))+210*B*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+266*A*cos(1/2*d*x+1/2*c)^4-240*B*cos(1/2*d*x+1/2*c)^4-135*
A*cos(1/2*d*x+1/2*c)^2+105*B*cos(1/2*d*x+1/2*c)^2+5*A-5*B)/cos(1/2*d*x+1/2*c)^3/(-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((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^2*sec(d*x + c)^(5/2)), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.98, size = 395, normalized size = 1.62 \begin {gather*} -\frac {25 \, {\left (\sqrt {2} {\left (-3 i \, A + 2 i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-3 i \, A + 2 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A + 2 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 25 \, {\left (\sqrt {2} {\left (3 i \, A - 2 i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (3 i \, A - 2 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A - 2 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, {\left (\sqrt {2} {\left (-8 i \, A + 5 i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-8 i \, A + 5 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-8 i \, A + 5 i \, B\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 ) + 21 \, {\left (\sqrt {2} {\left (8 i \, A - 5 i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (8 i \, A - 5 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (8 i \, A - 5 i \, B\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 (6 \, A \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{3} - {\left (94 \, A - 65 \, B\right )} \cos \left (d x + c\right )^{2} - 25 \, {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/30*(25*(sqrt(2)*(-3*I*A + 2*I*B)*cos(d*x + c)^2 + 2*sqrt(2)*(-3*I*A + 2*I*B)*cos(d*x + c) + sqrt(2)*(-3*I*A
+ 2*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 25*(sqrt(2)*(3*I*A - 2*I*B)*cos(d*x + c
)^2 + 2*sqrt(2)*(3*I*A - 2*I*B)*cos(d*x + c) + sqrt(2)*(3*I*A - 2*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c
) - I*sin(d*x + c)) + 21*(sqrt(2)*(-8*I*A + 5*I*B)*cos(d*x + c)^2 + 2*sqrt(2)*(-8*I*A + 5*I*B)*cos(d*x + c) +
sqrt(2)*(-8*I*A + 5*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) +
21*(sqrt(2)*(8*I*A - 5*I*B)*cos(d*x + c)^2 + 2*sqrt(2)*(8*I*A - 5*I*B)*cos(d*x + c) + sqrt(2)*(8*I*A - 5*I*B))
*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(6*A*cos(d*x + c)^4 - 2
*(4*A - 5*B)*cos(d*x + c)^3 - (94*A - 65*B)*cos(d*x + c)^2 - 25*(3*A - 2*B)*cos(d*x + c))*sin(d*x + c)/sqrt(co
s(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {A}{\sec ^{\frac {9}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {7}{2}}{\left (c + d x \right )} + \sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{\frac {9}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {7}{2}}{\left (c + d x \right )} + \sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c))**2,x)

[Out]

(Integral(A/(sec(c + d*x)**(9/2) + 2*sec(c + d*x)**(7/2) + sec(c + d*x)**(5/2)), x) + Integral(B*sec(c + d*x)/
(sec(c + d*x)**(9/2) + 2*sec(c + d*x)**(7/2) + sec(c + d*x)**(5/2)), x))/a**2

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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((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^2*sec(d*x + c)^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^2*(1/cos(c + d*x))^(5/2)),x)

[Out]

int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^2*(1/cos(c + d*x))^(5/2)), x)

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