3.6.65 \(\int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^2} \, dx\) [565]

Optimal. Leaf size=251 \[ \frac {(A-4 B+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {(2 A-5 B+10 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {(A-4 B+7 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac {(2 A-5 B+10 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}-\frac {(A-4 B+7 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

1/3*(2*A-5*B+10*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d-1/3*(A-4*B+7*C)*sec(d*x+c)^(5/2)*sin(d*x+c)/a^2/d/(1+sec(
d*x+c))-1/3*(A-B+C)*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^2-(A-4*B+7*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a
^2/d+(A-4*B+7*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^2/d+1/3*(2*A-5*B+10*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^2/d

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Rubi [A]
time = 0.27, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4169, 4104, 3872, 3853, 3856, 2719, 2720} \begin {gather*} -\frac {(A-4 B+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {(2 A-5 B+10 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a^2 d}-\frac {(A-4 B+7 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {(2 A-5 B+10 C) \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 {(A-4 B+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((A - 4*B + 7*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d) + ((2*A - 5*B + 10*C
)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) - ((A - 4*B + 7*C)*Sqrt[Sec[c + d
*x]]*Sin[c + d*x])/(a^2*d) + ((2*A - 5*B + 10*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a^2*d) - ((A - 4*B + 7*C)
*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A - B + C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])
/(3*d*(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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[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 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 4169

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + a*C))*Cot[e + f*x]*(a + b*C
sc[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[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(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, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=-\frac {(A-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (\frac {1}{2} a (A+5 B-5 C)+\frac {3}{2} a (A-B+3 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {(A-4 B+7 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \sec ^{\frac {3}{2}}(c+d x) \left (-\frac {3}{2} a^2 (A-4 B+7 C)+\frac {3}{2} a^2 (2 A-5 B+10 C) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {(A-4 B+7 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(A-4 B+7 C) \int \sec ^{\frac {3}{2}}(c+d x) \, dx}{2 a^2}+\frac {(2 A-5 B+10 C) \int \sec ^{\frac {5}{2}}(c+d x) \, dx}{2 a^2}\\ &=-\frac {(A-4 B+7 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac {(2 A-5 B+10 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}-\frac {(A-4 B+7 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(A-4 B+7 C) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a^2}+\frac {(2 A-5 B+10 C) \int \sqrt {\sec (c+d x)} \, dx}{6 a^2}\\ &=-\frac {(A-4 B+7 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac {(2 A-5 B+10 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}-\frac {(A-4 B+7 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\left ((A-4 B+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^2}+\frac {\left ((2 A-5 B+10 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^2}\\ &=\frac {(A-4 B+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {(2 A-5 B+10 C) \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 {(A-4 B+7 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac {(2 A-5 B+10 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}-\frac {(A-4 B+7 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (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 = 8.01, size = 1347, normalized size = 5.37 \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 ^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 ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2}+\frac {8 \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 ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2}-\frac {14 \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 ^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 ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2}+\frac {8 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 ) \sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sin (c)}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 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 ) \sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sin (c)}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2}+\frac {40 C \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 ) \sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sin (c)}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2}+\frac {\cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\frac {4 (A-4 B+7 C) \cos (d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )}{d}+\frac {4 \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 )+C \sin \left (\frac {d x}{2}\right )\right )}{3 d}+\frac {8 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (2 A \sin \left (\frac {d x}{2}\right )-5 B \sin \left (\frac {d x}{2}\right )+8 C \sin \left (\frac {d x}{2}\right )\right )}{3 d}+\frac {16 C \sec (c) \sec (c+d x) \sin (d x)}{3 d}+\frac {8 (2 C+2 A \cos (c)-5 B \cos (c)+10 C \cos (c)) \sec (c) \tan \left (\frac {c}{2}\right )}{3 d}+\frac {4 (A-B+C) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{3 d}\right )}{(A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^2,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]
^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]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*(A + 2*C + 2*B*Cos
[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) + (8*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))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + B*Sec[c + d*
x] + C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^
2) - (14*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]^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]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*(A + 2*C + 2*
B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) + (8*A*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Cs
c[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[c])/(
3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) - (20*B*Cos[c/2 + (d*x)/2]^4*Sqr
t[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c +
d*x]^2)*Sin[c])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) + (40*C*Cos[c/
2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*(A + B*Sec[c
+ d*x] + C*Sec[c + d*x]^2)*Sin[c])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])
^2) + (Cos[c/2 + (d*x)/2]^4*Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-4*(A - 4*B + 7*C)*Co
s[d*x]*Csc[c/2]*Sec[c/2])/d + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/
2]))/(3*d) + (8*Sec[c/2]*Sec[c/2 + (d*x)/2]*(2*A*Sin[(d*x)/2] - 5*B*Sin[(d*x)/2] + 8*C*Sin[(d*x)/2]))/(3*d) +
(16*C*Sec[c]*Sec[c + d*x]*Sin[d*x])/(3*d) + (8*(2*C + 2*A*Cos[c] - 5*B*Cos[c] + 10*C*Cos[c])*Sec[c]*Tan[c/2])/
(3*d) + (4*(A - B + C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3*d)))/((A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x
])*(a + a*Sec[c + d*x])^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(723\) vs. \(2(279)=558\).
time = 0.16, size = 724, normalized size = 2.88

method	result	size
default	\(\text {Expression too large to display}\)	\(724\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^2*((4*C-2*B)*(cos(1/2*d*x+1/2*c)*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x
+1/2*c),2^(1/2)))-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)/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)+4*C*(-1/6*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)/(co
s(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1
/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+1/3*(A-B+C)*(2*(sin(1/2*d*x+1/2*c)^
2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(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*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1
)^(1/2)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-1
2*sin(1/2*d*x+1/2*c)^6+20*sin(1/2*d*x+1/2*c)^4-7*sin(1/2*d*x+1/2*c)^2)/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)^2-1)+(-8*C+4*B)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c
)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^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)*EllipticE(cos(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)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,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 = 1.21, size = 464, normalized size = 1.85 \begin {gather*} \frac {{\left (\sqrt {2} {\left (-2 i \, A + 5 i \, B - 10 i \, C\right )} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} {\left (2 i \, A - 5 i \, B + 10 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-2 i \, A + 5 i \, B - 10 i \, C\right )} \cos \left (d x + 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 (2 i \, A - 5 i \, B + 10 i \, C\right )} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} {\left (-2 i \, A + 5 i \, B - 10 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (2 i \, A - 5 i \, B + 10 i \, C\right )} \cos \left (d x + 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 + 4 i \, B - 7 i \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (-i \, A + 4 i \, B - 7 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A + 4 i \, B - 7 i \, C\right )} \cos \left (d x + 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 - 4 i \, B + 7 i \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (i \, A - 4 i \, B + 7 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A - 4 i \, B + 7 i \, C\right )} \cos \left (d x + 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 - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (4 \, A - 19 \, B + 32 \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (3 \, B - 4 \, C\right )} \cos \left (d x + c\right ) - 2 \, C\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((sqrt(2)*(-2*I*A + 5*I*B - 10*I*C)*cos(d*x + c)^3 - 2*sqrt(2)*(2*I*A - 5*I*B + 10*I*C)*cos(d*x + c)^2 + s
qrt(2)*(-2*I*A + 5*I*B - 10*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sq
rt(2)*(2*I*A - 5*I*B + 10*I*C)*cos(d*x + c)^3 - 2*sqrt(2)*(-2*I*A + 5*I*B - 10*I*C)*cos(d*x + c)^2 + sqrt(2)*(
2*I*A - 5*I*B + 10*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*(sqrt(2)*(
-I*A + 4*I*B - 7*I*C)*cos(d*x + c)^3 + 2*sqrt(2)*(-I*A + 4*I*B - 7*I*C)*cos(d*x + c)^2 + sqrt(2)*(-I*A + 4*I*B
- 7*I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*
(sqrt(2)*(I*A - 4*I*B + 7*I*C)*cos(d*x + c)^3 + 2*sqrt(2)*(I*A - 4*I*B + 7*I*C)*cos(d*x + c)^2 + sqrt(2)*(I*A
- 4*I*B + 7*I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)
)) - 2*(3*(A - 4*B + 7*C)*cos(d*x + c)^3 + (4*A - 19*B + 32*C)*cos(d*x + c)^2 - 2*(3*B - 4*C)*cos(d*x + c) - 2
*C)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3063 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(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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