3.6.56 \(\int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx\) [556]

Optimal. Leaf size=307 \[ \frac {4 a^3 (15 A+17 B+21 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^3 (105 A+121 B+143 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{231 d}+\frac {4 a^3 (210 A+253 B+264 C) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (105 A+121 B+143 C) \sin (c+d x)}{231 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 (6 A+11 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (105 A+143 B+99 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x)} \]

[Out]

4/1155*a^3*(210*A+253*B+264*C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/11*A*(a+a*sec(d*x+c))^3*sin(d*x+c)/d/sec(d*x+c)
^(9/2)+2/99*(6*A+11*B)*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/a/d/sec(d*x+c)^(7/2)+2/693*(105*A+143*B+99*C)*(a^3+a^
3*sec(d*x+c))*sin(d*x+c)/d/sec(d*x+c)^(5/2)+4/231*a^3*(105*A+121*B+143*C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+4/15*a
^3*(15*A+17*B+21*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)/d+4/231*a^3*(105*A+121*B+143*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)/d

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Rubi [A]
time = 0.45, antiderivative size = 307, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4171, 4102, 4081, 3872, 3854, 3856, 2720, 2719} \begin {gather*} \frac {4 a^3 (210 A+253 B+264 C) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (105 A+143 B+99 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{693 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {4 a^3 (105 A+121 B+143 C) \sin (c+d x)}{231 d \sqrt {\sec (c+d x)}}+\frac {4 a^3 (105 A+121 B+143 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {4 a^3 (15 A+17 B+21 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 (6 A+11 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{99 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{11 d \sec ^{\frac {9}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(11/2),x]

[Out]

(4*a^3*(15*A + 17*B + 21*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (4*a^3*(
105*A + 121*B + 143*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(231*d) + (4*a^3*(210*
A + 253*B + 264*C)*Sin[c + d*x])/(1155*d*Sec[c + d*x]^(3/2)) + (4*a^3*(105*A + 121*B + 143*C)*Sin[c + d*x])/(2
31*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(11*d*Sec[c + d*x]^(9/2)) + (2*(6*A + 11*
B)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(99*a*d*Sec[c + d*x]^(7/2)) + (2*(105*A + 143*B + 99*C)*(a^3 + a^3
*Sec[c + d*x])*Sin[c + d*x])/(693*d*Sec[c + d*x]^(5/2))

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 4081

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

Rule 4102

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

Rule 4171

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*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*
Csc[e + f*x])^n/(f*n)), x] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])

Rubi steps

\begin {align*} \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx &=\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \sec (c+d x))^3 \left (\frac {1}{2} a (6 A+11 B)+\frac {1}{2} a (3 A+11 C) \sec (c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx}{11 a}\\ &=\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 (6 A+11 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {4 \int \frac {(a+a \sec (c+d x))^2 \left (\frac {1}{4} a^2 (105 A+143 B+99 C)+\frac {3}{4} a^2 (15 A+11 B+33 C) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx}{99 a}\\ &=\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 (6 A+11 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (105 A+143 B+99 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {8 \int \frac {(a+a \sec (c+d x)) \left (\frac {3}{4} a^3 (210 A+253 B+264 C)+\frac {15}{4} a^3 (21 A+22 B+33 C) \sec (c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{693 a}\\ &=\frac {4 a^3 (210 A+253 B+264 C) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 (6 A+11 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (105 A+143 B+99 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {16 \int \frac {-\frac {45}{8} a^4 (105 A+121 B+143 C)-\frac {231}{8} a^4 (15 A+17 B+21 C) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{3465 a}\\ &=\frac {4 a^3 (210 A+253 B+264 C) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 (6 A+11 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (105 A+143 B+99 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{15} \left (2 a^3 (15 A+17 B+21 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{77} \left (2 a^3 (105 A+121 B+143 C)\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {4 a^3 (210 A+253 B+264 C) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (105 A+121 B+143 C) \sin (c+d x)}{231 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 (6 A+11 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (105 A+143 B+99 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{231} \left (2 a^3 (105 A+121 B+143 C)\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{15} \left (2 a^3 (15 A+17 B+21 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {4 a^3 (15 A+17 B+21 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^3 (210 A+253 B+264 C) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (105 A+121 B+143 C) \sin (c+d x)}{231 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 (6 A+11 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (105 A+143 B+99 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{231} \left (2 a^3 (105 A+121 B+143 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {4 a^3 (15 A+17 B+21 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^3 (105 A+121 B+143 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{231 d}+\frac {4 a^3 (210 A+253 B+264 C) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (105 A+121 B+143 C) \sin (c+d x)}{231 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 (6 A+11 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (105 A+143 B+99 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 5.22, size = 246, normalized size = 0.80 \begin {gather*} \frac {a^3 \sqrt {\sec (c+d x)} \left (480 (105 A+121 B+143 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-2464 i (15 A+17 B+21 C) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+\cos (c+d x) (110880 i A+125664 i B+155232 i C+30 (1953 A+2134 B+2354 C) \sin (c+d x)+308 (75 A+73 B+54 C) \sin (2 (c+d x))+8505 A \sin (3 (c+d x))+5940 B \sin (3 (c+d x))+1980 C \sin (3 (c+d x))+2310 A \sin (4 (c+d x))+770 B \sin (4 (c+d x))+315 A \sin (5 (c+d x)))\right )}{27720 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(11/2),x]

[Out]

(a^3*Sqrt[Sec[c + d*x]]*(480*(105*A + 121*B + 143*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - (2464*I)*(
15*A + 17*B + 21*C)*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(
c + d*x))] + Cos[c + d*x]*((110880*I)*A + (125664*I)*B + (155232*I)*C + 30*(1953*A + 2134*B + 2354*C)*Sin[c +
d*x] + 308*(75*A + 73*B + 54*C)*Sin[2*(c + d*x)] + 8505*A*Sin[3*(c + d*x)] + 5940*B*Sin[3*(c + d*x)] + 1980*C*
Sin[3*(c + d*x)] + 2310*A*Sin[4*(c + d*x)] + 770*B*Sin[4*(c + d*x)] + 315*A*Sin[5*(c + d*x)])))/(27720*d)

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Maple [A]
time = 0.10, size = 545, normalized size = 1.78

method	result	size
default	\(-\frac {4 \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 )}\, a^{3} \left (10080 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-43680 A -6160 B \right ) \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (77280 A +24200 B +3960 C \right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-72240 A -37532 B -14256 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (39270 A +29722 B +19866 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-8820 A -8118 B -6864 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1575 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{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 )-3465 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{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 )+1815 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{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 )-3927 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{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 )+2145 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{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 )-4851 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{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 )\right )}{3465 \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}\)	\(545\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x,method=_RETURNVERBOSE)

[Out]

-4/3465*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(10080*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^12+(-43680*A-6160*B)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(77280*A+24200*B+3960*C)*sin(1/2*d*x+1/2*c)
^8*cos(1/2*d*x+1/2*c)+(-72240*A-37532*B-14256*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(39270*A+29722*B+1986
6*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-8820*A-8118*B-6864*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+1
575*A*(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))-3465
*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+1815*B*
(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))-3927*B*(si
n(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2145*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))-4851*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)*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)^(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((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),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 = 1.81, size = 272, normalized size = 0.89 \begin {gather*} -\frac {2 \, {\left (15 i \, \sqrt {2} {\left (105 \, A + 121 \, B + 143 \, C\right )} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 15 i \, \sqrt {2} {\left (105 \, A + 121 \, B + 143 \, C\right )} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 231 i \, \sqrt {2} {\left (15 \, A + 17 \, B + 21 \, C\right )} a^{3} {\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 ) + 231 i \, \sqrt {2} {\left (15 \, A + 17 \, B + 21 \, C\right )} a^{3} {\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 {{\left (315 \, A a^{3} \cos \left (d x + c\right )^{5} + 385 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{4} + 45 \, {\left (42 \, A + 33 \, B + 11 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 77 \, {\left (30 \, A + 34 \, B + 27 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 30 \, {\left (105 \, A + 121 \, B + 143 \, C\right )} a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{3465 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

-2/3465*(15*I*sqrt(2)*(105*A + 121*B + 143*C)*a^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) -
15*I*sqrt(2)*(105*A + 121*B + 143*C)*a^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 231*I*sqr
t(2)*(15*A + 17*B + 21*C)*a^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))
) + 231*I*sqrt(2)*(15*A + 17*B + 21*C)*a^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*
sin(d*x + c))) - (315*A*a^3*cos(d*x + c)^5 + 385*(3*A + B)*a^3*cos(d*x + c)^4 + 45*(42*A + 33*B + 11*C)*a^3*co
s(d*x + c)^3 + 77*(30*A + 34*B + 27*C)*a^3*cos(d*x + c)^2 + 30*(105*A + 121*B + 143*C)*a^3*cos(d*x + c))*sin(d
*x + c)/sqrt(cos(d*x + c)))/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((a+a*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(11/2),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((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^3/sec(d*x + c)^(11/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{11/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(11/2),x)

[Out]

int(((a + a/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(11/2), x)

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