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

Optimal. Leaf size=345 \[ \frac {2 b \left (22 a^2 B+27 a A b+7 b^2 B\right ) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{45 d}+\frac {2 \left (7 a^3 B+21 a^2 A b+15 a b^2 B+5 A b^3\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 \left (15 a^3 A+27 a^2 b B+27 a A b^2+7 b^3 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}+\frac {2 \left (7 a^3 B+21 a^2 A b+15 a b^2 B+5 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {2 \left (15 a^3 A+27 a^2 b B+27 a A b^2+7 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 b^2 (13 a B+9 A b) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 b B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2}{9 d} \]

[Out]

2/21*(21*A*a^2*b+5*A*b^3+7*B*a^3+15*B*a*b^2)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/45*b*(27*A*a*b+22*B*a^2+7*B*b^2)*
sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/63*b^2*(9*A*b+13*B*a)*sec(d*x+c)^(7/2)*sin(d*x+c)/d+2/9*b*B*sec(d*x+c)^(5/2)*(
a+b*sec(d*x+c))^2*sin(d*x+c)/d+2/15*(15*A*a^3+27*A*a*b^2+27*B*a^2*b+7*B*b^3)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/1
5*(15*A*a^3+27*A*a*b^2+27*B*a^2*b+7*B*b^3)*(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+2/21*(21*A*a^2*b+5*A*b^3+7*B*a^3+15*B*a*b^2)*(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 = 0.57, antiderivative size = 345, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {4026, 4076, 4047, 3768, 3771, 2641, 4046, 2639} \[ \frac {2 b \left (22 a^2 B+27 a A b+7 b^2 B\right ) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{45 d}+\frac {2 \left (21 a^2 A b+7 a^3 B+15 a b^2 B+5 A b^3\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 \left (15 a^3 A+27 a^2 b B+27 a A b^2+7 b^3 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}+\frac {2 \left (21 a^2 A b+7 a^3 B+15 a b^2 B+5 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {2 \left (15 a^3 A+27 a^2 b B+27 a A b^2+7 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 b^2 (13 a B+9 A b) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 b B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(15*a^3*A + 27*a*A*b^2 + 27*a^2*b*B + 7*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d
*x]])/(15*d) + (2*(21*a^2*A*b + 5*A*b^3 + 7*a^3*B + 15*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*S
qrt[Sec[c + d*x]])/(21*d) + (2*(15*a^3*A + 27*a*A*b^2 + 27*a^2*b*B + 7*b^3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])
/(15*d) + (2*(21*a^2*A*b + 5*A*b^3 + 7*a^3*B + 15*a*b^2*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (2*b*(27*
a*A*b + 22*a^2*B + 7*b^2*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(45*d) + (2*b^2*(9*A*b + 13*a*B)*Sec[c + d*x]^(7/
2)*Sin[c + d*x])/(63*d) + (2*b*B*Sec[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(9*d)

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 3768

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 3771

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 4026

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

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 4076

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

Rubi steps

\begin {align*} \int \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx &=\frac {2 b B \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {2}{9} \int \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) \left (\frac {3}{2} a (3 a A+b B)+\frac {1}{2} \left (7 b^2 B+9 a (2 A b+a B)\right ) \sec (c+d x)+\frac {1}{2} b (9 A b+13 a B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 b^2 (9 A b+13 a B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b B \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {4}{63} \int \sec ^{\frac {3}{2}}(c+d x) \left (\frac {21}{4} a^2 (3 a A+b B)+\frac {9}{4} \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sec (c+d x)+\frac {7}{4} b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 b^2 (9 A b+13 a B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b B \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {4}{63} \int \sec ^{\frac {3}{2}}(c+d x) \left (\frac {21}{4} a^2 (3 a A+b B)+\frac {7}{4} b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \sec ^2(c+d x)\right ) \, dx+\frac {1}{7} \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \int \sec ^{\frac {5}{2}}(c+d x) \, dx\\ &=\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 b^2 (9 A b+13 a B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b B \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {1}{21} \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{15} \left (15 a^3 A+27 a A b^2+27 a^2 b B+7 b^3 B\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {2 \left (15 a^3 A+27 a A b^2+27 a^2 b B+7 b^3 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 b^2 (9 A b+13 a B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b B \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {1}{15} \left (-15 a^3 A-27 a A b^2-27 a^2 b B-7 b^3 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{21} \left (\left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 \left (15 a^3 A+27 a A b^2+27 a^2 b B+7 b^3 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 b^2 (9 A b+13 a B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b B \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {1}{15} \left (\left (-15 a^3 A-27 a A b^2-27 a^2 b B-7 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 \left (15 a^3 A+27 a A b^2+27 a^2 b B+7 b^3 B\right ) \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 {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 \left (15 a^3 A+27 a A b^2+27 a^2 b B+7 b^3 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 b^2 (9 A b+13 a B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b B \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}\\ \end {align*}

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Mathematica [A]  time = 6.57, size = 452, normalized size = 1.31 \[ \frac {\cos ^4(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \left (2 \left (35 a^3 B+105 a^2 A b+75 a b^2 B+25 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\frac {2 \left (-105 a^3 A-189 a^2 b B-189 a A b^2-49 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}\right )}{105 d (a \cos (c+d x)+b)^3 (A \cos (c+d x)+B)}+\frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \left (\frac {2}{45} \sec ^2(c+d x) \left (27 a^2 b B \sin (c+d x)+27 a A b^2 \sin (c+d x)+7 b^3 B \sin (c+d x)\right )+\frac {2}{15} \left (15 a^3 A+27 a^2 b B+27 a A b^2+7 b^3 B\right ) \sin (c+d x)+\frac {2}{21} \sec (c+d x) \left (7 a^3 B \sin (c+d x)+21 a^2 A b \sin (c+d x)+15 a b^2 B \sin (c+d x)+5 A b^3 \sin (c+d x)\right )+\frac {2}{7} \sec ^3(c+d x) \left (3 a b^2 B \sin (c+d x)+A b^3 \sin (c+d x)\right )+\frac {2}{9} b^3 B \tan (c+d x) \sec ^3(c+d x)\right )}{d \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+b)^3 (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^4*((2*(-105*a^3*A - 189*a*A*b^2 - 189*a^2*b*B - 49*b^3*B)*EllipticE[(c + d*x)/2, 2])/(Sqrt[Cos[c
+ d*x]]*Sqrt[Sec[c + d*x]]) + 2*(105*a^2*A*b + 25*A*b^3 + 35*a^3*B + 75*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF
[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/(105*d*(b + a*Cos[c + d*x])^
3*(B + A*Cos[c + d*x])) + ((a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x])*((2*(15*a^3*A + 27*a*A*b^2 + 27*a^2*b*B
+ 7*b^3*B)*Sin[c + d*x])/15 + (2*Sec[c + d*x]^3*(A*b^3*Sin[c + d*x] + 3*a*b^2*B*Sin[c + d*x]))/7 + (2*Sec[c +
d*x]*(21*a^2*A*b*Sin[c + d*x] + 5*A*b^3*Sin[c + d*x] + 7*a^3*B*Sin[c + d*x] + 15*a*b^2*B*Sin[c + d*x]))/21 +
(2*Sec[c + d*x]^2*(27*a*A*b^2*Sin[c + d*x] + 27*a^2*b*B*Sin[c + d*x] + 7*b^3*B*Sin[c + d*x]))/45 + (2*b^3*B*Se
c[c + d*x]^3*Tan[c + d*x])/9))/(d*(b + a*Cos[c + d*x])^3*(B + A*Cos[c + d*x])*Sec[c + d*x]^(7/2))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B b^{3} \sec \left (d x + c\right )^{5} + A a^{3} \sec \left (d x + c\right ) + {\left (3 \, B a b^{2} + A b^{3}\right )} \sec \left (d x + c\right )^{4} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} \sec \left (d x + c\right )^{3} + {\left (B a^{3} + 3 \, A a^{2} b\right )} \sec \left (d x + c\right )^{2}\right )} \sqrt {\sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*b^3*sec(d*x + c)^5 + A*a^3*sec(d*x + c) + (3*B*a*b^2 + A*b^3)*sec(d*x + c)^4 + 3*(B*a^2*b + A*a*b^
2)*sec(d*x + c)^3 + (B*a^3 + 3*A*a^2*b)*sec(d*x + c)^2)*sqrt(sec(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 20.42, size = 1193, normalized size = 3.46 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*b^2*(A*b+3*B*a)*(-1/56*cos(1/2*d*x+1/2*c)*(-2*si
n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1/2+cos(1/2*d*x+1/2*c)^2)^4-5/42*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)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+5/21*(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)))+2*b^3*B*(-1/144*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)/(-1/2+co
s(1/2*d*x+1/2*c)^2)^5-7/180*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)/(-1/2+cos(
1/2*d*x+1/2*c)^2)^3-14/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/
2*c)^2)^(1/2)+7/15*(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))-7/15*(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
))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))))+2*a^2*(3*A*b+B*a)*(-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)/(-1/2+cos(1/2*d*x+1/2*c)^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)))+2*
A*a^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)^(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*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)-6/5*a*b*(A*b+B*a)/(8*sin(
1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(12*EllipticE(cos(1/2*
d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*cos(
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*(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))-8*sin(1/2*d*x+1/2*c)^2*co
s(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*cos(1/2*d*x+1/2*
c)^2-1)^(1/2)/d

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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