3.5.7 \(\int \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\) [407]

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

[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

________________________________________________________________________________________

Rubi [A]
time = 0.37, 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 = {4111, 4161, 4132, 3853, 3856, 2720, 4131, 2719} \begin {gather*} \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} \end {gather*}

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 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 4111

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 4131

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 4132

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 4161

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*}

________________________________________________________________________________________

Mathematica [A]
time = 6.56, size = 452, normalized size = 1.31 \begin {gather*} \frac {\cos ^4(c+d x) \left (\frac {2 \left (-105 a^3 A-189 a A b^2-189 a^2 b B-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)}}+2 \left (105 a^2 A b+25 A b^3+35 a^3 B+75 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)}\right ) (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))}+\frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \left (\frac {2}{15} \left (15 a^3 A+27 a A b^2+27 a^2 b B+7 b^3 B\right ) \sin (c+d x)+\frac {2}{7} \sec ^3(c+d x) \left (A b^3 \sin (c+d x)+3 a b^2 B \sin (c+d x)\right )+\frac {2}{21} \sec (c+d x) \left (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)\right )+\frac {2}{45} \sec ^2(c+d x) \left (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)\right )+\frac {2}{9} b^3 B \sec ^3(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))^3 (B+A \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)} \end {gather*}

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))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1165\) vs. \(2(365)=730\).
time = 9.21, size = 1166, normalized size = 3.38

method	result	size
default	\(\text {Expression too large to display}\)	\(1166\)

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,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(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)/(cos(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)))+6/5*b*a*(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*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x
+1/2*c)+12*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)
*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(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)))*(-2*sin(1/2*d*x+1/2*c)^4+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*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)^2-1/2)^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)/(cos(1/2*d*x+1/2*
c)^2-1/2)^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*A*a^3/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)))+2*B*b^3*(-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)/(cos(1/2*d*x+1/2*c)^2-1/2)^5-7/180*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^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)*Ellipt
icF(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)))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.20, size = 401, normalized size = 1.16 \begin {gather*} -\frac {15 \, \sqrt {2} {\left (7 i \, B a^{3} + 21 i \, A a^{2} b + 15 i \, B a b^{2} + 5 i \, A b^{3}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 \, \sqrt {2} {\left (-7 i \, B a^{3} - 21 i \, A a^{2} b - 15 i \, B a b^{2} - 5 i \, A b^{3}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (15 i \, A a^{3} + 27 i \, B a^{2} b + 27 i \, A a b^{2} + 7 i \, B b^{3}\right )} \cos \left (d x + c\right )^{4} {\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 \, \sqrt {2} {\left (-15 i \, A a^{3} - 27 i \, B a^{2} b - 27 i \, A a b^{2} - 7 i \, B b^{3}\right )} \cos \left (d x + c\right )^{4} {\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 (21 \, {\left (15 \, A a^{3} + 27 \, B a^{2} b + 27 \, A a b^{2} + 7 \, B b^{3}\right )} \cos \left (d x + c\right )^{4} + 35 \, B b^{3} + 15 \, {\left (7 \, B a^{3} + 21 \, A a^{2} b + 15 \, B a b^{2} + 5 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} + 7 \, {\left (27 \, B a^{2} b + 27 \, A a b^{2} + 7 \, B b^{3}\right )} \cos \left (d x + c\right )^{2} + 45 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{315 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

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]

-1/315*(15*sqrt(2)*(7*I*B*a^3 + 21*I*A*a^2*b + 15*I*B*a*b^2 + 5*I*A*b^3)*cos(d*x + c)^4*weierstrassPInverse(-4
, 0, cos(d*x + c) + I*sin(d*x + c)) + 15*sqrt(2)*(-7*I*B*a^3 - 21*I*A*a^2*b - 15*I*B*a*b^2 - 5*I*A*b^3)*cos(d*
x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*sqrt(2)*(15*I*A*a^3 + 27*I*B*a^2*b + 2
7*I*A*a*b^2 + 7*I*B*b^3)*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin
(d*x + c))) + 21*sqrt(2)*(-15*I*A*a^3 - 27*I*B*a^2*b - 27*I*A*a*b^2 - 7*I*B*b^3)*cos(d*x + c)^4*weierstrassZet
a(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(21*(15*A*a^3 + 27*B*a^2*b + 27*A*a*b^
2 + 7*B*b^3)*cos(d*x + c)^4 + 35*B*b^3 + 15*(7*B*a^3 + 21*A*a^2*b + 15*B*a*b^2 + 5*A*b^3)*cos(d*x + c)^3 + 7*(
27*B*a^2*b + 27*A*a*b^2 + 7*B*b^3)*cos(d*x + c)^2 + 45*(3*B*a*b^2 + A*b^3)*cos(d*x + c))*sin(d*x + c)/sqrt(cos
(d*x + c)))/(d*cos(d*x + c)^4)

________________________________________________________________________________________

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 7316 deep

________________________________________________________________________________________

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+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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

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)

________________________________________________________________________________________