3.5.8 \(\int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\) [408]

Optimal. Leaf size=295 \[ -\frac {2 \left (15 a^2 A b+3 A b^3+5 a^3 B+9 a b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (21 a^3 A+21 a A b^2+21 a^2 b B+5 b^3 B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 \left (15 a^2 A b+3 A b^3+5 a^3 B+9 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b \left (21 a A b+18 a^2 B+5 b^2 B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 b^2 (7 A b+11 a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b B \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d} \]

[Out]

2/21*b*(21*A*a*b+18*B*a^2+5*B*b^2)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/35*b^2*(7*A*b+11*B*a)*sec(d*x+c)^(5/2)*sin(
d*x+c)/d+2/7*b*B*sec(d*x+c)^(3/2)*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+2/5*(15*A*a^2*b+3*A*b^3+5*B*a^3+9*B*a*b^2)*s
in(d*x+c)*sec(d*x+c)^(1/2)/d-2/5*(15*A*a^2*b+3*A*b^3+5*B*a^3+9*B*a*b^2)*(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^3+21*A*a*b^2+
21*B*a^2*b+5*B*b^3)*(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 = 0.33, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 9, 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, 2719, 4131, 2720} \begin {gather*} \frac {2 b \left (18 a^2 B+21 a A b+5 b^2 B\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 \left (5 a^3 B+15 a^2 A b+9 a b^2 B+3 A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (21 a^3 A+21 a^2 b B+21 a A b^2+5 b^3 B\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 (5 a^3 B+15 a^2 A b+9 a b^2 B+3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b^2 (11 a B+7 A b) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2}{7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(15*a^2*A*b + 3*A*b^3 + 5*a^3*B + 9*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x
]])/(5*d) + (2*(21*a^3*A + 21*a*A*b^2 + 21*a^2*b*B + 5*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqr
t[Sec[c + d*x]])/(21*d) + (2*(15*a^2*A*b + 3*A*b^3 + 5*a^3*B + 9*a*b^2*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*
d) + (2*b*(21*a*A*b + 18*a^2*B + 5*b^2*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (2*b^2*(7*A*b + 11*a*B)*Se
c[c + d*x]^(5/2)*Sin[c + d*x])/(35*d) + (2*b*B*Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(7*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 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx &=\frac {2 b B \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac {2}{7} \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (\frac {1}{2} a (7 a A+b B)+\frac {1}{2} \left (5 b^2 B+7 a (2 A b+a B)\right ) \sec (c+d x)+\frac {1}{2} b (7 A b+11 a B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 b^2 (7 A b+11 a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b B \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac {4}{35} \int \sqrt {\sec (c+d x)} \left (\frac {5}{4} a^2 (7 a A+b B)+\frac {7}{4} \left (15 a^2 A b+3 A b^3+5 a^3 B+9 a b^2 B\right ) \sec (c+d x)+\frac {5}{4} b \left (21 a A b+18 a^2 B+5 b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 b^2 (7 A b+11 a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b B \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac {4}{35} \int \sqrt {\sec (c+d x)} \left (\frac {5}{4} a^2 (7 a A+b B)+\frac {5}{4} b \left (21 a A b+18 a^2 B+5 b^2 B\right ) \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} \left (15 a^2 A b+3 A b^3+5 a^3 B+9 a b^2 B\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {2 \left (15 a^2 A b+3 A b^3+5 a^3 B+9 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b \left (21 a A b+18 a^2 B+5 b^2 B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 b^2 (7 A b+11 a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b B \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac {1}{5} \left (-15 a^2 A b-3 A b^3-5 a^3 B-9 a b^2 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{21} \left (21 a^3 A+21 a A b^2+21 a^2 b B+5 b^3 B\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 \left (15 a^2 A b+3 A b^3+5 a^3 B+9 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b \left (21 a A b+18 a^2 B+5 b^2 B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 b^2 (7 A b+11 a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b B \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac {1}{5} \left (\left (-15 a^2 A b-3 A b^3-5 a^3 B-9 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} \left (\left (21 a^3 A+21 a A b^2+21 a^2 b B+5 b^3 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 (15 a^2 A b+3 A b^3+5 a^3 B+9 a b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (21 a^3 A+21 a A b^2+21 a^2 b B+5 b^3 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^2 A b+3 A b^3+5 a^3 B+9 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b \left (21 a A b+18 a^2 B+5 b^2 B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 b^2 (7 A b+11 a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b B \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]
time = 3.88, size = 225, normalized size = 0.76 \begin {gather*} \frac {2 \sqrt {\sec (c+d x)} \left (-21 \left (15 a^2 A b+3 A b^3+5 a^3 B+9 a b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 \left (21 a^3 A+21 a A b^2+21 a^2 b B+5 b^3 B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+21 \left (15 a^2 A b+3 A b^3+5 a^3 B+9 a b^2 B\right ) \sin (c+d x)+5 b \left (21 a A b+21 a^2 B+5 b^2 B\right ) \tan (c+d x)+21 b^2 (A b+3 a B) \sec (c+d x) \tan (c+d x)+15 b^3 B \sec ^2(c+d x) \tan (c+d x)\right )}{105 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[Sec[c + d*x]]*(-21*(15*a^2*A*b + 3*A*b^3 + 5*a^3*B + 9*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)
/2, 2] + 5*(21*a^3*A + 21*a*A*b^2 + 21*a^2*b*B + 5*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 21*(1
5*a^2*A*b + 3*A*b^3 + 5*a^3*B + 9*a*b^2*B)*Sin[c + d*x] + 5*b*(21*a*A*b + 21*a^2*B + 5*b^2*B)*Tan[c + d*x] + 2
1*b^2*(A*b + 3*a*B)*Sec[c + d*x]*Tan[c + d*x] + 15*b^3*B*Sec[c + d*x]^2*Tan[c + d*x]))/(105*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(916\) vs. \(2(319)=638\).
time = 6.65, size = 917, normalized size = 3.11

method	result	size
default	\(\text {Expression too large to display}\)	\(917\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(1/2)*(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*a^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*b*a*(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)))+2/5*b^2*(A*b+3*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*b^3*(-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^2*
(3*A*b+B*a)/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)*El
lipticE(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]
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(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.97, size = 364, normalized size = 1.23 \begin {gather*} -\frac {5 \, \sqrt {2} {\left (21 i \, A a^{3} + 21 i \, B a^{2} b + 21 i \, A a b^{2} + 5 i \, B b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-21 i \, A a^{3} - 21 i \, B a^{2} b - 21 i \, A a b^{2} - 5 i \, B b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (5 i \, B a^{3} + 15 i \, A a^{2} b + 9 i \, B a b^{2} + 3 i \, A b^{3}\right )} \cos \left (d x + c\right )^{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 ) + 21 \, \sqrt {2} {\left (-5 i \, B a^{3} - 15 i \, A a^{2} b - 9 i \, B a b^{2} - 3 i \, A b^{3}\right )} \cos \left (d x + c\right )^{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 {2 \, {\left (15 \, B b^{3} + 21 \, {\left (5 \, B a^{3} + 15 \, A a^{2} b + 9 \, B a b^{2} + 3 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (21 \, B a^{2} b + 21 \, A a b^{2} + 5 \, B b^{3}\right )} \cos \left (d x + c\right )^{2} + 21 \, {\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 )}}}{105 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

-1/105*(5*sqrt(2)*(21*I*A*a^3 + 21*I*B*a^2*b + 21*I*A*a*b^2 + 5*I*B*b^3)*cos(d*x + c)^3*weierstrassPInverse(-4
, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-21*I*A*a^3 - 21*I*B*a^2*b - 21*I*A*a*b^2 - 5*I*B*b^3)*cos(d*
x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*sqrt(2)*(5*I*B*a^3 + 15*I*A*a^2*b + 9*
I*B*a*b^2 + 3*I*A*b^3)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d
*x + c))) + 21*sqrt(2)*(-5*I*B*a^3 - 15*I*A*a^2*b - 9*I*B*a*b^2 - 3*I*A*b^3)*cos(d*x + c)^3*weierstrassZeta(-4
, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(15*B*b^3 + 21*(5*B*a^3 + 15*A*a^2*b + 9*B
*a*b^2 + 3*A*b^3)*cos(d*x + c)^3 + 5*(21*B*a^2*b + 21*A*a*b^2 + 5*B*b^3)*cos(d*x + c)^2 + 21*(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)^3)

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3061 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)^(1/2)*(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*sqrt(sec(d*x + c)), x)

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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\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}} \,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))^(1/2),x)

[Out]

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

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