3.5.9 \(\int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [409]

Optimal. Leaf size=244 \[ \frac {2 \left (5 a^3 A-15 a A b^2-15 a^2 b B-3 b^3 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 b \left (15 a A b+14 a^2 B+3 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b^2 (5 A b+9 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d} \]

[Out]

2/15*b^2*(5*A*b+9*B*a)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*b*(15*A*a*b+14*B*a^2+3*B*b^2)*sin(d*x+c)*sec(d*x+c)^(
1/2)/d+2/5*b*B*(a+b*sec(d*x+c))^2*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/5*(5*A*a^3-15*A*a*b^2-15*B*a^2*b-3*B*b^3)*(c
os(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+2/3*(9*A*a^2*b+A*b^3+3*B*a^3+3*B*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(si
n(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 = 0.31, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4111, 4161, 4132, 3856, 2720, 4131, 2719} \begin {gather*} \frac {2 b \left (14 a^2 B+15 a A b+3 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+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 )}{3 d}+\frac {2 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 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 )}{5 d}+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*a^3*A - 15*a*A*b^2 - 15*a^2*b*B - 3*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x
]])/(5*d) + (2*(9*a^2*A*b + A*b^3 + 3*a^3*B + 3*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec
[c + d*x]])/(3*d) + (2*b*(15*a*A*b + 14*a^2*B + 3*b^2*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*b^2*(5*A*
b + 9*a*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) + (2*b*B*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^2*Sin[c +
d*x])/(5*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 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 \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx &=\frac {2 b B \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2}{5} \int \frac {(a+b \sec (c+d x)) \left (\frac {1}{2} a (5 a A-b B)+\frac {1}{2} \left (3 b^2 B+5 a (2 A b+a B)\right ) \sec (c+d x)+\frac {1}{2} b (5 A b+9 a B) \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 b^2 (5 A b+9 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {3}{4} a^2 (5 a A-b B)+\frac {5}{4} \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \sec (c+d x)+\frac {3}{4} b \left (15 a A b+14 a^2 B+3 b^2 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 b^2 (5 A b+9 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {3}{4} a^2 (5 a A-b B)+\frac {3}{4} b \left (15 a A b+14 a^2 B+3 b^2 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 b \left (15 a A b+14 a^2 B+3 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b^2 (5 A b+9 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{5} \left (5 a^3 A-15 a A b^2-15 a^2 b B-3 b^3 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (\left (9 a^2 A b+A b^3+3 a^3 B+3 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 (9 a^2 A b+A b^3+3 a^3 B+3 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)}}{3 d}+\frac {2 b \left (15 a A b+14 a^2 B+3 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b^2 (5 A b+9 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{5} \left (\left (5 a^3 A-15 a A b^2-15 a^2 b B-3 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 (5 a^3 A-15 a A b^2-15 a^2 b B-3 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)}}{5 d}+\frac {2 \left (9 a^2 A b+A b^3+3 a^3 B+3 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)}}{3 d}+\frac {2 b \left (15 a A b+14 a^2 B+3 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b^2 (5 A b+9 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]
time = 2.46, size = 190, normalized size = 0.78 \begin {gather*} \frac {\sec ^{\frac {5}{2}}(c+d x) \left (12 \left (5 a^3 A-15 a A b^2-15 a^2 b B-3 b^3 B\right ) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 b \left (15 \left (3 a A b+3 a^2 B+b^2 B\right )+10 b (A b+3 a B) \cos (c+d x)+9 \left (5 a A b+5 a^2 B+b^2 B\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right )}{30 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^(5/2)*(12*(5*a^3*A - 15*a*A*b^2 - 15*a^2*b*B - 3*b^3*B)*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2
, 2] + 20*(9*a^2*A*b + A*b^3 + 3*a^3*B + 3*a*b^2*B)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 2*b*(15*(3*
a*A*b + 3*a^2*B + b^2*B) + 10*b*(A*b + 3*a*B)*Cos[c + d*x] + 9*(5*a*A*b + 5*a^2*B + b^2*B)*Cos[2*(c + d*x)])*S
in[c + d*x]))/(30*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(969\) vs. \(2(272)=544\).
time = 5.30, size = 970, normalized size = 3.98

method	result	size
default	\(\text {Expression too large to display}\)	\(970\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),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
))-EllipticE(cos(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*A*a^2*b*(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^3*B*(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^2*(A*b+3*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*b^3/(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)+6*b*a*(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)*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]
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((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),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 = 0.93, size = 326, normalized size = 1.34 \begin {gather*} -\frac {5 \, \sqrt {2} {\left (3 i \, B a^{3} + 9 i \, A a^{2} b + 3 i \, B a b^{2} + i \, A b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-3 i \, B a^{3} - 9 i \, A a^{2} b - 3 i \, B a b^{2} - i \, A b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, A a^{3} + 15 i \, B a^{2} b + 15 i \, A a b^{2} + 3 i \, B b^{3}\right )} \cos \left (d x + c\right )^{2} {\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 \, \sqrt {2} {\left (5 i \, A a^{3} - 15 i \, B a^{2} b - 15 i \, A a b^{2} - 3 i \, B b^{3}\right )} \cos \left (d x + c\right )^{2} {\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 \, B b^{3} + 9 \, {\left (5 \, B a^{2} b + 5 \, A a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\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 )}}}{15 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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