3.5.10 \(\int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [410]

Optimal. Leaf size=239 \[ \frac {2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 \left (a^3 A+9 a A b^2+9 a^2 b B+b^3 B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 b \left (2 a^2 A-3 A b^2-9 a b B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 (a A-b B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a A (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]

[Out]

-2/3*b^2*(A*a-B*b)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/3*a*A*(a+b*sec(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(1/2)-2/3*
b*(2*A*a^2-3*A*b^2-9*B*a*b)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2*(3*A*a^2*b-A*b^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)*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*
(A*a^3+9*A*a*b^2+9*B*a^2*b+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 = 239, 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 = {4110, 4161, 4132, 3856, 2720, 4131, 2719} \begin {gather*} -\frac {2 b \left (2 a^2 A-9 a b B-3 A b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 \left (a^3 A+9 a^2 b B+9 a A b^2+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 )}{3 d}+\frac {2 \left (a^3 B+3 a^2 A b-3 a b^2 B-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 )}{d}-\frac {2 b^2 (a A-b B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4110

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[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[a*(a*B*n - A*b*(m - n - 1)) + (
2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; Free
Q[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LeQ[n, -1]

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))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx &=\frac {2 a A (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2}{3} \int \frac {(a+b \sec (c+d x)) \left (-\frac {1}{2} a (7 A b+3 a B)-\frac {1}{2} \left (a^2 A+3 A b^2+6 a b B\right ) \sec (c+d x)+\frac {3}{2} b (a A-b B) \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\\ &=-\frac {2 b^2 (a A-b B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a A (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {4}{9} \int \frac {-\frac {3}{4} a^2 (7 A b+3 a B)-\frac {3}{4} \left (a^3 A+9 a A b^2+9 a^2 b B+b^3 B\right ) \sec (c+d x)+\frac {3}{4} b \left (2 a^2 A-3 A b^2-9 a b B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx\\ &=-\frac {2 b^2 (a A-b B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a A (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {4}{9} \int \frac {-\frac {3}{4} a^2 (7 A b+3 a B)+\frac {3}{4} b \left (2 a^2 A-3 A b^2-9 a b B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx-\frac {1}{3} \left (-a^3 A-9 a A b^2-9 a^2 b B-b^3 B\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=-\frac {2 b \left (2 a^2 A-3 A b^2-9 a b B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 (a A-b B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a A (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\left (-3 a^2 A b+A b^3-a^3 B+3 a b^2 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx-\frac {1}{3} \left (\left (-a^3 A-9 a A b^2-9 a^2 b B-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 (a^3 A+9 a A b^2+9 a^2 b B+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)}}{3 d}-\frac {2 b \left (2 a^2 A-3 A b^2-9 a b B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 (a A-b B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a A (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\left (\left (-3 a^2 A b+A b^3-a^3 B+3 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {2 \left (3 a^2 A b-A b^3+a^3 B-3 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)}}{d}+\frac {2 \left (a^3 A+9 a A b^2+9 a^2 b B+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)}}{3 d}-\frac {2 b \left (2 a^2 A-3 A b^2-9 a b B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 (a A-b B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a A (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.98, size = 166, normalized size = 0.69 \begin {gather*} \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (6 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \left (a^3 A+9 a A b^2+9 a^2 b B+b^3 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\frac {\left (a^3 A+2 b^3 B+6 b^2 (A b+3 a B) \cos (c+d x)+a^3 A \cos (2 (c+d x))\right ) \sin (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}\right )}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(6*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*EllipticE[(c + d*x)/2, 2] +
2*(a^3*A + 9*a*A*b^2 + 9*a^2*b*B + b^3*B)*EllipticF[(c + d*x)/2, 2] + ((a^3*A + 2*b^3*B + 6*b^2*(A*b + 3*a*B)*
Cos[c + d*x] + a^3*A*Cos[2*(c + d*x)])*Sin[c + d*x])/Cos[c + d*x]^(3/2)))/(3*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1211\) vs. \(2(269)=538\).
time = 3.90, size = 1212, normalized size = 5.07

method	result	size
default	\(\text {Expression too large to display}\)	\(1212\)

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)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1
)/sin(1/2*d*x+1/2*c)^3*(8*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*a^3-8*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^4*a^3-12*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*b^3+2*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(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*a^3+18*A*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*EllipticF(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*a*b^2-18*A*(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)^2-1)^(1/2)*sin(1/2*d*x+1/2
*c)^2*a^2*b+6*A*(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)^2-1)^
(1/2)*sin(1/2*d*x+1/2*c)^2*b^3-36*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a*b^2+18*B*(sin(1/2*d*x+1/2*c)^2)^
(1/2)*EllipticF(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*a^2*b+2*B*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(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*b^3-6*B*(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)^2-1)
^(1/2)*sin(1/2*d*x+1/2*c)^2*a^3+18*B*(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)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2*a*b^2+2*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^3+6*A*cos(
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^3-A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(
2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a^3-9*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*
sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a*b^2+9*A*(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)^2-1)^(1/2)*a^2*b-3*A*(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)^2-1)^(1/2)*b^3+18*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a*b^2+2*B*cos(1/2*d*x+1/2*c)*si
n(1/2*d*x+1/2*c)^2*b^3-9*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1
/2*c)^2-1)^(1/2)*a^2*b-B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2
*c)^2-1)^(1/2)*b^3+3*B*(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
)^2-1)^(1/2)*a^3-9*B*(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)^
2-1)^(1/2)*a*b^2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(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)^(3/2),x, algorithm="maxima")

[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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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.78, size = 298, normalized size = 1.25 \begin {gather*} \frac {\sqrt {2} {\left (-i \, A a^{3} - 9 i \, B a^{2} b - 9 i \, A a b^{2} - i \, B b^{3}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, A a^{3} + 9 i \, B a^{2} b + 9 i \, A a b^{2} + i \, B b^{3}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (-i \, B a^{3} - 3 i \, A a^{2} b + 3 i \, B a b^{2} + i \, A b^{3}\right )} \cos \left (d x + c\right ) {\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 (i \, B a^{3} + 3 i \, A a^{2} b - 3 i \, B a b^{2} - i \, A b^{3}\right )} \cos \left (d x + c\right ) {\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 (A a^{3} \cos \left (d x + c\right )^{2} + B b^{3} + 3 \, {\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 )}}}{3 \, d \cos \left (d x + c\right )} \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)^(3/2),x, algorithm="fricas")

[Out]

1/3*(sqrt(2)*(-I*A*a^3 - 9*I*B*a^2*b - 9*I*A*a*b^2 - I*B*b^3)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x
+ c) + I*sin(d*x + c)) + sqrt(2)*(I*A*a^3 + 9*I*B*a^2*b + 9*I*A*a*b^2 + I*B*b^3)*cos(d*x + c)*weierstrassPInve
rse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sqrt(2)*(-I*B*a^3 - 3*I*A*a^2*b + 3*I*B*a*b^2 + I*A*b^3)*cos(d*x
+ c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*sqrt(2)*(I*B*a^3 +
3*I*A*a^2*b - 3*I*B*a*b^2 - I*A*b^3)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x +
c) - I*sin(d*x + c))) + 2*(A*a^3*cos(d*x + c)^2 + B*b^3 + 3*(3*B*a*b^2 + A*b^3)*cos(d*x + c))*sin(d*x + c)/sq
rt(cos(d*x + c)))/(d*cos(d*x + c))

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

[Out]

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

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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)^(3/2),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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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}{{\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)

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