\(\int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\) [186]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [B] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F(-1)]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 33, antiderivative size = 234 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=-\frac {4 a^2 (4 A+3 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (7 A+6 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {4 a^2 (4 A+3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {4 a^2 (7 A+6 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 a^2 (7 A+9 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 B \sec ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{7 d} \]

[Out]

4/21*a^2*(7*A+6*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/35*a^2*(7*A+9*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/7*B*sec(d*
x+c)^(5/2)*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d+4/5*a^2*(4*A+3*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-4/5*a^2*(4*A+3*B)
*(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+4/21*a^2*(7*A+6*B)*(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] (verified)

Time = 0.38 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4103, 4082, 3872, 3853, 3856, 2719, 2720} \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {2 a^2 (7 A+9 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {4 a^2 (7 A+6 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {4 a^2 (4 A+3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (7 A+6 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}-\frac {4 a^2 (4 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{7 d} \]

[In]

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

[Out]

(-4*a^2*(4*A + 3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^2*(7*A + 6*B
)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (4*a^2*(4*A + 3*B)*Sqrt[Sec[c + d*
x]]*Sin[c + d*x])/(5*d) + (4*a^2*(7*A + 6*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (2*a^2*(7*A + 9*B)*Sec[
c + d*x]^(5/2)*Sin[c + d*x])/(35*d) + (2*B*Sec[c + d*x]^(5/2)*(a^2 + a^2*Sec[c + d*x])*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 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4082

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

Rule 4103

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 B \sec ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{7 d}+\frac {2}{7} \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x)) \left (\frac {1}{2} a (7 A+3 B)+\frac {1}{2} a (7 A+9 B) \sec (c+d x)\right ) \, dx \\ & = \frac {2 a^2 (7 A+9 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 B \sec ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{7 d}+\frac {4}{35} \int \sec ^{\frac {3}{2}}(c+d x) \left (\frac {7}{2} a^2 (4 A+3 B)+\frac {5}{2} a^2 (7 A+6 B) \sec (c+d x)\right ) \, dx \\ & = \frac {2 a^2 (7 A+9 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 B \sec ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{7 d}+\frac {1}{5} \left (2 a^2 (4 A+3 B)\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx+\frac {1}{7} \left (2 a^2 (7 A+6 B)\right ) \int \sec ^{\frac {5}{2}}(c+d x) \, dx \\ & = \frac {4 a^2 (4 A+3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {4 a^2 (7 A+6 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 a^2 (7 A+9 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 B \sec ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{7 d}-\frac {1}{5} \left (2 a^2 (4 A+3 B)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{21} \left (2 a^2 (7 A+6 B)\right ) \int \sqrt {\sec (c+d x)} \, dx \\ & = \frac {4 a^2 (4 A+3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {4 a^2 (7 A+6 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 a^2 (7 A+9 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 B \sec ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{7 d}-\frac {1}{5} \left (2 a^2 (4 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} \left (2 a^2 (7 A+6 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {4 a^2 (4 A+3 B) \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 {4 a^2 (7 A+6 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {4 a^2 (4 A+3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {4 a^2 (7 A+6 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 a^2 (7 A+9 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 B \sec ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{7 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 4.99 (sec) , antiderivative size = 463, normalized size of antiderivative = 1.98 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {a^2 e^{-i d x} \cos ^3(c+d x) \csc (c) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (7 \sqrt {2} (4 A+3 B) e^{2 i d x} \left (-1+e^{2 i c}\right ) \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )-\frac {e^{-i (c-d x)} \left (-1+e^{2 i c}\right ) \left (7 A \left (-5+9 e^{i (c+d x)}-5 e^{2 i (c+d x)}+36 e^{3 i (c+d x)}+5 e^{4 i (c+d x)}+39 e^{5 i (c+d x)}+5 e^{6 i (c+d x)}+12 e^{7 i (c+d x)}\right )+3 B \left (-10+7 e^{i (c+d x)}-20 e^{2 i (c+d x)}+63 e^{3 i (c+d x)}+20 e^{4 i (c+d x)}+77 e^{5 i (c+d x)}+10 e^{6 i (c+d x)}+21 e^{7 i (c+d x)}\right )+5 i (7 A+6 B) \left (1+e^{2 i (c+d x)}\right )^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )\right ) \sqrt {\sec (c+d x)}}{\left (1+e^{2 i (c+d x)}\right )^3}\right ) (1+\sec (c+d x))^2 (A+B \sec (c+d x))}{210 d (B+A \cos (c+d x))} \]

[In]

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

[Out]

(a^2*Cos[c + d*x]^3*Csc[c]*Sec[(c + d*x)/2]^4*(7*Sqrt[2]*(4*A + 3*B)*E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Sqrt[E^(
I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*
I)*(c + d*x))] - ((-1 + E^((2*I)*c))*(7*A*(-5 + 9*E^(I*(c + d*x)) - 5*E^((2*I)*(c + d*x)) + 36*E^((3*I)*(c + d
*x)) + 5*E^((4*I)*(c + d*x)) + 39*E^((5*I)*(c + d*x)) + 5*E^((6*I)*(c + d*x)) + 12*E^((7*I)*(c + d*x))) + 3*B*
(-10 + 7*E^(I*(c + d*x)) - 20*E^((2*I)*(c + d*x)) + 63*E^((3*I)*(c + d*x)) + 20*E^((4*I)*(c + d*x)) + 77*E^((5
*I)*(c + d*x)) + 10*E^((6*I)*(c + d*x)) + 21*E^((7*I)*(c + d*x))) + (5*I)*(7*A + 6*B)*(1 + E^((2*I)*(c + d*x))
)^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])*Sqrt[Sec[c + d*x]])/(E^(I*(c - d*x))*(1 + E^((2*I)*(c + d*x)
))^3))*(1 + Sec[c + d*x])^2*(A + B*Sec[c + d*x]))/(210*d*E^(I*d*x)*(B + A*Cos[c + d*x]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(824\) vs. \(2(258)=516\).

Time = 43.42 (sec) , antiderivative size = 825, normalized size of antiderivative = 3.53

method result size
default \(\text {Expression too large to display}\) \(825\)
parts \(\text {Expression too large to display}\) \(1026\)

[In]

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

[Out]

-a^2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*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*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)^(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))+2*B*(-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)*Elliptic
F(cos(1/2*d*x+1/2*c),2^(1/2)))+8/5*(1/4*A+1/2*B)/sin(1/2*d*x+1/2*c)^2/(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)*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(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)^4-24*sin(1/2*d*x+1/2
*c)^4*cos(1/2*d*x+1/2*c)+12*(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+8*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*(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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x
+1/2*c)^2)^(1/2)+8*(1/2*A+1/4*B)*(-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))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*
d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.12 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (7 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (7 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} {\left (4 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} {\left (4 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (42 \, {\left (4 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \, {\left (7 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 21 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right ) + 15 \, B a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{105 \, d \cos \left (d x + c\right )^{3}} \]

[In]

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

[Out]

-2/105*(5*I*sqrt(2)*(7*A + 6*B)*a^2*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) -
 5*I*sqrt(2)*(7*A + 6*B)*a^2*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*I*s
qrt(2)*(4*A + 3*B)*a^2*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d
*x + c))) - 21*I*sqrt(2)*(4*A + 3*B)*a^2*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(
d*x + c) - I*sin(d*x + c))) - (42*(4*A + 3*B)*a^2*cos(d*x + c)^3 + 10*(7*A + 6*B)*a^2*cos(d*x + c)^2 + 21*(A +
 2*B)*a^2*cos(d*x + c) + 15*B*a^2)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^3)

Sympy [F(-1)]

Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F]

\[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]

[In]

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

[Out]

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