\(\int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\) [193]

Optimal result

Integrand size = 33, antiderivative size = 277 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=-\frac {4 a^3 (21 A+17 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^3 (13 A+11 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {4 a^3 (21 A+17 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {4 a^3 (13 A+11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {4 a^3 (24 A+23 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 a B \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {2 (9 A+13 B) \sec ^{\frac {5}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d} \]

[Out]

4/21*a^3*(13*A+11*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+4/105*a^3*(24*A+23*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/9*a*B
*sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^2*sin(d*x+c)/d+2/63*(9*A+13*B)*sec(d*x+c)^(5/2)*(a^3+a^3*sec(d*x+c))*sin(d*
x+c)/d+4/15*a^3*(21*A+17*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-4/15*a^3*(21*A+17*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/co
s(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^3*(13*A+11*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 = 0.53 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4103, 4082, 3872, 3853, 3856, 2719, 2720} \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {4 a^3 (24 A+23 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{105 d}+\frac {4 a^3 (13 A+11 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 (9 A+13 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{63 d}+\frac {4 a^3 (21 A+17 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^3 (13 A+11 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}-\frac {4 a^3 (21 A+17 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d} \]

[In]

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

[Out]

(-4*a^3*(21*A + 17*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (4*a^3*(13*A +
11*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (4*a^3*(21*A + 17*B)*Sqrt[Sec
[c + d*x]]*Sin[c + d*x])/(15*d) + (4*a^3*(13*A + 11*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (4*a^3*(24*A
+ 23*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(105*d) + (2*a*B*Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2*Sin[c + d*
x])/(9*d) + (2*(9*A + 13*B)*Sec[c + d*x]^(5/2)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(63*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 a B \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {2}{9} \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (\frac {3}{2} a (3 A+B)+\frac {1}{2} a (9 A+13 B) \sec (c+d x)\right ) \, dx \\ & = \frac {2 a B \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {2 (9 A+13 B) \sec ^{\frac {5}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d}+\frac {4}{63} \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x)) \left (\frac {15}{2} a^2 (3 A+2 B)+\frac {3}{2} a^2 (24 A+23 B) \sec (c+d x)\right ) \, dx \\ & = \frac {4 a^3 (24 A+23 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 a B \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {2 (9 A+13 B) \sec ^{\frac {5}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d}+\frac {8}{315} \int \sec ^{\frac {3}{2}}(c+d x) \left (\frac {21}{4} a^3 (21 A+17 B)+\frac {45}{4} a^3 (13 A+11 B) \sec (c+d x)\right ) \, dx \\ & = \frac {4 a^3 (24 A+23 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 a B \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {2 (9 A+13 B) \sec ^{\frac {5}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d}+\frac {1}{7} \left (2 a^3 (13 A+11 B)\right ) \int \sec ^{\frac {5}{2}}(c+d x) \, dx+\frac {1}{15} \left (2 a^3 (21 A+17 B)\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx \\ & = \frac {4 a^3 (21 A+17 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {4 a^3 (13 A+11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {4 a^3 (24 A+23 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 a B \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {2 (9 A+13 B) \sec ^{\frac {5}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d}+\frac {1}{21} \left (2 a^3 (13 A+11 B)\right ) \int \sqrt {\sec (c+d x)} \, dx-\frac {1}{15} \left (2 a^3 (21 A+17 B)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {4 a^3 (21 A+17 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {4 a^3 (13 A+11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {4 a^3 (24 A+23 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 a B \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {2 (9 A+13 B) \sec ^{\frac {5}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d}+\frac {1}{21} \left (2 a^3 (13 A+11 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{15} \left (2 a^3 (21 A+17 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = -\frac {4 a^3 (21 A+17 B) \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 {4 a^3 (13 A+11 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^3 (21 A+17 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {4 a^3 (13 A+11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {4 a^3 (24 A+23 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 a B \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {2 (9 A+13 B) \sec ^{\frac {5}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 8.72 (sec) , antiderivative size = 793, normalized size of antiderivative = 2.86 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {7 A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^4(c+d x) \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{30 \sqrt {2} d (B+A \cos (c+d x))}+\frac {17 B e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^4(c+d x) \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{90 \sqrt {2} d (B+A \cos (c+d x))}+\frac {13 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{42 d (B+A \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}+\frac {11 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{42 d (B+A \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}+\frac {\sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \left (\frac {(21 A+17 B) \cos (d x) \csc (c)}{30 d}+\frac {B \sec (c) \sec ^4(c+d x) \sin (d x)}{36 d}+\frac {\sec (c) \sec ^3(c+d x) (7 B \sin (c)+9 A \sin (d x)+27 B \sin (d x))}{252 d}+\frac {\sec (c) \sec ^2(c+d x) (45 A \sin (c)+135 B \sin (c)+189 A \sin (d x)+238 B \sin (d x))}{1260 d}+\frac {\sec (c) \sec (c+d x) (189 A \sin (c)+238 B \sin (c)+390 A \sin (d x)+330 B \sin (d x))}{1260 d}+\frac {(13 A+11 B) \tan (c)}{42 d}\right )}{(B+A \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)} \]

[In]

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

[Out]

(7*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*x]^4*Csc[c]*(-3*S
qrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c
+ d*x))])*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/(30*Sqrt[2]*d*E^(I*d*x)*(B + A*Cos
[c + d*x])) + (17*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*x]
^4*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4
, -E^((2*I)*(c + d*x))])*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/(90*Sqrt[2]*d*E^(I*
d*x)*(B + A*Cos[c + d*x])) + (13*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sec[c/2 + (d*x)/2]^6*(a + a*Se
c[c + d*x])^3*(A + B*Sec[c + d*x]))/(42*d*(B + A*Cos[c + d*x])*Sec[c + d*x]^(7/2)) + (11*B*Sqrt[Cos[c + d*x]]*
EllipticF[(c + d*x)/2, 2]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/(42*d*(B + A*Cos[c
+ d*x])*Sec[c + d*x]^(7/2)) + (Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x])*(((21*A + 17*
B)*Cos[d*x]*Csc[c])/(30*d) + (B*Sec[c]*Sec[c + d*x]^4*Sin[d*x])/(36*d) + (Sec[c]*Sec[c + d*x]^3*(7*B*Sin[c] +
9*A*Sin[d*x] + 27*B*Sin[d*x]))/(252*d) + (Sec[c]*Sec[c + d*x]^2*(45*A*Sin[c] + 135*B*Sin[c] + 189*A*Sin[d*x] +
238*B*Sin[d*x]))/(1260*d) + (Sec[c]*Sec[c + d*x]*(189*A*Sin[c] + 238*B*Sin[c] + 390*A*Sin[d*x] + 330*B*Sin[d*
x]))/(1260*d) + ((13*A + 11*B)*Tan[c])/(42*d)))/((B + A*Cos[c + d*x])*Sec[c + d*x]^(7/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1152\) vs. \(2(297)=594\).

Time = 58.79 (sec) , antiderivative size = 1153, normalized size of antiderivative = 4.16

method	result	size
default	\(\text {Expression too large to display}\)	\(1153\)
parts	\(\text {Expression too large to display}\)	\(1421\)

[In]

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

[Out]

-a^3*(-(-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*(-1/144*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*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)^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)*EllipticF(co
s(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))))+16*(1/8*A+3/8*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+16/5*(3/8*A+3/8*B)/(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*(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)*si
n(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)+16*(3/8*A+1/8*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 = 283, normalized size of antiderivative = 1.02 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=-\frac {2 \, {\left (15 i \, \sqrt {2} {\left (13 \, A + 11 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 15 i \, \sqrt {2} {\left (13 \, A + 11 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} {\left (21 \, A + 17 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} {\left (21 \, A + 17 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (42 \, {\left (21 \, A + 17 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} + 30 \, {\left (13 \, A + 11 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 7 \, {\left (27 \, A + 34 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 45 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) + 35 \, B a^{3}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{315 \, d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

-2/315*(15*I*sqrt(2)*(13*A + 11*B)*a^3*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)
) - 15*I*sqrt(2)*(13*A + 11*B)*a^3*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) +
21*I*sqrt(2)*(21*A + 17*B)*a^3*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) +
I*sin(d*x + c))) - 21*I*sqrt(2)*(21*A + 17*B)*a^3*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-
4, 0, cos(d*x + c) - I*sin(d*x + c))) - (42*(21*A + 17*B)*a^3*cos(d*x + c)^4 + 30*(13*A + 11*B)*a^3*cos(d*x +
c)^3 + 7*(27*A + 34*B)*a^3*cos(d*x + c)^2 + 45*(A + 3*B)*a^3*cos(d*x + c) + 35*B*a^3)*sin(d*x + c)/sqrt(cos(d*
x + c)))/(d*cos(d*x + c)^4)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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