\(\int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) [1000]

Optimal result

Integrand size = 43, antiderivative size = 313 \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \left (15 a^2 b B-5 b^3 B+15 a b^2 (A-C)+a^3 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (a^3 B+9 a b^2 B+b^3 (3 A+C)+3 a^2 b (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 b \left (10 a^2 B-15 b^2 B+3 a b (7 A-15 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 (9 A b+5 a B-5 b C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 (6 A b+5 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

-2/15*b^2*(9*A*b+5*B*a-5*C*b)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*A*(a+b*sec(d*x+c))^3*sin(d*x+c)/d/sec(d*x+c)^(
3/2)+2/15*(6*A*b+5*B*a)*(a+b*sec(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(1/2)-2/15*b*(10*B*a^2-15*B*b^2+3*a*b*(7*A-
15*C))*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/5*(15*B*a^2*b-5*B*b^3+15*a*b^2*(A-C)+a^3*(3*A+5*C))*(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*(B*
a^3+9*B*a*b^2+b^3*(3*A+C)+3*a^2*b*(A+3*C))*(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 = 0.93 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4179, 4161, 4132, 3856, 2720, 4131, 2719} \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 b \sin (c+d x) \sqrt {\sec (c+d x)} \left (10 a^2 B+3 a b (7 A-15 C)-15 b^2 B\right )}{15 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^3 B+3 a^2 b (A+3 C)+9 a b^2 B+b^3 (3 A+C)\right )}{3 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 (3 A+5 C)+15 a^2 b B+15 a b^2 (A-C)-5 b^3 B\right )}{5 d}-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (5 a B+9 A b-5 b C)}{15 d}+\frac {2 (5 a B+6 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

[In]

Int[((a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(5/2),x]

[Out]

(2*(15*a^2*b*B - 5*b^3*B + 15*a*b^2*(A - C) + a^3*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sq
rt[Sec[c + d*x]])/(5*d) + (2*(a^3*B + 9*a*b^2*B + b^3*(3*A + C) + 3*a^2*b*(A + 3*C))*Sqrt[Cos[c + d*x]]*Ellipt
icF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (2*b*(10*a^2*B - 15*b^2*B + 3*a*b*(7*A - 15*C))*Sqrt[Sec[c + d
*x]]*Sin[c + d*x])/(15*d) - (2*b^2*(9*A*b + 5*a*B - 5*b*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) + (2*(6*A*b
+ 5*a*B)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + b*Sec[c + d*x])^3*Sin[c +
d*x])/(5*d*Sec[c + d*x]^(3/2))

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 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]

Rule 4179

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_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*
Csc[e + f*x])^n/(f*n)), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2}{5} \int \frac {(a+b \sec (c+d x))^2 \left (\frac {1}{2} (6 A b+5 a B)+\frac {1}{2} (3 a A+5 b B+5 a C) \sec (c+d x)-\frac {1}{2} b (3 A-5 C) \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 (6 A b+5 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4}{15} \int \frac {(a+b \sec (c+d x)) \left (\frac {1}{4} \left (24 A b^2+35 a b B+3 a^2 (3 A+5 C)\right )+\frac {1}{4} \left (5 a^2 B+15 b^2 B+6 a b (A+5 C)\right ) \sec (c+d x)-\frac {3}{4} b (9 A b+5 a B-5 b C) \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx \\ & = -\frac {2 b^2 (9 A b+5 a B-5 b C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 (6 A b+5 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {8}{45} \int \frac {\frac {3}{8} a \left (24 A b^2+35 a b B+3 a^2 (3 A+5 C)\right )+\frac {15}{8} \left (a^3 B+9 a b^2 B+b^3 (3 A+C)+3 a^2 b (A+3 C)\right ) \sec (c+d x)-\frac {3}{8} b \left (21 a A b+10 a^2 B-15 b^2 B-45 a b C\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx \\ & = -\frac {2 b^2 (9 A b+5 a B-5 b C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 (6 A b+5 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {8}{45} \int \frac {\frac {3}{8} a \left (24 A b^2+35 a b B+3 a^2 (3 A+5 C)\right )-\frac {3}{8} b \left (21 a A b+10 a^2 B-15 b^2 B-45 a b C\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (a^3 B+9 a b^2 B+b^3 (3 A+C)+3 a^2 b (A+3 C)\right ) \int \sqrt {\sec (c+d x)} \, dx \\ & = -\frac {2 b \left (10 a^2 B-15 b^2 B+3 a b (7 A-15 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 (9 A b+5 a B-5 b C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 (6 A b+5 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{5} \left (15 a^2 b B-5 b^3 B+15 a b^2 (A-C)+a^3 (3 A+5 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (\left (a^3 B+9 a b^2 B+b^3 (3 A+C)+3 a^2 b (A+3 C)\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 B+9 a b^2 B+b^3 (3 A+C)+3 a^2 b (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 b \left (10 a^2 B-15 b^2 B+3 a b (7 A-15 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 (9 A b+5 a B-5 b C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 (6 A b+5 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{5} \left (\left (15 a^2 b B-5 b^3 B+15 a b^2 (A-C)+a^3 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = \frac {2 \left (15 a^2 b B-5 b^3 B+15 a b^2 (A-C)+a^3 (3 A+5 C)\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 (a^3 B+9 a b^2 B+b^3 (3 A+C)+3 a^2 b (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 b \left (10 a^2 B-15 b^2 B+3 a b (7 A-15 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 (9 A b+5 a B-5 b C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 (6 A b+5 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.20 (sec) , antiderivative size = 295, normalized size of antiderivative = 0.94 \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (12 \left (15 a^2 b B-5 b^3 B+15 a b^2 (A-C)+a^3 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 \left (a^3 B+9 a b^2 B+b^3 (3 A+C)+3 a^2 b (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+3 a^3 A \sin (c+d x)+60 b^3 B \sin (c+d x)+180 a b^2 C \sin (c+d x)+30 a^2 A b \sin (2 (c+d x))+10 a^3 B \sin (2 (c+d x))+3 a^3 A \sin (3 (c+d x))+20 b^3 C \tan (c+d x)\right )}{15 d (b+a \cos (c+d x))^3 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) \sec ^{\frac {9}{2}}(c+d x)} \]

[In]

Integrate[((a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(5/2),x]

[Out]

((a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(12*(15*a^2*b*B - 5*b^3*B + 15*a*b^2*(A - C) +
a^3*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 20*(a^3*B + 9*a*b^2*B + b^3*(3*A + C) + 3*a^2
*b*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 3*a^3*A*Sin[c + d*x] + 60*b^3*B*Sin[c + d*x] + 18
0*a*b^2*C*Sin[c + d*x] + 30*a^2*A*b*Sin[2*(c + d*x)] + 10*a^3*B*Sin[2*(c + d*x)] + 3*a^3*A*Sin[3*(c + d*x)] +
20*b^3*C*Tan[c + d*x]))/(15*d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sec[c +
d*x]^(9/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1111\) vs. \(2(337)=674\).

Time = 6.32 (sec) , antiderivative size = 1112, normalized size of antiderivative = 3.55

method	result	size
parts	\(\text {Expression too large to display}\)	\(1112\)
default	\(\text {Expression too large to display}\)	\(1837\)

[In]

int((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/3*(3*A*a^2*b+B*a^3)*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*cos(1/2*d*x+1/2*c)*sin(1/2*d
*x+1/2*c)^4-2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^
(1/2)*EllipticF(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)/sin(1/2*d*x+
1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d-2*(B*b^3+3*C*a*b^2)*(-2*(-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)*sin(1/2*d*x+1/2*c)^2+(sin(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)*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)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d-2*(A*b^3+3*B*a*b^2+3
*C*a^2*b)*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(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+2*(3*A*a*b^2+3*B*a^2*b+C*a^3)*((2*cos(1/2*d*x+1/2*c)^2-1
)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(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)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/
2*c)^2-1)^(1/2)/d-2/5*A*a^3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-8*sin(1/2*d*x+1/2*c)^6*c
os(1/2*d*x+1/2*c)+8*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*Ellipt
icE(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)^(1/2))/(-2*sin(1/2*d*x
+1/2*c)^4+sin(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-2/3*C*b^3*(-2*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/
2*c)^2-2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)
*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(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*cos(1/2*d*x+1/2*c)^2-1)^(3/2)/sin(1/2*d*x+1/2*c)/d

Fricas [C] (verification not implemented)

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

Time = 0.14 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (i \, B a^{3} + 3 i \, {\left (A + 3 \, C\right )} a^{2} b + 9 i \, B a b^{2} + i \, {\left (3 \, A + C\right )} b^{3}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, B a^{3} - 3 i \, {\left (A + 3 \, C\right )} a^{2} b - 9 i \, B a b^{2} - i \, {\left (3 \, A + C\right )} b^{3}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (-i \, {\left (3 \, A + 5 \, C\right )} a^{3} - 15 i \, B a^{2} b - 15 i \, {\left (A - C\right )} a b^{2} + 5 i \, B b^{3}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (i \, {\left (3 \, A + 5 \, C\right )} a^{3} + 15 i \, B a^{2} b + 15 i \, {\left (A - C\right )} a b^{2} - 5 i \, B b^{3}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, A a^{3} \cos \left (d x + c\right )^{3} + 5 \, C b^{3} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )} \]

[In]

integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (a + b \sec {\left (c + d x \right )}\right )^{3} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

[In]

integrate((a+b*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(5/2),x)

[Out]

Integral((a + b*sec(c + d*x))**3*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)/sec(c + d*x)**(5/2), x)

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^3/sec(d*x + c)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int(((a + b/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(5/2),x)

[Out]

int(((a + b/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(5/2), x)