\(\int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) [553]

Optimal result

Integrand size = 43, antiderivative size = 270 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {4 a^3 (9 A+5 B-5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (3 A+5 (B+C)) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {4 a^3 (6 A-5 B-20 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (6 A+5 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d \sqrt {\sec (c+d x)}}-\frac {2 (9 A+5 B-5 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d} \]

[Out]

2/5*A*(a+a*sec(d*x+c))^3*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/15*(6*A+5*B)*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/a/d/se
c(d*x+c)^(1/2)-4/15*a^3*(6*A-5*B-20*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/15*(9*A+5*B-5*C)*(a^3+a^3*sec(d*x+c))*s
in(d*x+c)*sec(d*x+c)^(1/2)/d+4/5*a^3*(9*A+5*B-5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(s
in(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+4/3*a^3*(3*A+5*B+5*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 = 0.70 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4171, 4102, 4103, 4082, 3872, 3856, 2719, 2720} \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {4 a^3 (6 A-5 B-20 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}-\frac {2 (9 A+5 B-5 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac {4 a^3 (3 A+5 (B+C)) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {4 a^3 (9 A+5 B-5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{15 a d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

[In]

Int[((a + a*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]

(4*a^3*(9*A + 5*B - 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*(3*A
+ 5*(B + C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (4*a^3*(6*A - 5*B - 20*C
)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d) + (2*A*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))
+ (2*(6*A + 5*B)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(15*a*d*Sqrt[Sec[c + d*x]]) - (2*(9*A + 5*B - 5*C)*
Sqrt[Sec[c + d*x]]*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(15*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 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 4102

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

Rule 4171

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \sec (c+d x))^3 \left (\frac {1}{2} a (6 A+5 B)-\frac {1}{2} a (3 A-5 C) \sec (c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{5 a} \\ & = \frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (6 A+5 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d \sqrt {\sec (c+d x)}}+\frac {4 \int \frac {(a+a \sec (c+d x))^2 \left (\frac {1}{4} a^2 (33 A+35 B+15 C)-\frac {3}{4} a^2 (9 A+5 B-5 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{15 a} \\ & = \frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (6 A+5 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d \sqrt {\sec (c+d x)}}-\frac {2 (9 A+5 B-5 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {8 \int \frac {(a+a \sec (c+d x)) \left (\frac {3}{4} a^3 (21 A+20 B+5 C)-\frac {3}{4} a^3 (6 A-5 B-20 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{45 a} \\ & = -\frac {4 a^3 (6 A-5 B-20 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (6 A+5 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d \sqrt {\sec (c+d x)}}-\frac {2 (9 A+5 B-5 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {16 \int \frac {\frac {9}{8} a^4 (9 A+5 B-5 C)+\frac {15}{8} a^4 (3 A+5 (B+C)) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{45 a} \\ & = -\frac {4 a^3 (6 A-5 B-20 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (6 A+5 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d \sqrt {\sec (c+d x)}}-\frac {2 (9 A+5 B-5 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{5} \left (2 a^3 (9 A+5 B-5 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (2 a^3 (3 A+5 (B+C))\right ) \int \sqrt {\sec (c+d x)} \, dx \\ & = -\frac {4 a^3 (6 A-5 B-20 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (6 A+5 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d \sqrt {\sec (c+d x)}}-\frac {2 (9 A+5 B-5 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{5} \left (2 a^3 (9 A+5 B-5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (2 a^3 (3 A+5 (B+C)) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {4 a^3 (9 A+5 B-5 C) \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^3 (3 A+5 (B+C)) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {4 a^3 (6 A-5 B-20 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (6 A+5 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d \sqrt {\sec (c+d x)}}-\frac {2 (9 A+5 B-5 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 5.15 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.02 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a^3 e^{-i d x} \sec ^{\frac {3}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (216 i A+120 i B-120 i C+216 i A \cos (2 (c+d x))+120 i B \cos (2 (c+d x))-120 i C \cos (2 (c+d x))+80 (3 A+5 (B+C)) \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-8 i (9 A+5 B-5 C) \left (1+e^{2 i (c+d x)}\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+30 A \sin (c+d x)+10 B \sin (c+d x)+40 C \sin (c+d x)+6 A \sin (2 (c+d x))+60 B \sin (2 (c+d x))+180 C \sin (2 (c+d x))+30 A \sin (3 (c+d x))+10 B \sin (3 (c+d x))+3 A \sin (4 (c+d x))\right )}{60 d} \]

[In]

Integrate[((a + a*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^3*Sec[c + d*x]^(3/2)*(Cos[d*x] + I*Sin[d*x])*((216*I)*A + (120*I)*B - (120*I)*C + (216*I)*A*Cos[2*(c + d*x)
] + (120*I)*B*Cos[2*(c + d*x)] - (120*I)*C*Cos[2*(c + d*x)] + 80*(3*A + 5*(B + C))*Cos[c + d*x]^(3/2)*Elliptic
F[(c + d*x)/2, 2] - (8*I)*(9*A + 5*B - 5*C)*(1 + E^((2*I)*(c + d*x)))^(3/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -
E^((2*I)*(c + d*x))] + 30*A*Sin[c + d*x] + 10*B*Sin[c + d*x] + 40*C*Sin[c + d*x] + 6*A*Sin[2*(c + d*x)] + 60*B
*Sin[2*(c + d*x)] + 180*C*Sin[2*(c + d*x)] + 30*A*Sin[3*(c + d*x)] + 10*B*Sin[3*(c + d*x)] + 3*A*Sin[4*(c + d*
x)]))/(60*d*E^(I*d*x))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(949\) vs. \(2(295)=590\).

Time = 5.95 (sec) , antiderivative size = 950, normalized size of antiderivative = 3.52

method	result	size
default	\(\text {Expression too large to display}\)	\(950\)
parts	\(\text {Expression too large to display}\)	\(1106\)

[In]

int((a+a*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]

-4/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3/(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*(24*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-96*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^6-20*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+78*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-30*A*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2
*c)^2+54*A*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)^(1/2)
*sin(1/2*d*x+1/2*c)^2+50*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-50*B*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)^(1/2)*sin(1/2*d*x+1/2*c)^2+30*B*(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+90*C*cos(1/
2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-50*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s
in(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2-30*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*
c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2-18*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A
+15*A*(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))-27*A
*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(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))-20*B*cos(
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+25*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipti
cF(cos(1/2*d*x+1/2*c),2^(1/2))-15*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(co
s(1/2*d*x+1/2*c),2^(1/2))-50*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*C+25*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+15*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)*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)/(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 = 0.12 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.96 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (3 \, A + 5 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (3 \, A + 5 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (9 \, A + 5 \, B - 5 \, C\right )} a^{3} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (9 \, A + 5 \, B - 5 \, C\right )} a^{3} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (3 \, A a^{3} \cos \left (d x + c\right )^{3} + 5 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + 15 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 5 \, C a^{3}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )} \]

[In]

integrate((a+a*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]

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

Sympy [F]

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

[In]

integrate((a+a*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]

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

Maxima [F(-1)]

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

[In]

integrate((a+a*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]

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

[In]

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

Mupad [F(-1)]

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

[In]

int(((a + a/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 + a/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)