3.6.52 \(\int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [552]

Optimal. Leaf size=271 \[ \frac {4 a^3 (5 A-5 B-9 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (5 A+5 B+3 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^3 (5 A+20 B+21 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (5 A-3 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d}-\frac {2 (5 A-5 B-9 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d} \]

[Out]

2/3*A*(a+a*sec(d*x+c))^3*sin(d*x+c)/d/sec(d*x+c)^(1/2)+4/15*a^3*(5*A+20*B+21*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-
2/15*(5*A-3*C)*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d-2/15*(5*A-5*B-9*C)*(a^3+a^3*sec(d*x+c))*
sin(d*x+c)*sec(d*x+c)^(1/2)/d+4/5*a^3*(5*A-5*B-9*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+4/3*a^3*(5*A+5*B+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]
time = 0.42, antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4171, 4103, 4082, 3872, 3856, 2719, 2720} \begin {gather*} \frac {4 a^3 (5 A+20 B+21 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}-\frac {2 (5 A-5 B-9 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 (5 A+5 B+3 C) \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 {4 a^3 (5 A-5 B-9 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 (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{15 a d}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d \sqrt {\sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[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]^(3/2),x]

[Out]

(4*a^3*(5*A - 5*B - 9*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*(5*A
+ 5*B + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (4*a^3*(5*A + 20*B + 21*
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])/(3*d*Sqrt[Sec[c + d*x]]
) - (2*(5*A - 3*C)*Sqrt[Sec[c + d*x]]*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(15*a*d) - (2*(5*A - 5*B - 9*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 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*} \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 {3}{2}}(c+d x)} \, dx &=\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \int \frac {(a+a \sec (c+d x))^3 \left (\frac {3}{2} a (2 A+B)-\frac {1}{2} a (5 A-3 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{3 a}\\ &=\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (5 A-3 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d}+\frac {4 \int \frac {(a+a \sec (c+d x))^2 \left (\frac {1}{4} a^2 (35 A+15 B-3 C)-\frac {3}{4} a^2 (5 A-5 B-9 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)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (5 A-3 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d}-\frac {2 (5 A-5 B-9 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 (20 A+5 B-6 C)+\frac {3}{4} a^3 (5 A+20 B+21 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{45 a}\\ &=\frac {4 a^3 (5 A+20 B+21 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)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (5 A-3 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d}-\frac {2 (5 A-5 B-9 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 (5 A-5 B-9 C)+\frac {15}{8} a^4 (5 A+5 B+3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{45 a}\\ &=\frac {4 a^3 (5 A+20 B+21 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)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (5 A-3 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d}-\frac {2 (5 A-5 B-9 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 (5 A-5 B-9 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (2 a^3 (5 A+5 B+3 C)\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {4 a^3 (5 A+20 B+21 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)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (5 A-3 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d}-\frac {2 (5 A-5 B-9 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 (5 A-5 B-9 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 (5 A+5 B+3 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 (5 A-5 B-9 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 (5 A+5 B+3 C) \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 {4 a^3 (5 A+20 B+21 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)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (5 A-3 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 a d}-\frac {2 (5 A-5 B-9 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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 4.85, size = 316, normalized size = 1.17 \begin {gather*} \frac {a^3 e^{-i d x} \sec ^{\frac {5}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (180 i A \cos (c+d x)-180 i B \cos (c+d x)-324 i C \cos (c+d x)+60 i A \cos (3 (c+d x))-60 i B \cos (3 (c+d x))-108 i C \cos (3 (c+d x))+80 (5 A+5 B+3 C) \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-4 i (5 A-5 B-9 C) e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+30 A \sin (c+d x)+90 B \sin (c+d x)+132 C \sin (c+d x)+10 A \sin (2 (c+d x))+20 B \sin (2 (c+d x))+60 C \sin (2 (c+d x))+30 A \sin (3 (c+d x))+90 B \sin (3 (c+d x))+108 C \sin (3 (c+d x))+5 A \sin (4 (c+d x))\right )}{60 d} \end {gather*}

Antiderivative was successfully verified.

[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]^(3/2),x]

[Out]

(a^3*Sec[c + d*x]^(5/2)*(Cos[d*x] + I*Sin[d*x])*((180*I)*A*Cos[c + d*x] - (180*I)*B*Cos[c + d*x] - (324*I)*C*C
os[c + d*x] + (60*I)*A*Cos[3*(c + d*x)] - (60*I)*B*Cos[3*(c + d*x)] - (108*I)*C*Cos[3*(c + d*x)] + 80*(5*A + 5
*B + 3*C)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] - ((4*I)*(5*A - 5*B - 9*C)*(1 + E^((2*I)*(c + d*x)))^(5
/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(I*(c + d*x)) + 30*A*Sin[c + d*x] + 90*B*Sin[c +
d*x] + 132*C*Sin[c + d*x] + 10*A*Sin[2*(c + d*x)] + 20*B*Sin[2*(c + d*x)] + 60*C*Sin[2*(c + d*x)] + 30*A*Sin[
3*(c + d*x)] + 90*B*Sin[3*(c + d*x)] + 108*C*Sin[3*(c + d*x)] + 5*A*Sin[4*(c + d*x)]))/(60*d*E^(I*d*x))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1327\) vs. \(2(295)=590\).
time = 0.17, size = 1328, normalized size = 4.90

method	result	size
default	\(\text {Expression too large to display}\)	\(1328\)

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

4/15*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)^6-12*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)^3*(-100*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(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+60*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(
sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-100*B*(2*sin(1/2*d*x+1/
2*c)^2-1)^(1/2)*(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)^2-60*C*(
2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(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)^2-108*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(
1/2))*sin(1/2*d*x+1/2*c)^2+40*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+90*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^4-20*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A-72*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*C+100*A*(2*sin(
1/2*d*x+1/2*c)^2-1)^(1/2)*(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
)^4-60*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*s
in(1/2*d*x+1/2*c)^4+100*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(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)^4+60*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipt
icE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4+60*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(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)^4+108*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-60*B*(2*sin(1/2*d*x+1/2*
c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-120*A*co
s(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+25*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)*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))-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)*EllipticE(
cos(1/2*d*x+1/2*c),2^(1/2))+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)*EllipticF(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(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)*EllipticF(cos(1/2*d*x+1/2
*c),2^(1/2))+27*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))-180*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-216*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+190*B*cos(
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+246*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-50*B*cos(1/2*d*x+1/2*c)*sin(
1/2*d*x+1/2*c)^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

________________________________________________________________________________________

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

[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)^(3/2), x)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.04, size = 272, normalized size = 1.00 \begin {gather*} -\frac {2 \, {\left (5 i \, \sqrt {2} {\left (5 \, A + 5 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (5 \, A + 5 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, A - 5 \, B - 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\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 (5 \, A - 5 \, B - 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\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 (5 \, A a^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 15 \, B + 18 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 5 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 3 \, 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 )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)^(3/2),x, algorithm="fricas")

[Out]

-2/15*(5*I*sqrt(2)*(5*A + 5*B + 3*C)*a^3*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x +
c)) - 5*I*sqrt(2)*(5*A + 5*B + 3*C)*a^3*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c
)) - 3*I*sqrt(2)*(5*A - 5*B - 9*C)*a^3*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*
x + c) + I*sin(d*x + c))) + 3*I*sqrt(2)*(5*A - 5*B - 9*C)*a^3*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstras
sPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (5*A*a^3*cos(d*x + c)^3 + 3*(5*A + 15*B + 18*C)*a^3*cos(d*x
+ c)^2 + 5*(B + 3*C)*a^3*cos(d*x + c) + 3*C*a^3)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3061 deep

________________________________________________________________________________________

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))^(3/2),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))^(3/2), x)

________________________________________________________________________________________