3.1000 \(\int \frac{(a+b \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=313 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (3 a^2 b (A+3 C)+a^3 B+9 a b^2 B+b^3 (3 A+C)\right )}{3 d}-\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)} \left (10 a^2 B+3 a b (7 A-15 C)-15 b^2 B\right )}{15 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^3 (3 A+5 C)+15 a^2 b B+15 a b^2 (A-C)-5 b^3 B\right )}{5 d}-\frac{2 b^2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (5 a B+9 A b-5 b C)}{15 d}+\frac{2 (5 a B+6 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^3}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]
[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))
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Rubi [A] time = 0.833526, antiderivative size = 313, normalized size of antiderivative = 1.,
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 =
{4094, 4076, 4047, 3771, 2641, 4046, 2639} \[ -\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)} \left (10 a^2 B+3 a b (7 A-15 C)-15 b^2 B\right )}{15 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b (A+3 C)+a^3 B+9 a b^2 B+b^3 (3 A+C)\right )}{3 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^3 (3 A+5 C)+15 a^2 b B+15 a b^2 (A-C)-5 b^3 B\right )}{5 d}-\frac{2 b^2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (5 a B+9 A b-5 b C)}{15 d}+\frac{2 (5 a B+6 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^3}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[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 4094
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]
Rule 4076
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 4047
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 3771
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 2641
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]
Rule 4046
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 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rubi steps
\begin{align*} \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 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)} F\left (\left .\frac{1}{2} (c+d x)\right |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)} F\left (\left .\frac{1}{2} (c+d x)\right |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] time = 2.69708, size = 295, normalized size = 0.94 \[ \frac{(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (20 \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (3 a^2 b (A+3 C)+a^3 B+9 a b^2 B+b^3 (3 A+C)\right )+12 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^3 (3 A+5 C)+15 a^2 b B+15 a b^2 (A-C)-5 b^3 B\right )+30 a^2 A b \sin (2 (c+d x))+3 a^3 A \sin (c+d x)+3 a^3 A \sin (3 (c+d x))+10 a^3 B \sin (2 (c+d x))+180 a b^2 C \sin (c+d x)+60 b^3 B \sin (c+d x)+20 b^3 C \tan (c+d x)\right )}{15 d \sec ^{\frac{9}{2}}(c+d x) (a \cos (c+d x)+b)^3 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]
Antiderivative was successfully verified.
[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))
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Maple [B] time = 7.42, size = 1837, normalized size = 5.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
[Out]
2/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(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*(-45*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*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)*a^2*b+30*A*a^2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+90*C*a*b^2*cos(1/2*d*x+1/2*c)*si
n(1/2*d*x+1/2*c)^2+120*A*a^2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-120*A*a^2*b*cos(1/2*d*x+1/2*c)*sin(1/2*
d*x+1/2*c)^4-180*C*a*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+9*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(co
s(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a^3-5*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*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)*a^3-15*B*(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)*b^3-15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*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)*b^3-5*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*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)*b^3+15*C*(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)*a^3-36*A*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-40*B*
a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-60*B*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6*A*a^3*cos(1/2*d
*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+10*B*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+30*B*b^3*cos(1/2*d*x+1/2*c)*si
n(1/2*d*x+1/2*c)^2+10*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-48*A*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^8+72*A*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+40*B*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+30*A*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*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)*a^2*b*sin(
1/2*d*x+1/2*c)^2-90*A*(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)*a*b^2*sin(1/2*d*x+1/2*c)^2+90*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*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)*a*b^2*sin(1/2*d*x+1/2*c)^2-90*B*(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)*a^2*b*sin(1/2*d*x+1/2*c)^2+90*C*(sin(1/2*d*x+1/2*c)^2)
^(1/2)*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)*a^2*b*sin(1/2*d*x+1/2*c)^2+90*C*
(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)*a*b^2*sin(
1/2*d*x+1/2*c)^2-45*C*(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)*a*b^2+30*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*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)*b^3*sin(1/2*d*x+1/2*c)^2-18*A*(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)*a^3*sin(1/2*d*x+1/2*c)^2+10*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*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)*a^3*sin(1/2*d*x+1/2*c)^2+30*B*(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)*b^3*sin(1/2*d*x+1/2*c)^2+10*C*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*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)*b^3*sin(1/2*d*x+1
/2*c)^2-30*C*(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)*a^3*sin(1/2*d*x+1/2*c)^2-15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*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)*a^2*b+45*A*(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)*a*b^2-45*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*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)*a*b^2+45*B*(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)*a^2*b)*(-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
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C b^{3} \sec \left (d x + c\right )^{5} +{\left (3 \, C a b^{2} + B b^{3}\right )} \sec \left (d x + c\right )^{4} + A a^{3} +{\left (3 \, C a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} \sec \left (d x + c\right )^{3} +{\left (C a^{3} + 3 \, B a^{2} b + 3 \, A a b^{2}\right )} \sec \left (d x + c\right )^{2} +{\left (B a^{3} + 3 \, A a^{2} b\right )} \sec \left (d x + c\right )}{\sec \left (d x + c\right )^{\frac{5}{2}}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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]
integral((C*b^3*sec(d*x + c)^5 + (3*C*a*b^2 + B*b^3)*sec(d*x + c)^4 + A*a^3 + (3*C*a^2*b + 3*B*a*b^2 + A*b^3)*
sec(d*x + c)^3 + (C*a^3 + 3*B*a^2*b + 3*A*a*b^2)*sec(d*x + c)^2 + (B*a^3 + 3*A*a^2*b)*sec(d*x + c))/sec(d*x +
c)^(5/2), x)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)