3.1001 \(\int \frac{(a+b \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=317 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (a^3 (5 A+7 C)+21 a^2 b B+21 a b^2 (A+3 C)+21 b^3 B\right )}{21 d}+\frac{2 a \sin (c+d x) \left (5 a^2 (5 A+7 C)+63 a b B+24 A b^2\right )}{105 d \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b (3 A+5 C)+3 a^3 B+15 a b^2 B+5 b^3 (A-C)\right )}{5 d}-\frac{2 b^2 \sin (c+d x) \sqrt{\sec (c+d x)} (7 a B+11 A b-35 b C)}{35 d}+\frac{2 (7 a B+6 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^3}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

[Out]

(2*(3*a^3*B + 15*a*b^2*B + 5*b^3*(A - C) + 3*a^2*b*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*S
qrt[Sec[c + d*x]])/(5*d) + (2*(21*a^2*b*B + 21*b^3*B + 21*a*b^2*(A + 3*C) + a^3*(5*A + 7*C))*Sqrt[Cos[c + d*x]
]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a*(24*A*b^2 + 63*a*b*B + 5*a^2*(5*A + 7*C))*Sin[c
+ d*x])/(105*d*Sqrt[Sec[c + d*x]]) - (2*b^2*(11*A*b + 7*a*B - 35*b*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(35*d)
+ (2*(6*A*b + 7*a*B)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2)) + (2*A*(a + b*Sec[c + d*x]
)^3*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2))

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Rubi [A]  time = 0.831875, antiderivative size = 317, 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, 4074, 4047, 3771, 2641, 4046, 2639} \[ \frac{2 a \sin (c+d x) \left (5 a^2 (5 A+7 C)+63 a b B+24 A b^2\right )}{105 d \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^3 (5 A+7 C)+21 a^2 b B+21 a b^2 (A+3 C)+21 b^3 B\right )}{21 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b (3 A+5 C)+3 a^3 B+15 a b^2 B+5 b^3 (A-C)\right )}{5 d}-\frac{2 b^2 \sin (c+d x) \sqrt{\sec (c+d x)} (7 a B+11 A b-35 b C)}{35 d}+\frac{2 (7 a B+6 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^3}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

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]^(7/2),x]

[Out]

(2*(3*a^3*B + 15*a*b^2*B + 5*b^3*(A - C) + 3*a^2*b*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*S
qrt[Sec[c + d*x]])/(5*d) + (2*(21*a^2*b*B + 21*b^3*B + 21*a*b^2*(A + 3*C) + a^3*(5*A + 7*C))*Sqrt[Cos[c + d*x]
]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a*(24*A*b^2 + 63*a*b*B + 5*a^2*(5*A + 7*C))*Sin[c
+ d*x])/(105*d*Sqrt[Sec[c + d*x]]) - (2*b^2*(11*A*b + 7*a*B - 35*b*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(35*d)
+ (2*(6*A*b + 7*a*B)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2)) + (2*A*(a + b*Sec[c + d*x]
)^3*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2))

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 4074

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[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, 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{7}{2}}(c+d x)} \, dx &=\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2}{7} \int \frac{(a+b \sec (c+d x))^2 \left (\frac{1}{2} (6 A b+7 a B)+\frac{1}{2} (5 a A+7 b B+7 a C) \sec (c+d x)-\frac{1}{2} b (A-7 C) \sec ^2(c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 (6 A b+7 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4}{35} \int \frac{(a+b \sec (c+d x)) \left (\frac{1}{4} \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right )+\frac{1}{4} \left (38 a A b+21 a^2 B+35 b^2 B+70 a b C\right ) \sec (c+d x)-\frac{1}{4} b (11 A b+7 a B-35 b C) \sec ^2(c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 (6 A b+7 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}-\frac{8}{105} \int \frac{-\frac{3}{8} \left (24 A b^3+21 a^3 B+98 a b^2 B+21 a^2 b (3 A+5 C)\right )-\frac{5}{8} \left (21 a^2 b B+21 b^3 B+21 a b^2 (A+3 C)+a^3 (5 A+7 C)\right ) \sec (c+d x)+\frac{3}{8} b^2 (11 A b+7 a B-35 b C) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 (6 A b+7 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}-\frac{8}{105} \int \frac{-\frac{3}{8} \left (24 A b^3+21 a^3 B+98 a b^2 B+21 a^2 b (3 A+5 C)\right )+\frac{3}{8} b^2 (11 A b+7 a B-35 b C) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx-\frac{1}{21} \left (-21 a^2 b B-21 b^3 B-21 a b^2 (A+3 C)-a^3 (5 A+7 C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}-\frac{2 b^2 (11 A b+7 a B-35 b C) \sqrt{\sec (c+d x)} \sin (c+d x)}{35 d}+\frac{2 (6 A b+7 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}-\frac{1}{5} \left (-3 a^3 B-15 a b^2 B-5 b^3 (A-C)-3 a^2 b (3 A+5 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx-\frac{1}{21} \left (\left (-21 a^2 b B-21 b^3 B-21 a b^2 (A+3 C)-a^3 (5 A+7 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 (21 a^2 b B+21 b^3 B+21 a b^2 (A+3 C)+a^3 (5 A+7 C)\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{21 d}+\frac{2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}-\frac{2 b^2 (11 A b+7 a B-35 b C) \sqrt{\sec (c+d x)} \sin (c+d x)}{35 d}+\frac{2 (6 A b+7 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}-\frac{1}{5} \left (\left (-3 a^3 B-15 a b^2 B-5 b^3 (A-C)-3 a^2 b (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 (3 a^3 B+15 a b^2 B+5 b^3 (A-C)+3 a^2 b (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 (21 a^2 b B+21 b^3 B+21 a b^2 (A+3 C)+a^3 (5 A+7 C)\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{21 d}+\frac{2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}-\frac{2 b^2 (11 A b+7 a B-35 b C) \sqrt{\sec (c+d x)} \sin (c+d x)}{35 d}+\frac{2 (6 A b+7 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 5.06322, size = 234, normalized size = 0.74 \[ \frac{\sqrt{\sec (c+d x)} \left (40 \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (a^3 (5 A+7 C)+21 a^2 b B+21 a b^2 (A+3 C)+21 b^3 B\right )+2 \sin (c+d x) \left (5 a \cos (c+d x) \left (a^2 (29 A+28 C)+84 a b B+84 A b^2\right )+42 \left (3 a^2 A b+a^3 B+10 b^3 C\right )+42 a^2 (a B+3 A b) \cos (2 (c+d x))+15 a^3 A \cos (3 (c+d x))\right )+168 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b (3 A+5 C)+3 a^3 B+15 a b^2 B+5 b^3 (A-C)\right )\right )}{420 d} \]

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]^(7/2),x]

[Out]

(Sqrt[Sec[c + d*x]]*(168*(3*a^3*B + 15*a*b^2*B + 5*b^3*(A - C) + 3*a^2*b*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*Ellip
ticE[(c + d*x)/2, 2] + 40*(21*a^2*b*B + 21*b^3*B + 21*a*b^2*(A + 3*C) + a^3*(5*A + 7*C))*Sqrt[Cos[c + d*x]]*El
lipticF[(c + d*x)/2, 2] + 2*(42*(3*a^2*A*b + a^3*B + 10*b^3*C) + 5*a*(84*A*b^2 + 84*a*b*B + a^2*(29*A + 28*C))
*Cos[c + d*x] + 42*a^2*(3*A*b + a*B)*Cos[2*(c + d*x)] + 15*a^3*A*Cos[3*(c + d*x)])*Sin[c + d*x]))/(420*d)

________________________________________________________________________________________

Maple [B]  time = 2.813, size = 1278, normalized size = 4. \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)^(7/2),x)

[Out]

-2/105*(240*A*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8
-24*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(15*A*a+21*A*b+7*B*a)*sin(1/2*d*x+1/2*c)^6*cos(1/
2*d*x+1/2*c)+28*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(10*A*a^2+18*A*a*b+15*A*b^2+6*B*a^2+15*
B*a*b+5*C*a^2)*sin(1/2*d*x+1/2*c)^4*cos(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)*
(40*A*a^3+63*A*a^2*b+105*A*a*b^2+21*B*a^3+105*B*a^2*b+35*C*a^3+105*C*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2
*c)+25*A*a^3*(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)+105*A*a*b^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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)
-189*A*(-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)^(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))*a^2*b-105*A*(-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)^(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))*b^3+1
05*B*a^2*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))
*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+105*B*b^3*(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)-63*
B*(-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)^(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))*a^3-315*B*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(s
in(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))*a*b^2+35*a^3
*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)+315*C*a*b^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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-315*C*(-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)^(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))*a^2*b+105*C*(-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)^(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))*b^3)/(-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

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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)^(7/2),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{7}{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)^(7/2),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)^(7/2), 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)**(7/2),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{7}{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)^(7/2),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)^(7/2), x)