\(\int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [950]

Optimal result

Integrand size = 43, antiderivative size = 610 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (a-b) \sqrt {a+b} \left (110 a^4 b B-3069 a^2 b^3 B-1617 b^5 B-40 a^5 C-15 a^3 b^2 (33 A+17 C)-15 a b^4 (319 A+247 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3465 b^4 d}+\frac {2 (a-b) \sqrt {a+b} \left (10 a^3 b (11 B-3 C)-40 a^4 C-15 a^2 b^2 (33 A-121 B+19 C)-3 b^4 (275 A-539 B+225 C)+6 a b^3 (660 A-209 B+505 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3465 b^3 d}-\frac {2 \left (110 a^3 b B-1254 a b^3 B-40 a^4 C-75 b^4 (11 A+9 C)-15 a^2 b^2 (33 A+19 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3465 b^2 d}-\frac {2 \left (110 a^2 b B-539 b^3 B-40 a^3 C-5 a b^2 (99 A+67 C)\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{3465 b^2 d}+\frac {2 \left (99 A b^2-22 a b B+8 a^2 C+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}+\frac {2 (11 b B-4 a C) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d} \]

[Out]

2/3465*(a-b)*(110*B*a^4*b-3069*B*a^2*b^3-1617*B*b^5-40*a^5*C-15*a^3*b^2*(33*A+17*C)-15*a*b^4*(319*A+247*C))*co
t(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b)
)^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^4/d+2/3465*(a-b)*(10*a^3*b*(11*B-3*C)-40*a^4*C-15*a^2*b^2*(33*A-121*
B+19*C)-3*b^4*(275*A-539*B+225*C)+6*a*b^3*(660*A-209*B+505*C))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+
b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^3/d
-2/3465*(110*B*a^2*b-539*B*b^3-40*a^3*C-5*a*b^2*(99*A+67*C))*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b^2/d+2/693*(99
*A*b^2-22*B*a*b+8*C*a^2+81*C*b^2)*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b^2/d+2/99*(11*B*b-4*C*a)*(a+b*sec(d*x+c))
^(7/2)*tan(d*x+c)/b^2/d+2/11*C*sec(d*x+c)*(a+b*sec(d*x+c))^(7/2)*tan(d*x+c)/b/d-2/3465*(110*B*a^3*b-1254*B*a*b
^3-40*a^4*C-75*b^4*(11*A+9*C)-15*a^2*b^2*(33*A+19*C))*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/d

Rubi [A] (verified)

Time = 2.39 (sec) , antiderivative size = 610, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {4177, 4167, 4087, 4090, 3917, 4089} \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \tan (c+d x) \left (8 a^2 C-22 a b B+99 A b^2+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2}}{693 b^2 d}-\frac {2 \tan (c+d x) \left (-40 a^3 C+110 a^2 b B-5 a b^2 (99 A+67 C)-539 b^3 B\right ) (a+b \sec (c+d x))^{3/2}}{3465 b^2 d}+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-40 a^4 C+10 a^3 b (11 B-3 C)-15 a^2 b^2 (33 A-121 B+19 C)+6 a b^3 (660 A-209 B+505 C)-3 b^4 (275 A-539 B+225 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{3465 b^3 d}-\frac {2 \tan (c+d x) \left (-40 a^4 C+110 a^3 b B-15 a^2 b^2 (33 A+19 C)-1254 a b^3 B-75 b^4 (11 A+9 C)\right ) \sqrt {a+b \sec (c+d x)}}{3465 b^2 d}+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-40 a^5 C+110 a^4 b B-15 a^3 b^2 (33 A+17 C)-3069 a^2 b^3 B-15 a b^4 (319 A+247 C)-1617 b^5 B\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3465 b^4 d}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{99 b^2 d}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d} \]

[In]

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

[Out]

(2*(a - b)*Sqrt[a + b]*(110*a^4*b*B - 3069*a^2*b^3*B - 1617*b^5*B - 40*a^5*C - 15*a^3*b^2*(33*A + 17*C) - 15*a
*b^4*(319*A + 247*C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sq
rt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3465*b^4*d) + (2*(a - b)*Sqrt[a +
b]*(10*a^3*b*(11*B - 3*C) - 40*a^4*C - 15*a^2*b^2*(33*A - 121*B + 19*C) - 3*b^4*(275*A - 539*B + 225*C) + 6*a
*b^3*(660*A - 209*B + 505*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a
- b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3465*b^3*d) - (2*(110*a^3
*b*B - 1254*a*b^3*B - 40*a^4*C - 75*b^4*(11*A + 9*C) - 15*a^2*b^2*(33*A + 19*C))*Sqrt[a + b*Sec[c + d*x]]*Tan[
c + d*x])/(3465*b^2*d) - (2*(110*a^2*b*B - 539*b^3*B - 40*a^3*C - 5*a*b^2*(99*A + 67*C))*(a + b*Sec[c + d*x])^
(3/2)*Tan[c + d*x])/(3465*b^2*d) + (2*(99*A*b^2 - 22*a*b*B + 8*a^2*C + 81*b^2*C)*(a + b*Sec[c + d*x])^(5/2)*Ta
n[c + d*x])/(693*b^2*d) + (2*(11*b*B - 4*a*C)*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(99*b^2*d) + (2*C*Sec[c
+ d*x]*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(11*b*d)

Rule 3917

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin
[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4087

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[(-B)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[1/(m + 1), Int[Csc[e + f
*x]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /;
FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 4089

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4090

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, Int[Csc[e + f*x]*((1 +
Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]

Rule 4167

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m
+ 2))), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*
B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 4177

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^
(m + 1)/(b*f*(m + 3))), x] + Dist[1/(b*(m + 3)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m +
2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B*(m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C,
m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac {2 \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (a C+\frac {1}{2} b (11 A+9 C) \sec (c+d x)+\frac {1}{2} (11 b B-4 a C) \sec ^2(c+d x)\right ) \, dx}{11 b} \\ & = \frac {2 (11 b B-4 a C) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac {4 \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (\frac {1}{4} b (77 b B-10 a C)+\frac {1}{4} \left (99 A b^2-22 a b B+8 a^2 C+81 b^2 C\right ) \sec (c+d x)\right ) \, dx}{99 b^2} \\ & = \frac {2 \left (99 A b^2-22 a b B+8 a^2 C+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}+\frac {2 (11 b B-4 a C) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac {8 \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (\frac {3}{8} b \left (165 A b^2+143 a b B-10 a^2 C+135 b^2 C\right )-\frac {1}{8} \left (110 a^2 b B-539 b^3 B-40 a^3 C-5 a b^2 (99 A+67 C)\right ) \sec (c+d x)\right ) \, dx}{693 b^2} \\ & = -\frac {2 \left (110 a^2 b B-539 b^3 B-40 a^3 C-5 a b^2 (99 A+67 C)\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{3465 b^2 d}+\frac {2 \left (99 A b^2-22 a b B+8 a^2 C+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}+\frac {2 (11 b B-4 a C) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac {16 \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (\frac {3}{16} b \left (605 a^2 b B+539 b^3 B-10 a^3 C+10 a b^2 (132 A+101 C)\right )-\frac {3}{16} \left (110 a^3 b B-1254 a b^3 B-40 a^4 C-75 b^4 (11 A+9 C)-15 a^2 b^2 (33 A+19 C)\right ) \sec (c+d x)\right ) \, dx}{3465 b^2} \\ & = -\frac {2 \left (110 a^3 b B-1254 a b^3 B-40 a^4 C-75 b^4 (11 A+9 C)-15 a^2 b^2 (33 A+19 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3465 b^2 d}-\frac {2 \left (110 a^2 b B-539 b^3 B-40 a^3 C-5 a b^2 (99 A+67 C)\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{3465 b^2 d}+\frac {2 \left (99 A b^2-22 a b B+8 a^2 C+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}+\frac {2 (11 b B-4 a C) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac {32 \int \frac {\sec (c+d x) \left (\frac {3}{32} b \left (1705 a^3 b B+2871 a b^3 B+10 a^4 C+75 b^4 (11 A+9 C)+15 a^2 b^2 (297 A+221 C)\right )-\frac {3}{32} \left (110 a^4 b B-3069 a^2 b^3 B-1617 b^5 B-40 a^5 C-15 a^3 b^2 (33 A+17 C)-15 a b^4 (319 A+247 C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{10395 b^2} \\ & = -\frac {2 \left (110 a^3 b B-1254 a b^3 B-40 a^4 C-75 b^4 (11 A+9 C)-15 a^2 b^2 (33 A+19 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3465 b^2 d}-\frac {2 \left (110 a^2 b B-539 b^3 B-40 a^3 C-5 a b^2 (99 A+67 C)\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{3465 b^2 d}+\frac {2 \left (99 A b^2-22 a b B+8 a^2 C+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}+\frac {2 (11 b B-4 a C) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}-\frac {\left (110 a^4 b B-3069 a^2 b^3 B-1617 b^5 B-40 a^5 C-15 a^3 b^2 (33 A+17 C)-15 a b^4 (319 A+247 C)\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{3465 b^2}+\frac {\left (32 \left (\frac {3}{32} b \left (1705 a^3 b B+2871 a b^3 B+10 a^4 C+75 b^4 (11 A+9 C)+15 a^2 b^2 (297 A+221 C)\right )+\frac {3}{32} \left (110 a^4 b B-3069 a^2 b^3 B-1617 b^5 B-40 a^5 C-15 a^3 b^2 (33 A+17 C)-15 a b^4 (319 A+247 C)\right )\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{10395 b^2} \\ & = \frac {2 (a-b) \sqrt {a+b} \left (110 a^4 b B-3069 a^2 b^3 B-1617 b^5 B-40 a^5 C-15 a^3 b^2 (33 A+17 C)-15 a b^4 (319 A+247 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3465 b^4 d}+\frac {2 (a-b) \sqrt {a+b} \left (a^3 b (110 B-30 C)-40 a^4 C-15 a^2 b^2 (33 A-121 B+19 C)-3 b^4 (275 A-539 B+225 C)+6 a b^3 (660 A-209 B+505 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3465 b^3 d}-\frac {2 \left (110 a^3 b B-1254 a b^3 B-40 a^4 C-75 b^4 (11 A+9 C)-15 a^2 b^2 (33 A+19 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3465 b^2 d}-\frac {2 \left (110 a^2 b B-539 b^3 B-40 a^3 C-5 a b^2 (99 A+67 C)\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{3465 b^2 d}+\frac {2 \left (99 A b^2-22 a b B+8 a^2 C+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}+\frac {2 (11 b B-4 a C) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(5369\) vs. \(2(610)=1220\).

Time = 32.94 (sec) , antiderivative size = 5369, normalized size of antiderivative = 8.80 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]

[In]

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

[Out]

Result too large to show

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(9226\) vs. \(2(568)=1136\).

Time = 134.14 (sec) , antiderivative size = 9227, normalized size of antiderivative = 15.13

method	result	size
parts	\(\text {Expression too large to display}\)	\(9227\)
default	\(\text {Expression too large to display}\)	\(9339\)

[In]

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

[Out]

result too large to display

Fricas [F]

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

[In]

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

[Out]

integral((C*b^2*sec(d*x + c)^6 + (2*C*a*b + B*b^2)*sec(d*x + c)^5 + A*a^2*sec(d*x + c)^2 + (C*a^2 + 2*B*a*b +
A*b^2)*sec(d*x + c)^4 + (B*a^2 + 2*A*a*b)*sec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a), x)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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