\(\int \frac {\sec (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [972]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (warning: unable to verify)
   Maple [B] (warning: unable to verify)
   Fricas [F]
   Sympy [F]
   Maxima [F(-1)]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 41, antiderivative size = 416 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 \left (4 a A b^2-a^2 b B-3 b^3 B-2 a^3 C+6 a b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^3 (a+b)^{3/2} d}+\frac {2 \left (2 a^2 C+a b (3 A+B+3 C)-b^2 (A+3 (B+C))\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^2 \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (a^2 b B+3 b^3 B+2 a^3 C-2 a b^2 (2 A+3 C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]

[Out]

-2/3*(4*A*a*b^2-B*a^2*b-3*B*b^3-2*C*a^3+6*C*a*b^2)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a
+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/(a-b)/b^3/(a+b)^(3/2)/d+2/3*(
2*C*a^2+a*b*(3*A+B+3*C)-b^2*(A+3*B+3*C))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(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^2/(a^2-b^2)/d/(a+b)^(1/2)-2/3*(A*b^2-
a*(B*b-C*a))*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)+2/3*(B*a^2*b+3*B*b^3+2*a^3*C-2*a*b^2*(2*A+3*C))*t
an(d*x+c)/b/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {4165, 4088, 4090, 3917, 4089} \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {2 \cot (c+d x) \left (2 a^2 C+a b (3 A+B+3 C)-b^2 (A+3 (B+C))\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{3 b^2 d \sqrt {a+b} \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {2 \cot (c+d x) \left (-2 a^3 C-a^2 b B+4 a A b^2+6 a b^2 C-3 b^3 B\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^3 d (a-b) (a+b)^{3/2}}+\frac {2 \tan (c+d x) \left (2 a^3 C+a^2 b B-2 a b^2 (2 A+3 C)+3 b^3 B\right )}{3 b d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}} \]

[In]

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

[Out]

(-2*(4*a*A*b^2 - a^2*b*B - 3*b^3*B - 2*a^3*C + 6*a*b^2*C)*Cot[c + d*x]*EllipticE[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))]
)/(3*(a - b)*b^3*(a + b)^(3/2)*d) + (2*(2*a^2*C + a*b*(3*A + B + 3*C) - b^2*(A + 3*(B + C)))*Cot[c + d*x]*Elli
pticF[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))])/(3*b^2*Sqrt[a + b]*(a^2 - b^2)*d) - (2*(A*b^2 - a*(b*B - a*C))*Tan[c + d*
x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) + (2*(a^2*b*B + 3*b^3*B + 2*a^3*C - 2*a*b^2*(2*A + 3*C))*Ta
n[c + d*x])/(3*b*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]])

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 4088

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[(a*A - b*B)*(m + 1) - (A*b - a
*B)*(m + 2)*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]
 && LtQ[m, -1]

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 4165

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \int \frac {\sec (c+d x) \left (\frac {3}{2} b (b B-a (A+C))+\frac {1}{2} \left (A b^2-a b B-2 a^2 C+3 b^2 C\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )} \\ & = -\frac {2 \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (a^2 b B+3 b^3 B+2 a^3 C-2 a b^2 (2 A+3 C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {4 \int \frac {\sec (c+d x) \left (-\frac {1}{4} b \left (4 a b B-a^2 (3 A+C)-b^2 (A+3 C)\right )-\frac {1}{4} \left (a^2 b B+3 b^3 B+2 a^3 C-2 a b^2 (2 A+3 C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 b \left (a^2-b^2\right )^2} \\ & = -\frac {2 \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (a^2 b B+3 b^3 B+2 a^3 C-2 a b^2 (2 A+3 C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {\left (a^2 b B+3 b^3 B+2 a^3 C-2 a b^2 (2 A+3 C)\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 b \left (a^2-b^2\right )^2}+\frac {\left (2 a^2 C+a b (3 A+B+3 C)-b^2 (A+3 (B+C))\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 (a-b) b (a+b)^2} \\ & = \frac {2 \left (a^2 b B+3 b^3 B+2 a^3 C-2 a b^2 (2 A+3 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}}}{3 (a-b) b^3 (a+b)^{3/2} d}+\frac {2 \left (2 a^2 C+a b (3 A+B+3 C)-b^2 (A+3 (B+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}}}{3 (a-b) b^2 (a+b)^{3/2} d}-\frac {2 \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (a^2 b B+3 b^3 B+2 a^3 C-2 a b^2 (2 A+3 C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(3980\) vs. \(2(416)=832\).

Time = 30.66 (sec) , antiderivative size = 3980, normalized size of antiderivative = 9.57 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

((b + a*Cos[c + d*x])^3*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-4*(-4*a*A*b^2 + a^2*b*B + 3*b^
3*B + 2*a^3*C - 6*a*b^2*C)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)^2) + (4*(A*b^2*Sin[c + d*x] - a*b*B*Sin[c + d*x] +
 a^2*C*Sin[c + d*x]))/(3*a*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) + (4*(-5*a^2*A*b^2*Sin[c + d*x] + A*b^4*Sin[c +
 d*x] + 2*a^3*b*B*Sin[c + d*x] + 2*a*b^3*B*Sin[c + d*x] + a^4*C*Sin[c + d*x] - 5*a^2*b^2*C*Sin[c + d*x]))/(3*a
*b*(-a^2 + b^2)^2*(b + a*Cos[c + d*x]))))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c +
d*x])^(5/2)) + (4*(b + a*Cos[c + d*x])^2*((-8*a*A*b)/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d
*x]]) + (2*a^2*B)/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*b^2*B)/((-a^2 + b^2)^2*S
qrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (4*a^3*C)/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[
c + d*x]]) - (4*a*b*C)/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*a^2*A*Sqrt[Sec[c + d*
x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (2*A*b^2*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*C
os[c + d*x]]) + (2*a^3*B*Sqrt[Sec[c + d*x]])/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (2*a*b*B*Sqrt[Sec
[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (10*a^2*C*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt
[b + a*Cos[c + d*x]]) + (4*a^4*C*Sqrt[Sec[c + d*x]])/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (2*b^2*
C*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (8*a^2*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]]
)/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (2*a^3*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*b*(-a^2 + b^2
)^2*Sqrt[b + a*Cos[c + d*x]]) + (2*a*b*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c
 + d*x]]) - (4*a^2*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (4*a^4*C
*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]))*Sqrt[Sec[c + d*x]]*Sqrt
[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(2*(a + b)*(a^2*b*B + 3*b^3*B + 2*a^
3*C - 2*a*b^2*(2*A + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c
 + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(-2*a^2*C + b^2*(A - 3*B + 3*C)
+ a*b*(3*A - B + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d
*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (a^2*b*B + 3*b^3*B + 2*a^3*C - 2*a*b^2*(2*A + 3*
C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*(-(a^2*b) + b^3)^2*d*(A + 2*C +
 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[Sec[(c + d*x)/2]^2]*(a + b*Sec[c + d*x])^(5/2)*((2*a*Sqrt[Cos[(c
+ d*x)/2]^2*Sec[c + d*x]]*Sin[c + d*x]*(2*(a + b)*(a^2*b*B + 3*b^3*B + 2*a^3*C - 2*a*b^2*(2*A + 3*C))*Sqrt[Cos
[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c
+ d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(-2*a^2*C + b^2*(A - 3*B + 3*C) + a*b*(3*A - B + 3*C))*Sqrt[Cos[c +
 d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*
x)/2]], (a - b)/(a + b)] + (a^2*b*B + 3*b^3*B + 2*a^3*C - 2*a*b^2*(2*A + 3*C))*Cos[c + d*x]*(b + a*Cos[c + d*x
])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*(-(a^2*b) + b^3)^2*(b + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[(c + d*x)/2
]^2]) - (2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Tan[(c + d*x)/2]*(2*(a + b)*(a^2*b*B + 3*b^3*B + 2*a^3*C - 2*
a*b^2*(2*A + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x])
)]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(-2*a^2*C + b^2*(A - 3*B + 3*C) + a*b*(3
*A - B + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*E
llipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (a^2*b*B + 3*b^3*B + 2*a^3*C - 2*a*b^2*(2*A + 3*C))*Cos[
c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*(-(a^2*b) + b^3)^2*Sqrt[b + a*Cos[c + d
*x]]*Sqrt[Sec[(c + d*x)/2]^2]) + (4*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(((a^2*b*B + 3*b^3*B + 2*a^3*C - 2*a
*b^2*(2*A + 3*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^4)/2 + ((a + b)*(a^2*b*B + 3*b^3*B + 2*a^
3*C - 2*a*b^2*(2*A + 3*C))*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d
*x)/2]], (a - b)/(a + b)]*((Cos[c + d*x]*Sin[c + d*x])/(1 + Cos[c + d*x])^2 - Sin[c + d*x]/(1 + Cos[c + d*x]))
)/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])] + (b*(a + b)*(-2*a^2*C + b^2*(A - 3*B + 3*C) + a*b*(3*A - B + 3*C))*Sq
rt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*((C
os[c + d*x]*Sin[c + d*x])/(1 + Cos[c + d*x])^2 - Sin[c + d*x]/(1 + Cos[c + d*x])))/Sqrt[Cos[c + d*x]/(1 + Cos[
c + d*x])] + ((a + b)*(a^2*b*B + 3*b^3*B + 2*a^3*C - 2*a*b^2*(2*A + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])
]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*(-((a*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x]))) + ((b
 + a*Cos[c + d*x])*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x])^2)))/Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[
c + d*x]))] + (b*(a + b)*(-2*a^2*C + b^2*(A - 3*B + 3*C) + a*b*(3*A - B + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c +
 d*x])]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*(-((a*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x])))
 + ((b + a*Cos[c + d*x])*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x])^2)))/Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1
+ Cos[c + d*x]))] - a*(a^2*b*B + 3*b^3*B + 2*a^3*C - 2*a*b^2*(2*A + 3*C))*Cos[c + d*x]*Sec[(c + d*x)/2]^2*Sin[
c + d*x]*Tan[(c + d*x)/2] - (a^2*b*B + 3*b^3*B + 2*a^3*C - 2*a*b^2*(2*A + 3*C))*(b + a*Cos[c + d*x])*Sec[(c +
d*x)/2]^2*Sin[c + d*x]*Tan[(c + d*x)/2] + (a^2*b*B + 3*b^3*B + 2*a^3*C - 2*a*b^2*(2*A + 3*C))*Cos[c + d*x]*(b
+ a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]^2 + (b*(a + b)*(-2*a^2*C + b^2*(A - 3*B + 3*C) + a*b*(3*
A - B + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*Se
c[(c + d*x)/2]^2)/(Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[1 - ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)]) + ((a + b)*(a^
2*b*B + 3*b^3*B + 2*a^3*C - 2*a*b^2*(2*A + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x
])/((a + b)*(1 + Cos[c + d*x]))]*Sec[(c + d*x)/2]^2*Sqrt[1 - ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])/Sqrt[1 - T
an[(c + d*x)/2]^2]))/(3*(-(a^2*b) + b^3)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) + (2*(2*(a + b)*
(a^2*b*B + 3*b^3*B + 2*a^3*C - 2*a*b^2*(2*A + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c +
d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(-2*a^2
*C + b^2*(A - 3*B + 3*C) + a*b*(3*A - B + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x]
)/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (a^2*b*B + 3*b^3*B + 2*
a^3*C - 2*a*b^2*(2*A + 3*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])*(-(Cos[(c
+ d*x)/2]*Sec[c + d*x]*Sin[(c + d*x)/2]) + Cos[(c + d*x)/2]^2*Sec[c + d*x]*Tan[c + d*x]))/(3*(-(a^2*b) + b^3)^
2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]])))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(8563\) vs. \(2(387)=774\).

Time = 21.39 (sec) , antiderivative size = 8564, normalized size of antiderivative = 20.59

method result size
parts \(\text {Expression too large to display}\) \(8564\)
default \(\text {Expression too large to display}\) \(8624\)

[In]

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

[Out]

result too large to display

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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