\(\int \frac {\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\) [742]

Optimal result

Integrand size = 35, antiderivative size = 460 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \left (2 a^2 b^2 (5 A-4 C)+16 a^4 C-b^4 (5 A+3 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 b^5 \sqrt {a+b} d}-\frac {2 \left (16 a^3 C+12 a^2 b C+2 a b^2 (5 A+2 C)+b^3 (5 A+3 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 b^4 \sqrt {a+b} d}-\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 a \left (5 A b^2+8 a^2 C-3 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d} \]

[Out]

-2/5*(2*a^2*b^2*(5*A-4*C)+16*a^4*C-b^4*(5*A+3*C))*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)/b^5/d/(a+b)^(1/2)-2/5*(16*a^3*
C+12*a^2*b*C+2*a*b^2*(5*A+2*C)+b^3*(5*A+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^4/d/(a+b)^(1/2)-2*(A*b^2+C*a^2)*
sec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)-2/5*a*(5*A*b^2+8*C*a^2-3*C*b^2)*(a+b*sec(d*x+c))^
(1/2)*tan(d*x+c)/b^3/(a^2-b^2)/d+2/5*(5*A*b^2+6*C*a^2-C*b^2)*sec(d*x+c)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/
(a^2-b^2)/d

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 460, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4184, 4177, 4167, 4090, 3917, 4089} \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b^2 d \left (a^2-b^2\right )}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b^3 d \left (a^2-b^2\right )}-\frac {2 \left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{5 b^5 d \sqrt {a+b}}-\frac {2 \left (16 a^3 C+12 a^2 b C+2 a b^2 (5 A+2 C)+b^3 (5 A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{5 b^4 d \sqrt {a+b}} \]

[In]

Int[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(-2*(2*a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*A + 3*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))])
/(5*b^5*Sqrt[a + b]*d) - (2*(16*a^3*C + 12*a^2*b*C + 2*a*b^2*(5*A + 2*C) + b^3*(5*A + 3*C))*Cot[c + d*x]*Ellip
ticF[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))])/(5*b^4*Sqrt[a + b]*d) - (2*(A*b^2 + a^2*C)*Sec[c + d*x]^2*Tan[c + d*x])/(b
*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) - (2*a*(5*A*b^2 + 8*a^2*C - 3*b^2*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c +
d*x])/(5*b^3*(a^2 - b^2)*d) + (2*(5*A*b^2 + 6*a^2*C - b^2*C)*Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*
x])/(5*b^2*(a^2 - b^2)*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 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]

Rule 4184

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 \int \frac {\sec ^2(c+d x) \left (2 \left (A b^2+a^2 C\right )-\frac {1}{2} a b (A+C) \sec (c+d x)-\frac {1}{2} \left (5 A b^2+6 a^2 C-b^2 C\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{b \left (a^2-b^2\right )} \\ & = -\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (5 A b^2+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {4 \int \frac {\sec (c+d x) \left (-\frac {1}{2} a \left (b^2 (5 A-C)+6 a^2 C\right )+\frac {1}{4} b \left (5 A b^2+2 a^2 C+3 b^2 C\right ) \sec (c+d x)+\frac {3}{4} a \left (5 A b^2+8 a^2 C-3 b^2 C\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{5 b^2 \left (a^2-b^2\right )} \\ & = -\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 a \left (5 A b^2+8 a^2 C-3 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {8 \int \frac {\sec (c+d x) \left (-\frac {3}{8} a b \left (5 A b^2+\left (4 a^2+b^2\right ) C\right )-\frac {3}{8} \left (2 a^2 b^2 (5 A-4 C)+16 a^4 C-b^4 (5 A+3 C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 b^3 \left (a^2-b^2\right )} \\ & = -\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 a \left (5 A b^2+8 a^2 C-3 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left (16 a^3 C+12 a^2 b C+2 a b^2 (5 A+2 C)+b^3 (5 A+3 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{5 b^3 (a+b)}+\frac {\left (2 a^2 b^2 (5 A-4 C)+16 a^4 C-b^4 (5 A+3 C)\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{5 b^3 \left (a^2-b^2\right )} \\ & = -\frac {2 \left (2 a^2 b^2 (5 A-4 C)+16 a^4 C-b^4 (5 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}}}{5 b^5 \sqrt {a+b} d}-\frac {2 \left (16 a^3 C+12 a^2 b C+2 a b^2 (5 A+2 C)+b^3 (5 A+3 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}}}{5 b^4 \sqrt {a+b} d}-\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 a \left (5 A b^2+8 a^2 C-3 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(3853\) vs. \(2(460)=920\).

Time = 29.57 (sec) , antiderivative size = 3853, normalized size of antiderivative = 8.38 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Result too large to show} \]

[In]

Integrate[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

((b + a*Cos[c + d*x])^2*(A + C*Sec[c + d*x]^2)*((4*(-10*a^2*A*b^2 + 5*A*b^4 - 16*a^4*C + 8*a^2*b^2*C + 3*b^4*C
)*Sin[c + d*x])/(5*b^4*(-a^2 + b^2)) + (4*(a^2*A*b^2*Sin[c + d*x] + a^4*C*Sin[c + d*x]))/(b^3*(-a^2 + b^2)*(b
+ a*Cos[c + d*x])) - (12*a*C*Tan[c + d*x])/(5*b^3) + (4*C*Sec[c + d*x]*Tan[c + d*x])/(5*b^2)))/(d*(A + 2*C + A
*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(3/2)) + (4*(b + a*Cos[c + d*x])*((4*a^2*A)/(b*(-a^2 + b^2)*Sqrt[b + a
*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*A*b)/((-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (32*
a^4*C)/(5*b^3*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (16*a^2*C)/(5*b*(-a^2 + b^2)*Sqrt[b
+ a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (6*b*C)/(5*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) -
(4*a*A*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) + (4*a^3*A*Sqrt[Sec[c + d*x]])/(b^2*(-a^2
+ b^2)*Sqrt[b + a*Cos[c + d*x]]) - (8*a*C*Sqrt[Sec[c + d*x]])/(5*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) + (32*
a^5*C*Sqrt[Sec[c + d*x]])/(5*b^4*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) - (24*a^3*C*Sqrt[Sec[c + d*x]])/(5*b^2
*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) - (2*a*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)*Sqrt[b + a
*Cos[c + d*x]]) + (4*a^3*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(b^2*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) -
(6*a*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(5*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) + (32*a^5*C*Cos[2*(c + d
*x)]*Sqrt[Sec[c + d*x]])/(5*b^4*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) - (16*a^3*C*Cos[2*(c + d*x)]*Sqrt[Sec[c
+ d*x]])/(5*b^2*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]))*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(A + C*Sec[c +
d*x]^2)*(2*(a + b)*(2*a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*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)*(-16*a^3*C + 12*a^2*b*C - 2*a*b^2*(5*A + 2*C) + b^3*(5*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]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a +
b)] + (2*a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*A + 3*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]
^2*Tan[(c + d*x)/2]))/(5*b^4*(-a^2 + b^2)*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[Sec[(c + d*x)/2]^2]*Sqrt[Sec[c
+ d*x]]*(a + b*Sec[c + d*x])^(3/2)*((2*a*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sin[c + d*x]*(2*(a + b)*(2*a^2
*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*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)*(-16*a^3*C +
12*a^2*b*C - 2*a*b^2*(5*A + 2*C) + b^3*(5*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]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (2*a^2*b^2*(5*A -
4*C) + 16*a^4*C - b^4*(5*A + 3*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(5*
b^4*(-a^2 + b^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)*(2*a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*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)*(-16*a^3*C + 12*a^2*b*C - 2*a*b^2*(5*A + 2*C) + b^3*(5*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]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]],
(a - b)/(a + b)] + (2*a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*A + 3*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[
(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(5*b^4*(-a^2 + b^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]]*(((2*a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*A + 3*C))*Cos[c + d*x]*(b +
a*Cos[c + d*x])*Sec[(c + d*x)/2]^4)/2 + ((a + b)*(2*a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*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)*(-16*a^3*C + 12*a^2*b*C - 2*a*b^2*(5*A + 2*C) + b^3*(5*A + 3*C))*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)]*((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])] + ((a + b)*(2
*a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*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
)*(-16*a^3*C + 12*a^2*b*C - 2*a*b^2*(5*A + 2*C) + b^3*(5*A + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Ellip
ticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*(-((a*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x]))) + ((b + a*Co
s[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*(2*a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*A + 3*C))*Cos[c + d*x]*Sec[(c + d*x)/2]^2*Sin[c + d*x]*Tan
[(c + d*x)/2] - (2*a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*A + 3*C))*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*S
in[c + d*x]*Tan[(c + d*x)/2] + (2*a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*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)*(-16*a^3*C + 12*a^2*b*C - 2*a*b^2*(5*A + 2*C) + b^3
*(5*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]))]*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)*(2*
a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*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 -
Tan[(c + d*x)/2]^2]))/(5*b^4*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) + (2*(2*(a + b)*(
2*a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*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)*(-16*a^
3*C + 12*a^2*b*C - 2*a*b^2*(5*A + 2*C) + b^3*(5*A + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Co
s[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (2*a^2*b^2*(5
*A - 4*C) + 16*a^4*C - b^4*(5*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]))/(5*b^4*(-
a^2 + b^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] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(5108\) vs. \(2(428)=856\).

Time = 29.12 (sec) , antiderivative size = 5109, normalized size of antiderivative = 11.11

method	result	size
parts	\(\text {Expression too large to display}\)	\(5109\)
default	\(\text {Expression too large to display}\)	\(5163\)

[In]

int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F]

\[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(sec(d*x+c)**3*(A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(3/2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**3/(a + b*sec(c + d*x))**(3/2), x)

Maxima [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F]

\[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x))^(3/2)),x)

[Out]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x))^(3/2)), x)