\(\int \frac {\sec (c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [750]

Optimal result

Integrand size = 33, antiderivative size = 378 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {4 a \left (2 A b^2-a^2 C+3 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^3 (a+b)^{3/2} d}+\frac {2 \left (2 a^2 C+3 a b (A+C)-b^2 (A+3 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^2 \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {4 a \left (2 A b^2-a^2 C+3 b^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4166, 4088, 4090, 3917, 4089} \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {2 \left (2 a^2 C+3 a b (A+C)-b^2 (A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{3 b^2 d \sqrt {a+b} \left (a^2-b^2\right )}-\frac {4 a \left (a^2 (-C)+2 A b^2+3 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^3 d (a-b) (a+b)^{3/2}}-\frac {4 a \left (a^2 (-C)+2 A b^2+3 b^2 C\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}} \]

[In]

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

[Out]

(-4*a*(2*A*b^2 - a^2*C + 3*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 + 3*a*b*(A + C) - b^2*(A + 3*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))])
/(3*b^2*Sqrt[a + b]*(a^2 - b^2)*d) - (2*(A*b^2 + a^2*C)*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^
(3/2)) - (4*a*(2*A*b^2 - a^2*C + 3*b^2*C)*Tan[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 4166

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (A b^2+a^2 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} a b (A+C)+\frac {1}{2} \left (A b^2-2 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^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {4 a \left (2 A b^2-a^2 C+3 b^2 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 (a^2 (3 A+C)+b^2 (A+3 C)\right )+\frac {1}{2} a \left (2 A b^2-\left (a^2-3 b^2\right ) 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^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {4 a \left (2 A b^2-a^2 C+3 b^2 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 (2 a \left (2 A b^2-a^2 C+3 b^2 C\right )\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+3 a b (A+C)-b^2 (A+3 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 (a-b) b (a+b)^2} \\ & = -\frac {4 a \left (2 A b^2-\left (a^2-3 b^2\right ) 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+3 a b (A+C)-b^2 (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}}}{3 (a-b) b^2 (a+b)^{3/2} d}-\frac {2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {4 a \left (2 A b^2-a^2 C+3 b^2 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. \(3369\) vs. \(2(378)=756\).

Time = 25.23 (sec) , antiderivative size = 3369, normalized size of antiderivative = 8.91 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Result too large to show} \]

[In]

Integrate[(Sec[c + d*x]*(A + 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 + C*Sec[c + d*x]^2)*((-8*a*(-2*A*b^2 + a^2*C - 3*b^2*C)*Sin[c + d*x])/
(3*b^2*(a^2 - b^2)^2) + (4*(A*b^2*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] + 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 + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(5/2)) + (8*(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]]) + (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*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]]) - (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 + C*Sec[c + d*x]^2)*(2*a*(a
+ b)*(-2*A*b^2 + (a^2 - 3*b^2)*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)] + b*(a + b)*(-2*a^2*C + 3*a*b*(A + C) +
b^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]))]*
EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + a*(-2*A*b^2 + (a^2 - 3*b^2)*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 + A*Cos[2*c + 2*d*x])*Sqrt[Se
c[(c + d*x)/2]^2]*(a + b*Sec[c + d*x])^(5/2)*((4*a*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sin[c + d*x]*(2*a*(a
+ b)*(-2*A*b^2 + (a^2 - 3*b^2)*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)] + b*(a + b)*(-2*a^2*C + 3*a*b*(A + C) +
b^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]))]*
EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + a*(-2*A*b^2 + (a^2 - 3*b^2)*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]) - (4*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Tan[(c + d*x)/2]*(2*a*(a + b)*(-2*A*b^2 + (a^2 - 3*b^2
)*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[A
rcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + b*(a + b)*(-2*a^2*C + 3*a*b*(A + C) + b^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]))]*EllipticF[ArcSin[Tan[(c + d*x
)/2]], (a - b)/(a + b)] + a*(-2*A*b^2 + (a^2 - 3*b^2)*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]) + (8*Sqrt[Cos[(c +
d*x)/2]^2*Sec[c + d*x]]*((a*(-2*A*b^2 + (a^2 - 3*b^2)*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^4
)/2 + (a*(a + b)*(-2*A*b^2 + (a^2 - 3*b^2)*C)*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*Elliptic
E[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 + 3*a*b*(A + C) + b^2*(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])))/(2*Sqrt[Cos[c + d*x]
/(1 + Cos[c + d*x])]) + (a*(a + b)*(-2*A*b^2 + (a^2 - 3*b^2)*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Elliptic
E[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 + 3*a*b*(A + C) + b^2*(A + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*EllipticF[ArcS
in[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)))/(2*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))])
- a^2*(-2*A*b^2 + (a^2 - 3*b^2)*C)*Cos[c + d*x]*Sec[(c + d*x)/2]^2*Sin[c + d*x]*Tan[(c + d*x)/2] - a*(-2*A*b^2
+ (a^2 - 3*b^2)*C)*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Sin[c + d*x]*Tan[(c + d*x)/2] + a*(-2*A*b^2 + (a^2
- 3*b^2)*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 +
3*a*b*(A + C) + b^2*(A + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + C
os[c + d*x]))]*Sec[(c + d*x)/2]^2)/(2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[1 - ((a - b)*Tan[(c + d*x)/2]^2)/(a +
b)]) + (a*(a + b)*(-2*A*b^2 + (a^2 - 3*b^2)*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]))/(3*(-(a^2*b) + b^3)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) + (4*(2*a*(a + b)*
(-2*A*b^2 + (a^2 - 3*b^2)*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)] + b*(a + b)*(-2*a^2*C + 3*a*b*(A + C) + b^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]))]*Ellip
ticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + a*(-2*A*b^2 + (a^2 - 3*b^2)*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[C
os[(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. \(5563\) vs. \(2(348)=696\).

Time = 17.29 (sec) , antiderivative size = 5564, normalized size of antiderivative = 14.72

method	result	size
parts	\(\text {Expression too large to display}\)	\(5564\)
default	\(\text {Expression too large to display}\)	\(5624\)

[In]

int(sec(d*x+c)*(A+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]

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

[In]

integrate(sec(d*x+c)*(A+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 + 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]

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

[In]

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

[Out]

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

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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