\(\int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [388]

Optimal result

Integrand size = 31, antiderivative size = 353 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 \left (4 a A b-a^2 B-3 b^2 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^2 (a+b)^{3/2} d}+\frac {2 (3 a A-A b+a B-3 b B) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b (a+b)^{3/2} d}-\frac {2 (A b-a B) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (4 a A b-a^2 B-3 b^2 B\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {4088, 4090, 3917, 4089} \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 \left (a^2 (-B)+4 a A b-3 b^2 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^2 d (a-b) (a+b)^{3/2}}-\frac {2 \left (a^2 (-B)+4 a A b-3 b^2 B\right ) \tan (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}-\frac {2 (A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {2 (3 a A+a B-A b-3 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{3 b d (a-b) (a+b)^{3/2}} \]

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (A b-a B) \tan (c+d x)}{3 \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 A-b B)+\frac {1}{2} (A b-a B) \sec (c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 \left (a^2-b^2\right )} \\ & = -\frac {2 (A b-a B) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (4 a A b-a^2 B-3 b^2 B\right ) \tan (c+d x)}{3 \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} \left (3 a^2 A+A b^2-4 a b B\right )+\frac {1}{4} \left (4 a A b-a^2 B-3 b^2 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^2} \\ & = -\frac {2 (A b-a B) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (4 a A b-a^2 B-3 b^2 B\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {\left (4 a A b-a^2 B-3 b^2 B\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^2}+\frac {(a (3 A+B)-b (A+3 B)) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 (a-b) (a+b)^2} \\ & = -\frac {2 \left (4 a A b-a^2 B-3 b^2 B\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^2 (a+b)^{3/2} d}+\frac {2 (a (3 A+B)-b (A+3 B)) \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 (a+b)^{3/2} d}-\frac {2 (A b-a B) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (4 a A b-a^2 B-3 b^2 B\right ) \tan (c+d x)}{3 \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. \(3225\) vs. \(2(353)=706\).

Time = 23.83 (sec) , antiderivative size = 3225, normalized size of antiderivative = 9.14 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

((b + a*Cos[c + d*x])^3*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*((-2*(-4*a*A*b + a^2*B + 3*b^2*B)*Sin[c + d*x])/(3
*b*(-a^2 + b^2)^2) + (2*(A*b^2*Sin[c + d*x] - a*b*B*Sin[c + d*x]))/(3*a*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) +
(2*(-5*a^2*A*b*Sin[c + d*x] + A*b^3*Sin[c + d*x] + 2*a^3*B*Sin[c + d*x] + 2*a*b^2*B*Sin[c + d*x]))/(3*a*(a^2 -
b^2)^2*(b + a*Cos[c + d*x]))))/(d*(B + A*Cos[c + d*x])*(a + b*Sec[c + d*x])^(5/2)) + (2*(b + a*Cos[c + d*x])^
2*((-4*a*A*b)/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (a^2*B)/(3*(-a^2 + b^2)^2*Sqrt[
b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (b^2*B)/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]])
- (a^2*A*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (A*b^2*Sqrt[Sec[c + d*x]])/(3*(-a^
2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (a^3*B*Sqrt[Sec[c + d*x]])/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]
) - (a*b*B*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (4*a^2*A*Cos[2*(c + d*x)]*Sqrt[Se
c[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (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]]) + (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]]))*Sec[c + d*x]^(3/2)*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(A + B*Sec[c + d*x])*(2*(a + b)*(-4
*a*A*b + a^2*B + 3*b^2*B)*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)*(3*a*A + A*b - a*B - 3*b*B)*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)] + (-4*a*A*b + a^2*B + 3*b^2*B)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)
/2]^2*Tan[(c + d*x)/2]))/(3*b*(a^2 - b^2)^2*d*(B + A*Cos[c + d*x])*Sqrt[Sec[(c + d*x)/2]^2]*(a + b*Sec[c + d*x
])^(5/2)*((a*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sin[c + d*x]*(2*(a + b)*(-4*a*A*b + a^2*B + 3*b^2*B)*Sqrt[C
os[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)*(3*a*A + A*b - a*B - 3*b*B)*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)]
+ (-4*a*A*b + a^2*B + 3*b^2*B)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b*(a
^2 - b^2)^2*(b + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[(c + d*x)/2]^2]) - (Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Tan[
(c + d*x)/2]*(2*(a + b)*(-4*a*A*b + a^2*B + 3*b^2*B)*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)*(3*a*A
+ A*b - a*B - 3*b*B)*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)] + (-4*a*A*b + a^2*B + 3*b^2*B)*Cos[c + d*x]*(b + a*
Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b*(a^2 - b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c +
d*x)/2]^2]) + (2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(((-4*a*A*b + a^2*B + 3*b^2*B)*Cos[c + d*x]*(b + a*Cos[
c + d*x])*Sec[(c + d*x)/2]^4)/2 + ((a + b)*(-4*a*A*b + a^2*B + 3*b^2*B)*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)*(3*a*A + A*b
- a*B - 3*b*B)*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)*(-4*a*A*b + a^2*B + 3*b^2*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Ell
ipticE[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)*(3*a*A + A*b - a*B - 3*b*B)*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*(-4*a*A*b
+ a^2*B + 3*b^2*B)*Cos[c + d*x]*Sec[(c + d*x)/2]^2*Sin[c + d*x]*Tan[(c + d*x)/2] - (-4*a*A*b + a^2*B + 3*b^2*B
)*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Sin[c + d*x]*Tan[(c + d*x)/2] + (-4*a*A*b + a^2*B + 3*b^2*B)*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)*(3*a*A + A*b - a*B - 3*b*B)*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 - Tan[(c + d*x)/2]^2]*Sqrt[1 - ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)]) + ((a + b)*(-4*a*A*b + a^2*B + 3
*b^2*B)*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*b*(a^2 - b^2)^2*S
qrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) + ((2*(a + b)*(-4*a*A*b + a^2*B + 3*b^2*B)*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)*(3*a*A + A*b - a*B - 3*b*B)*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)] + (-4*a*A*b
+ a^2*B + 3*b^2*B)*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*b*(a^2 - b^2)^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. \(5259\) vs. \(2(323)=646\).

Time = 14.82 (sec) , antiderivative size = 5260, normalized size of antiderivative = 14.90

method	result	size
parts	\(\text {Expression too large to display}\)	\(5260\)
default	\(\text {Expression too large to display}\)	\(5303\)

[In]

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

[Out]

result too large to display

Fricas [F]

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

[In]

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

[Out]

integral((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]

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

[In]

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

[Out]

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

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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