\(\int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^3} \, dx\) [153]

Optimal result

Integrand size = 27, antiderivative size = 287 \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^3} \, dx=\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{c^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {a^{3/2} \sqrt {d} \left (15 c^2+20 c d+8 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{4 c^3 (c+d)^{5/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {a d \tan (e+f x)}{2 c (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2}-\frac {a d (7 c+4 d) \tan (e+f x)}{4 c^2 (c+d)^2 f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \]

[Out]

-1/2*a*d*tan(f*x+e)/c/(c+d)/f/(c+d*sec(f*x+e))^2/(a+a*sec(f*x+e))^(1/2)-1/4*a*d*(7*c+4*d)*tan(f*x+e)/c^2/(c+d)
^2/f/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)+2*a^(3/2)*arctanh((a-a*sec(f*x+e))^(1/2)/a^(1/2))*tan(f*x+e)/c^3/
f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-1/4*a^(3/2)*(15*c^2+20*c*d+8*d^2)*arctanh(d^(1/2)*(a-a*sec(f*x
+e))^(1/2)/a^(1/2)/(c+d)^(1/2))*d^(1/2)*tan(f*x+e)/c^3/(c+d)^(5/2)/f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(
1/2)

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {4025, 105, 156, 162, 65, 212, 214} \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^3} \, dx=\frac {2 a^{3/2} \tan (e+f x) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{c^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}-\frac {a^{3/2} \sqrt {d} \left (15 c^2+20 c d+8 d^2\right ) \tan (e+f x) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{4 c^3 f (c+d)^{5/2} \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}-\frac {a d (7 c+4 d) \tan (e+f x)}{4 c^2 f (c+d)^2 \sqrt {a \sec (e+f x)+a} (c+d \sec (e+f x))}-\frac {a d \tan (e+f x)}{2 c f (c+d) \sqrt {a \sec (e+f x)+a} (c+d \sec (e+f x))^2} \]

[In]

Int[Sqrt[a + a*Sec[e + f*x]]/(c + d*Sec[e + f*x])^3,x]

[Out]

(2*a^(3/2)*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]]*Tan[e + f*x])/(c^3*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*
Sec[e + f*x]]) - (a^(3/2)*Sqrt[d]*(15*c^2 + 20*c*d + 8*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a
]*Sqrt[c + d])]*Tan[e + f*x])/(4*c^3*(c + d)^(5/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (a*d
*Tan[e + f*x])/(2*c*(c + d)*f*Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])^2) - (a*d*(7*c + 4*d)*Tan[e + f*x]
)/(4*c^2*(c + d)^2*f*Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x]))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 4025

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[a^2*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])), Subst[Int[(a + b*x)^(m - 1/2)*((c
+ d*x)^n/(x*Sqrt[a - b*x])), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && IntegerQ[m - 1/2]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a^2 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a-a x} (c+d x)^3} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = -\frac {a d \tan (e+f x)}{2 c (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2}-\frac {(a \tan (e+f x)) \text {Subst}\left (\int \frac {2 a (c+d)-\frac {3 a d x}{2}}{x \sqrt {a-a x} (c+d x)^2} \, dx,x,\sec (e+f x)\right )}{2 c (c+d) f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = -\frac {a d \tan (e+f x)}{2 c (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2}-\frac {a d (7 c+4 d) \tan (e+f x)}{4 c^2 (c+d)^2 f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))}-\frac {\tan (e+f x) \text {Subst}\left (\int \frac {2 a^2 (c+d)^2-\frac {1}{4} a^2 d (7 c+4 d) x}{x \sqrt {a-a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{2 c^2 (c+d)^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = -\frac {a d \tan (e+f x)}{2 c (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2}-\frac {a d (7 c+4 d) \tan (e+f x)}{4 c^2 (c+d)^2 f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))}-\frac {\left (a^2 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a-a x}} \, dx,x,\sec (e+f x)\right )}{c^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {\left (a^2 d \left (15 c^2+20 c d+8 d^2\right ) \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a-a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{8 c^3 (c+d)^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = -\frac {a d \tan (e+f x)}{2 c (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2}-\frac {a d (7 c+4 d) \tan (e+f x)}{4 c^2 (c+d)^2 f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))}+\frac {(2 a \tan (e+f x)) \text {Subst}\left (\int \frac {1}{1-\frac {x^2}{a}} \, dx,x,\sqrt {a-a \sec (e+f x)}\right )}{c^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\left (a d \left (15 c^2+20 c d+8 d^2\right ) \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{c+d-\frac {d x^2}{a}} \, dx,x,\sqrt {a-a \sec (e+f x)}\right )}{4 c^3 (c+d)^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = \frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{c^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {a^{3/2} \sqrt {d} \left (15 c^2+20 c d+8 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{4 c^3 (c+d)^{5/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {a d \tan (e+f x)}{2 c (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2}-\frac {a d (7 c+4 d) \tan (e+f x)}{4 c^2 (c+d)^2 f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 7.57 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^3} \, dx=\frac {(d+c \cos (e+f x))^3 \sec \left (\frac {1}{2} (e+f x)\right ) \sec ^{\frac {5}{2}}(e+f x) \sqrt {a (1+\sec (e+f x))} \left (\frac {\left (8 (c+d)^{5/2} \arctan \left (\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )-\sqrt {d} \left (15 c^2+20 c d+8 d^2\right ) \arctan \left (\frac {\sqrt {d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d} \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )\right ) \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}} \sec \left (\frac {1}{2} (e+f x)\right ) \sqrt {1+\sec (e+f x)}}{(c+d)^{5/2} \sqrt {\frac {1}{1+\cos (e+f x)}}}-\frac {2 c d (d (7 c+4 d)+3 c (3 c+2 d) \cos (e+f x)) \sec ^{\frac {3}{2}}(e+f x) \sin \left (\frac {1}{2} (e+f x)\right )}{(c+d)^2 (c+d \sec (e+f x))^2}\right )}{8 c^3 f (c+d \sec (e+f x))^3} \]

[In]

Integrate[Sqrt[a + a*Sec[e + f*x]]/(c + d*Sec[e + f*x])^3,x]

[Out]

((d + c*Cos[e + f*x])^3*Sec[(e + f*x)/2]*Sec[e + f*x]^(5/2)*Sqrt[a*(1 + Sec[e + f*x])]*(((8*(c + d)^(5/2)*ArcT
an[Tan[(e + f*x)/2]/Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])]] - Sqrt[d]*(15*c^2 + 20*c*d + 8*d^2)*ArcTan[(Sqrt[d]
*Tan[(e + f*x)/2])/(Sqrt[c + d]*Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])])])*Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])]
*Sec[(e + f*x)/2]*Sqrt[1 + Sec[e + f*x]])/((c + d)^(5/2)*Sqrt[(1 + Cos[e + f*x])^(-1)]) - (2*c*d*(d*(7*c + 4*d
) + 3*c*(3*c + 2*d)*Cos[e + f*x])*Sec[e + f*x]^(3/2)*Sin[(e + f*x)/2])/((c + d)^2*(c + d*Sec[e + f*x])^2)))/(8
*c^3*f*(c + d*Sec[e + f*x])^3)

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(76268\) vs. \(2(249)=498\).

Time = 17.66 (sec) , antiderivative size = 76269, normalized size of antiderivative = 265.75

method	result	size
default	\(\text {Expression too large to display}\)	\(76269\)

[In]

int((a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 545 vs. \(2 (249) = 498\).

Time = 4.36 (sec) , antiderivative size = 2368, normalized size of antiderivative = 8.25 \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

[1/8*((15*c^2*d^2 + 20*c*d^3 + 8*d^4 + (15*c^4 + 20*c^3*d + 8*c^2*d^2)*cos(f*x + e)^3 + (15*c^4 + 50*c^3*d + 4
8*c^2*d^2 + 16*c*d^3)*cos(f*x + e)^2 + (30*c^3*d + 55*c^2*d^2 + 36*c*d^3 + 8*d^4)*cos(f*x + e))*sqrt(-a*d/(c +
d))*log((2*(c + d)*sqrt(-a*d/(c + d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + (a*
c + 2*a*d)*cos(f*x + e)^2 - a*d + (a*c + a*d)*cos(f*x + e))/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e) + d)) + 8
*(c^2*d^2 + 2*c*d^3 + d^4 + (c^4 + 2*c^3*d + c^2*d^2)*cos(f*x + e)^3 + (c^4 + 4*c^3*d + 5*c^2*d^2 + 2*c*d^3)*c
os(f*x + e)^2 + (2*c^3*d + 5*c^2*d^2 + 4*c*d^3 + d^4)*cos(f*x + e))*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(
-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)
) - 2*(3*(3*c^3*d + 2*c^2*d^2)*cos(f*x + e)^2 + (7*c^2*d^2 + 4*c*d^3)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/
cos(f*x + e))*sin(f*x + e))/((c^7 + 2*c^6*d + c^5*d^2)*f*cos(f*x + e)^3 + (c^7 + 4*c^6*d + 5*c^5*d^2 + 2*c^4*d
^3)*f*cos(f*x + e)^2 + (2*c^6*d + 5*c^5*d^2 + 4*c^4*d^3 + c^3*d^4)*f*cos(f*x + e) + (c^5*d^2 + 2*c^4*d^3 + c^3
*d^4)*f), -1/8*(16*(c^2*d^2 + 2*c*d^3 + d^4 + (c^4 + 2*c^3*d + c^2*d^2)*cos(f*x + e)^3 + (c^4 + 4*c^3*d + 5*c^
2*d^2 + 2*c*d^3)*cos(f*x + e)^2 + (2*c^3*d + 5*c^2*d^2 + 4*c*d^3 + d^4)*cos(f*x + e))*sqrt(a)*arctan(sqrt((a*c
os(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - (15*c^2*d^2 + 20*c*d^3 + 8*d^4 + (15*c^4
+ 20*c^3*d + 8*c^2*d^2)*cos(f*x + e)^3 + (15*c^4 + 50*c^3*d + 48*c^2*d^2 + 16*c*d^3)*cos(f*x + e)^2 + (30*c^3
*d + 55*c^2*d^2 + 36*c*d^3 + 8*d^4)*cos(f*x + e))*sqrt(-a*d/(c + d))*log((2*(c + d)*sqrt(-a*d/(c + d))*sqrt((a
*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + (a*c + 2*a*d)*cos(f*x + e)^2 - a*d + (a*c + a*d)*
cos(f*x + e))/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e) + d)) + 2*(3*(3*c^3*d + 2*c^2*d^2)*cos(f*x + e)^2 + (7*
c^2*d^2 + 4*c*d^3)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/((c^7 + 2*c^6*d + c^5*d
^2)*f*cos(f*x + e)^3 + (c^7 + 4*c^6*d + 5*c^5*d^2 + 2*c^4*d^3)*f*cos(f*x + e)^2 + (2*c^6*d + 5*c^5*d^2 + 4*c^4
*d^3 + c^3*d^4)*f*cos(f*x + e) + (c^5*d^2 + 2*c^4*d^3 + c^3*d^4)*f), 1/4*((15*c^2*d^2 + 20*c*d^3 + 8*d^4 + (15
*c^4 + 20*c^3*d + 8*c^2*d^2)*cos(f*x + e)^3 + (15*c^4 + 50*c^3*d + 48*c^2*d^2 + 16*c*d^3)*cos(f*x + e)^2 + (30
*c^3*d + 55*c^2*d^2 + 36*c*d^3 + 8*d^4)*cos(f*x + e))*sqrt(a*d/(c + d))*arctan((c + d)*sqrt(a*d/(c + d))*sqrt(
(a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(a*d*sin(f*x + e))) + 4*(c^2*d^2 + 2*c*d^3 + d^4 + (c^4 + 2*c^
3*d + c^2*d^2)*cos(f*x + e)^3 + (c^4 + 4*c^3*d + 5*c^2*d^2 + 2*c*d^3)*cos(f*x + e)^2 + (2*c^3*d + 5*c^2*d^2 +
4*c*d^3 + d^4)*cos(f*x + e))*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x +
e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) - (3*(3*c^3*d + 2*c^2*d^2)*cos(f*x +
e)^2 + (7*c^2*d^2 + 4*c*d^3)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/((c^7 + 2*c^6
*d + c^5*d^2)*f*cos(f*x + e)^3 + (c^7 + 4*c^6*d + 5*c^5*d^2 + 2*c^4*d^3)*f*cos(f*x + e)^2 + (2*c^6*d + 5*c^5*d
^2 + 4*c^4*d^3 + c^3*d^4)*f*cos(f*x + e) + (c^5*d^2 + 2*c^4*d^3 + c^3*d^4)*f), -1/4*(8*(c^2*d^2 + 2*c*d^3 + d^
4 + (c^4 + 2*c^3*d + c^2*d^2)*cos(f*x + e)^3 + (c^4 + 4*c^3*d + 5*c^2*d^2 + 2*c*d^3)*cos(f*x + e)^2 + (2*c^3*d
+ 5*c^2*d^2 + 4*c*d^3 + d^4)*cos(f*x + e))*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e
)/(sqrt(a)*sin(f*x + e))) - (15*c^2*d^2 + 20*c*d^3 + 8*d^4 + (15*c^4 + 20*c^3*d + 8*c^2*d^2)*cos(f*x + e)^3 +
(15*c^4 + 50*c^3*d + 48*c^2*d^2 + 16*c*d^3)*cos(f*x + e)^2 + (30*c^3*d + 55*c^2*d^2 + 36*c*d^3 + 8*d^4)*cos(f*
x + e))*sqrt(a*d/(c + d))*arctan((c + d)*sqrt(a*d/(c + d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e
)/(a*d*sin(f*x + e))) + (3*(3*c^3*d + 2*c^2*d^2)*cos(f*x + e)^2 + (7*c^2*d^2 + 4*c*d^3)*cos(f*x + e))*sqrt((a*
cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/((c^7 + 2*c^6*d + c^5*d^2)*f*cos(f*x + e)^3 + (c^7 + 4*c^6*d + 5
*c^5*d^2 + 2*c^4*d^3)*f*cos(f*x + e)^2 + (2*c^6*d + 5*c^5*d^2 + 4*c^4*d^3 + c^3*d^4)*f*cos(f*x + e) + (c^5*d^2
+ 2*c^4*d^3 + c^3*d^4)*f)]

Sympy [F]

\[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^3} \, dx=\int \frac {\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )}}{\left (c + d \sec {\left (e + f x \right )}\right )^{3}}\, dx \]

[In]

integrate((a+a*sec(f*x+e))**(1/2)/(c+d*sec(f*x+e))**3,x)

[Out]

Integral(sqrt(a*(sec(e + f*x) + 1))/(c + d*sec(e + f*x))**3, x)

Maxima [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^3} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^3} \, dx=\int { \frac {\sqrt {a \sec \left (f x + e\right ) + a}}{{\left (d \sec \left (f x + e\right ) + c\right )}^{3}} \,d x } \]

[In]

integrate((a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e))^3,x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^3} \, dx=\int \frac {\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}}{{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^3} \,d x \]

[In]

int((a + a/cos(e + f*x))^(1/2)/(c + d/cos(e + f*x))^3,x)

[Out]

int((a + a/cos(e + f*x))^(1/2)/(c + d/cos(e + f*x))^3, x)