3.170 \(\int \frac{1}{\sqrt{a+a \sec (e+f x)} (c+d \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=416 \[ \frac{d^2 \tan (e+f x)}{c f \left (c^2-d^2\right ) \sqrt{a \sec (e+f x)+a} (c+d \sec (e+f x))}+\frac{2 \sqrt{a} d^{3/2} (2 c-d) \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a-a \sec (e+f x)}}{\sqrt{a} \sqrt{c+d}}\right )}{c^2 f (c-d)^2 \sqrt{c+d} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{2 \sqrt{a} \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{c^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{\sqrt{a} d^{3/2} \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a-a \sec (e+f x)}}{\sqrt{a} \sqrt{c+d}}\right )}{c f (c-d) (c+d)^{3/2} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{\sqrt{2} \sqrt{a} \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right )}{f (c-d)^2 \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}} \]

[Out]

(2*Sqrt[a]*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]]*Tan[e + f*x])/(c^2*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*
Sec[e + f*x]]) - (Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])]*Tan[e + f*x])/((c - d)^2
*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + (Sqrt[a]*d^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f
*x]])/(Sqrt[a]*Sqrt[c + d])]*Tan[e + f*x])/(c*(c - d)*(c + d)^(3/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[
e + f*x]]) + (2*Sqrt[a]*(2*c - d)*d^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])]*Ta
n[e + f*x])/(c^2*(c - d)^2*Sqrt[c + d]*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + (d^2*Tan[e + f*x
])/(c*(c^2 - d^2)*f*Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x]))

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Rubi [A]  time = 0.422065, antiderivative size = 416, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3940, 180, 63, 206, 51, 208} \[ \frac{d^2 \tan (e+f x)}{c f \left (c^2-d^2\right ) \sqrt{a \sec (e+f x)+a} (c+d \sec (e+f x))}+\frac{2 \sqrt{a} d^{3/2} (2 c-d) \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a-a \sec (e+f x)}}{\sqrt{a} \sqrt{c+d}}\right )}{c^2 f (c-d)^2 \sqrt{c+d} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{2 \sqrt{a} \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{c^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{\sqrt{a} d^{3/2} \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a-a \sec (e+f x)}}{\sqrt{a} \sqrt{c+d}}\right )}{c f (c-d) (c+d)^{3/2} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{\sqrt{2} \sqrt{a} \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right )}{f (c-d)^2 \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])^2),x]

[Out]

(2*Sqrt[a]*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]]*Tan[e + f*x])/(c^2*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*
Sec[e + f*x]]) - (Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])]*Tan[e + f*x])/((c - d)^2
*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + (Sqrt[a]*d^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f
*x]])/(Sqrt[a]*Sqrt[c + d])]*Tan[e + f*x])/(c*(c - d)*(c + d)^(3/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[
e + f*x]]) + (2*Sqrt[a]*(2*c - d)*d^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])]*Ta
n[e + f*x])/(c^2*(c - d)^2*Sqrt[c + d]*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + (d^2*Tan[e + f*x
])/(c*(c^2 - d^2)*f*Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x]))

Rule 3940

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]

Rule 180

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

Rule 63

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 206

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a+a \sec (e+f x)} (c+d \sec (e+f x))^2} \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a-a x} (a+a x) (c+d x)^2} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \left (\frac{1}{a c^2 x \sqrt{a-a x}}-\frac{1}{a (c-d)^2 (1+x) \sqrt{a-a x}}+\frac{d^2}{a c (c-d) \sqrt{a-a x} (c+d x)^2}+\frac{(2 c-d) d^2}{a c^2 (c-d)^2 \sqrt{a-a x} (c+d x)}\right ) \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{(a \tan (e+f x)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{c^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{(a \tan (e+f x)) \operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{(c-d)^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\left (a d^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} (c+d x)^2} \, dx,x,\sec (e+f x)\right )}{c (c-d) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\left (a (2 c-d) d^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{c^2 (c-d)^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{d^2 \tan (e+f x)}{c \left (c^2-d^2\right ) f \sqrt{a+a \sec (e+f x)} (c+d \sec (e+f x))}+\frac{(2 \tan (e+f x)) \operatorname{Subst}\left (\int \frac{1}{1-\frac{x^2}{a}} \, dx,x,\sqrt{a-a \sec (e+f x)}\right )}{c^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{(2 \tan (e+f x)) \operatorname{Subst}\left (\int \frac{1}{2-\frac{x^2}{a}} \, dx,x,\sqrt{a-a \sec (e+f x)}\right )}{(c-d)^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{\left (2 (2 c-d) d^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{c+d-\frac{d x^2}{a}} \, dx,x,\sqrt{a-a \sec (e+f x)}\right )}{c^2 (c-d)^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\left (a d^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{2 c (c-d) (c+d) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right ) \tan (e+f x)}{c^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right ) \tan (e+f x)}{(c-d)^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{2 \sqrt{a} (2 c-d) d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a-a \sec (e+f x)}}{\sqrt{a} \sqrt{c+d}}\right ) \tan (e+f x)}{c^2 (c-d)^2 \sqrt{c+d} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{d^2 \tan (e+f x)}{c \left (c^2-d^2\right ) f \sqrt{a+a \sec (e+f x)} (c+d \sec (e+f x))}+\frac{\left (d^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{c+d-\frac{d x^2}{a}} \, dx,x,\sqrt{a-a \sec (e+f x)}\right )}{c (c-d) (c+d) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right ) \tan (e+f x)}{c^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right ) \tan (e+f x)}{(c-d)^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{\sqrt{a} d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a-a \sec (e+f x)}}{\sqrt{a} \sqrt{c+d}}\right ) \tan (e+f x)}{c (c-d) (c+d)^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{2 \sqrt{a} (2 c-d) d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a-a \sec (e+f x)}}{\sqrt{a} \sqrt{c+d}}\right ) \tan (e+f x)}{c^2 (c-d)^2 \sqrt{c+d} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{d^2 \tan (e+f x)}{c \left (c^2-d^2\right ) f \sqrt{a+a \sec (e+f x)} (c+d \sec (e+f x))}\\ \end{align*}

Mathematica [C]  time = 35.8607, size = 472069, normalized size = 1134.78 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])^2),x]

[Out]

Result too large to show

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Maple [B]  time = 2.335, size = 117715, normalized size = 283. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \sec \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right ) + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 96.0913, size = 6564, normalized size = 15.78 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*(c^2*d^2 - c*d^3)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + sqrt(2)*(a*c^3*d
+ a*c^2*d^2 + (a*c^4 + a*c^3*d)*cos(f*x + e)^2 + (a*c^4 + 2*a*c^3*d + a*c^2*d^2)*cos(f*x + e))*sqrt(-1/a)*log
((4*sqrt(2)*(3*cos(f*x + e)^2 - cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(-1/a)*sin(f*x + e)
+ 17*cos(f*x + e)^3 + 3*cos(f*x + e)^2 - 13*cos(f*x + e) + 1)/(cos(f*x + e)^3 + 3*cos(f*x + e)^2 + 3*cos(f*x +
e) + 1)) - (5*a*c^2*d^2 + a*c*d^3 - 2*a*d^4 + (5*a*c^3*d + a*c^2*d^2 - 2*a*c*d^3)*cos(f*x + e)^2 + (5*a*c^3*d
+ 6*a*c^2*d^2 - a*c*d^3 - 2*a*d^4)*cos(f*x + e))*sqrt(-d/(a*c + a*d))*log(((c^2 + 8*c*d + 8*d^2)*cos(f*x + e)
^3 + (c^2 + 2*c*d)*cos(f*x + e)^2 + 4*((c^2 + 3*c*d + 2*d^2)*cos(f*x + e)^2 - (c*d + d^2)*cos(f*x + e))*sqrt(-
d/(a*c + a*d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) + d^2 - (6*c*d + 7*d^2)*cos(f*x + e))/(c^2
*cos(f*x + e)^3 + (c^2 + 2*c*d)*cos(f*x + e)^2 + d^2 + (2*c*d + d^2)*cos(f*x + e))) - 2*(c^3*d - c^2*d^2 - c*d
^3 + d^4 + (c^4 - c^3*d - c^2*d^2 + c*d^3)*cos(f*x + e)^2 + (c^4 - 2*c^2*d^2 + d^4)*cos(f*x + e))*sqrt(-a)*log
((8*a*cos(f*x + e)^3 + 4*(2*cos(f*x + e)^2 - cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*si
n(f*x + e) - 7*a*cos(f*x + e) + a)/(cos(f*x + e) + 1)))/((a*c^6 - a*c^5*d - a*c^4*d^2 + a*c^3*d^3)*f*cos(f*x +
e)^2 + (a*c^6 - 2*a*c^4*d^2 + a*c^2*d^4)*f*cos(f*x + e) + (a*c^5*d - a*c^4*d^2 - a*c^3*d^3 + a*c^2*d^4)*f), 1
/4*(4*(c^2*d^2 - c*d^3)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + sqrt(2)*(a*c^3*d +
a*c^2*d^2 + (a*c^4 + a*c^3*d)*cos(f*x + e)^2 + (a*c^4 + 2*a*c^3*d + a*c^2*d^2)*cos(f*x + e))*sqrt(-1/a)*log((
4*sqrt(2)*(3*cos(f*x + e)^2 - cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(-1/a)*sin(f*x + e) +
17*cos(f*x + e)^3 + 3*cos(f*x + e)^2 - 13*cos(f*x + e) + 1)/(cos(f*x + e)^3 + 3*cos(f*x + e)^2 + 3*cos(f*x + e
) + 1)) - 2*(5*a*c^2*d^2 + a*c*d^3 - 2*a*d^4 + (5*a*c^3*d + a*c^2*d^2 - 2*a*c*d^3)*cos(f*x + e)^2 + (5*a*c^3*d
+ 6*a*c^2*d^2 - a*c*d^3 - 2*a*d^4)*cos(f*x + e))*sqrt(d/(a*c + a*d))*arctan(1/2*((c + 2*d)*cos(f*x + e) - d)*
sqrt(d/(a*c + a*d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))/(d*sin(f*x + e))) - 2*(c^3*d - c^2*d^2 - c*d^3 + d
^4 + (c^4 - c^3*d - c^2*d^2 + c*d^3)*cos(f*x + e)^2 + (c^4 - 2*c^2*d^2 + d^4)*cos(f*x + e))*sqrt(-a)*log((8*a*
cos(f*x + e)^3 + 4*(2*cos(f*x + e)^2 - cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x
+ e) - 7*a*cos(f*x + e) + a)/(cos(f*x + e) + 1)))/((a*c^6 - a*c^5*d - a*c^4*d^2 + a*c^3*d^3)*f*cos(f*x + e)^2
+ (a*c^6 - 2*a*c^4*d^2 + a*c^2*d^4)*f*cos(f*x + e) + (a*c^5*d - a*c^4*d^2 - a*c^3*d^3 + a*c^2*d^4)*f), 1/4*(4*
(c^2*d^2 - c*d^3)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - 4*(c^3*d - c^2*d^2 - c*d
^3 + d^4 + (c^4 - c^3*d - c^2*d^2 + c*d^3)*cos(f*x + e)^2 + (c^4 - 2*c^2*d^2 + d^4)*cos(f*x + e))*sqrt(a)*arct
an(1/2*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*(2*cos(f*x + e) - 1)/(sqrt(a)*sin(f*x + e))) - (5*a*c^2*d^2 + a
*c*d^3 - 2*a*d^4 + (5*a*c^3*d + a*c^2*d^2 - 2*a*c*d^3)*cos(f*x + e)^2 + (5*a*c^3*d + 6*a*c^2*d^2 - a*c*d^3 - 2
*a*d^4)*cos(f*x + e))*sqrt(-d/(a*c + a*d))*log(((c^2 + 8*c*d + 8*d^2)*cos(f*x + e)^3 + (c^2 + 2*c*d)*cos(f*x +
e)^2 + 4*((c^2 + 3*c*d + 2*d^2)*cos(f*x + e)^2 - (c*d + d^2)*cos(f*x + e))*sqrt(-d/(a*c + a*d))*sqrt((a*cos(f
*x + e) + a)/cos(f*x + e))*sin(f*x + e) + d^2 - (6*c*d + 7*d^2)*cos(f*x + e))/(c^2*cos(f*x + e)^3 + (c^2 + 2*c
*d)*cos(f*x + e)^2 + d^2 + (2*c*d + d^2)*cos(f*x + e))) + 2*sqrt(2)*(a*c^3*d + a*c^2*d^2 + (a*c^4 + a*c^3*d)*c
os(f*x + e)^2 + (a*c^4 + 2*a*c^3*d + a*c^2*d^2)*cos(f*x + e))*arctan(1/4*sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos
(f*x + e))*(3*cos(f*x + e) - 1)/(sqrt(a)*sin(f*x + e)))/sqrt(a))/((a*c^6 - a*c^5*d - a*c^4*d^2 + a*c^3*d^3)*f*
cos(f*x + e)^2 + (a*c^6 - 2*a*c^4*d^2 + a*c^2*d^4)*f*cos(f*x + e) + (a*c^5*d - a*c^4*d^2 - a*c^3*d^3 + a*c^2*d
^4)*f), 1/2*(2*(c^2*d^2 - c*d^3)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - (5*a*c^2*
d^2 + a*c*d^3 - 2*a*d^4 + (5*a*c^3*d + a*c^2*d^2 - 2*a*c*d^3)*cos(f*x + e)^2 + (5*a*c^3*d + 6*a*c^2*d^2 - a*c*
d^3 - 2*a*d^4)*cos(f*x + e))*sqrt(d/(a*c + a*d))*arctan(1/2*((c + 2*d)*cos(f*x + e) - d)*sqrt(d/(a*c + a*d))*s
qrt((a*cos(f*x + e) + a)/cos(f*x + e))/(d*sin(f*x + e))) - 2*(c^3*d - c^2*d^2 - c*d^3 + d^4 + (c^4 - c^3*d - c
^2*d^2 + c*d^3)*cos(f*x + e)^2 + (c^4 - 2*c^2*d^2 + d^4)*cos(f*x + e))*sqrt(a)*arctan(1/2*sqrt((a*cos(f*x + e)
+ a)/cos(f*x + e))*(2*cos(f*x + e) - 1)/(sqrt(a)*sin(f*x + e))) + sqrt(2)*(a*c^3*d + a*c^2*d^2 + (a*c^4 + a*c
^3*d)*cos(f*x + e)^2 + (a*c^4 + 2*a*c^3*d + a*c^2*d^2)*cos(f*x + e))*arctan(1/4*sqrt(2)*sqrt((a*cos(f*x + e) +
a)/cos(f*x + e))*(3*cos(f*x + e) - 1)/(sqrt(a)*sin(f*x + e)))/sqrt(a))/((a*c^6 - a*c^5*d - a*c^4*d^2 + a*c^3*
d^3)*f*cos(f*x + e)^2 + (a*c^6 - 2*a*c^4*d^2 + a*c^2*d^4)*f*cos(f*x + e) + (a*c^5*d - a*c^4*d^2 - a*c^3*d^3 +
a*c^2*d^4)*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \left (\sec{\left (e + f x \right )} + 1\right )} \left (c + d \sec{\left (e + f x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(a*(sec(e + f*x) + 1))*(c + d*sec(e + f*x))**2), x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out