3.199 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c+d \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=130 \[ \frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{f \sqrt {c-d} (c+d)^{5/2}}+\frac {3 a^2 \tan (e+f x)}{2 f (c+d)^2 (c+d \sec (e+f x))}+\frac {\tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{2 f (c+d) (c+d \sec (e+f x))^2} \]

[Out]

3*a^2*arctanh((c-d)^(1/2)*tan(1/2*e+1/2*f*x)/(c+d)^(1/2))/(c+d)^(5/2)/f/(c-d)^(1/2)+1/2*(a^2+a^2*sec(f*x+e))*t
an(f*x+e)/(c+d)/f/(c+d*sec(f*x+e))^2+3/2*a^2*tan(f*x+e)/(c+d)^2/f/(c+d*sec(f*x+e))

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Rubi [A]  time = 0.20, antiderivative size = 184, normalized size of antiderivative = 1.42, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3987, 94, 93, 205} \[ -\frac {3 a^3 \tan (e+f x) \tan ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a \sec (e+f x)+a}}{\sqrt {c-d} \sqrt {a-a \sec (e+f x)}}\right )}{f \sqrt {c-d} (c+d)^{5/2} \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {3 a^2 \tan (e+f x)}{2 f (c+d)^2 (c+d \sec (e+f x))}+\frac {\tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{2 f (c+d) (c+d \sec (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^2)/(c + d*Sec[e + f*x])^3,x]

[Out]

(-3*a^3*ArcTan[(Sqrt[c + d]*Sqrt[a + a*Sec[e + f*x]])/(Sqrt[c - d]*Sqrt[a - a*Sec[e + f*x]])]*Tan[e + f*x])/(S
qrt[c - d]*(c + d)^(5/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + ((a^2 + a^2*Sec[e + f*x])*Tan[
e + f*x])/(2*(c + d)*f*(c + d*Sec[e + f*x])^2) + (3*a^2*Tan[e + f*x])/(2*(c + d)^2*f*(c + d*Sec[e + f*x]))

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 94

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

Rule 205

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

Rule 3987

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

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c+d \sec (e+f x))^3} \, dx &=-\frac {\left (a^2 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{3/2}}{\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 {\left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{2 (c+d) f (c+d \sec (e+f x))^2}-\frac {\left (3 a^3 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+a x}}{\sqrt {a-a x} (c+d x)^2} \, dx,x,\sec (e+f x)\right )}{2 (c+d) f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {\left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{2 (c+d) f (c+d \sec (e+f x))^2}+\frac {3 a^2 \tan (e+f x)}{2 (c+d)^2 f (c+d \sec (e+f x))}-\frac {\left (3 a^4 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-a x} \sqrt {a+a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{2 (c+d)^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {\left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{2 (c+d) f (c+d \sec (e+f x))^2}+\frac {3 a^2 \tan (e+f x)}{2 (c+d)^2 f (c+d \sec (e+f x))}-\frac {\left (3 a^4 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{a c-a d-(-a c-a d) x^2} \, dx,x,\frac {\sqrt {a+a \sec (e+f 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 {3 a^3 \tan ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a+a \sec (e+f x)}}{\sqrt {c-d} \sqrt {a-a \sec (e+f x)}}\right ) \tan (e+f x)}{\sqrt {c-d} (c+d)^{5/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {\left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{2 (c+d) f (c+d \sec (e+f x))^2}+\frac {3 a^2 \tan (e+f x)}{2 (c+d)^2 f (c+d \sec (e+f x))}\\ \end {align*}

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Mathematica [C]  time = 1.25, size = 249, normalized size = 1.92 \[ \frac {a^2 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) (\sec (e+f x)+1)^2 (c \cos (e+f x)+d) \left (-\frac {6 i (\cos (e)-i \sin (e)) (c \cos (e+f x)+d)^2 \tan ^{-1}\left (\frac {(\sin (e)+i \cos (e)) \left (\tan \left (\frac {f x}{2}\right ) (c \cos (e)-d)+c \sin (e)\right )}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}\right )}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}+\frac {\sec (e) \left (\left (c^2-4 c d-2 d^2\right ) \sin (e)+c (4 c+d) \sin (f x)\right ) (c \cos (e+f x)+d)}{c^2}+\frac {(c-d) (c+d) \sec (e) (c \sin (f x)-d \sin (e))}{c^2}\right )}{8 f (c+d)^2 (c+d \sec (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^2)/(c + d*Sec[e + f*x])^3,x]

[Out]

(a^2*(d + c*Cos[e + f*x])*Sec[(e + f*x)/2]^4*Sec[e + f*x]*(1 + Sec[e + f*x])^2*(((-6*I)*ArcTan[((I*Cos[e] + Si
n[e])*(c*Sin[e] + (-d + c*Cos[e])*Tan[(f*x)/2]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e])^2])]*(d + c*Cos[e +
f*x])^2*(Cos[e] - I*Sin[e]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e])^2]) + ((c - d)*(c + d)*Sec[e]*(-(d*Sin
[e]) + c*Sin[f*x]))/c^2 + ((d + c*Cos[e + f*x])*Sec[e]*((c^2 - 4*c*d - 2*d^2)*Sin[e] + c*(4*c + d)*Sin[f*x]))/
c^2))/(8*(c + d)^2*f*(c + d*Sec[e + f*x])^3)

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fricas [B]  time = 0.50, size = 622, normalized size = 4.78 \[ \left [\frac {3 \, {\left (a^{2} c^{2} \cos \left (f x + e\right )^{2} + 2 \, a^{2} c d \cos \left (f x + e\right ) + a^{2} d^{2}\right )} \sqrt {c^{2} - d^{2}} \log \left (\frac {2 \, c d \cos \left (f x + e\right ) - {\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {c^{2} - d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) + 2 \, {\left (a^{2} c^{3} + 4 \, a^{2} c^{2} d - a^{2} c d^{2} - 4 \, a^{2} d^{3} + {\left (4 \, a^{2} c^{3} + a^{2} c^{2} d - 4 \, a^{2} c d^{2} - a^{2} d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, {\left ({\left (c^{6} + 2 \, c^{5} d - 2 \, c^{3} d^{3} - c^{2} d^{4}\right )} f \cos \left (f x + e\right )^{2} + 2 \, {\left (c^{5} d + 2 \, c^{4} d^{2} - 2 \, c^{2} d^{4} - c d^{5}\right )} f \cos \left (f x + e\right ) + {\left (c^{4} d^{2} + 2 \, c^{3} d^{3} - 2 \, c d^{5} - d^{6}\right )} f\right )}}, \frac {3 \, {\left (a^{2} c^{2} \cos \left (f x + e\right )^{2} + 2 \, a^{2} c d \cos \left (f x + e\right ) + a^{2} d^{2}\right )} \sqrt {-c^{2} + d^{2}} \arctan \left (-\frac {\sqrt {-c^{2} + d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )}}{{\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}\right ) + {\left (a^{2} c^{3} + 4 \, a^{2} c^{2} d - a^{2} c d^{2} - 4 \, a^{2} d^{3} + {\left (4 \, a^{2} c^{3} + a^{2} c^{2} d - 4 \, a^{2} c d^{2} - a^{2} d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (c^{6} + 2 \, c^{5} d - 2 \, c^{3} d^{3} - c^{2} d^{4}\right )} f \cos \left (f x + e\right )^{2} + 2 \, {\left (c^{5} d + 2 \, c^{4} d^{2} - 2 \, c^{2} d^{4} - c d^{5}\right )} f \cos \left (f x + e\right ) + {\left (c^{4} d^{2} + 2 \, c^{3} d^{3} - 2 \, c d^{5} - d^{6}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(3*(a^2*c^2*cos(f*x + e)^2 + 2*a^2*c*d*cos(f*x + e) + a^2*d^2)*sqrt(c^2 - d^2)*log((2*c*d*cos(f*x + e) -
(c^2 - 2*d^2)*cos(f*x + e)^2 + 2*sqrt(c^2 - d^2)*(d*cos(f*x + e) + c)*sin(f*x + e) + 2*c^2 - d^2)/(c^2*cos(f*x
+ e)^2 + 2*c*d*cos(f*x + e) + d^2)) + 2*(a^2*c^3 + 4*a^2*c^2*d - a^2*c*d^2 - 4*a^2*d^3 + (4*a^2*c^3 + a^2*c^2
*d - 4*a^2*c*d^2 - a^2*d^3)*cos(f*x + e))*sin(f*x + e))/((c^6 + 2*c^5*d - 2*c^3*d^3 - c^2*d^4)*f*cos(f*x + e)^
2 + 2*(c^5*d + 2*c^4*d^2 - 2*c^2*d^4 - c*d^5)*f*cos(f*x + e) + (c^4*d^2 + 2*c^3*d^3 - 2*c*d^5 - d^6)*f), 1/2*(
3*(a^2*c^2*cos(f*x + e)^2 + 2*a^2*c*d*cos(f*x + e) + a^2*d^2)*sqrt(-c^2 + d^2)*arctan(-sqrt(-c^2 + d^2)*(d*cos
(f*x + e) + c)/((c^2 - d^2)*sin(f*x + e))) + (a^2*c^3 + 4*a^2*c^2*d - a^2*c*d^2 - 4*a^2*d^3 + (4*a^2*c^3 + a^2
*c^2*d - 4*a^2*c*d^2 - a^2*d^3)*cos(f*x + e))*sin(f*x + e))/((c^6 + 2*c^5*d - 2*c^3*d^3 - c^2*d^4)*f*cos(f*x +
e)^2 + 2*(c^5*d + 2*c^4*d^2 - 2*c^2*d^4 - c*d^5)*f*cos(f*x + e) + (c^4*d^2 + 2*c^3*d^3 - 2*c*d^5 - d^6)*f)]

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giac [A]  time = 3.14, size = 220, normalized size = 1.69 \[ -\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c^{2} + d^{2}}}\right )\right )} a^{2}}{{\left (c^{2} + 2 \, c d + d^{2}\right )} \sqrt {-c^{2} + d^{2}}} + \frac {3 \, a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 5 \, a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 5 \, a^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )}^{2} {\left (c^{2} + 2 \, c d + d^{2}\right )}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(2*c - 2*d) + arctan((c*tan(1/2*f*x + 1/2*e) - d*tan(1/2*f*x + 1/2*e)
)/sqrt(-c^2 + d^2)))*a^2/((c^2 + 2*c*d + d^2)*sqrt(-c^2 + d^2)) + (3*a^2*c*tan(1/2*f*x + 1/2*e)^3 - 3*a^2*d*ta
n(1/2*f*x + 1/2*e)^3 - 5*a^2*c*tan(1/2*f*x + 1/2*e) - 5*a^2*d*tan(1/2*f*x + 1/2*e))/((c*tan(1/2*f*x + 1/2*e)^2
- d*tan(1/2*f*x + 1/2*e)^2 - c - d)^2*(c^2 + 2*c*d + d^2)))/f

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maple [A]  time = 0.77, size = 167, normalized size = 1.28 \[ \frac {8 a^{2} \left (\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{4 \left (c +d \right ) \left (\left (\tan ^{2}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c -\left (\tan ^{2}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) d -c -d \right )^{2}}+\frac {-\frac {3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{8 \left (c +d \right ) \left (\left (\tan ^{2}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c -\left (\tan ^{2}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) d -c -d \right )}+\frac {3 \arctanh \left (\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \left (c -d \right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{8 \left (c +d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}}{c +d}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/f*a^2*(1/4*tan(1/2*e+1/2*f*x)/(c+d)/(tan(1/2*e+1/2*f*x)^2*c-tan(1/2*e+1/2*f*x)^2*d-c-d)^2+3/4/(c+d)*(-1/2*ta
n(1/2*e+1/2*f*x)/(c+d)/(tan(1/2*e+1/2*f*x)^2*c-tan(1/2*e+1/2*f*x)^2*d-c-d)+1/2/(c+d)/((c+d)*(c-d))^(1/2)*arcta
nh(tan(1/2*e+1/2*f*x)*(c-d)/((c+d)*(c-d))^(1/2))))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*c^2-4*d^2>0)', see `assume?`
for more details)Is 4*c^2-4*d^2 positive or negative?

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mupad [B]  time = 3.57, size = 158, normalized size = 1.22 \[ \frac {\frac {5\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{c+d}-\frac {3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (a^2\,c-a^2\,d\right )}{{\left (c+d\right )}^2}}{f\,\left (2\,c\,d-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,c^2-2\,d^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (c^2-2\,c\,d+d^2\right )+c^2+d^2\right )}+\frac {3\,a^2\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c-d}}{\sqrt {c+d}}\right )}{f\,{\left (c+d\right )}^{5/2}\,\sqrt {c-d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((5*a^2*tan(e/2 + (f*x)/2))/(c + d) - (3*tan(e/2 + (f*x)/2)^3*(a^2*c - a^2*d))/(c + d)^2)/(f*(2*c*d - tan(e/2
+ (f*x)/2)^2*(2*c^2 - 2*d^2) + tan(e/2 + (f*x)/2)^4*(c^2 - 2*c*d + d^2) + c^2 + d^2)) + (3*a^2*atanh((tan(e/2
+ (f*x)/2)*(c - d)^(1/2))/(c + d)^(1/2)))/(f*(c + d)^(5/2)*(c - d)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int \frac {\sec {\left (e + f x \right )}}{c^{3} + 3 c^{2} d \sec {\left (e + f x \right )} + 3 c d^{2} \sec ^{2}{\left (e + f x \right )} + d^{3} \sec ^{3}{\left (e + f x \right )}}\, dx + \int \frac {2 \sec ^{2}{\left (e + f x \right )}}{c^{3} + 3 c^{2} d \sec {\left (e + f x \right )} + 3 c d^{2} \sec ^{2}{\left (e + f x \right )} + d^{3} \sec ^{3}{\left (e + f x \right )}}\, dx + \int \frac {\sec ^{3}{\left (e + f x \right )}}{c^{3} + 3 c^{2} d \sec {\left (e + f x \right )} + 3 c d^{2} \sec ^{2}{\left (e + f x \right )} + d^{3} \sec ^{3}{\left (e + f x \right )}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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