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

Optimal. Leaf size=129 \[ \frac {2 d^2 \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{a^2 f (c-d)^{5/2} \sqrt {c+d}}+\frac {(c-4 d) \tan (e+f x)}{3 f (c-d)^2 \left (a^2 \sec (e+f x)+a^2\right )}+\frac {\tan (e+f x)}{3 f (c-d) (a \sec (e+f x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 183, normalized size of antiderivative = 1.42, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3987, 104, 152, 12, 93, 205} \[ \frac {(c-4 d) \tan (e+f x)}{3 f (c-d)^2 \left (a^2 \sec (e+f x)+a^2\right )}-\frac {2 d^2 \tan (e+f x) \tan ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a \sec (e+f x)+a}}{\sqrt {c-d} \sqrt {a-a \sec (e+f x)}}\right )}{a f (c-d)^{5/2} \sqrt {c+d} \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {\tan (e+f x)}{3 f (c-d) (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 104

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] && LtQ[m, -1] &&
IntegersQ[2*m, 2*n, 2*p]

Rule 152

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] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

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

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Mathematica [C]  time = 1.65, size = 209, normalized size = 1.62 \[ \frac {\cos \left (\frac {1}{2} (e+f x)\right ) \left (\sec \left (\frac {e}{2}\right ) \left (-3 (c-2 d) \sin \left (e+\frac {f x}{2}\right )+(2 c-5 d) \sin \left (e+\frac {3 f x}{2}\right )+3 (c-3 d) \sin \left (\frac {f x}{2}\right )\right )-\frac {24 i d^2 (\cos (e)-i \sin (e)) \cos ^3\left (\frac {1}{2} (e+f x)\right ) \tan ^{-1}\left (\frac {(\sin (e)+i \cos (e)) \left (\tan \left (\frac {f x}{2}\right ) (c \cos (e)-d)+c \sin (e)\right )}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}\right )}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}\right )}{3 a^2 f (c-d)^2 (\cos (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(e + f*x)/2]*(((-24*I)*d^2*ArcTan[((I*Cos[e] + Sin[e])*(c*Sin[e] + (-d + c*Cos[e])*Tan[(f*x)/2]))/(Sqrt[c
^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e])^2])]*Cos[(e + f*x)/2]^3*(Cos[e] - I*Sin[e]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e]
- I*Sin[e])^2]) + Sec[e/2]*(3*(c - 3*d)*Sin[(f*x)/2] - 3*(c - 2*d)*Sin[e + (f*x)/2] + (2*c - 5*d)*Sin[e + (3*f
*x)/2])))/(3*a^2*(c - d)^2*f*(1 + Cos[e + f*x])^2)

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fricas [B]  time = 0.48, size = 598, normalized size = 4.64 \[ \left [\frac {3 \, {\left (d^{2} \cos \left (f x + e\right )^{2} + 2 \, d^{2} \cos \left (f x + e\right ) + d^{2}\right )} \sqrt {c^{2} - d^{2}} \log \left (\frac {2 \, c d \cos \left (f x + e\right ) - {\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {c^{2} - d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) + 2 \, {\left (c^{3} - 4 \, c^{2} d - c d^{2} + 4 \, d^{3} + {\left (2 \, c^{3} - 5 \, c^{2} d - 2 \, c d^{2} + 5 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{6 \, {\left ({\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right )^{2} + 2 \, {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right ) + {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f\right )}}, \frac {3 \, {\left (d^{2} \cos \left (f x + e\right )^{2} + 2 \, d^{2} \cos \left (f x + e\right ) + d^{2}\right )} \sqrt {-c^{2} + d^{2}} \arctan \left (-\frac {\sqrt {-c^{2} + d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )}}{{\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}\right ) + {\left (c^{3} - 4 \, c^{2} d - c d^{2} + 4 \, d^{3} + {\left (2 \, c^{3} - 5 \, c^{2} d - 2 \, c d^{2} + 5 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left ({\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right )^{2} + 2 \, {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right ) + {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f\right )}}\right ] \]

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)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.59, size = 258, normalized size = 2.00 \[ -\frac {\frac {12 \, {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c^{2} + d^{2}}}\right )\right )} d^{2}}{{\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} \sqrt {-c^{2} + d^{2}}} + \frac {a^{4} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a^{4} c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + a^{4} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a^{4} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 12 \, a^{4} c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 9 \, a^{4} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{6} c^{3} - 3 \, a^{6} c^{2} d + 3 \, a^{6} c d^{2} - a^{6} d^{3}}}{6 \, f} \]

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)),x, algorithm="giac")

[Out]

-1/6*(12*(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)))*d^2/((a^2*c^2 - 2*a^2*c*d + a^2*d^2)*sqrt(-c^2 + d^2)) + (a^4*c^2*tan(1/2*f*x + 1/2*
e)^3 - 2*a^4*c*d*tan(1/2*f*x + 1/2*e)^3 + a^4*d^2*tan(1/2*f*x + 1/2*e)^3 - 3*a^4*c^2*tan(1/2*f*x + 1/2*e) + 12
*a^4*c*d*tan(1/2*f*x + 1/2*e) - 9*a^4*d^2*tan(1/2*f*x + 1/2*e))/(a^6*c^3 - 3*a^6*c^2*d + 3*a^6*c*d^2 - a^6*d^3
))/f

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maple [A]  time = 0.74, size = 122, normalized size = 0.95 \[ \frac {-\frac {\frac {\left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c}{3}-\frac {\left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) d}{3}-\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) c +3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right ) d}{\left (c -d \right )^{2}}+\frac {4 d^{2} \arctanh \left (\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \left (c -d \right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (c -d \right )^{2} \sqrt {\left (c +d \right ) \left (c -d \right )}}}{2 f \,a^{2}} \]

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)),x)

[Out]

1/2/f/a^2*(-1/(c-d)^2*(1/3*tan(1/2*e+1/2*f*x)^3*c-1/3*tan(1/2*e+1/2*f*x)^3*d-tan(1/2*e+1/2*f*x)*c+3*tan(1/2*e+
1/2*f*x)*d)+4*d^2/(c-d)^2/((c+d)*(c-d))^(1/2)*arctanh(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)),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 = 1.92, size = 168, normalized size = 1.30 \[ \frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {1}{a^2\,\left (c-d\right )}-\frac {c+d}{2\,a^2\,{\left (c-d\right )}^2}\right )}{f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{6\,a^2\,f\,\left (c-d\right )}-\frac {d^2\,\mathrm {atan}\left (\frac {1{}\mathrm {i}\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c^3-3{}\mathrm {i}\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c^2\,d+3{}\mathrm {i}\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c\,d^2-1{}\mathrm {i}\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,d^3}{\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}\right )\,2{}\mathrm {i}}{a^2\,f\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(e/2 + (f*x)/2)*(1/(a^2*(c - d)) - (c + d)/(2*a^2*(c - d)^2)))/f - tan(e/2 + (f*x)/2)^3/(6*a^2*f*(c - d))
- (d^2*atan((c^3*tan(e/2 + (f*x)/2)*1i - d^3*tan(e/2 + (f*x)/2)*1i + c*d^2*tan(e/2 + (f*x)/2)*3i - c^2*d*tan(e
/2 + (f*x)/2)*3i)/((c + d)^(1/2)*(c - d)^(5/2)))*2i)/(a^2*f*(c + d)^(1/2)*(c - d)^(5/2))

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

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)),x)

[Out]

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

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