\(\int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^2} \, dx\) [220]

Optimal result

Integrand size = 31, antiderivative size = 89 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^2} \, dx=\frac {d^2 \text {arctanh}(\sin (e+f x))}{a^2 f}+\frac {(c-d)^2 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {(c-d) (c+5 d) \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.67, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4072, 91, 79, 65, 223, 209} \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^2} \, dx=\frac {(c+5 d) (c-d) \tan (e+f x)}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac {2 d^2 \tan (e+f x) \arctan \left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a (\sec (e+f x)+1)}}\right )}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {(c-d)^2 \tan (e+f x)}{3 f (a \sec (e+f x)+a)^2} \]

[In]

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

[Out]

((c - d)^2*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2) + (2*d^2*ArcTan[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a*(1 + Sec
[e + f*x])]]*Tan[e + f*x])/(a*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + ((c - d)*(c + 5*d)*Tan[e
+ f*x])/(3*f*(a^2 + a^2*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 79

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

Rule 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 209

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 4072

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(181\) vs. \(2(89)=178\).

Time = 2.41 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.03 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^2} \, dx=-\frac {2 \cos \left (\frac {1}{2} (e+f x)\right ) \left (6 d^2 \cos ^3\left (\frac {1}{2} (e+f x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+(c-d)^2 \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right )-4 \left (c^2+c d-2 d^2\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right )+(c-d)^2 \cos \left (\frac {1}{2} (e+f x)\right ) \tan \left (\frac {e}{2}\right )\right )}{3 a^2 f (1+\cos (e+f x))^2} \]

[In]

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

[Out]

(-2*Cos[(e + f*x)/2]*(6*d^2*Cos[(e + f*x)/2]^3*(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2
] + Sin[(e + f*x)/2]]) + (c - d)^2*Sec[e/2]*Sin[(f*x)/2] - 4*(c^2 + c*d - 2*d^2)*Cos[(e + f*x)/2]^2*Sec[e/2]*S
in[(f*x)/2] + (c - d)^2*Cos[(e + f*x)/2]*Tan[e/2]))/(3*a^2*f*(1 + Cos[e + f*x])^2)

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94

method	result	size
parallelrisch	\(\frac {-6 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) d^{2}+6 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) d^{2}-\left (c -d \right ) \left (\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-3 c -9 d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{6 a^{2} f}\)	\(84\)
derivativedivides	\(\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c d +\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} c d}{3}+\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c^{2}-3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d^{2}-2 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) d^{2}-\frac {c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} d^{2}}{3}+2 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) d^{2}}{2 f \,a^{2}}\)	\(131\)
default	\(\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c d +\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} c d}{3}+\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c^{2}-3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d^{2}-2 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) d^{2}-\frac {c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} d^{2}}{3}+2 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) d^{2}}{2 f \,a^{2}}\)	\(131\)
risch	\(\frac {2 i \left (3 c^{2} {\mathrm e}^{2 i \left (f x +e \right )}-3 d^{2} {\mathrm e}^{2 i \left (f x +e \right )}+3 c^{2} {\mathrm e}^{i \left (f x +e \right )}+6 d \,{\mathrm e}^{i \left (f x +e \right )} c -9 d^{2} {\mathrm e}^{i \left (f x +e \right )}+2 c^{2}+2 c d -4 d^{2}\right )}{3 f \,a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3}}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{a^{2} f}-\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{a^{2} f}\)	\(155\)
norman	\(\frac {-\frac {\left (c^{2}-2 c d +d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{6 a f}+\frac {\left (c^{2}+2 c d -3 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 a f}+\frac {\left (5 c^{2}+2 c d -7 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{6 a f}-\frac {\left (7 c^{2}+10 c d -17 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{6 a f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{2} a}+\frac {d^{2} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{a^{2} f}-\frac {d^{2} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{a^{2} f}\)	\(195\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.74 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^2} \, dx=\frac {3 \, {\left (d^{2} \cos \left (f x + e\right )^{2} + 2 \, d^{2} \cos \left (f x + e\right ) + d^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, {\left (d^{2} \cos \left (f x + e\right )^{2} + 2 \, d^{2} \cos \left (f x + e\right ) + d^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (c^{2} + 4 \, c d - 5 \, d^{2} + 2 \, {\left (c^{2} + c d - 2 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{6 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^2} \, dx=\frac {\int \frac {c^{2} \sec {\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{2} \sec ^{3}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {2 c d \sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (85) = 170\).

Time = 0.22 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.19 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^2} \, dx=-\frac {d^{2} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} - \frac {2 \, c d {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}} - \frac {c^{2} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, f} \]

[In]

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

[Out]

-1/6*(d^2*((9*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 6*log(sin(f*x + e)/
(cos(f*x + e) + 1) + 1)/a^2 + 6*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^2) - 2*c*d*(3*sin(f*x + e)/(cos(f*x
+ e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - c^2*(3*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^
3/(cos(f*x + e) + 1)^3)/a^2)/f

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.78 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^2} \, dx=\frac {\frac {6 \, d^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, d^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}} - \frac {a^{4} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a^{4} c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + a^{4} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a^{4} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 6 \, a^{4} c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 9 \, a^{4} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{6}}}{6 \, f} \]

[In]

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

[Out]

1/6*(6*d^2*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^2 - 6*d^2*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^2 - (a^4*c^2*ta
n(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) - 6*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)/f

Mupad [B] (verification not implemented)

Time = 13.42 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^2} \, dx=\frac {2\,d^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{a^2\,f}-\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {{\left (c-d\right )}^2}{2\,a^2}-\frac {c^2-d^2}{a^2}\right )}{f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,{\left (c-d\right )}^2}{6\,a^2\,f} \]

[In]

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

[Out]

(2*d^2*atanh(tan(e/2 + (f*x)/2)))/(a^2*f) - (tan(e/2 + (f*x)/2)*((c - d)^2/(2*a^2) - (c^2 - d^2)/a^2))/f - (ta
n(e/2 + (f*x)/2)^3*(c - d)^2)/(6*a^2*f)