3.1.2 \(\int (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^4 \, dx\) [2]

Optimal. Leaf size=140 \[ a^2 c^4 x-\frac {3 a^2 c^4 \tanh ^{-1}(\sin (e+f x))}{4 f}-\frac {a^2 c^4 \tan (e+f x)}{f}+\frac {3 a^2 c^4 \sec (e+f x) \tan (e+f x)}{4 f}+\frac {a^2 c^4 \tan ^3(e+f x)}{3 f}-\frac {a^2 c^4 \sec (e+f x) \tan ^3(e+f x)}{2 f}+\frac {a^2 c^4 \tan ^5(e+f x)}{5 f} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3989, 3971, 3554, 8, 2691, 3855, 2687, 30} \begin {gather*} \frac {a^2 c^4 \tan ^5(e+f x)}{5 f}+\frac {a^2 c^4 \tan ^3(e+f x)}{3 f}-\frac {a^2 c^4 \tan (e+f x)}{f}-\frac {3 a^2 c^4 \tanh ^{-1}(\sin (e+f x))}{4 f}-\frac {a^2 c^4 \tan ^3(e+f x) \sec (e+f x)}{2 f}+\frac {3 a^2 c^4 \tan (e+f x) \sec (e+f x)}{4 f}+a^2 c^4 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^4,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3971

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

Rule 3989

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

Rubi steps

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

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Mathematica [A]
time = 1.19, size = 146, normalized size = 1.04 \begin {gather*} \frac {a^2 c^4 \sec ^5(e+f x) \left (600 (e+f x) \cos (e+f x)-720 \tanh ^{-1}(\sin (e+f x)) \cos ^5(e+f x)+300 e \cos (3 (e+f x))+300 f x \cos (3 (e+f x))+60 e \cos (5 (e+f x))+60 f x \cos (5 (e+f x))+40 \sin (e+f x)+60 \sin (2 (e+f x))-220 \sin (3 (e+f x))+150 \sin (4 (e+f x))-68 \sin (5 (e+f x))\right )}{960 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^4,x]

[Out]

(a^2*c^4*Sec[e + f*x]^5*(600*(e + f*x)*Cos[e + f*x] - 720*ArcTanh[Sin[e + f*x]]*Cos[e + f*x]^5 + 300*e*Cos[3*(
e + f*x)] + 300*f*x*Cos[3*(e + f*x)] + 60*e*Cos[5*(e + f*x)] + 60*f*x*Cos[5*(e + f*x)] + 40*Sin[e + f*x] + 60*
Sin[2*(e + f*x)] - 220*Sin[3*(e + f*x)] + 150*Sin[4*(e + f*x)] - 68*Sin[5*(e + f*x)]))/(960*f)

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Maple [A]
time = 0.10, size = 206, normalized size = 1.47

method	result	size
risch	\(a^{2} c^{4} x -\frac {i c^{4} a^{2} \left (75 \,{\mathrm e}^{9 i \left (f x +e \right )}+60 \,{\mathrm e}^{8 i \left (f x +e \right )}+30 \,{\mathrm e}^{7 i \left (f x +e \right )}+360 \,{\mathrm e}^{6 i \left (f x +e \right )}+320 \,{\mathrm e}^{4 i \left (f x +e \right )}-30 \,{\mathrm e}^{3 i \left (f x +e \right )}+280 \,{\mathrm e}^{2 i \left (f x +e \right )}-75 \,{\mathrm e}^{i \left (f x +e \right )}+68\right )}{30 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{5}}-\frac {3 c^{4} a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{4 f}+\frac {3 c^{4} a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{4 f}\)	\(173\)
derivativedivides	\(\frac {-c^{4} a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (f x +e \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (f x +e \right )\right )}{15}\right ) \tan \left (f x +e \right )-2 c^{4} a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (f x +e \right )\right )}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )+c^{4} a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )+4 c^{4} a^{2} \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )-c^{4} a^{2} \tan \left (f x +e \right )-2 c^{4} a^{2} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+c^{4} a^{2} \left (f x +e \right )}{f}\)	\(206\)
default	\(\frac {-c^{4} a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (f x +e \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (f x +e \right )\right )}{15}\right ) \tan \left (f x +e \right )-2 c^{4} a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (f x +e \right )\right )}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )+c^{4} a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )+4 c^{4} a^{2} \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )-c^{4} a^{2} \tan \left (f x +e \right )-2 c^{4} a^{2} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+c^{4} a^{2} \left (f x +e \right )}{f}\)	\(206\)
norman	\(\frac {a^{2} c^{4} x \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-a^{2} c^{4} x +5 a^{2} c^{4} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-10 a^{2} c^{4} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+10 a^{2} c^{4} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-5 a^{2} c^{4} x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {c^{4} a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 f}-\frac {11 c^{4} a^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}+\frac {164 c^{4} a^{2} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 f}-\frac {53 c^{4} a^{2} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}+\frac {7 c^{4} a^{2} \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}+\frac {3 c^{4} a^{2} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4 f}-\frac {3 c^{4} a^{2} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4 f}\)	\(281\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 259, normalized size = 1.85 \begin {gather*} \frac {8 \, {\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} a^{2} c^{4} - 40 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c^{4} + 120 \, {\left (f x + e\right )} a^{2} c^{4} + 15 \, a^{2} c^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 120 \, a^{2} c^{4} {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 240 \, a^{2} c^{4} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) - 120 \, a^{2} c^{4} \tan \left (f x + e\right )}{120 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^4,x, algorithm="maxima")

[Out]

1/120*(8*(3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^2*c^4 - 40*(tan(f*x + e)^3 + 3*tan(f*x + e
))*a^2*c^4 + 120*(f*x + e)*a^2*c^4 + 15*a^2*c^4*(2*(3*sin(f*x + e)^3 - 5*sin(f*x + e))/(sin(f*x + e)^4 - 2*sin
(f*x + e)^2 + 1) - 3*log(sin(f*x + e) + 1) + 3*log(sin(f*x + e) - 1)) - 120*a^2*c^4*(2*sin(f*x + e)/(sin(f*x +
e)^2 - 1) - log(sin(f*x + e) + 1) + log(sin(f*x + e) - 1)) - 240*a^2*c^4*log(sec(f*x + e) + tan(f*x + e)) - 1
20*a^2*c^4*tan(f*x + e))/f

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Fricas [A]
time = 2.65, size = 174, normalized size = 1.24 \begin {gather*} \frac {120 \, a^{2} c^{4} f x \cos \left (f x + e\right )^{5} - 45 \, a^{2} c^{4} \cos \left (f x + e\right )^{5} \log \left (\sin \left (f x + e\right ) + 1\right ) + 45 \, a^{2} c^{4} \cos \left (f x + e\right )^{5} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (68 \, a^{2} c^{4} \cos \left (f x + e\right )^{4} - 75 \, a^{2} c^{4} \cos \left (f x + e\right )^{3} + 4 \, a^{2} c^{4} \cos \left (f x + e\right )^{2} + 30 \, a^{2} c^{4} \cos \left (f x + e\right ) - 12 \, a^{2} c^{4}\right )} \sin \left (f x + e\right )}{120 \, f \cos \left (f x + e\right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^4,x, algorithm="fricas")

[Out]

1/120*(120*a^2*c^4*f*x*cos(f*x + e)^5 - 45*a^2*c^4*cos(f*x + e)^5*log(sin(f*x + e) + 1) + 45*a^2*c^4*cos(f*x +
e)^5*log(-sin(f*x + e) + 1) - 2*(68*a^2*c^4*cos(f*x + e)^4 - 75*a^2*c^4*cos(f*x + e)^3 + 4*a^2*c^4*cos(f*x +
e)^2 + 30*a^2*c^4*cos(f*x + e) - 12*a^2*c^4)*sin(f*x + e))/(f*cos(f*x + e)^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} c^{4} \left (\int 1\, dx + \int \left (- 2 \sec {\left (e + f x \right )}\right )\, dx + \int \left (- \sec ^{2}{\left (e + f x \right )}\right )\, dx + \int 4 \sec ^{3}{\left (e + f x \right )}\, dx + \int \left (- \sec ^{4}{\left (e + f x \right )}\right )\, dx + \int \left (- 2 \sec ^{5}{\left (e + f x \right )}\right )\, dx + \int \sec ^{6}{\left (e + f x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))**2*(c-c*sec(f*x+e))**4,x)

[Out]

a**2*c**4*(Integral(1, x) + Integral(-2*sec(e + f*x), x) + Integral(-sec(e + f*x)**2, x) + Integral(4*sec(e +
f*x)**3, x) + Integral(-sec(e + f*x)**4, x) + Integral(-2*sec(e + f*x)**5, x) + Integral(sec(e + f*x)**6, x))

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Giac [A]
time = 0.52, size = 172, normalized size = 1.23 \begin {gather*} \frac {60 \, {\left (f x + e\right )} a^{2} c^{4} - 45 \, a^{2} c^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right ) + 45 \, a^{2} c^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right ) + \frac {2 \, {\left (105 \, a^{2} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} - 530 \, a^{2} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 328 \, a^{2} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 110 \, a^{2} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, a^{2} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{5}}}{60 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^4,x, algorithm="giac")

[Out]

1/60*(60*(f*x + e)*a^2*c^4 - 45*a^2*c^4*log(abs(tan(1/2*f*x + 1/2*e) + 1)) + 45*a^2*c^4*log(abs(tan(1/2*f*x +
1/2*e) - 1)) + 2*(105*a^2*c^4*tan(1/2*f*x + 1/2*e)^9 - 530*a^2*c^4*tan(1/2*f*x + 1/2*e)^7 + 328*a^2*c^4*tan(1/
2*f*x + 1/2*e)^5 - 110*a^2*c^4*tan(1/2*f*x + 1/2*e)^3 + 15*a^2*c^4*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)
^2 - 1)^5)/f

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Mupad [B]
time = 2.30, size = 195, normalized size = 1.39 \begin {gather*} a^2\,c^4\,x+\frac {\frac {7\,a^2\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{2}-\frac {53\,a^2\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{3}+\frac {164\,a^2\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{15}-\frac {11\,a^2\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3}+\frac {a^2\,c^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{2}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}-\frac {3\,a^2\,c^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{2\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^2*(c - c/cos(e + f*x))^4,x)

[Out]

a^2*c^4*x + ((164*a^2*c^4*tan(e/2 + (f*x)/2)^5)/15 - (11*a^2*c^4*tan(e/2 + (f*x)/2)^3)/3 - (53*a^2*c^4*tan(e/2
+ (f*x)/2)^7)/3 + (7*a^2*c^4*tan(e/2 + (f*x)/2)^9)/2 + (a^2*c^4*tan(e/2 + (f*x)/2))/2)/(f*(5*tan(e/2 + (f*x)/
2)^2 - 10*tan(e/2 + (f*x)/2)^4 + 10*tan(e/2 + (f*x)/2)^6 - 5*tan(e/2 + (f*x)/2)^8 + tan(e/2 + (f*x)/2)^10 - 1)
) - (3*a^2*c^4*atanh(tan(e/2 + (f*x)/2)))/(2*f)

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