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

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

[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)

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Rubi [A]  time = 0.198988, antiderivative size = 140, normalized size of antiderivative = 1., 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 = {3904, 3886, 3473, 8, 2611, 3770, 2607, 30} \[ \frac{a^2 c^4 \tan ^5(e+f x)}{5 f}+\frac{a^2 c^4 \tan ^3(e+f x)}{3 f}-\frac{a^2 c^4 \tan (e+f x)}{f}-\frac{3 a^2 c^4 \tanh ^{-1}(\sin (e+f x))}{4 f}-\frac{a^2 c^4 \tan ^3(e+f x) \sec (e+f x)}{2 f}+\frac{3 a^2 c^4 \tan (e+f x) \sec (e+f x)}{4 f}+a^2 c^4 x \]

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 3904

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])

Rule 3886

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 3473

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 8

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

Rule 2611

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 3770

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

Rule 2607

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 30

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

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 ) \operatorname{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*}

Mathematica [A]  time = 1.15177, size = 146, normalized size = 1.04 \[ \frac{a^2 c^4 \sec ^5(e+f x) \left (40 \sin (e+f x)+60 \sin (2 (e+f x))-220 \sin (3 (e+f x))+150 \sin (4 (e+f x))-68 \sin (5 (e+f x))+600 (e+f x) \cos (e+f x)+300 e \cos (3 (e+f x))+300 f x \cos (3 (e+f x))+60 e \cos (5 (e+f x))+60 f x \cos (5 (e+f x))-720 \cos ^5(e+f x) \tanh ^{-1}(\sin (e+f x))\right )}{960 f} \]

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 = 0.028, size = 161, normalized size = 1.2 \begin{align*} -{\frac{17\,{c}^{4}{a}^{2}\tan \left ( fx+e \right ) }{15\,f}}-{\frac{{c}^{4}{a}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{15\,f}}+{\frac{5\,{c}^{4}{a}^{2}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{4\,f}}-{\frac{3\,{c}^{4}{a}^{2}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{4\,f}}+{a}^{2}{c}^{4}x+{\frac{{a}^{2}{c}^{4}e}{f}}-{\frac{{c}^{4}{a}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{2\,f}}+{\frac{{c}^{4}{a}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{4}}{5\,f}} \end{align*}

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)

[Out]

-17/15*a^2*c^4*tan(f*x+e)/f-1/15/f*c^4*a^2*tan(f*x+e)*sec(f*x+e)^2+5/4*a^2*c^4*sec(f*x+e)*tan(f*x+e)/f-3/4/f*c
^4*a^2*ln(sec(f*x+e)+tan(f*x+e))+a^2*c^4*x+1/f*a^2*c^4*e-1/2/f*c^4*a^2*tan(f*x+e)*sec(f*x+e)^3+1/5/f*c^4*a^2*t
an(f*x+e)*sec(f*x+e)^4

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Maxima [A]  time = 1.03812, size = 324, normalized size = 2.31 \begin{align*} \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{align*}

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 = 1.13531, size = 404, normalized size = 2.89 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} a^{2} c^{4} \left (\int 1\, dx + \int - 2 \sec{\left (e + f x \right )}\, dx + \int - \sec ^{2}{\left (e + f x \right )}\, dx + \int 4 \sec ^{3}{\left (e + f x \right )}\, dx + \int - \sec ^{4}{\left (e + f x \right )}\, dx + \int - 2 \sec ^{5}{\left (e + f x \right )}\, dx + \int \sec ^{6}{\left (e + f x \right )}\, dx\right ) \end{align*}

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 = 1.44588, size = 244, normalized size = 1.74 \begin{align*} \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{align*}

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