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

Optimal. Leaf size=61 \[ -\frac{a^2 c \tan ^3(e+f x)}{3 f}+\frac{a^2 c \tanh ^{-1}(\sin (e+f x))}{2 f}-\frac{a^2 c \tan (e+f x) \sec (e+f x)}{2 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.102527, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3958, 2611, 3770, 2607, 30} \[ -\frac{a^2 c \tan ^3(e+f x)}{3 f}+\frac{a^2 c \tanh ^{-1}(\sin (e+f x))}{2 f}-\frac{a^2 c \tan (e+f x) \sec (e+f x)}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3958

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)
)^(n_.), x_Symbol] :> Dist[(-(a*c))^m, Int[ExpandTrig[csc[e + f*x]*cot[e + f*x]^(2*m), (c + d*csc[e + f*x])^(n
- m), x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegersQ[m,
n] && GeQ[n - m, 0] && GtQ[m*n, 0]

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

Mathematica [A]  time = 0.102027, size = 45, normalized size = 0.74 \[ \frac{a^2 c \left (-2 \tan ^3(e+f x)+3 \tanh ^{-1}(\sin (e+f x))-3 \tan (e+f x) \sec (e+f x)\right )}{6 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c*(3*ArcTanh[Sin[e + f*x]] - 3*Sec[e + f*x]*Tan[e + f*x] - 2*Tan[e + f*x]^3))/(6*f)

________________________________________________________________________________________

Maple [A]  time = 0.02, size = 84, normalized size = 1.4 \begin{align*}{\frac{{a}^{2}c\tan \left ( fx+e \right ) }{3\,f}}+{\frac{{a}^{2}c\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{2\,f}}-{\frac{{a}^{2}c\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{2\,f}}-{\frac{{a}^{2}c\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{3\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e)),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.00824, size = 146, normalized size = 2.39 \begin{align*} -\frac{4 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c - 3 \, a^{2} c{\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 )} - 12 \, a^{2} c \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) - 12 \, a^{2} c \tan \left (f x + e\right )}{12 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(4*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*c - 3*a^2*c*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x
+ e) + 1) + log(sin(f*x + e) - 1)) - 12*a^2*c*log(sec(f*x + e) + tan(f*x + e)) - 12*a^2*c*tan(f*x + e))/f

________________________________________________________________________________________

Fricas [A]  time = 0.481272, size = 263, normalized size = 4.31 \begin{align*} \frac{3 \, a^{2} c \cos \left (f x + e\right )^{3} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, a^{2} c \cos \left (f x + e\right )^{3} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (2 \, a^{2} c \cos \left (f x + e\right )^{2} - 3 \, a^{2} c \cos \left (f x + e\right ) - 2 \, a^{2} c\right )} \sin \left (f x + e\right )}{12 \, f \cos \left (f x + e\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a^2*c*cos(f*x + e)^3*log(sin(f*x + e) + 1) - 3*a^2*c*cos(f*x + e)^3*log(-sin(f*x + e) + 1) + 2*(2*a^2*
c*cos(f*x + e)^2 - 3*a^2*c*cos(f*x + e) - 2*a^2*c)*sin(f*x + e))/(f*cos(f*x + e)^3)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.33076, size = 158, normalized size = 2.59 \begin{align*} \frac{3 \, a^{2} c \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right ) - 3 \, a^{2} c \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right ) - \frac{2 \,{\left (3 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 8 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 3 \, a^{2} c \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 )}^{3}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*a^2*c*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 3*a^2*c*log(abs(tan(1/2*f*x + 1/2*e) - 1)) - 2*(3*a^2*c*tan(
1/2*f*x + 1/2*e)^5 - 8*a^2*c*tan(1/2*f*x + 1/2*e)^3 - 3*a^2*c*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 -
1)^3)/f