3.12 \(\int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3 \, dx\)

Optimal. Leaf size=94 \[ -\frac{a^2 c^3 \tan ^5(e+f x)}{5 f}+\frac{3 a^2 c^3 \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{a^2 c^3 \tan ^3(e+f x) \sec (e+f x)}{4 f}-\frac{3 a^2 c^3 \tan (e+f x) \sec (e+f x)}{8 f} \]

[Out]

(3*a^2*c^3*ArcTanh[Sin[e + f*x]])/(8*f) - (3*a^2*c^3*Sec[e + f*x]*Tan[e + f*x])/(8*f) + (a^2*c^3*Sec[e + f*x]*
Tan[e + f*x]^3)/(4*f) - (a^2*c^3*Tan[e + f*x]^5)/(5*f)

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Rubi [A]  time = 0.148851, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {3958, 2611, 3770, 2607, 30} \[ -\frac{a^2 c^3 \tan ^5(e+f x)}{5 f}+\frac{3 a^2 c^3 \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{a^2 c^3 \tan ^3(e+f x) \sec (e+f x)}{4 f}-\frac{3 a^2 c^3 \tan (e+f x) \sec (e+f x)}{8 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^3,x]

[Out]

(3*a^2*c^3*ArcTanh[Sin[e + f*x]])/(8*f) - (3*a^2*c^3*Sec[e + f*x]*Tan[e + f*x])/(8*f) + (a^2*c^3*Sec[e + f*x]*
Tan[e + f*x]^3)/(4*f) - (a^2*c^3*Tan[e + f*x]^5)/(5*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))^3 \, dx &=\left (a^2 c^2\right ) \int \left (c \sec (e+f x) \tan ^4(e+f x)-c \sec ^2(e+f x) \tan ^4(e+f x)\right ) \, dx\\ &=\left (a^2 c^3\right ) \int \sec (e+f x) \tan ^4(e+f x) \, dx-\left (a^2 c^3\right ) \int \sec ^2(e+f x) \tan ^4(e+f x) \, dx\\ &=\frac{a^2 c^3 \sec (e+f x) \tan ^3(e+f x)}{4 f}-\frac{1}{4} \left (3 a^2 c^3\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx-\frac{\left (a^2 c^3\right ) \operatorname{Subst}\left (\int x^4 \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{3 a^2 c^3 \sec (e+f x) \tan (e+f x)}{8 f}+\frac{a^2 c^3 \sec (e+f x) \tan ^3(e+f x)}{4 f}-\frac{a^2 c^3 \tan ^5(e+f x)}{5 f}+\frac{1}{8} \left (3 a^2 c^3\right ) \int \sec (e+f x) \, dx\\ &=\frac{3 a^2 c^3 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac{3 a^2 c^3 \sec (e+f x) \tan (e+f x)}{8 f}+\frac{a^2 c^3 \sec (e+f x) \tan ^3(e+f x)}{4 f}-\frac{a^2 c^3 \tan ^5(e+f x)}{5 f}\\ \end{align*}

Mathematica [A]  time = 0.741826, size = 82, normalized size = 0.87 \[ \frac{a^2 c^3 \left (120 \tanh ^{-1}(\sin (e+f x))-(40 \sin (e+f x)+10 \sin (2 (e+f x))-20 \sin (3 (e+f x))+25 \sin (4 (e+f x))+4 \sin (5 (e+f x))) \sec ^5(e+f x)\right )}{320 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^3,x]

[Out]

(a^2*c^3*(120*ArcTanh[Sin[e + f*x]] - Sec[e + f*x]^5*(40*Sin[e + f*x] + 10*Sin[2*(e + f*x)] - 20*Sin[3*(e + f*
x)] + 25*Sin[4*(e + f*x)] + 4*Sin[5*(e + f*x)])))/(320*f)

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Maple [A]  time = 0.026, size = 142, normalized size = 1.5 \begin{align*} -{\frac{{a}^{2}{c}^{3}\tan \left ( fx+e \right ) }{5\,f}}+{\frac{2\,{a}^{2}{c}^{3}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{5\,f}}-{\frac{5\,{a}^{2}{c}^{3}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{8\,f}}+{\frac{3\,{a}^{2}{c}^{3}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{8\,f}}+{\frac{{a}^{2}{c}^{3}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{4\,f}}-{\frac{{a}^{2}{c}^{3}\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(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^3,x)

[Out]

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

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Maxima [B]  time = 0.977232, size = 306, normalized size = 3.26 \begin{align*} -\frac{16 \,{\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^{3} - 160 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c^{3} + 15 \, a^{2} c^{3}{\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^{3}{\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^{3} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 240 \, a^{2} c^{3} \tan \left (f x + e\right )}{240 \, 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))^3,x, algorithm="maxima")

[Out]

-1/240*(16*(3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^2*c^3 - 160*(tan(f*x + e)^3 + 3*tan(f*x
+ e))*a^2*c^3 + 15*a^2*c^3*(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^3*(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^3*log(sec(f*x + e) + tan(f*x + e)) + 240*a^2*c^3*tan(f*x +
e))/f

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Fricas [A]  time = 0.49673, size = 356, normalized size = 3.79 \begin{align*} \frac{15 \, a^{2} c^{3} \cos \left (f x + e\right )^{5} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, a^{2} c^{3} \cos \left (f x + e\right )^{5} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (8 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} + 25 \, a^{2} c^{3} \cos \left (f x + e\right )^{3} - 16 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} - 10 \, a^{2} c^{3} \cos \left (f x + e\right ) + 8 \, a^{2} c^{3}\right )} \sin \left (f x + e\right )}{80 \, f \cos \left (f x + e\right )^{5}} \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))^3,x, algorithm="fricas")

[Out]

1/80*(15*a^2*c^3*cos(f*x + e)^5*log(sin(f*x + e) + 1) - 15*a^2*c^3*cos(f*x + e)^5*log(-sin(f*x + e) + 1) - 2*(
8*a^2*c^3*cos(f*x + e)^4 + 25*a^2*c^3*cos(f*x + e)^3 - 16*a^2*c^3*cos(f*x + e)^2 - 10*a^2*c^3*cos(f*x + e) + 8
*a^2*c^3)*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^{3} \left (\int - \sec{\left (e + f x \right )}\, dx + \int \sec ^{2}{\left (e + f x \right )}\, dx + \int 2 \sec ^{3}{\left (e + f x \right )}\, dx + \int - 2 \sec ^{4}{\left (e + f x \right )}\, dx + \int - \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(sec(f*x+e)*(a+a*sec(f*x+e))**2*(c-c*sec(f*x+e))**3,x)

[Out]

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

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Giac [A]  time = 1.26778, size = 225, normalized size = 2.39 \begin{align*} \frac{15 \, a^{2} c^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right ) - 15 \, a^{2} c^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right ) - \frac{2 \,{\left (15 \, a^{2} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9} - 70 \, a^{2} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} - 128 \, a^{2} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 70 \, a^{2} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 15 \, a^{2} c^{3} \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}}}{40 \, 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))^3,x, algorithm="giac")

[Out]

1/40*(15*a^2*c^3*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 15*a^2*c^3*log(abs(tan(1/2*f*x + 1/2*e) - 1)) - 2*(15*a^
2*c^3*tan(1/2*f*x + 1/2*e)^9 - 70*a^2*c^3*tan(1/2*f*x + 1/2*e)^7 - 128*a^2*c^3*tan(1/2*f*x + 1/2*e)^5 + 70*a^2
*c^3*tan(1/2*f*x + 1/2*e)^3 - 15*a^2*c^3*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^5)/f