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

Optimal. Leaf size=73 \[ \frac{3 a^2 c^2 \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{a^2 c^2 \tan ^3(e+f x) \sec (e+f x)}{4 f}-\frac{3 a^2 c^2 \tan (e+f x) \sec (e+f x)}{8 f} \]

[Out]

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

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Rubi [A]  time = 0.108332, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {3958, 2611, 3770} \[ \frac{3 a^2 c^2 \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{a^2 c^2 \tan ^3(e+f x) \sec (e+f x)}{4 f}-\frac{3 a^2 c^2 \tan (e+f x) \sec (e+f x)}{8 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin{align*} \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^2 \, dx &=\left (a^2 c^2\right ) \int \sec (e+f x) \tan ^4(e+f x) \, dx\\ &=\frac{a^2 c^2 \sec (e+f x) \tan ^3(e+f x)}{4 f}-\frac{1}{4} \left (3 a^2 c^2\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx\\ &=-\frac{3 a^2 c^2 \sec (e+f x) \tan (e+f x)}{8 f}+\frac{a^2 c^2 \sec (e+f x) \tan ^3(e+f x)}{4 f}+\frac{1}{8} \left (3 a^2 c^2\right ) \int \sec (e+f x) \, dx\\ &=\frac{3 a^2 c^2 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac{3 a^2 c^2 \sec (e+f x) \tan (e+f x)}{8 f}+\frac{a^2 c^2 \sec (e+f x) \tan ^3(e+f x)}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.149303, size = 51, normalized size = 0.7 \[ \frac{a^2 c^2 \left (6 \tanh ^{-1}(\sin (e+f x))-(5 \cos (2 (e+f x))+1) \tan (e+f x) \sec ^3(e+f x)\right )}{16 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c^2*(6*ArcTanh[Sin[e + f*x]] - (1 + 5*Cos[2*(e + f*x)])*Sec[e + f*x]^3*Tan[e + f*x]))/(16*f)

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Maple [A]  time = 0.02, size = 75, normalized size = 1. \begin{align*} -{\frac{5\,{a}^{2}{c}^{2}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{8\,f}}+{\frac{3\,{a}^{2}{c}^{2}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{8\,f}}+{\frac{{a}^{2}{c}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{4\,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))^2,x)

[Out]

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

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Maxima [B]  time = 0.967191, size = 203, normalized size = 2.78 \begin{align*} -\frac{a^{2} c^{2}{\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 )} - 8 \, a^{2} c^{2}{\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 )} - 16 \, a^{2} c^{2} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right )}{16 \, 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))^2,x, algorithm="maxima")

[Out]

-1/16*(a^2*c^2*(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)) - 8*a^2*c^2*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1)
+ log(sin(f*x + e) - 1)) - 16*a^2*c^2*log(sec(f*x + e) + tan(f*x + e)))/f

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Fricas [A]  time = 0.48342, size = 243, normalized size = 3.33 \begin{align*} \frac{3 \, a^{2} c^{2} \cos \left (f x + e\right )^{4} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, a^{2} c^{2} \cos \left (f x + e\right )^{4} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (5 \, a^{2} c^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} c^{2}\right )} \sin \left (f x + e\right )}{16 \, f \cos \left (f x + e\right )^{4}} \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))^2,x, algorithm="fricas")

[Out]

1/16*(3*a^2*c^2*cos(f*x + e)^4*log(sin(f*x + e) + 1) - 3*a^2*c^2*cos(f*x + e)^4*log(-sin(f*x + e) + 1) - 2*(5*
a^2*c^2*cos(f*x + e)^2 - 2*a^2*c^2)*sin(f*x + e))/(f*cos(f*x + e)^4)

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

[Out]

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

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Giac [A]  time = 1.24941, size = 124, normalized size = 1.7 \begin{align*} \frac{3 \, a^{2} c^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, a^{2} c^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) + \frac{2 \,{\left (5 \, a^{2} c^{2} \sin \left (f x + e\right )^{3} - 3 \, a^{2} c^{2} \sin \left (f x + e\right )\right )}}{{\left (\sin \left (f x + e\right )^{2} - 1\right )}^{2}}}{16 \, 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))^2,x, algorithm="giac")

[Out]

1/16*(3*a^2*c^2*log(sin(f*x + e) + 1) - 3*a^2*c^2*log(-sin(f*x + e) + 1) + 2*(5*a^2*c^2*sin(f*x + e)^3 - 3*a^2
*c^2*sin(f*x + e))/(sin(f*x + e)^2 - 1)^2)/f