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

Optimal. Leaf size=100 \[ \frac{5 a^3 c^3 \tanh ^{-1}(\sin (e+f x))}{16 f}-\frac{a^3 c^3 \tan ^5(e+f x) \sec (e+f x)}{6 f}+\frac{5 a^3 c^3 \tan ^3(e+f x) \sec (e+f x)}{24 f}-\frac{5 a^3 c^3 \tan (e+f x) \sec (e+f x)}{16 f} \]

[Out]

(5*a^3*c^3*ArcTanh[Sin[e + f*x]])/(16*f) - (5*a^3*c^3*Sec[e + f*x]*Tan[e + f*x])/(16*f) + (5*a^3*c^3*Sec[e + f
*x]*Tan[e + f*x]^3)/(24*f) - (a^3*c^3*Sec[e + f*x]*Tan[e + f*x]^5)/(6*f)

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Rubi [A]  time = 0.131113, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {3958, 2611, 3770} \[ \frac{5 a^3 c^3 \tanh ^{-1}(\sin (e+f x))}{16 f}-\frac{a^3 c^3 \tan ^5(e+f x) \sec (e+f x)}{6 f}+\frac{5 a^3 c^3 \tan ^3(e+f x) \sec (e+f x)}{24 f}-\frac{5 a^3 c^3 \tan (e+f x) \sec (e+f x)}{16 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^3*c^3*ArcTanh[Sin[e + f*x]])/(16*f) - (5*a^3*c^3*Sec[e + f*x]*Tan[e + f*x])/(16*f) + (5*a^3*c^3*Sec[e + f
*x]*Tan[e + f*x]^3)/(24*f) - (a^3*c^3*Sec[e + f*x]*Tan[e + f*x]^5)/(6*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))^3 (c-c \sec (e+f x))^3 \, dx &=-\left (\left (a^3 c^3\right ) \int \sec (e+f x) \tan ^6(e+f x) \, dx\right )\\ &=-\frac{a^3 c^3 \sec (e+f x) \tan ^5(e+f x)}{6 f}+\frac{1}{6} \left (5 a^3 c^3\right ) \int \sec (e+f x) \tan ^4(e+f x) \, dx\\ &=\frac{5 a^3 c^3 \sec (e+f x) \tan ^3(e+f x)}{24 f}-\frac{a^3 c^3 \sec (e+f x) \tan ^5(e+f x)}{6 f}-\frac{1}{8} \left (5 a^3 c^3\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx\\ &=-\frac{5 a^3 c^3 \sec (e+f x) \tan (e+f x)}{16 f}+\frac{5 a^3 c^3 \sec (e+f x) \tan ^3(e+f x)}{24 f}-\frac{a^3 c^3 \sec (e+f x) \tan ^5(e+f x)}{6 f}+\frac{1}{16} \left (5 a^3 c^3\right ) \int \sec (e+f x) \, dx\\ &=\frac{5 a^3 c^3 \tanh ^{-1}(\sin (e+f x))}{16 f}-\frac{5 a^3 c^3 \sec (e+f x) \tan (e+f x)}{16 f}+\frac{5 a^3 c^3 \sec (e+f x) \tan ^3(e+f x)}{24 f}-\frac{a^3 c^3 \sec (e+f x) \tan ^5(e+f x)}{6 f}\\ \end{align*}

Mathematica [A]  time = 0.248216, size = 60, normalized size = 0.6 \[ -\frac{a^3 c^3 \left ((28 \cos (2 (e+f x))+33 \cos (4 (e+f x))+59) \tan (e+f x) \sec ^5(e+f x)-120 \tanh ^{-1}(\sin (e+f x))\right )}{384 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*c^3*(-120*ArcTanh[Sin[e + f*x]] + (59 + 28*Cos[2*(e + f*x)] + 33*Cos[4*(e + f*x)])*Sec[e + f*x]^5*Tan[e
+ f*x]))/(384*f)

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Maple [A]  time = 0.024, size = 100, normalized size = 1. \begin{align*} -{\frac{11\,{a}^{3}{c}^{3}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{16\,f}}+{\frac{5\,{a}^{3}{c}^{3}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{16\,f}}+{\frac{13\,{a}^{3}{c}^{3}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{24\,f}}-{\frac{{a}^{3}{c}^{3}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{5}}{6\,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))^3*(c-c*sec(f*x+e))^3,x)

[Out]

-11/16*a^3*c^3*sec(f*x+e)*tan(f*x+e)/f+5/16/f*a^3*c^3*ln(sec(f*x+e)+tan(f*x+e))+13/24/f*a^3*c^3*tan(f*x+e)*sec
(f*x+e)^3-1/6/f*a^3*c^3*tan(f*x+e)*sec(f*x+e)^5

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Maxima [B]  time = 0.984666, size = 329, normalized size = 3.29 \begin{align*} \frac{a^{3} c^{3}{\left (\frac{2 \,{\left (15 \, \sin \left (f x + e\right )^{5} - 40 \, \sin \left (f x + e\right )^{3} + 33 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1} - 15 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 15 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 18 \, a^{3} 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 )} + 72 \, a^{3} 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 )} + 96 \, a^{3} c^{3} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right )}{96 \, 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))^3*(c-c*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

1/96*(a^3*c^3*(2*(15*sin(f*x + e)^5 - 40*sin(f*x + e)^3 + 33*sin(f*x + e))/(sin(f*x + e)^6 - 3*sin(f*x + e)^4
+ 3*sin(f*x + e)^2 - 1) - 15*log(sin(f*x + e) + 1) + 15*log(sin(f*x + e) - 1)) - 18*a^3*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)) + 72*a^3*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)) + 96*a^
3*c^3*log(sec(f*x + e) + tan(f*x + e)))/f

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Fricas [A]  time = 0.495505, size = 285, normalized size = 2.85 \begin{align*} \frac{15 \, a^{3} c^{3} \cos \left (f x + e\right )^{6} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, a^{3} c^{3} \cos \left (f x + e\right )^{6} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (33 \, a^{3} c^{3} \cos \left (f x + e\right )^{4} - 26 \, a^{3} c^{3} \cos \left (f x + e\right )^{2} + 8 \, a^{3} c^{3}\right )} \sin \left (f x + e\right )}{96 \, f \cos \left (f x + e\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/96*(15*a^3*c^3*cos(f*x + e)^6*log(sin(f*x + e) + 1) - 15*a^3*c^3*cos(f*x + e)^6*log(-sin(f*x + e) + 1) - 2*(
33*a^3*c^3*cos(f*x + e)^4 - 26*a^3*c^3*cos(f*x + e)^2 + 8*a^3*c^3)*sin(f*x + e))/(f*cos(f*x + e)^6)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - a^{3} c^{3} \left (\int - \sec{\left (e + f x \right )}\, dx + \int 3 \sec ^{3}{\left (e + f x \right )}\, dx + \int - 3 \sec ^{5}{\left (e + f x \right )}\, dx + \int \sec ^{7}{\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))**3*(c-c*sec(f*x+e))**3,x)

[Out]

-a**3*c**3*(Integral(-sec(e + f*x), x) + Integral(3*sec(e + f*x)**3, x) + Integral(-3*sec(e + f*x)**5, x) + In
tegral(sec(e + f*x)**7, x))

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Giac [A]  time = 1.36111, size = 147, normalized size = 1.47 \begin{align*} \frac{15 \, a^{3} c^{3} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, a^{3} c^{3} \log \left (-\sin \left (f x + e\right ) + 1\right ) + \frac{2 \,{\left (33 \, a^{3} c^{3} \sin \left (f x + e\right )^{5} - 40 \, a^{3} c^{3} \sin \left (f x + e\right )^{3} + 15 \, a^{3} c^{3} \sin \left (f x + e\right )\right )}}{{\left (\sin \left (f x + e\right )^{2} - 1\right )}^{3}}}{96 \, 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))^3*(c-c*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/96*(15*a^3*c^3*log(sin(f*x + e) + 1) - 15*a^3*c^3*log(-sin(f*x + e) + 1) + 2*(33*a^3*c^3*sin(f*x + e)^5 - 40
*a^3*c^3*sin(f*x + e)^3 + 15*a^3*c^3*sin(f*x + e))/(sin(f*x + e)^2 - 1)^3)/f