3.23 \(\int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4 \, dx\)

Optimal. Leaf size=121 \[ \frac{a^3 c^4 \tan ^7(e+f x)}{7 f}+\frac{5 a^3 c^4 \tanh ^{-1}(\sin (e+f x))}{16 f}-\frac{a^3 c^4 \tan ^5(e+f x) \sec (e+f x)}{6 f}+\frac{5 a^3 c^4 \tan ^3(e+f x) \sec (e+f x)}{24 f}-\frac{5 a^3 c^4 \tan (e+f x) \sec (e+f x)}{16 f} \]

[Out]

(5*a^3*c^4*ArcTanh[Sin[e + f*x]])/(16*f) - (5*a^3*c^4*Sec[e + f*x]*Tan[e + f*x])/(16*f) + (5*a^3*c^4*Sec[e + f
*x]*Tan[e + f*x]^3)/(24*f) - (a^3*c^4*Sec[e + f*x]*Tan[e + f*x]^5)/(6*f) + (a^3*c^4*Tan[e + f*x]^7)/(7*f)

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Rubi [A]  time = 0.178182, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{a^3 c^4 \tan ^7(e+f x)}{7 f}+\frac{5 a^3 c^4 \tanh ^{-1}(\sin (e+f x))}{16 f}-\frac{a^3 c^4 \tan ^5(e+f x) \sec (e+f x)}{6 f}+\frac{5 a^3 c^4 \tan ^3(e+f x) \sec (e+f x)}{24 f}-\frac{5 a^3 c^4 \tan (e+f x) \sec (e+f x)}{16 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*(a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^4,x]

[Out]

(5*a^3*c^4*ArcTanh[Sin[e + f*x]])/(16*f) - (5*a^3*c^4*Sec[e + f*x]*Tan[e + f*x])/(16*f) + (5*a^3*c^4*Sec[e + f
*x]*Tan[e + f*x]^3)/(24*f) - (a^3*c^4*Sec[e + f*x]*Tan[e + f*x]^5)/(6*f) + (a^3*c^4*Tan[e + f*x]^7)/(7*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))^3 (c-c \sec (e+f x))^4 \, dx &=-\left (\left (a^3 c^3\right ) \int \left (c \sec (e+f x) \tan ^6(e+f x)-c \sec ^2(e+f x) \tan ^6(e+f x)\right ) \, dx\right )\\ &=-\left (\left (a^3 c^4\right ) \int \sec (e+f x) \tan ^6(e+f x) \, dx\right )+\left (a^3 c^4\right ) \int \sec ^2(e+f x) \tan ^6(e+f x) \, dx\\ &=-\frac{a^3 c^4 \sec (e+f x) \tan ^5(e+f x)}{6 f}+\frac{1}{6} \left (5 a^3 c^4\right ) \int \sec (e+f x) \tan ^4(e+f x) \, dx+\frac{\left (a^3 c^4\right ) \operatorname{Subst}\left (\int x^6 \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{5 a^3 c^4 \sec (e+f x) \tan ^3(e+f x)}{24 f}-\frac{a^3 c^4 \sec (e+f x) \tan ^5(e+f x)}{6 f}+\frac{a^3 c^4 \tan ^7(e+f x)}{7 f}-\frac{1}{8} \left (5 a^3 c^4\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx\\ &=-\frac{5 a^3 c^4 \sec (e+f x) \tan (e+f x)}{16 f}+\frac{5 a^3 c^4 \sec (e+f x) \tan ^3(e+f x)}{24 f}-\frac{a^3 c^4 \sec (e+f x) \tan ^5(e+f x)}{6 f}+\frac{a^3 c^4 \tan ^7(e+f x)}{7 f}+\frac{1}{16} \left (5 a^3 c^4\right ) \int \sec (e+f x) \, dx\\ &=\frac{5 a^3 c^4 \tanh ^{-1}(\sin (e+f x))}{16 f}-\frac{5 a^3 c^4 \sec (e+f x) \tan (e+f x)}{16 f}+\frac{5 a^3 c^4 \sec (e+f x) \tan ^3(e+f x)}{24 f}-\frac{a^3 c^4 \sec (e+f x) \tan ^5(e+f x)}{6 f}+\frac{a^3 c^4 \tan ^7(e+f x)}{7 f}\\ \end{align*}

Mathematica [A]  time = 1.71463, size = 102, normalized size = 0.84 \[ \frac{a^3 c^4 \left (3360 \tanh ^{-1}(\sin (e+f x))-(-840 \sin (e+f x)+595 \sin (2 (e+f x))+504 \sin (3 (e+f x))+196 \sin (4 (e+f x))-168 \sin (5 (e+f x))+231 \sin (6 (e+f x))+24 \sin (7 (e+f x))) \sec ^7(e+f x)\right )}{10752 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^4,x]

[Out]

(a^3*c^4*(3360*ArcTanh[Sin[e + f*x]] - Sec[e + f*x]^7*(-840*Sin[e + f*x] + 595*Sin[2*(e + f*x)] + 504*Sin[3*(e
+ f*x)] + 196*Sin[4*(e + f*x)] - 168*Sin[5*(e + f*x)] + 231*Sin[6*(e + f*x)] + 24*Sin[7*(e + f*x)])))/(10752*
f)

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Maple [A]  time = 0.029, size = 192, normalized size = 1.6 \begin{align*}{\frac{13\,{a}^{3}{c}^{4}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{24\,f}}-{\frac{11\,{a}^{3}{c}^{4}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{16\,f}}+{\frac{5\,{a}^{3}{c}^{4}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{16\,f}}-{\frac{{a}^{3}{c}^{4}\tan \left ( fx+e \right ) }{7\,f}}+{\frac{3\,{a}^{3}{c}^{4}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{7\,f}}-{\frac{3\,{a}^{3}{c}^{4}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{4}}{7\,f}}-{\frac{{a}^{3}{c}^{4}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{5}}{6\,f}}+{\frac{{a}^{3}{c}^{4}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{6}}{7\,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))^4,x)

[Out]

13/24/f*a^3*c^4*tan(f*x+e)*sec(f*x+e)^3-11/16*a^3*c^4*sec(f*x+e)*tan(f*x+e)/f+5/16/f*a^3*c^4*ln(sec(f*x+e)+tan
(f*x+e))-1/7/f*a^3*c^4*tan(f*x+e)+3/7/f*a^3*c^4*tan(f*x+e)*sec(f*x+e)^2-3/7/f*a^3*c^4*tan(f*x+e)*sec(f*x+e)^4-
1/6/f*a^3*c^4*tan(f*x+e)*sec(f*x+e)^5+1/7/f*a^3*c^4*tan(f*x+e)*sec(f*x+e)^6

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Maxima [B]  time = 1.02888, size = 497, normalized size = 4.11 \begin{align*} \frac{96 \,{\left (5 \, \tan \left (f x + e\right )^{7} + 21 \, \tan \left (f x + e\right )^{5} + 35 \, \tan \left (f x + e\right )^{3} + 35 \, \tan \left (f x + e\right )\right )} a^{3} c^{4} - 672 \,{\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^{3} c^{4} + 3360 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{3} c^{4} + 35 \, a^{3} c^{4}{\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 )} - 630 \, a^{3} 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 )} + 2520 \, a^{3} 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 )} + 3360 \, a^{3} c^{4} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) - 3360 \, a^{3} c^{4} \tan \left (f x + e\right )}{3360 \, 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))^4,x, algorithm="maxima")

[Out]

1/3360*(96*(5*tan(f*x + e)^7 + 21*tan(f*x + e)^5 + 35*tan(f*x + e)^3 + 35*tan(f*x + e))*a^3*c^4 - 672*(3*tan(f
*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^3*c^4 + 3360*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^3*c^4 + 35
*a^3*c^4*(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*s
in(f*x + e)^2 - 1) - 15*log(sin(f*x + e) + 1) + 15*log(sin(f*x + e) - 1)) - 630*a^3*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))
+ 2520*a^3*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)) + 3360*a^
3*c^4*log(sec(f*x + e) + tan(f*x + e)) - 3360*a^3*c^4*tan(f*x + e))/f

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Fricas [A]  time = 0.511063, size = 444, normalized size = 3.67 \begin{align*} \frac{105 \, a^{3} c^{4} \cos \left (f x + e\right )^{7} \log \left (\sin \left (f x + e\right ) + 1\right ) - 105 \, a^{3} c^{4} \cos \left (f x + e\right )^{7} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (48 \, a^{3} c^{4} \cos \left (f x + e\right )^{6} + 231 \, a^{3} c^{4} \cos \left (f x + e\right )^{5} - 144 \, a^{3} c^{4} \cos \left (f x + e\right )^{4} - 182 \, a^{3} c^{4} \cos \left (f x + e\right )^{3} + 144 \, a^{3} c^{4} \cos \left (f x + e\right )^{2} + 56 \, a^{3} c^{4} \cos \left (f x + e\right ) - 48 \, a^{3} c^{4}\right )} \sin \left (f x + e\right )}{672 \, f \cos \left (f x + e\right )^{7}} \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))^4,x, algorithm="fricas")

[Out]

1/672*(105*a^3*c^4*cos(f*x + e)^7*log(sin(f*x + e) + 1) - 105*a^3*c^4*cos(f*x + e)^7*log(-sin(f*x + e) + 1) -
2*(48*a^3*c^4*cos(f*x + e)^6 + 231*a^3*c^4*cos(f*x + e)^5 - 144*a^3*c^4*cos(f*x + e)^4 - 182*a^3*c^4*cos(f*x +
e)^3 + 144*a^3*c^4*cos(f*x + e)^2 + 56*a^3*c^4*cos(f*x + e) - 48*a^3*c^4)*sin(f*x + e))/(f*cos(f*x + e)^7)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} c^{4} \left (\int \sec{\left (e + f x \right )}\, dx + \int - \sec ^{2}{\left (e + f x \right )}\, dx + \int - 3 \sec ^{3}{\left (e + f x \right )}\, dx + \int 3 \sec ^{4}{\left (e + f x \right )}\, dx + \int 3 \sec ^{5}{\left (e + f x \right )}\, dx + \int - 3 \sec ^{6}{\left (e + f x \right )}\, dx + \int - \sec ^{7}{\left (e + f x \right )}\, dx + \int \sec ^{8}{\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))**4,x)

[Out]

a**3*c**4*(Integral(sec(e + f*x), x) + Integral(-sec(e + f*x)**2, x) + Integral(-3*sec(e + f*x)**3, x) + Integ
ral(3*sec(e + f*x)**4, x) + Integral(3*sec(e + f*x)**5, x) + Integral(-3*sec(e + f*x)**6, x) + Integral(-sec(e
+ f*x)**7, x) + Integral(sec(e + f*x)**8, x))

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Giac [A]  time = 1.37582, size = 279, normalized size = 2.31 \begin{align*} \frac{105 \, a^{3} c^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right ) - 105 \, a^{3} 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^{3} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{13} - 700 \, a^{3} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{11} + 1981 \, a^{3} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9} + 3072 \, a^{3} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} - 1981 \, a^{3} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 700 \, a^{3} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 105 \, a^{3} 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 )}^{7}}}{336 \, 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))^4,x, algorithm="giac")

[Out]

1/336*(105*a^3*c^4*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 105*a^3*c^4*log(abs(tan(1/2*f*x + 1/2*e) - 1)) - 2*(10
5*a^3*c^4*tan(1/2*f*x + 1/2*e)^13 - 700*a^3*c^4*tan(1/2*f*x + 1/2*e)^11 + 1981*a^3*c^4*tan(1/2*f*x + 1/2*e)^9
+ 3072*a^3*c^4*tan(1/2*f*x + 1/2*e)^7 - 1981*a^3*c^4*tan(1/2*f*x + 1/2*e)^5 + 700*a^3*c^4*tan(1/2*f*x + 1/2*e)
^3 - 105*a^3*c^4*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^7)/f