3.11 \(\int (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^5 \, dx\)

Optimal. Leaf size=188 \[ -\frac{a^3 c^5 \tan ^7(e+f x)}{7 f}-\frac{a^3 c^5 \tan ^5(e+f x)}{5 f}+\frac{a^3 c^5 \tan ^3(e+f x)}{3 f}-\frac{a^3 c^5 \tan (e+f x)}{f}-\frac{5 a^3 c^5 \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{a^3 c^5 \tan ^5(e+f x) \sec (e+f x)}{3 f}-\frac{5 a^3 c^5 \tan ^3(e+f x) \sec (e+f x)}{12 f}+\frac{5 a^3 c^5 \tan (e+f x) \sec (e+f x)}{8 f}+a^3 c^5 x \]

[Out]

a^3*c^5*x - (5*a^3*c^5*ArcTanh[Sin[e + f*x]])/(8*f) - (a^3*c^5*Tan[e + f*x])/f + (5*a^3*c^5*Sec[e + f*x]*Tan[e
+ f*x])/(8*f) + (a^3*c^5*Tan[e + f*x]^3)/(3*f) - (5*a^3*c^5*Sec[e + f*x]*Tan[e + f*x]^3)/(12*f) - (a^3*c^5*Ta
n[e + f*x]^5)/(5*f) + (a^3*c^5*Sec[e + f*x]*Tan[e + f*x]^5)/(3*f) - (a^3*c^5*Tan[e + f*x]^7)/(7*f)

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Rubi [A]  time = 0.23721, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {3904, 3886, 3473, 8, 2611, 3770, 2607, 30} \[ -\frac{a^3 c^5 \tan ^7(e+f x)}{7 f}-\frac{a^3 c^5 \tan ^5(e+f x)}{5 f}+\frac{a^3 c^5 \tan ^3(e+f x)}{3 f}-\frac{a^3 c^5 \tan (e+f x)}{f}-\frac{5 a^3 c^5 \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{a^3 c^5 \tan ^5(e+f x) \sec (e+f x)}{3 f}-\frac{5 a^3 c^5 \tan ^3(e+f x) \sec (e+f x)}{12 f}+\frac{5 a^3 c^5 \tan (e+f x) \sec (e+f x)}{8 f}+a^3 c^5 x \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^5,x]

[Out]

a^3*c^5*x - (5*a^3*c^5*ArcTanh[Sin[e + f*x]])/(8*f) - (a^3*c^5*Tan[e + f*x])/f + (5*a^3*c^5*Sec[e + f*x]*Tan[e
+ f*x])/(8*f) + (a^3*c^5*Tan[e + f*x]^3)/(3*f) - (5*a^3*c^5*Sec[e + f*x]*Tan[e + f*x]^3)/(12*f) - (a^3*c^5*Ta
n[e + f*x]^5)/(5*f) + (a^3*c^5*Sec[e + f*x]*Tan[e + f*x]^5)/(3*f) - (a^3*c^5*Tan[e + f*x]^7)/(7*f)

Rule 3904

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[(-(a*c))^m, Int[Cot[e + f*x]^(2*m)*(c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] &&  !(IntegerQ[n] && GtQ[m - n, 0])

Rule 3886

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^5 \, dx &=-\left (\left (a^3 c^3\right ) \int (c-c \sec (e+f x))^2 \tan ^6(e+f x) \, dx\right )\\ &=-\left (\left (a^3 c^3\right ) \int \left (c^2 \tan ^6(e+f x)-2 c^2 \sec (e+f x) \tan ^6(e+f x)+c^2 \sec ^2(e+f x) \tan ^6(e+f x)\right ) \, dx\right )\\ &=-\left (\left (a^3 c^5\right ) \int \tan ^6(e+f x) \, dx\right )-\left (a^3 c^5\right ) \int \sec ^2(e+f x) \tan ^6(e+f x) \, dx+\left (2 a^3 c^5\right ) \int \sec (e+f x) \tan ^6(e+f x) \, dx\\ &=-\frac{a^3 c^5 \tan ^5(e+f x)}{5 f}+\frac{a^3 c^5 \sec (e+f x) \tan ^5(e+f x)}{3 f}+\left (a^3 c^5\right ) \int \tan ^4(e+f x) \, dx-\frac{1}{3} \left (5 a^3 c^5\right ) \int \sec (e+f x) \tan ^4(e+f x) \, dx-\frac{\left (a^3 c^5\right ) \operatorname{Subst}\left (\int x^6 \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{a^3 c^5 \tan ^3(e+f x)}{3 f}-\frac{5 a^3 c^5 \sec (e+f x) \tan ^3(e+f x)}{12 f}-\frac{a^3 c^5 \tan ^5(e+f x)}{5 f}+\frac{a^3 c^5 \sec (e+f x) \tan ^5(e+f x)}{3 f}-\frac{a^3 c^5 \tan ^7(e+f x)}{7 f}-\left (a^3 c^5\right ) \int \tan ^2(e+f x) \, dx+\frac{1}{4} \left (5 a^3 c^5\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx\\ &=-\frac{a^3 c^5 \tan (e+f x)}{f}+\frac{5 a^3 c^5 \sec (e+f x) \tan (e+f x)}{8 f}+\frac{a^3 c^5 \tan ^3(e+f x)}{3 f}-\frac{5 a^3 c^5 \sec (e+f x) \tan ^3(e+f x)}{12 f}-\frac{a^3 c^5 \tan ^5(e+f x)}{5 f}+\frac{a^3 c^5 \sec (e+f x) \tan ^5(e+f x)}{3 f}-\frac{a^3 c^5 \tan ^7(e+f x)}{7 f}-\frac{1}{8} \left (5 a^3 c^5\right ) \int \sec (e+f x) \, dx+\left (a^3 c^5\right ) \int 1 \, dx\\ &=a^3 c^5 x-\frac{5 a^3 c^5 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac{a^3 c^5 \tan (e+f x)}{f}+\frac{5 a^3 c^5 \sec (e+f x) \tan (e+f x)}{8 f}+\frac{a^3 c^5 \tan ^3(e+f x)}{3 f}-\frac{5 a^3 c^5 \sec (e+f x) \tan ^3(e+f x)}{12 f}-\frac{a^3 c^5 \tan ^5(e+f x)}{5 f}+\frac{a^3 c^5 \sec (e+f x) \tan ^5(e+f x)}{3 f}-\frac{a^3 c^5 \tan ^7(e+f x)}{7 f}\\ \end{align*}

Mathematica [A]  time = 2.2194, size = 189, normalized size = 1.01 \[ \frac{a^3 c^5 \sec ^7(e+f x) \left (-4200 \sin (e+f x)+2975 \sin (2 (e+f x))-2184 \sin (3 (e+f x))+980 \sin (4 (e+f x))-2408 \sin (5 (e+f x))+1155 \sin (6 (e+f x))-584 \sin (7 (e+f x))+14700 (e+f x) \cos (e+f x)+8820 e \cos (3 (e+f x))+8820 f x \cos (3 (e+f x))+2940 e \cos (5 (e+f x))+2940 f x \cos (5 (e+f x))+420 e \cos (7 (e+f x))+420 f x \cos (7 (e+f x))-16800 \cos ^7(e+f x) \tanh ^{-1}(\sin (e+f x))\right )}{26880 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^5,x]

[Out]

(a^3*c^5*Sec[e + f*x]^7*(14700*(e + f*x)*Cos[e + f*x] - 16800*ArcTanh[Sin[e + f*x]]*Cos[e + f*x]^7 + 8820*e*Co
s[3*(e + f*x)] + 8820*f*x*Cos[3*(e + f*x)] + 2940*e*Cos[5*(e + f*x)] + 2940*f*x*Cos[5*(e + f*x)] + 420*e*Cos[7
*(e + f*x)] + 420*f*x*Cos[7*(e + f*x)] - 4200*Sin[e + f*x] + 2975*Sin[2*(e + f*x)] - 2184*Sin[3*(e + f*x)] + 9
80*Sin[4*(e + f*x)] - 2408*Sin[5*(e + f*x)] + 1155*Sin[6*(e + f*x)] - 584*Sin[7*(e + f*x)]))/(26880*f)

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Maple [A]  time = 0.035, size = 211, normalized size = 1.1 \begin{align*} -{\frac{13\,{c}^{5}{a}^{3}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{12\,f}}+{\frac{11\,{c}^{5}{a}^{3}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{8\,f}}-{\frac{5\,{c}^{5}{a}^{3}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{8\,f}}-{\frac{146\,{c}^{5}{a}^{3}\tan \left ( fx+e \right ) }{105\,f}}+{a}^{3}{c}^{5}x+{\frac{{a}^{3}{c}^{5}e}{f}}+{\frac{8\,{c}^{5}{a}^{3}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{4}}{35\,f}}+{\frac{32\,{c}^{5}{a}^{3}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{105\,f}}+{\frac{{c}^{5}{a}^{3}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{5}}{3\,f}}-{\frac{{c}^{5}{a}^{3}\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((a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^5,x)

[Out]

-13/12/f*c^5*a^3*tan(f*x+e)*sec(f*x+e)^3+11/8*a^3*c^5*sec(f*x+e)*tan(f*x+e)/f-5/8/f*c^5*a^3*ln(sec(f*x+e)+tan(
f*x+e))-146/105*a^3*c^5*tan(f*x+e)/f+a^3*c^5*x+1/f*a^3*c^5*e+8/35/f*c^5*a^3*tan(f*x+e)*sec(f*x+e)^4+32/105/f*c
^5*a^3*tan(f*x+e)*sec(f*x+e)^2+1/3/f*c^5*a^3*tan(f*x+e)*sec(f*x+e)^5-1/7/f*c^5*a^3*tan(f*x+e)*sec(f*x+e)^6

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Maxima [B]  time = 1.00956, size = 481, normalized size = 2.56 \begin{align*} -\frac{48 \,{\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^{5} - 224 \,{\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^{5} - 1680 \,{\left (f x + e\right )} a^{3} c^{5} + 35 \, a^{3} c^{5}{\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^{5}{\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^{5}{\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^{5} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 3360 \, a^{3} c^{5} \tan \left (f x + e\right )}{1680 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^5,x, algorithm="maxima")

[Out]

-1/1680*(48*(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^5 - 224*(3*tan(
f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^3*c^5 - 1680*(f*x + e)*a^3*c^5 + 35*a^3*c^5*(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)) - 630*a^3*c^5*(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^5*(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^5*log(sec(f*x + e)
+ tan(f*x + e)) + 3360*a^3*c^5*tan(f*x + e))/f

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Fricas [A]  time = 1.21925, size = 498, normalized size = 2.65 \begin{align*} \frac{1680 \, a^{3} c^{5} f x \cos \left (f x + e\right )^{7} - 525 \, a^{3} c^{5} \cos \left (f x + e\right )^{7} \log \left (\sin \left (f x + e\right ) + 1\right ) + 525 \, a^{3} c^{5} \cos \left (f x + e\right )^{7} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (1168 \, a^{3} c^{5} \cos \left (f x + e\right )^{6} - 1155 \, a^{3} c^{5} \cos \left (f x + e\right )^{5} - 256 \, a^{3} c^{5} \cos \left (f x + e\right )^{4} + 910 \, a^{3} c^{5} \cos \left (f x + e\right )^{3} - 192 \, a^{3} c^{5} \cos \left (f x + e\right )^{2} - 280 \, a^{3} c^{5} \cos \left (f x + e\right ) + 120 \, a^{3} c^{5}\right )} \sin \left (f x + e\right )}{1680 \, f \cos \left (f x + e\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^5,x, algorithm="fricas")

[Out]

1/1680*(1680*a^3*c^5*f*x*cos(f*x + e)^7 - 525*a^3*c^5*cos(f*x + e)^7*log(sin(f*x + e) + 1) + 525*a^3*c^5*cos(f
*x + e)^7*log(-sin(f*x + e) + 1) - 2*(1168*a^3*c^5*cos(f*x + e)^6 - 1155*a^3*c^5*cos(f*x + e)^5 - 256*a^3*c^5*
cos(f*x + e)^4 + 910*a^3*c^5*cos(f*x + e)^3 - 192*a^3*c^5*cos(f*x + e)^2 - 280*a^3*c^5*cos(f*x + e) + 120*a^3*
c^5)*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^{5} \left (\int \left (-1\right )\, dx + \int 2 \sec{\left (e + f x \right )}\, dx + \int 2 \sec ^{2}{\left (e + f x \right )}\, dx + \int - 6 \sec ^{3}{\left (e + f x \right )}\, dx + \int 6 \sec ^{5}{\left (e + f x \right )}\, dx + \int - 2 \sec ^{6}{\left (e + f x \right )}\, dx + \int - 2 \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((a+a*sec(f*x+e))**3*(c-c*sec(f*x+e))**5,x)

[Out]

-a**3*c**5*(Integral(-1, x) + Integral(2*sec(e + f*x), x) + Integral(2*sec(e + f*x)**2, x) + Integral(-6*sec(e
+ f*x)**3, x) + Integral(6*sec(e + f*x)**5, x) + Integral(-2*sec(e + f*x)**6, x) + Integral(-2*sec(e + f*x)**
7, x) + Integral(sec(e + f*x)**8, x))

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Giac [A]  time = 1.42914, size = 298, normalized size = 1.59 \begin{align*} \frac{840 \,{\left (f x + e\right )} a^{3} c^{5} - 525 \, a^{3} c^{5} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right ) + 525 \, a^{3} c^{5} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right ) + \frac{2 \,{\left (1365 \, a^{3} c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{13} - 9660 \, a^{3} c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{11} + 29673 \, a^{3} c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9} - 21216 \, a^{3} c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 9863 \, a^{3} c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 2660 \, a^{3} c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 315 \, a^{3} c^{5} \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}}}{840 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^5,x, algorithm="giac")

[Out]

1/840*(840*(f*x + e)*a^3*c^5 - 525*a^3*c^5*log(abs(tan(1/2*f*x + 1/2*e) + 1)) + 525*a^3*c^5*log(abs(tan(1/2*f*
x + 1/2*e) - 1)) + 2*(1365*a^3*c^5*tan(1/2*f*x + 1/2*e)^13 - 9660*a^3*c^5*tan(1/2*f*x + 1/2*e)^11 + 29673*a^3*
c^5*tan(1/2*f*x + 1/2*e)^9 - 21216*a^3*c^5*tan(1/2*f*x + 1/2*e)^7 + 9863*a^3*c^5*tan(1/2*f*x + 1/2*e)^5 - 2660
*a^3*c^5*tan(1/2*f*x + 1/2*e)^3 + 315*a^3*c^5*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^7)/f