3.36 \(\int \frac{1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))} \, dx\)

Optimal. Leaf size=126 \[ \frac{2 \cot ^5(e+f x)}{5 a^3 c f}-\frac{\cot ^3(e+f x)}{3 a^3 c f}+\frac{\cot (e+f x)}{a^3 c f}-\frac{2 \csc ^5(e+f x)}{5 a^3 c f}+\frac{4 \csc ^3(e+f x)}{3 a^3 c f}-\frac{2 \csc (e+f x)}{a^3 c f}+\frac{x}{a^3 c} \]

[Out]

x/(a^3*c) + Cot[e + f*x]/(a^3*c*f) - Cot[e + f*x]^3/(3*a^3*c*f) + (2*Cot[e + f*x]^5)/(5*a^3*c*f) - (2*Csc[e +
f*x])/(a^3*c*f) + (4*Csc[e + f*x]^3)/(3*a^3*c*f) - (2*Csc[e + f*x]^5)/(5*a^3*c*f)

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Rubi [A]  time = 0.193262, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {3904, 3886, 3473, 8, 2606, 194, 2607, 30} \[ \frac{2 \cot ^5(e+f x)}{5 a^3 c f}-\frac{\cot ^3(e+f x)}{3 a^3 c f}+\frac{\cot (e+f x)}{a^3 c f}-\frac{2 \csc ^5(e+f x)}{5 a^3 c f}+\frac{4 \csc ^3(e+f x)}{3 a^3 c f}-\frac{2 \csc (e+f x)}{a^3 c f}+\frac{x}{a^3 c} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])),x]

[Out]

x/(a^3*c) + Cot[e + f*x]/(a^3*c*f) - Cot[e + f*x]^3/(3*a^3*c*f) + (2*Cot[e + f*x]^5)/(5*a^3*c*f) - (2*Csc[e +
f*x])/(a^3*c*f) + (4*Csc[e + f*x]^3)/(3*a^3*c*f) - (2*Csc[e + f*x]^5)/(5*a^3*c*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 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

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

Mathematica [A]  time = 0.929026, size = 197, normalized size = 1.56 \[ -\frac{\csc \left (\frac{e}{2}\right ) \sec \left (\frac{e}{2}\right ) \csc \left (\frac{1}{2} (e+f x)\right ) \sec ^5\left (\frac{1}{2} (e+f x)\right ) (-445 \sin (e+f x)-356 \sin (2 (e+f x))-89 \sin (3 (e+f x))+240 \sin (2 e+f x)+296 \sin (e+2 f x)+120 \sin (3 e+2 f x)+104 \sin (2 e+3 f x)+150 f x \cos (2 e+f x)-120 f x \cos (e+2 f x)+120 f x \cos (3 e+2 f x)-30 f x \cos (2 e+3 f x)+30 f x \cos (4 e+3 f x)+80 \sin (e)+280 \sin (f x)-150 f x \cos (f x))}{3840 a^3 c f} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])),x]

[Out]

-(Csc[e/2]*Csc[(e + f*x)/2]*Sec[e/2]*Sec[(e + f*x)/2]^5*(-150*f*x*Cos[f*x] + 150*f*x*Cos[2*e + f*x] - 120*f*x*
Cos[e + 2*f*x] + 120*f*x*Cos[3*e + 2*f*x] - 30*f*x*Cos[2*e + 3*f*x] + 30*f*x*Cos[4*e + 3*f*x] + 80*Sin[e] + 28
0*Sin[f*x] - 445*Sin[e + f*x] - 356*Sin[2*(e + f*x)] - 89*Sin[3*(e + f*x)] + 240*Sin[2*e + f*x] + 296*Sin[e +
2*f*x] + 120*Sin[3*e + 2*f*x] + 104*Sin[2*e + 3*f*x]))/(3840*a^3*c*f)

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Maple [A]  time = 0.061, size = 109, normalized size = 0.9 \begin{align*} -{\frac{1}{40\,f{a}^{3}c} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}+{\frac{5}{24\,f{a}^{3}c} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-{\frac{11}{8\,f{a}^{3}c}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{a}^{3}c}}+{\frac{1}{8\,f{a}^{3}c} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e)),x)

[Out]

-1/40/f/a^3/c*tan(1/2*f*x+1/2*e)^5+5/24/f/a^3/c*tan(1/2*f*x+1/2*e)^3-11/8/f/a^3/c*tan(1/2*f*x+1/2*e)+2/f/a^3/c
*arctan(tan(1/2*f*x+1/2*e))+1/8/f/a^3/c/tan(1/2*f*x+1/2*e)

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Maxima [A]  time = 1.55291, size = 165, normalized size = 1.31 \begin{align*} -\frac{\frac{\frac{165 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{25 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3} c} - \frac{240 \, \arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{3} c} - \frac{15 \,{\left (\cos \left (f x + e\right ) + 1\right )}}{a^{3} c \sin \left (f x + e\right )}}{120 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e)),x, algorithm="maxima")

[Out]

-1/120*((165*sin(f*x + e)/(cos(f*x + e) + 1) - 25*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos(
f*x + e) + 1)^5)/(a^3*c) - 240*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/(a^3*c) - 15*(cos(f*x + e) + 1)/(a^3*c*
sin(f*x + e)))/f

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Fricas [A]  time = 1.03937, size = 284, normalized size = 2.25 \begin{align*} \frac{26 \, \cos \left (f x + e\right )^{3} + 22 \, \cos \left (f x + e\right )^{2} + 15 \,{\left (f x \cos \left (f x + e\right )^{2} + 2 \, f x \cos \left (f x + e\right ) + f x\right )} \sin \left (f x + e\right ) - 17 \, \cos \left (f x + e\right ) - 16}{15 \,{\left (a^{3} c f \cos \left (f x + e\right )^{2} + 2 \, a^{3} c f \cos \left (f x + e\right ) + a^{3} c f\right )} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e)),x, algorithm="fricas")

[Out]

1/15*(26*cos(f*x + e)^3 + 22*cos(f*x + e)^2 + 15*(f*x*cos(f*x + e)^2 + 2*f*x*cos(f*x + e) + f*x)*sin(f*x + e)
- 17*cos(f*x + e) - 16)/((a^3*c*f*cos(f*x + e)^2 + 2*a^3*c*f*cos(f*x + e) + a^3*c*f)*sin(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{1}{\sec ^{4}{\left (e + f x \right )} + 2 \sec ^{3}{\left (e + f x \right )} - 2 \sec{\left (e + f x \right )} - 1}\, dx}{a^{3} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e)),x)

[Out]

-Integral(1/(sec(e + f*x)**4 + 2*sec(e + f*x)**3 - 2*sec(e + f*x) - 1), x)/(a**3*c)

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Giac [A]  time = 1.33161, size = 144, normalized size = 1.14 \begin{align*} \frac{\frac{120 \,{\left (f x + e\right )}}{a^{3} c} + \frac{15}{a^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )} - \frac{3 \, a^{12} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 25 \, a^{12} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 165 \, a^{12} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{15} c^{5}}}{120 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e)),x, algorithm="giac")

[Out]

1/120*(120*(f*x + e)/(a^3*c) + 15/(a^3*c*tan(1/2*f*x + 1/2*e)) - (3*a^12*c^4*tan(1/2*f*x + 1/2*e)^5 - 25*a^12*
c^4*tan(1/2*f*x + 1/2*e)^3 + 165*a^12*c^4*tan(1/2*f*x + 1/2*e))/(a^15*c^5))/f