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

Optimal. Leaf size=78 \[ -\frac{2 \cot ^5(e+f x)}{5 a^3 c f}+\frac{2 \csc ^5(e+f x)}{5 a^3 c f}-\frac{\csc ^3(e+f x)}{a^3 c f}+\frac{\csc (e+f x)}{a^3 c f} \]

[Out]

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

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Rubi [A]  time = 0.180993, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3958, 2606, 194, 2607, 30, 14} \[ -\frac{2 \cot ^5(e+f x)}{5 a^3 c f}+\frac{2 \csc ^5(e+f x)}{5 a^3 c f}-\frac{\csc ^3(e+f x)}{a^3 c f}+\frac{\csc (e+f x)}{a^3 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.773441, size = 109, normalized size = 1.4 \[ -\frac{\csc (e) \sin ^4\left (\frac{1}{2} (e+f x)\right ) (65 \sin (e+f x)+52 \sin (2 (e+f x))+13 \sin (3 (e+f x))-40 \sin (2 e+f x)-12 \sin (e+2 f x)-20 \sin (3 e+2 f x)-8 \sin (2 e+3 f x)-40 \sin (e)) \csc ^5(e+f x)}{20 a^3 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[e]*Csc[e + f*x]^5*Sin[(e + f*x)/2]^4*(-40*Sin[e] + 65*Sin[e + f*x] + 52*Sin[2*(e + f*x)] + 13*Sin[3*(e +
f*x)] - 40*Sin[2*e + f*x] - 12*Sin[e + 2*f*x] - 20*Sin[3*e + 2*f*x] - 8*Sin[2*e + 3*f*x]))/(20*a^3*c*f)

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Maple [A]  time = 0.053, size = 61, normalized size = 0.8 \begin{align*}{\frac{1}{8\,f{a}^{3}c} \left ({\frac{1}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}- \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}+3\,\tan \left ( 1/2\,fx+e/2 \right ) + \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-1} \right ) } \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)),x)

[Out]

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

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Maxima [A]  time = 0.971455, size = 128, normalized size = 1.64 \begin{align*} \frac{\frac{\frac{15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3} c} + \frac{5 \,{\left (\cos \left (f x + e\right ) + 1\right )}}{a^{3} c \sin \left (f x + e\right )}}{40 \, 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)),x, algorithm="maxima")

[Out]

1/40*((15*sin(f*x + e)/(cos(f*x + e) + 1) - 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + sin(f*x + e)^5/(cos(f*x +
e) + 1)^5)/(a^3*c) + 5*(cos(f*x + e) + 1)/(a^3*c*sin(f*x + e)))/f

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Fricas [A]  time = 0.443794, size = 186, normalized size = 2.38 \begin{align*} -\frac{2 \, \cos \left (f x + e\right )^{3} - \cos \left (f x + e\right )^{2} - 4 \, \cos \left (f x + e\right ) - 2}{5 \,{\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(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e)),x, algorithm="fricas")

[Out]

-1/5*(2*cos(f*x + e)^3 - cos(f*x + e)^2 - 4*cos(f*x + e) - 2)/((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{\sec{\left (e + f x \right )}}{\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(sec(f*x+e)/(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e)),x)

[Out]

-Integral(sec(e + f*x)/(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 = 1.28059, size = 123, normalized size = 1.58 \begin{align*} \frac{\frac{5}{a^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )} + \frac{a^{12} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 5 \, a^{12} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 15 \, a^{12} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{15} c^{5}}}{40 \, 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)),x, algorithm="giac")

[Out]

1/40*(5/(a^3*c*tan(1/2*f*x + 1/2*e)) + (a^12*c^4*tan(1/2*f*x + 1/2*e)^5 - 5*a^12*c^4*tan(1/2*f*x + 1/2*e)^3 +
15*a^12*c^4*tan(1/2*f*x + 1/2*e))/(a^15*c^5))/f