3.49 \(\int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=80 \[ \frac{\cot ^5(e+f x)}{5 a^2 c^3 f}+\frac{\csc ^5(e+f x)}{5 a^2 c^3 f}-\frac{2 \csc ^3(e+f x)}{3 a^2 c^3 f}+\frac{\csc (e+f x)}{a^2 c^3 f} \]

[Out]

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

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Rubi [A]  time = 0.146201, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {3958, 2606, 194, 2607, 30} \[ \frac{\cot ^5(e+f x)}{5 a^2 c^3 f}+\frac{\csc ^5(e+f x)}{5 a^2 c^3 f}-\frac{2 \csc ^3(e+f x)}{3 a^2 c^3 f}+\frac{\csc (e+f x)}{a^2 c^3 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^3),x]

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.983502, size = 147, normalized size = 1.84 \[ -\frac{\csc (e) (534 \sin (e+f x)-178 \sin (2 (e+f x))-178 \sin (3 (e+f x))+89 \sin (4 (e+f x))+40 \sin (2 e+f x)-168 \sin (e+2 f x)+120 \sin (3 e+2 f x)+72 \sin (2 e+3 f x)-120 \sin (4 e+3 f x)+24 \sin (3 e+4 f x)-200 \sin (e)+104 \sin (f x)) \csc ^2\left (\frac{1}{2} (e+f x)\right ) \csc ^3(e+f x)}{1920 a^2 c^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^3),x]

[Out]

-(Csc[e]*Csc[(e + f*x)/2]^2*Csc[e + f*x]^3*(-200*Sin[e] + 104*Sin[f*x] + 534*Sin[e + f*x] - 178*Sin[2*(e + f*x
)] - 178*Sin[3*(e + f*x)] + 89*Sin[4*(e + f*x)] + 40*Sin[2*e + f*x] - 168*Sin[e + 2*f*x] + 120*Sin[3*e + 2*f*x
] + 72*Sin[2*e + 3*f*x] - 120*Sin[4*e + 3*f*x] + 24*Sin[3*e + 4*f*x]))/(1920*a^2*c^3*f)

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Maple [A]  time = 0.055, size = 76, normalized size = 1. \begin{align*}{\frac{1}{16\,f{a}^{2}{c}^{3}} \left ( -{\frac{1}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+4\,\tan \left ( 1/2\,fx+e/2 \right ) -{\frac{4}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-3}}+6\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{-1}+{\frac{1}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-5}} \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))^2/(c-c*sec(f*x+e))^3,x)

[Out]

1/16/f/a^2/c^3*(-1/3*tan(1/2*f*x+1/2*e)^3+4*tan(1/2*f*x+1/2*e)-4/3/tan(1/2*f*x+1/2*e)^3+6/tan(1/2*f*x+1/2*e)+1
/5/tan(1/2*f*x+1/2*e)^5)

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Maxima [A]  time = 0.983762, size = 163, normalized size = 2.04 \begin{align*} \frac{\frac{5 \,{\left (\frac{12 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2} c^{3}} - \frac{{\left (\frac{20 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{90 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 3\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{a^{2} c^{3} \sin \left (f x + e\right )^{5}}}{240 \, 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))^2/(c-c*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

1/240*(5*(12*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^2*c^3) - (20*sin(f*x +
e)^2/(cos(f*x + e) + 1)^2 - 90*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 3)*(cos(f*x + e) + 1)^5/(a^2*c^3*sin(f*x
+ e)^5))/f

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Fricas [A]  time = 0.452245, size = 259, normalized size = 3.24 \begin{align*} \frac{3 \, \cos \left (f x + e\right )^{4} + 12 \, \cos \left (f x + e\right )^{3} - 12 \, \cos \left (f x + e\right )^{2} - 8 \, \cos \left (f x + e\right ) + 8}{15 \,{\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} - a^{2} c^{3} f \cos \left (f x + e\right )^{2} - a^{2} c^{3} f \cos \left (f x + e\right ) + a^{2} c^{3} 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))^2/(c-c*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(3*cos(f*x + e)^4 + 12*cos(f*x + e)^3 - 12*cos(f*x + e)^2 - 8*cos(f*x + e) + 8)/((a^2*c^3*f*cos(f*x + e)^
3 - a^2*c^3*f*cos(f*x + e)^2 - a^2*c^3*f*cos(f*x + e) + a^2*c^3*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 ^{5}{\left (e + f x \right )} - \sec ^{4}{\left (e + f x \right )} - 2 \sec ^{3}{\left (e + f x \right )} + 2 \sec ^{2}{\left (e + f x \right )} + \sec{\left (e + f x \right )} - 1}\, dx}{a^{2} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+a*sec(f*x+e))**2/(c-c*sec(f*x+e))**3,x)

[Out]

-Integral(sec(e + f*x)/(sec(e + f*x)**5 - sec(e + f*x)**4 - 2*sec(e + f*x)**3 + 2*sec(e + f*x)**2 + sec(e + f*
x) - 1), x)/(a**2*c**3)

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Giac [A]  time = 1.28363, size = 136, normalized size = 1.7 \begin{align*} \frac{\frac{90 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 20 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3}{a^{2} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5}} - \frac{5 \,{\left (a^{4} c^{6} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 12 \, a^{4} c^{6} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{a^{6} c^{9}}}{240 \, 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))^2/(c-c*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/240*((90*tan(1/2*f*x + 1/2*e)^4 - 20*tan(1/2*f*x + 1/2*e)^2 + 3)/(a^2*c^3*tan(1/2*f*x + 1/2*e)^5) - 5*(a^4*c
^6*tan(1/2*f*x + 1/2*e)^3 - 12*a^4*c^6*tan(1/2*f*x + 1/2*e))/(a^6*c^9))/f