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

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

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 80, normalized size of antiderivative = 1.00, 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} \[ -\frac {\cot ^5(e+f x)}{5 a^3 c^2 f}+\frac {\csc ^5(e+f x)}{5 a^3 c^2 f}-\frac {2 \csc ^3(e+f x)}{3 a^3 c^2 f}+\frac {\csc (e+f x)}{a^3 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.82, size = 147, normalized size = 1.84 \[ \frac {\csc (e) \sin ^2\left (\frac {1}{2} (e+f x)\right ) (-534 \sin (e+f x)-178 \sin (2 (e+f x))+178 \sin (3 (e+f x))+89 \sin (4 (e+f x))+40 \sin (2 e+f x)+168 \sin (e+2 f x)-120 \sin (3 e+2 f x)+72 \sin (2 e+3 f x)-120 \sin (4 e+3 f x)-24 \sin (3 e+4 f x)+200 \sin (e)+104 \sin (f x)) \csc ^5(e+f x)}{480 a^3 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[e]*Csc[e + f*x]^5*Sin[(e + f*x)/2]^2*(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]))/(480*a^3*c^2*f)

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fricas [A]  time = 0.44, size = 109, normalized size = 1.36 \[ -\frac {3 \, \cos \left (f x + e\right )^{4} - 12 \, \cos \left (f x + e\right )^{3} - 12 \, \cos \left (f x + e\right )^{2} + 8 \, \cos \left (f x + e\right ) + 8}{15 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} + a^{3} c^{2} f \cos \left (f x + e\right )^{2} - a^{3} c^{2} f \cos \left (f x + e\right ) - a^{3} c^{2} f\right )} \sin \left (f x + e\right )} \]

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))^2,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^3*c^2*f*cos(f*x + e)
^3 + a^3*c^2*f*cos(f*x + e)^2 - a^3*c^2*f*cos(f*x + e) - a^3*c^2*f)*sin(f*x + e))

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giac [A]  time = 0.39, size = 108, normalized size = 1.35 \[ \frac {\frac {5 \, {\left (12 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}}{a^{3} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}} + \frac {3 \, a^{12} c^{8} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 20 \, a^{12} c^{8} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 90 \, a^{12} c^{8} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{15} c^{10}}}{240 \, f} \]

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))^2,x, algorithm="giac")

[Out]

1/240*(5*(12*tan(1/2*f*x + 1/2*e)^2 - 1)/(a^3*c^2*tan(1/2*f*x + 1/2*e)^3) + (3*a^12*c^8*tan(1/2*f*x + 1/2*e)^5
- 20*a^12*c^8*tan(1/2*f*x + 1/2*e)^3 + 90*a^12*c^8*tan(1/2*f*x + 1/2*e))/(a^15*c^10))/f

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maple [A]  time = 0.87, size = 76, normalized size = 0.95 \[ \frac {\frac {\left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{5}-\frac {4 \left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{3}+6 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )-\frac {1}{3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{3}}+\frac {4}{\tan \left (\frac {e}{2}+\frac {f x}{2}\right )}}{16 f \,a^{3} c^{2}} \]

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))^2,x)

[Out]

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

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maxima [A]  time = 0.34, size = 120, normalized size = 1.50 \[ \frac {\frac {\frac {90 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3} c^{2}} + \frac {5 \, {\left (\frac {12 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{a^{3} c^{2} \sin \left (f x + e\right )^{3}}}{240 \, f} \]

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))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.71, size = 111, normalized size = 1.39 \[ \frac {48\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-192\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+168\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-32\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+3}{240\,a^3\,c^2\,f\,\left ({\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(e + f*x)*(a + a/cos(e + f*x))^3*(c - c/cos(e + f*x))^2),x)

[Out]

(168*cos(e/2 + (f*x)/2)^4 - 32*cos(e/2 + (f*x)/2)^2 - 192*cos(e/2 + (f*x)/2)^6 + 48*cos(e/2 + (f*x)/2)^8 + 3)/
(240*a^3*c^2*f*(cos(e/2 + (f*x)/2)^5*sin(e/2 + (f*x)/2) - cos(e/2 + (f*x)/2)^7*sin(e/2 + (f*x)/2)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec {\left (e + f x \right )}}{\sec ^{5}{\left (e + f x \right )} + \sec ^{4}{\left (e + f x \right )} - 2 \sec ^{3}{\left (e + f x \right )} - 2 \sec ^{2}{\left (e + f x \right )} + \sec {\left (e + f x \right )} + 1}\, dx}{a^{3} c^{2}} \]

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))**2,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**3*c**2)

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