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

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

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 78, normalized size of antiderivative = 1.00, 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} \[ \frac {2 \cot ^5(e+f x)}{5 a c^3 f}+\frac {2 \csc ^5(e+f x)}{5 a c^3 f}-\frac {\csc ^3(e+f x)}{a c^3 f}+\frac {\csc (e+f x)}{a c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]]

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

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Mathematica [A]  time = 0.85, size = 107, normalized size = 1.37 \[ -\frac {\csc (e) (65 \sin (e+f x)-52 \sin (2 (e+f x))+13 \sin (3 (e+f x))+40 \sin (2 e+f x)-12 \sin (e+2 f x)-20 \sin (3 e+2 f x)+8 \sin (2 e+3 f x)-40 \sin (e)) \csc ^4\left (\frac {1}{2} (e+f x)\right ) \csc (e+f x)}{320 a c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/320*(Csc[e]*Csc[(e + f*x)/2]^4*Csc[e + f*x]*(-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]))/(a*c^3*f)

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fricas [A]  time = 0.44, size = 74, normalized size = 0.95 \[ \frac {2 \, \cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} - 4 \, \cos \left (f x + e\right ) + 2}{5 \, {\left (a c^{3} f \cos \left (f x + e\right )^{2} - 2 \, a c^{3} f \cos \left (f x + e\right ) + a c^{3} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.36, size = 73, normalized size = 0.94 \[ \frac {\frac {5 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a c^{3}} + \frac {15 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 5 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1}{a c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5}}}{40 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.03, size = 61, normalized size = 0.78 \[ \frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-\frac {1}{\tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{3}}+\frac {1}{5 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{5}}+\frac {3}{\tan \left (\frac {e}{2}+\frac {f x}{2}\right )}}{8 f a \,c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 97, normalized size = 1.24 \[ -\frac {\frac {{\left (\frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{a c^{3} \sin \left (f x + e\right )^{5}} - \frac {5 \, \sin \left (f x + e\right )}{a c^{3} {\left (\cos \left (f x + e\right ) + 1\right )}}}{40 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.72, size = 63, normalized size = 0.81 \[ \frac {5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1}{40\,a\,c^3\,f\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*tan(e/2 + (f*x)/2)^4 - 5*tan(e/2 + (f*x)/2)^2 + 5*tan(e/2 + (f*x)/2)^6 + 1)/(40*a*c^3*f*tan(e/2 + (f*x)/2)
^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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