3.50 \(\int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^4} \, dx\)

Optimal. Leaf size=98 \[ -\frac {2 \cot ^7(e+f x)}{7 a^2 c^4 f}-\frac {2 \csc ^7(e+f x)}{7 a^2 c^4 f}+\frac {\csc ^5(e+f x)}{a^2 c^4 f}-\frac {4 \csc ^3(e+f x)}{3 a^2 c^4 f}+\frac {\csc (e+f x)}{a^2 c^4 f} \]

[Out]

-2/7*cot(f*x+e)^7/a^2/c^4/f+csc(f*x+e)/a^2/c^4/f-4/3*csc(f*x+e)^3/a^2/c^4/f+csc(f*x+e)^5/a^2/c^4/f-2/7*csc(f*x
+e)^7/a^2/c^4/f

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Rubi [A]  time = 0.19, antiderivative size = 98, 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, 270} \[ -\frac {2 \cot ^7(e+f x)}{7 a^2 c^4 f}-\frac {2 \csc ^7(e+f x)}{7 a^2 c^4 f}+\frac {\csc ^5(e+f x)}{a^2 c^4 f}-\frac {4 \csc ^3(e+f x)}{3 a^2 c^4 f}+\frac {\csc (e+f x)}{a^2 c^4 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^4),x]

[Out]

(-2*Cot[e + f*x]^7)/(7*a^2*c^4*f) + Csc[e + f*x]/(a^2*c^4*f) - (4*Csc[e + f*x]^3)/(3*a^2*c^4*f) + Csc[e + f*x]
^5/(a^2*c^4*f) - (2*Csc[e + f*x]^7)/(7*a^2*c^4*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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && 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))^2 (c-c \sec (e+f x))^4} \, dx &=\frac {\int \left (a^2 \cot ^7(e+f x) \csc (e+f x)+2 a^2 \cot ^6(e+f x) \csc ^2(e+f x)+a^2 \cot ^5(e+f x) \csc ^3(e+f x)\right ) \, dx}{a^4 c^4}\\ &=\frac {\int \cot ^7(e+f x) \csc (e+f x) \, dx}{a^2 c^4}+\frac {\int \cot ^5(e+f x) \csc ^3(e+f x) \, dx}{a^2 c^4}+\frac {2 \int \cot ^6(e+f x) \csc ^2(e+f x) \, dx}{a^2 c^4}\\ &=-\frac {\operatorname {Subst}\left (\int x^2 \left (-1+x^2\right )^2 \, dx,x,\csc (e+f x)\right )}{a^2 c^4 f}-\frac {\operatorname {Subst}\left (\int \left (-1+x^2\right )^3 \, dx,x,\csc (e+f x)\right )}{a^2 c^4 f}+\frac {2 \operatorname {Subst}\left (\int x^6 \, dx,x,-\cot (e+f x)\right )}{a^2 c^4 f}\\ &=-\frac {2 \cot ^7(e+f x)}{7 a^2 c^4 f}-\frac {\operatorname {Subst}\left (\int \left (-1+3 x^2-3 x^4+x^6\right ) \, dx,x,\csc (e+f x)\right )}{a^2 c^4 f}-\frac {\operatorname {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\csc (e+f x)\right )}{a^2 c^4 f}\\ &=-\frac {2 \cot ^7(e+f x)}{7 a^2 c^4 f}+\frac {\csc (e+f x)}{a^2 c^4 f}-\frac {4 \csc ^3(e+f x)}{3 a^2 c^4 f}+\frac {\csc ^5(e+f x)}{a^2 c^4 f}-\frac {2 \csc ^7(e+f x)}{7 a^2 c^4 f}\\ \end {align*}

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Mathematica [A]  time = 0.93, size = 179, normalized size = 1.83 \[ \frac {\csc (e) (-182 \sin (e+f x)+104 \sin (2 (e+f x))+39 \sin (3 (e+f x))-52 \sin (4 (e+f x))+13 \sin (5 (e+f x))-56 \sin (2 e+f x)+76 \sin (e+2 f x)-28 \sin (3 e+2 f x)-24 \sin (2 e+3 f x)+42 \sin (4 e+3 f x)-3 \sin (3 e+4 f x)-21 \sin (5 e+4 f x)+6 \sin (4 e+5 f x)+42 \sin (e)-28 \sin (f x)) \csc ^4\left (\frac {1}{2} (e+f x)\right ) \csc ^3(e+f x)}{1344 a^2 c^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^4),x]

[Out]

(Csc[e]*Csc[(e + f*x)/2]^4*Csc[e + f*x]^3*(42*Sin[e] - 28*Sin[f*x] - 182*Sin[e + f*x] + 104*Sin[2*(e + f*x)] +
39*Sin[3*(e + f*x)] - 52*Sin[4*(e + f*x)] + 13*Sin[5*(e + f*x)] - 56*Sin[2*e + f*x] + 76*Sin[e + 2*f*x] - 28*
Sin[3*e + 2*f*x] - 24*Sin[2*e + 3*f*x] + 42*Sin[4*e + 3*f*x] - 3*Sin[3*e + 4*f*x] - 21*Sin[5*e + 4*f*x] + 6*Si
n[4*e + 5*f*x]))/(1344*a^2*c^4*f)

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fricas [A]  time = 0.45, size = 120, normalized size = 1.22 \[ \frac {6 \, \cos \left (f x + e\right )^{5} + 9 \, \cos \left (f x + e\right )^{4} - 24 \, \cos \left (f x + e\right )^{3} + 4 \, \cos \left (f x + e\right )^{2} + 16 \, \cos \left (f x + e\right ) - 8}{21 \, {\left (a^{2} c^{4} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} c^{4} f \cos \left (f x + e\right )^{3} + 2 \, a^{2} c^{4} f \cos \left (f x + e\right ) - a^{2} c^{4} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^4,x, algorithm="fricas")

[Out]

1/21*(6*cos(f*x + e)^5 + 9*cos(f*x + e)^4 - 24*cos(f*x + e)^3 + 4*cos(f*x + e)^2 + 16*cos(f*x + e) - 8)/((a^2*
c^4*f*cos(f*x + e)^4 - 2*a^2*c^4*f*cos(f*x + e)^3 + 2*a^2*c^4*f*cos(f*x + e) - a^2*c^4*f)*sin(f*x + e))

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giac [A]  time = 0.44, size = 115, normalized size = 1.17 \[ \frac {\frac {210 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 70 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 21 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3}{a^{2} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7}} - \frac {7 \, {\left (a^{4} c^{8} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 15 \, a^{4} c^{8} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{6} c^{12}}}{672 \, f} \]

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))^4,x, algorithm="giac")

[Out]

1/672*((210*tan(1/2*f*x + 1/2*e)^6 - 70*tan(1/2*f*x + 1/2*e)^4 + 21*tan(1/2*f*x + 1/2*e)^2 - 3)/(a^2*c^4*tan(1
/2*f*x + 1/2*e)^7) - 7*(a^4*c^8*tan(1/2*f*x + 1/2*e)^3 - 15*a^4*c^8*tan(1/2*f*x + 1/2*e))/(a^6*c^12))/f

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maple [A]  time = 0.84, size = 87, normalized size = 0.89 \[ \frac {-\frac {\left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{3}+5 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )-\frac {1}{7 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{7}}-\frac {10}{3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{3}}+\frac {1}{\tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{5}}+\frac {10}{\tan \left (\frac {e}{2}+\frac {f x}{2}\right )}}{32 f \,a^{2} c^{4}} \]

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))^4,x)

[Out]

1/32/f/a^2/c^4*(-1/3*tan(1/2*e+1/2*f*x)^3+5*tan(1/2*e+1/2*f*x)-1/7/tan(1/2*e+1/2*f*x)^7-10/3/tan(1/2*e+1/2*f*x
)^3+1/tan(1/2*e+1/2*f*x)^5+10/tan(1/2*e+1/2*f*x))

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maxima [A]  time = 0.34, size = 140, normalized size = 1.43 \[ \frac {\frac {7 \, {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2} c^{4}} + \frac {{\left (\frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {70 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {210 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{a^{2} c^{4} \sin \left (f x + e\right )^{7}}}{672 \, f} \]

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))^4,x, algorithm="maxima")

[Out]

1/672*(7*(15*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^2*c^4) + (21*sin(f*x +
e)^2/(cos(f*x + e) + 1)^2 - 70*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 210*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 -
3)*(cos(f*x + e) + 1)^7/(a^2*c^4*sin(f*x + e)^7))/f

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mupad [B]  time = 2.67, size = 89, normalized size = 0.91 \[ \frac {-7\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+105\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+210\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-70\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+21\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-3}{672\,a^2\,c^4\,f\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(e + f*x)*(a + a/cos(e + f*x))^2*(c - c/cos(e + f*x))^4),x)

[Out]

(21*tan(e/2 + (f*x)/2)^2 - 70*tan(e/2 + (f*x)/2)^4 + 210*tan(e/2 + (f*x)/2)^6 + 105*tan(e/2 + (f*x)/2)^8 - 7*t
an(e/2 + (f*x)/2)^10 - 3)/(672*a^2*c^4*f*tan(e/2 + (f*x)/2)^7)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec {\left (e + f x \right )}}{\sec ^{6}{\left (e + f x \right )} - 2 \sec ^{5}{\left (e + f x \right )} - \sec ^{4}{\left (e + f x \right )} + 4 \sec ^{3}{\left (e + f x \right )} - \sec ^{2}{\left (e + f x \right )} - 2 \sec {\left (e + f x \right )} + 1}\, dx}{a^{2} c^{4}} \]

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))**4,x)

[Out]

Integral(sec(e + f*x)/(sec(e + f*x)**6 - 2*sec(e + f*x)**5 - sec(e + f*x)**4 + 4*sec(e + f*x)**3 - sec(e + f*x
)**2 - 2*sec(e + f*x) + 1), x)/(a**2*c**4)

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