3.61 \(\int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^4} \, dx\)

Optimal. Leaf size=99 \[ -\frac {\cot ^7(e+f x)}{7 a^3 c^4 f}-\frac {\csc ^7(e+f x)}{7 a^3 c^4 f}+\frac {3 \csc ^5(e+f x)}{5 a^3 c^4 f}-\frac {\csc ^3(e+f x)}{a^3 c^4 f}+\frac {\csc (e+f x)}{a^3 c^4 f} \]

[Out]

-1/7*cot(f*x+e)^7/a^3/c^4/f+csc(f*x+e)/a^3/c^4/f-csc(f*x+e)^3/a^3/c^4/f+3/5*csc(f*x+e)^5/a^3/c^4/f-1/7*csc(f*x
+e)^7/a^3/c^4/f

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Rubi [A]  time = 0.15, antiderivative size = 99, 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} \[ -\frac {\cot ^7(e+f x)}{7 a^3 c^4 f}-\frac {\csc ^7(e+f x)}{7 a^3 c^4 f}+\frac {3 \csc ^5(e+f x)}{5 a^3 c^4 f}-\frac {\csc ^3(e+f x)}{a^3 c^4 f}+\frac {\csc (e+f x)}{a^3 c^4 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^4),x]

[Out]

-Cot[e + f*x]^7/(7*a^3*c^4*f) + Csc[e + f*x]/(a^3*c^4*f) - Csc[e + f*x]^3/(a^3*c^4*f) + (3*Csc[e + f*x]^5)/(5*
a^3*c^4*f) - Csc[e + f*x]^7/(7*a^3*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 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))^4} \, dx &=\frac {\int \left (a \cot ^7(e+f x) \csc (e+f x)+a \cot ^6(e+f x) \csc ^2(e+f x)\right ) \, dx}{a^4 c^4}\\ &=\frac {\int \cot ^7(e+f x) \csc (e+f x) \, dx}{a^3 c^4}+\frac {\int \cot ^6(e+f x) \csc ^2(e+f x) \, dx}{a^3 c^4}\\ &=\frac {\operatorname {Subst}\left (\int x^6 \, dx,x,-\cot (e+f x)\right )}{a^3 c^4 f}-\frac {\operatorname {Subst}\left (\int \left (-1+x^2\right )^3 \, dx,x,\csc (e+f x)\right )}{a^3 c^4 f}\\ &=-\frac {\cot ^7(e+f x)}{7 a^3 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^3 c^4 f}\\ &=-\frac {\cot ^7(e+f x)}{7 a^3 c^4 f}+\frac {\csc (e+f x)}{a^3 c^4 f}-\frac {\csc ^3(e+f x)}{a^3 c^4 f}+\frac {3 \csc ^5(e+f x)}{5 a^3 c^4 f}-\frac {\csc ^7(e+f x)}{7 a^3 c^4 f}\\ \end {align*}

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Mathematica [B]  time = 1.41, size = 211, normalized size = 2.13 \[ \frac {\csc (e) (-7620 \sin (e+f x)+1905 \sin (2 (e+f x))+3810 \sin (3 (e+f x))-1524 \sin (4 (e+f x))-762 \sin (5 (e+f x))+381 \sin (6 (e+f x))-2016 \sin (2 e+f x)+2080 \sin (e+2 f x)-1680 \sin (3 e+2 f x)+240 \sin (2 e+3 f x)+560 \sin (4 e+3 f x)-880 \sin (3 e+4 f x)+560 \sin (5 e+4 f x)+400 \sin (4 e+5 f x)-560 \sin (6 e+5 f x)+80 \sin (5 e+6 f x)+2912 \sin (e)+416 \sin (f x)) \csc ^2\left (\frac {1}{2} (e+f x)\right ) \csc ^5(e+f x)}{35840 a^3 c^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^4),x]

[Out]

(Csc[e]*Csc[(e + f*x)/2]^2*Csc[e + f*x]^5*(2912*Sin[e] + 416*Sin[f*x] - 7620*Sin[e + f*x] + 1905*Sin[2*(e + f*
x)] + 3810*Sin[3*(e + f*x)] - 1524*Sin[4*(e + f*x)] - 762*Sin[5*(e + f*x)] + 381*Sin[6*(e + f*x)] - 2016*Sin[2
*e + f*x] + 2080*Sin[e + 2*f*x] - 1680*Sin[3*e + 2*f*x] + 240*Sin[2*e + 3*f*x] + 560*Sin[4*e + 3*f*x] - 880*Si
n[3*e + 4*f*x] + 560*Sin[5*e + 4*f*x] + 400*Sin[4*e + 5*f*x] - 560*Sin[6*e + 5*f*x] + 80*Sin[5*e + 6*f*x]))/(3
5840*a^3*c^4*f)

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fricas [A]  time = 0.42, size = 163, normalized size = 1.65 \[ \frac {5 \, \cos \left (f x + e\right )^{6} + 30 \, \cos \left (f x + e\right )^{5} - 30 \, \cos \left (f x + e\right )^{4} - 40 \, \cos \left (f x + e\right )^{3} + 40 \, \cos \left (f x + e\right )^{2} + 16 \, \cos \left (f x + e\right ) - 16}{35 \, {\left (a^{3} c^{4} f \cos \left (f x + e\right )^{5} - a^{3} c^{4} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} c^{4} f \cos \left (f x + e\right )^{3} + 2 \, a^{3} c^{4} f \cos \left (f x + e\right )^{2} + a^{3} c^{4} f \cos \left (f x + e\right ) - a^{3} c^{4} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^4,x, algorithm="fricas")

[Out]

1/35*(5*cos(f*x + e)^6 + 30*cos(f*x + e)^5 - 30*cos(f*x + e)^4 - 40*cos(f*x + e)^3 + 40*cos(f*x + e)^2 + 16*co
s(f*x + e) - 16)/((a^3*c^4*f*cos(f*x + e)^5 - a^3*c^4*f*cos(f*x + e)^4 - 2*a^3*c^4*f*cos(f*x + e)^3 + 2*a^3*c^
4*f*cos(f*x + e)^2 + a^3*c^4*f*cos(f*x + e) - a^3*c^4*f)*sin(f*x + e))

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giac [A]  time = 1.01, size = 135, normalized size = 1.36 \[ \frac {\frac {700 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 175 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 42 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 5}{a^{3} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7}} + \frac {7 \, {\left (a^{12} c^{16} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 10 \, a^{12} c^{16} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 75 \, a^{12} c^{16} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{15} c^{20}}}{2240 \, f} \]

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))^4,x, algorithm="giac")

[Out]

1/2240*((700*tan(1/2*f*x + 1/2*e)^6 - 175*tan(1/2*f*x + 1/2*e)^4 + 42*tan(1/2*f*x + 1/2*e)^2 - 5)/(a^3*c^4*tan
(1/2*f*x + 1/2*e)^7) + 7*(a^12*c^16*tan(1/2*f*x + 1/2*e)^5 - 10*a^12*c^16*tan(1/2*f*x + 1/2*e)^3 + 75*a^12*c^1
6*tan(1/2*f*x + 1/2*e))/(a^15*c^20))/f

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maple [A]  time = 0.93, size = 102, normalized size = 1.03 \[ \frac {\frac {\left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{5}-2 \left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )+15 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )-\frac {1}{7 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{7}}-\frac {5}{\tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{3}}+\frac {6}{5 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{5}}+\frac {20}{\tan \left (\frac {e}{2}+\frac {f x}{2}\right )}}{64 f \,a^{3} c^{4}} \]

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))^4,x)

[Out]

1/64/f/a^3/c^4*(1/5*tan(1/2*e+1/2*f*x)^5-2*tan(1/2*e+1/2*f*x)^3+15*tan(1/2*e+1/2*f*x)-1/7/tan(1/2*e+1/2*f*x)^7
-5/tan(1/2*e+1/2*f*x)^3+6/5/tan(1/2*e+1/2*f*x)^5+20/tan(1/2*e+1/2*f*x))

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maxima [A]  time = 0.35, size = 159, normalized size = 1.61 \[ \frac {\frac {7 \, {\left (\frac {75 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3} c^{4}} + \frac {{\left (\frac {42 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {175 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {700 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 5\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{a^{3} c^{4} \sin \left (f x + e\right )^{7}}}{2240 \, f} \]

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))^4,x, algorithm="maxima")

[Out]

1/2240*(7*(75*sin(f*x + e)/(cos(f*x + e) + 1) - 10*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^4) + (42*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 175*sin(f*x + e)^4/(cos(f*x + e) + 1)^4
+ 700*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 5)*(cos(f*x + e) + 1)^7/(a^3*c^4*sin(f*x + e)^7))/f

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mupad [B]  time = 2.54, size = 129, normalized size = 1.30 \[ \frac {\left (2\,{\sin \left (\frac {e}{4}+\frac {f\,x}{4}\right )}^2-1\right )\,\left (\frac {235\,{\sin \left (e+f\,x\right )}^2}{16}-\frac {45\,{\sin \left (2\,e+2\,f\,x\right )}^2}{8}+\frac {19\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{2}+\frac {5\,{\sin \left (3\,e+3\,f\,x\right )}^2}{16}-\frac {5\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2}{4}+\frac {15\,{\sin \left (\frac {5\,e}{2}+\frac {5\,f\,x}{2}\right )}^2}{4}-5\right )}{2240\,a^3\,c^4\,f\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,{\left ({\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}^3} \]

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))^4),x)

[Out]

((2*sin(e/4 + (f*x)/4)^2 - 1)*((19*sin(e/2 + (f*x)/2)^2)/2 - (45*sin(2*e + 2*f*x)^2)/8 + (5*sin(3*e + 3*f*x)^2
)/16 - (5*sin((3*e)/2 + (3*f*x)/2)^2)/4 + (15*sin((5*e)/2 + (5*f*x)/2)^2)/4 + (235*sin(e + f*x)^2)/16 - 5))/(2
240*a^3*c^4*f*sin(e/2 + (f*x)/2)^7*(sin(e/2 + (f*x)/2)^2 - 1)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec {\left (e + f x \right )}}{\sec ^{7}{\left (e + f x \right )} - \sec ^{6}{\left (e + f x \right )} - 3 \sec ^{5}{\left (e + f x \right )} + 3 \sec ^{4}{\left (e + f x \right )} + 3 \sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} - \sec {\left (e + f x \right )} + 1}\, dx}{a^{3} c^{4}} \]

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))**4,x)

[Out]

Integral(sec(e + f*x)/(sec(e + f*x)**7 - sec(e + f*x)**6 - 3*sec(e + f*x)**5 + 3*sec(e + f*x)**4 + 3*sec(e + f
*x)**3 - 3*sec(e + f*x)**2 - sec(e + f*x) + 1), x)/(a**3*c**4)

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