3.51 \(\int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^5} \, dx\)

Optimal. Leaf size=141 \[ \frac {4 \cot ^9(e+f x)}{9 a^2 c^5 f}+\frac {\cot ^7(e+f x)}{7 a^2 c^5 f}+\frac {4 \csc ^9(e+f x)}{9 a^2 c^5 f}-\frac {13 \csc ^7(e+f x)}{7 a^2 c^5 f}+\frac {3 \csc ^5(e+f x)}{a^2 c^5 f}-\frac {7 \csc ^3(e+f x)}{3 a^2 c^5 f}+\frac {\csc (e+f x)}{a^2 c^5 f} \]

[Out]

1/7*cot(f*x+e)^7/a^2/c^5/f+4/9*cot(f*x+e)^9/a^2/c^5/f+csc(f*x+e)/a^2/c^5/f-7/3*csc(f*x+e)^3/a^2/c^5/f+3*csc(f*
x+e)^5/a^2/c^5/f-13/7*csc(f*x+e)^7/a^2/c^5/f+4/9*csc(f*x+e)^9/a^2/c^5/f

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Rubi [A]  time = 0.25, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3958, 2606, 194, 2607, 30, 270, 14} \[ \frac {4 \cot ^9(e+f x)}{9 a^2 c^5 f}+\frac {\cot ^7(e+f x)}{7 a^2 c^5 f}+\frac {4 \csc ^9(e+f x)}{9 a^2 c^5 f}-\frac {13 \csc ^7(e+f x)}{7 a^2 c^5 f}+\frac {3 \csc ^5(e+f x)}{a^2 c^5 f}-\frac {7 \csc ^3(e+f x)}{3 a^2 c^5 f}+\frac {\csc (e+f x)}{a^2 c^5 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^5),x]

[Out]

Cot[e + f*x]^7/(7*a^2*c^5*f) + (4*Cot[e + f*x]^9)/(9*a^2*c^5*f) + Csc[e + f*x]/(a^2*c^5*f) - (7*Csc[e + f*x]^3
)/(3*a^2*c^5*f) + (3*Csc[e + f*x]^5)/(a^2*c^5*f) - (13*Csc[e + f*x]^7)/(7*a^2*c^5*f) + (4*Csc[e + f*x]^9)/(9*a
^2*c^5*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 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))^5} \, dx &=-\frac {\int \left (a^3 \cot ^9(e+f x) \csc (e+f x)+3 a^3 \cot ^8(e+f x) \csc ^2(e+f x)+3 a^3 \cot ^7(e+f x) \csc ^3(e+f x)+a^3 \cot ^6(e+f x) \csc ^4(e+f x)\right ) \, dx}{a^5 c^5}\\ &=-\frac {\int \cot ^9(e+f x) \csc (e+f x) \, dx}{a^2 c^5}-\frac {\int \cot ^6(e+f x) \csc ^4(e+f x) \, dx}{a^2 c^5}-\frac {3 \int \cot ^8(e+f x) \csc ^2(e+f x) \, dx}{a^2 c^5}-\frac {3 \int \cot ^7(e+f x) \csc ^3(e+f x) \, dx}{a^2 c^5}\\ &=\frac {\operatorname {Subst}\left (\int \left (-1+x^2\right )^4 \, dx,x,\csc (e+f x)\right )}{a^2 c^5 f}-\frac {\operatorname {Subst}\left (\int x^6 \left (1+x^2\right ) \, dx,x,-\cot (e+f x)\right )}{a^2 c^5 f}-\frac {3 \operatorname {Subst}\left (\int x^8 \, dx,x,-\cot (e+f x)\right )}{a^2 c^5 f}+\frac {3 \operatorname {Subst}\left (\int x^2 \left (-1+x^2\right )^3 \, dx,x,\csc (e+f x)\right )}{a^2 c^5 f}\\ &=\frac {\cot ^9(e+f x)}{3 a^2 c^5 f}+\frac {\operatorname {Subst}\left (\int \left (1-4 x^2+6 x^4-4 x^6+x^8\right ) \, dx,x,\csc (e+f x)\right )}{a^2 c^5 f}-\frac {\operatorname {Subst}\left (\int \left (x^6+x^8\right ) \, dx,x,-\cot (e+f x)\right )}{a^2 c^5 f}+\frac {3 \operatorname {Subst}\left (\int \left (-x^2+3 x^4-3 x^6+x^8\right ) \, dx,x,\csc (e+f x)\right )}{a^2 c^5 f}\\ &=\frac {\cot ^7(e+f x)}{7 a^2 c^5 f}+\frac {4 \cot ^9(e+f x)}{9 a^2 c^5 f}+\frac {\csc (e+f x)}{a^2 c^5 f}-\frac {7 \csc ^3(e+f x)}{3 a^2 c^5 f}+\frac {3 \csc ^5(e+f x)}{a^2 c^5 f}-\frac {13 \csc ^7(e+f x)}{7 a^2 c^5 f}+\frac {4 \csc ^9(e+f x)}{9 a^2 c^5 f}\\ \end {align*}

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Mathematica [A]  time = 1.43, size = 211, normalized size = 1.50 \[ -\frac {\csc (e) (36252 \sin (e+f x)-27189 \sin (2 (e+f x))-2014 \sin (3 (e+f x))+12084 \sin (4 (e+f x))-6042 \sin (5 (e+f x))+1007 \sin (6 (e+f x))+12096 \sin (2 e+f x)-14400 \sin (e+2 f x)-2016 \sin (3 e+2 f x)+7520 \sin (2 e+3 f x)-8736 \sin (4 e+3 f x)+1248 \sin (3 e+4 f x)+6048 \sin (5 e+4 f x)-1632 \sin (4 e+5 f x)-2016 \sin (6 e+5 f x)+608 \sin (5 e+6 f x)-9408 \sin (e)+9792 \sin (f x)) \csc ^6\left (\frac {1}{2} (e+f x)\right ) \csc ^3(e+f x)}{516096 a^2 c^5 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^5),x]

[Out]

-1/516096*(Csc[e]*Csc[(e + f*x)/2]^6*Csc[e + f*x]^3*(-9408*Sin[e] + 9792*Sin[f*x] + 36252*Sin[e + f*x] - 27189
*Sin[2*(e + f*x)] - 2014*Sin[3*(e + f*x)] + 12084*Sin[4*(e + f*x)] - 6042*Sin[5*(e + f*x)] + 1007*Sin[6*(e + f
*x)] + 12096*Sin[2*e + f*x] - 14400*Sin[e + 2*f*x] - 2016*Sin[3*e + 2*f*x] + 7520*Sin[2*e + 3*f*x] - 8736*Sin[
4*e + 3*f*x] + 1248*Sin[3*e + 4*f*x] + 6048*Sin[5*e + 4*f*x] - 1632*Sin[4*e + 5*f*x] - 2016*Sin[6*e + 5*f*x] +
608*Sin[5*e + 6*f*x]))/(a^2*c^5*f)

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fricas [A]  time = 0.46, size = 163, normalized size = 1.16 \[ \frac {19 \, \cos \left (f x + e\right )^{6} + 6 \, \cos \left (f x + e\right )^{5} - 66 \, \cos \left (f x + e\right )^{4} + 56 \, \cos \left (f x + e\right )^{3} + 24 \, \cos \left (f x + e\right )^{2} - 48 \, \cos \left (f x + e\right ) + 16}{63 \, {\left (a^{2} c^{5} f \cos \left (f x + e\right )^{5} - 3 \, a^{2} c^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{2} c^{5} f \cos \left (f x + e\right )^{3} + 2 \, a^{2} c^{5} f \cos \left (f x + e\right )^{2} - 3 \, a^{2} c^{5} f \cos \left (f x + e\right ) + a^{2} c^{5} f\right )} \sin \left (f x + e\right )} \]

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))^5,x, algorithm="fricas")

[Out]

1/63*(19*cos(f*x + e)^6 + 6*cos(f*x + e)^5 - 66*cos(f*x + e)^4 + 56*cos(f*x + e)^3 + 24*cos(f*x + e)^2 - 48*co
s(f*x + e) + 16)/((a^2*c^5*f*cos(f*x + e)^5 - 3*a^2*c^5*f*cos(f*x + e)^4 + 2*a^2*c^5*f*cos(f*x + e)^3 + 2*a^2*
c^5*f*cos(f*x + e)^2 - 3*a^2*c^5*f*cos(f*x + e) + a^2*c^5*f)*sin(f*x + e))

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giac [A]  time = 1.38, size = 129, normalized size = 0.91 \[ \frac {\frac {945 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 420 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 189 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 54 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 7}{a^{2} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9}} - \frac {21 \, {\left (a^{4} c^{10} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 18 \, a^{4} c^{10} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{6} c^{15}}}{4032 \, f} \]

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))^5,x, algorithm="giac")

[Out]

1/4032*((945*tan(1/2*f*x + 1/2*e)^8 - 420*tan(1/2*f*x + 1/2*e)^6 + 189*tan(1/2*f*x + 1/2*e)^4 - 54*tan(1/2*f*x
+ 1/2*e)^2 + 7)/(a^2*c^5*tan(1/2*f*x + 1/2*e)^9) - 21*(a^4*c^10*tan(1/2*f*x + 1/2*e)^3 - 18*a^4*c^10*tan(1/2*
f*x + 1/2*e))/(a^6*c^15))/f

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maple [A]  time = 0.98, size = 102, normalized size = 0.72 \[ \frac {-\frac {\left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{3}+6 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )-\frac {6}{7 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{7}}+\frac {1}{9 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{9}}-\frac {20}{3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{3}}+\frac {3}{\tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{5}}+\frac {15}{\tan \left (\frac {e}{2}+\frac {f x}{2}\right )}}{64 f \,a^{2} c^{5}} \]

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))^5,x)

[Out]

1/64/f/a^2/c^5*(-1/3*tan(1/2*e+1/2*f*x)^3+6*tan(1/2*e+1/2*f*x)-6/7/tan(1/2*e+1/2*f*x)^7+1/9/tan(1/2*e+1/2*f*x)
^9-20/3/tan(1/2*e+1/2*f*x)^3+3/tan(1/2*e+1/2*f*x)^5+15/tan(1/2*e+1/2*f*x))

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maxima [A]  time = 0.34, size = 161, normalized size = 1.14 \[ \frac {\frac {21 \, {\left (\frac {18 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2} c^{5}} - \frac {{\left (\frac {54 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {189 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {420 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {945 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - 7\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{a^{2} c^{5} \sin \left (f x + e\right )^{9}}}{4032 \, f} \]

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))^5,x, algorithm="maxima")

[Out]

1/4032*(21*(18*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^2*c^5) - (54*sin(f*x
+ e)^2/(cos(f*x + e) + 1)^2 - 189*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 420*sin(f*x + e)^6/(cos(f*x + e) + 1)^
6 - 945*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - 7)*(cos(f*x + e) + 1)^9/(a^2*c^5*sin(f*x + e)^9))/f

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mupad [B]  time = 4.24, size = 102, normalized size = 0.72 \[ \frac {-21\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}+378\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+945\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-420\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+189\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-54\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+7}{4032\,a^2\,c^5\,f\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9} \]

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))^5),x)

[Out]

(189*tan(e/2 + (f*x)/2)^4 - 54*tan(e/2 + (f*x)/2)^2 - 420*tan(e/2 + (f*x)/2)^6 + 945*tan(e/2 + (f*x)/2)^8 + 37
8*tan(e/2 + (f*x)/2)^10 - 21*tan(e/2 + (f*x)/2)^12 + 7)/(4032*a^2*c^5*f*tan(e/2 + (f*x)/2)^9)

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

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))**5,x)

[Out]

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

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