3.1.62 \(\int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^5} \, dx\) [62]

Optimal. Leaf size=120 \[ \frac {2 \cot ^9(e+f x)}{9 a^3 c^5 f}+\frac {\csc (e+f x)}{a^3 c^5 f}-\frac {5 \csc ^3(e+f x)}{3 a^3 c^5 f}+\frac {9 \csc ^5(e+f x)}{5 a^3 c^5 f}-\frac {\csc ^7(e+f x)}{a^3 c^5 f}+\frac {2 \csc ^9(e+f x)}{9 a^3 c^5 f} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 120, 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 = {4043, 2686, 200, 2687, 30, 276} \begin {gather*} \frac {2 \cot ^9(e+f x)}{9 a^3 c^5 f}+\frac {2 \csc ^9(e+f x)}{9 a^3 c^5 f}-\frac {\csc ^7(e+f x)}{a^3 c^5 f}+\frac {9 \csc ^5(e+f x)}{5 a^3 c^5 f}-\frac {5 \csc ^3(e+f x)}{3 a^3 c^5 f}+\frac {\csc (e+f x)}{a^3 c^5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^5),x]

[Out]

(2*Cot[e + f*x]^9)/(9*a^3*c^5*f) + Csc[e + f*x]/(a^3*c^5*f) - (5*Csc[e + f*x]^3)/(3*a^3*c^5*f) + (9*Csc[e + f*
x]^5)/(5*a^3*c^5*f) - Csc[e + f*x]^7/(a^3*c^5*f) + (2*Csc[e + f*x]^9)/(9*a^3*c^5*f)

Rule 30

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

Rule 200

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 276

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 2686

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 2687

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 4043

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(257\) vs. \(2(120)=240\).
time = 1.50, size = 257, normalized size = 2.14 \begin {gather*} -\frac {\csc (e) \sec ^7(e+f x) (-33024 \sin (e)+6144 \sin (f x)+76455 \sin (e+f x)-33980 \sin (2 (e+f x))-32281 \sin (3 (e+f x))+27184 \sin (4 (e+f x))+1699 \sin (5 (e+f x))-6796 \sin (6 (e+f x))+1699 \sin (7 (e+f x))+22656 \sin (2 e+f x)-17216 \sin (e+2 f x)+4416 \sin (3 e+2 f x)+3200 \sin (2 e+3 f x)-15360 \sin (4 e+3 f x)+12160 \sin (3 e+4 f x)-1920 \sin (5 e+4 f x)-5120 \sin (4 e+5 f x)+5760 \sin (6 e+5 f x)+320 \sin (5 e+6 f x)-2880 \sin (7 e+6 f x)+640 \sin (6 e+7 f x)) \tan (e+f x)}{184320 a^3 c^5 f (-1+\sec (e+f x))^5 (1+\sec (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^5),x]

[Out]

-1/184320*(Csc[e]*Sec[e + f*x]^7*(-33024*Sin[e] + 6144*Sin[f*x] + 76455*Sin[e + f*x] - 33980*Sin[2*(e + f*x)]
- 32281*Sin[3*(e + f*x)] + 27184*Sin[4*(e + f*x)] + 1699*Sin[5*(e + f*x)] - 6796*Sin[6*(e + f*x)] + 1699*Sin[7
*(e + f*x)] + 22656*Sin[2*e + f*x] - 17216*Sin[e + 2*f*x] + 4416*Sin[3*e + 2*f*x] + 3200*Sin[2*e + 3*f*x] - 15
360*Sin[4*e + 3*f*x] + 12160*Sin[3*e + 4*f*x] - 1920*Sin[5*e + 4*f*x] - 5120*Sin[4*e + 5*f*x] + 5760*Sin[6*e +
5*f*x] + 320*Sin[5*e + 6*f*x] - 2880*Sin[7*e + 6*f*x] + 640*Sin[6*e + 7*f*x])*Tan[e + f*x])/(a^3*c^5*f*(-1 +
Sec[e + f*x])^5*(1 + Sec[e + f*x])^3)

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Maple [A]
time = 0.19, size = 115, normalized size = 0.96

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5}-\frac {7 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+21 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {21}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}+\frac {1}{9 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}+\frac {35}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}-\frac {35}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}}{128 f \,c^{5} a^{3}}\)	\(115\)
default	\(\frac {\frac {\left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5}-\frac {7 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+21 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {21}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}+\frac {1}{9 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}+\frac {35}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}-\frac {35}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}}{128 f \,c^{5} a^{3}}\)	\(115\)
risch	\(\frac {2 i \left (45 \,{\mathrm e}^{13 i \left (f x +e \right )}-90 \,{\mathrm e}^{12 i \left (f x +e \right )}+30 \,{\mathrm e}^{11 i \left (f x +e \right )}+240 \,{\mathrm e}^{10 i \left (f x +e \right )}-69 \,{\mathrm e}^{9 i \left (f x +e \right )}-354 \,{\mathrm e}^{8 i \left (f x +e \right )}+516 \,{\mathrm e}^{7 i \left (f x +e \right )}+96 \,{\mathrm e}^{6 i \left (f x +e \right )}-269 \,{\mathrm e}^{5 i \left (f x +e \right )}+50 \,{\mathrm e}^{4 i \left (f x +e \right )}+190 \,{\mathrm e}^{3 i \left (f x +e \right )}-80 \,{\mathrm e}^{2 i \left (f x +e \right )}+5 \,{\mathrm e}^{i \left (f x +e \right )}+10\right )}{45 f \,c^{5} a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{9}}\)	\(184\)
norman	\(\frac {\frac {1}{1152 a c f}-\frac {\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}{128 a c f}+\frac {21 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{640 a c f}-\frac {35 \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{384 a c f}+\frac {35 \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{128 a c f}+\frac {21 \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{128 a c f}-\frac {7 \left (\tan ^{12}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{384 a c f}+\frac {\tan ^{14}\left (\frac {f x}{2}+\frac {e}{2}\right )}{640 a c f}}{a^{2} c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}\)	\(185\)

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))^5,x,method=_RETURNVERBOSE)

[Out]

1/128/f/c^5/a^3*(1/5*tan(1/2*f*x+1/2*e)^5-7/3*tan(1/2*f*x+1/2*e)^3+21*tan(1/2*f*x+1/2*e)+21/5/tan(1/2*f*x+1/2*
e)^5+1/9/tan(1/2*f*x+1/2*e)^9-1/tan(1/2*f*x+1/2*e)^7+35/tan(1/2*f*x+1/2*e)-35/3/tan(1/2*f*x+1/2*e)^3)

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Maxima [A]
time = 0.32, size = 197, normalized size = 1.64 \begin {gather*} \frac {\frac {3 \, {\left (\frac {315 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {35 \, \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}}\right )}}{a^{3} c^{5}} - \frac {{\left (\frac {45 \, \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 {525 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {1575 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - 5\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{a^{3} c^{5} \sin \left (f x + e\right )^{9}}}{5760 \, f} \end {gather*}

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))^5,x, algorithm="maxima")

[Out]

1/5760*(3*(315*sin(f*x + e)/(cos(f*x + e) + 1) - 35*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(co
s(f*x + e) + 1)^5)/(a^3*c^5) - (45*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 189*sin(f*x + e)^4/(cos(f*x + e) + 1)
^4 + 525*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 1575*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - 5)*(cos(f*x + e) + 1
)^9/(a^3*c^5*sin(f*x + e)^9))/f

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Fricas [A]
time = 3.67, size = 204, normalized size = 1.70 \begin {gather*} \frac {10 \, \cos \left (f x + e\right )^{7} + 25 \, \cos \left (f x + e\right )^{6} - 60 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{4} + 80 \, \cos \left (f x + e\right )^{3} - 24 \, \cos \left (f x + e\right )^{2} - 32 \, \cos \left (f x + e\right ) + 16}{45 \, {\left (a^{3} c^{5} f \cos \left (f x + e\right )^{6} - 2 \, a^{3} c^{5} f \cos \left (f x + e\right )^{5} - a^{3} c^{5} f \cos \left (f x + e\right )^{4} + 4 \, a^{3} c^{5} f \cos \left (f x + e\right )^{3} - a^{3} c^{5} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} c^{5} f \cos \left (f x + e\right ) + a^{3} c^{5} f\right )} \sin \left (f x + e\right )} \end {gather*}

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))^5,x, algorithm="fricas")

[Out]

1/45*(10*cos(f*x + e)^7 + 25*cos(f*x + e)^6 - 60*cos(f*x + e)^5 - 10*cos(f*x + e)^4 + 80*cos(f*x + e)^3 - 24*c
os(f*x + e)^2 - 32*cos(f*x + e) + 16)/((a^3*c^5*f*cos(f*x + e)^6 - 2*a^3*c^5*f*cos(f*x + e)^5 - a^3*c^5*f*cos(
f*x + e)^4 + 4*a^3*c^5*f*cos(f*x + e)^3 - a^3*c^5*f*cos(f*x + e)^2 - 2*a^3*c^5*f*cos(f*x + e) + a^3*c^5*f)*sin
(f*x + e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {\sec {\left (e + f x \right )}}{\sec ^{8}{\left (e + f x \right )} - 2 \sec ^{7}{\left (e + f x \right )} - 2 \sec ^{6}{\left (e + f x \right )} + 6 \sec ^{5}{\left (e + f x \right )} - 6 \sec ^{3}{\left (e + f x \right )} + 2 \sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} - 1}\, dx}{a^{3} c^{5}} \end {gather*}

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))**5,x)

[Out]

-Integral(sec(e + f*x)/(sec(e + f*x)**8 - 2*sec(e + f*x)**7 - 2*sec(e + f*x)**6 + 6*sec(e + f*x)**5 - 6*sec(e
+ f*x)**3 + 2*sec(e + f*x)**2 + 2*sec(e + f*x) - 1), x)/(a**3*c**5)

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Giac [A]
time = 0.74, size = 142, normalized size = 1.18 \begin {gather*} \frac {\frac {1575 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 525 \, \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} - 45 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5}{a^{3} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9}} + \frac {3 \, {\left (3 \, a^{12} c^{20} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 35 \, a^{12} c^{20} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 315 \, a^{12} c^{20} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{15} c^{25}}}{5760 \, f} \end {gather*}

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))^5,x, algorithm="giac")

[Out]

1/5760*((1575*tan(1/2*f*x + 1/2*e)^8 - 525*tan(1/2*f*x + 1/2*e)^6 + 189*tan(1/2*f*x + 1/2*e)^4 - 45*tan(1/2*f*
x + 1/2*e)^2 + 5)/(a^3*c^5*tan(1/2*f*x + 1/2*e)^9) + 3*(3*a^12*c^20*tan(1/2*f*x + 1/2*e)^5 - 35*a^12*c^20*tan(
1/2*f*x + 1/2*e)^3 + 315*a^12*c^20*tan(1/2*f*x + 1/2*e))/(a^15*c^25))/f

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Mupad [B]
time = 2.88, size = 109, normalized size = 0.91 \begin {gather*} \frac {\frac {145\,\cos \left (3\,e+3\,f\,x\right )}{32}-\frac {169\,\cos \left (2\,e+2\,f\,x\right )}{32}-\frac {129\,\cos \left (e+f\,x\right )}{32}+\frac {55\,\cos \left (4\,e+4\,f\,x\right )}{16}-\frac {85\,\cos \left (5\,e+5\,f\,x\right )}{32}+\frac {25\,\cos \left (6\,e+6\,f\,x\right )}{32}+\frac {5\,\cos \left (7\,e+7\,f\,x\right )}{32}+\frac {129}{16}}{5760\,a^3\,c^5\,f\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9} \end {gather*}

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))^5),x)

[Out]

((145*cos(3*e + 3*f*x))/32 - (169*cos(2*e + 2*f*x))/32 - (129*cos(e + f*x))/32 + (55*cos(4*e + 4*f*x))/16 - (8
5*cos(5*e + 5*f*x))/32 + (25*cos(6*e + 6*f*x))/32 + (5*cos(7*e + 7*f*x))/32 + 129/16)/(5760*a^3*c^5*f*cos(e/2
+ (f*x)/2)^5*sin(e/2 + (f*x)/2)^9)

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