3.1.49 \(\int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3} \, dx\) [49]

Optimal. Leaf size=80 \[ \frac {\cot ^5(e+f x)}{5 a^2 c^3 f}+\frac {\csc (e+f x)}{a^2 c^3 f}-\frac {2 \csc ^3(e+f x)}{3 a^2 c^3 f}+\frac {\csc ^5(e+f x)}{5 a^2 c^3 f} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 80, 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 = {4043, 2686, 200, 2687, 30} \begin {gather*} \frac {\cot ^5(e+f x)}{5 a^2 c^3 f}+\frac {\csc ^5(e+f x)}{5 a^2 c^3 f}-\frac {2 \csc ^3(e+f x)}{3 a^2 c^3 f}+\frac {\csc (e+f x)}{a^2 c^3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^3),x]

[Out]

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

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Mathematica [A]
time = 0.99, size = 147, normalized size = 1.84 \begin {gather*} -\frac {\csc (e) \csc ^2\left (\frac {1}{2} (e+f x)\right ) \csc ^3(e+f x) (-200 \sin (e)+104 \sin (f x)+534 \sin (e+f x)-178 \sin (2 (e+f x))-178 \sin (3 (e+f x))+89 \sin (4 (e+f x))+40 \sin (2 e+f x)-168 \sin (e+2 f x)+120 \sin (3 e+2 f x)+72 \sin (2 e+3 f x)-120 \sin (4 e+3 f x)+24 \sin (3 e+4 f x))}{1920 a^2 c^3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^3),x]

[Out]

-1/1920*(Csc[e]*Csc[(e + f*x)/2]^2*Csc[e + f*x]^3*(-200*Sin[e] + 104*Sin[f*x] + 534*Sin[e + f*x] - 178*Sin[2*(
e + f*x)] - 178*Sin[3*(e + f*x)] + 89*Sin[4*(e + f*x)] + 40*Sin[2*e + f*x] - 168*Sin[e + 2*f*x] + 120*Sin[3*e
+ 2*f*x] + 72*Sin[2*e + 3*f*x] - 120*Sin[4*e + 3*f*x] + 24*Sin[3*e + 4*f*x]))/(a^2*c^3*f)

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Maple [A]
time = 0.17, size = 76, normalized size = 0.95

method	result	size
derivativedivides	\(\frac {-\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {6}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}-\frac {4}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}}{16 f \,c^{3} a^{2}}\)	\(76\)
default	\(\frac {-\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {6}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}-\frac {4}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}}{16 f \,c^{3} a^{2}}\)	\(76\)
risch	\(\frac {2 i \left (15 \,{\mathrm e}^{7 i \left (f x +e \right )}-15 \,{\mathrm e}^{6 i \left (f x +e \right )}-5 \,{\mathrm e}^{5 i \left (f x +e \right )}+25 \,{\mathrm e}^{4 i \left (f x +e \right )}+13 \,{\mathrm e}^{3 i \left (f x +e \right )}-21 \,{\mathrm e}^{2 i \left (f x +e \right )}+9 \,{\mathrm e}^{i \left (f x +e \right )}+3\right )}{15 f \,c^{3} a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{5}}\)	\(118\)
norman	\(\frac {\frac {1}{80 a c f}-\frac {\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}{12 a c f}+\frac {3 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8 a c f}+\frac {\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )}{4 a c f}-\frac {\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )}{48 a c f}}{a \,c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}\)	\(119\)

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))^3,x,method=_RETURNVERBOSE)

[Out]

1/16/f/c^3/a^2*(-1/3*tan(1/2*f*x+1/2*e)^3+4*tan(1/2*f*x+1/2*e)+6/tan(1/2*f*x+1/2*e)-4/3/tan(1/2*f*x+1/2*e)^3+1
/5/tan(1/2*f*x+1/2*e)^5)

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Maxima [A]
time = 0.30, size = 131, normalized size = 1.64 \begin {gather*} \frac {\frac {5 \, {\left (\frac {12 \, \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^{3}} - \frac {{\left (\frac {20 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {90 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{a^{2} c^{3} \sin \left (f x + e\right )^{5}}}{240 \, 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))^2/(c-c*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

1/240*(5*(12*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^2*c^3) - (20*sin(f*x +
e)^2/(cos(f*x + e) + 1)^2 - 90*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 3)*(cos(f*x + e) + 1)^5/(a^2*c^3*sin(f*x
+ e)^5))/f

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Fricas [A]
time = 2.85, size = 117, normalized size = 1.46 \begin {gather*} \frac {3 \, \cos \left (f x + e\right )^{4} + 12 \, \cos \left (f x + e\right )^{3} - 12 \, \cos \left (f x + e\right )^{2} - 8 \, \cos \left (f x + e\right ) + 8}{15 \, {\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} - a^{2} c^{3} f \cos \left (f x + e\right )^{2} - a^{2} c^{3} f \cos \left (f x + e\right ) + a^{2} c^{3} 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))^2/(c-c*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(3*cos(f*x + e)^4 + 12*cos(f*x + e)^3 - 12*cos(f*x + e)^2 - 8*cos(f*x + e) + 8)/((a^2*c^3*f*cos(f*x + e)^
3 - a^2*c^3*f*cos(f*x + e)^2 - a^2*c^3*f*cos(f*x + e) + a^2*c^3*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 ^{5}{\left (e + f x \right )} - \sec ^{4}{\left (e + f x \right )} - 2 \sec ^{3}{\left (e + f x \right )} + 2 \sec ^{2}{\left (e + f x \right )} + \sec {\left (e + f x \right )} - 1}\, dx}{a^{2} c^{3}} \end {gather*}

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))**3,x)

[Out]

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

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Giac [A]
time = 0.50, size = 96, normalized size = 1.20 \begin {gather*} \frac {\frac {90 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 20 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3}{a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5}} - \frac {5 \, {\left (a^{4} c^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 12 \, a^{4} c^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{6} c^{9}}}{240 \, 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))^2/(c-c*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/240*((90*tan(1/2*f*x + 1/2*e)^4 - 20*tan(1/2*f*x + 1/2*e)^2 + 3)/(a^2*c^3*tan(1/2*f*x + 1/2*e)^5) - 5*(a^4*c
^6*tan(1/2*f*x + 1/2*e)^3 - 12*a^4*c^6*tan(1/2*f*x + 1/2*e))/(a^6*c^9))/f

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Mupad [B]
time = 2.01, size = 76, normalized size = 0.95 \begin {gather*} \frac {-5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+60\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+90\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-20\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+3}{240\,a^2\,c^3\,f\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5} \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))^2*(c - c/cos(e + f*x))^3),x)

[Out]

(90*tan(e/2 + (f*x)/2)^4 - 20*tan(e/2 + (f*x)/2)^2 + 60*tan(e/2 + (f*x)/2)^6 - 5*tan(e/2 + (f*x)/2)^8 + 3)/(24
0*a^2*c^3*f*tan(e/2 + (f*x)/2)^5)

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