3.28 \(\int \frac {1}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=98 \[ \frac {\cot ^5(e+f x) (\sec (e+f x)+1)}{5 a^2 c^3 f}-\frac {\cot ^3(e+f x) (4 \sec (e+f x)+5)}{15 a^2 c^3 f}+\frac {\cot (e+f x) (8 \sec (e+f x)+15)}{15 a^2 c^3 f}+\frac {x}{a^2 c^3} \]

[Out]

x/a^2/c^3+1/5*cot(f*x+e)^5*(1+sec(f*x+e))/a^2/c^3/f-1/15*cot(f*x+e)^3*(5+4*sec(f*x+e))/a^2/c^3/f+1/15*cot(f*x+
e)*(15+8*sec(f*x+e))/a^2/c^3/f

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Rubi [A]  time = 0.14, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3904, 3882, 8} \[ \frac {\cot ^5(e+f x) (\sec (e+f x)+1)}{5 a^2 c^3 f}-\frac {\cot ^3(e+f x) (4 \sec (e+f x)+5)}{15 a^2 c^3 f}+\frac {\cot (e+f x) (8 \sec (e+f x)+15)}{15 a^2 c^3 f}+\frac {x}{a^2 c^3} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^3),x]

[Out]

x/(a^2*c^3) + (Cot[e + f*x]^5*(1 + Sec[e + f*x]))/(5*a^2*c^3*f) - (Cot[e + f*x]^3*(5 + 4*Sec[e + f*x]))/(15*a^
2*c^3*f) + (Cot[e + f*x]*(15 + 8*Sec[e + f*x]))/(15*a^2*c^3*f)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3882

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[((e*Cot[c
+ d*x])^(m + 1)*(a + b*Csc[c + d*x]))/(d*e*(m + 1)), x] - Dist[1/(e^2*(m + 1)), Int[(e*Cot[c + d*x])^(m + 2)*(
a*(m + 1) + b*(m + 2)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[m, -1]

Rule 3904

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[(-(a*c))^m, Int[Cot[e + f*x]^(2*m)*(c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] &&  !(IntegerQ[n] && GtQ[m - n, 0])

Rubi steps

\begin {align*} \int \frac {1}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3} \, dx &=-\frac {\int \cot ^6(e+f x) (a+a \sec (e+f x)) \, dx}{a^3 c^3}\\ &=\frac {\cot ^5(e+f x) (1+\sec (e+f x))}{5 a^2 c^3 f}-\frac {\int \cot ^4(e+f x) (-5 a-4 a \sec (e+f x)) \, dx}{5 a^3 c^3}\\ &=\frac {\cot ^5(e+f x) (1+\sec (e+f x))}{5 a^2 c^3 f}-\frac {\cot ^3(e+f x) (5+4 \sec (e+f x))}{15 a^2 c^3 f}-\frac {\int \cot ^2(e+f x) (15 a+8 a \sec (e+f x)) \, dx}{15 a^3 c^3}\\ &=\frac {\cot ^5(e+f x) (1+\sec (e+f x))}{5 a^2 c^3 f}-\frac {\cot ^3(e+f x) (5+4 \sec (e+f x))}{15 a^2 c^3 f}+\frac {\cot (e+f x) (15+8 \sec (e+f x))}{15 a^2 c^3 f}-\frac {\int -15 a \, dx}{15 a^3 c^3}\\ &=\frac {x}{a^2 c^3}+\frac {\cot ^5(e+f x) (1+\sec (e+f x))}{5 a^2 c^3 f}-\frac {\cot ^3(e+f x) (5+4 \sec (e+f x))}{15 a^2 c^3 f}+\frac {\cot (e+f x) (15+8 \sec (e+f x))}{15 a^2 c^3 f}\\ \end {align*}

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Mathematica [B]  time = 1.41, size = 257, normalized size = 2.62 \[ \frac {\csc \left (\frac {e}{2}\right ) \sec \left (\frac {e}{2}\right ) \csc ^5\left (\frac {1}{2} (e+f x)\right ) \sec ^3\left (\frac {1}{2} (e+f x)\right ) (-534 \sin (e+f x)+178 \sin (2 (e+f x))+178 \sin (3 (e+f x))-89 \sin (4 (e+f x))-520 \sin (2 e+f x)+248 \sin (e+2 f x)+120 \sin (3 e+2 f x)+248 \sin (2 e+3 f x)+120 \sin (4 e+3 f x)-184 \sin (3 e+4 f x)-360 f x \cos (2 e+f x)-120 f x \cos (e+2 f x)+120 f x \cos (3 e+2 f x)-120 f x \cos (2 e+3 f x)+120 f x \cos (4 e+3 f x)+60 f x \cos (3 e+4 f x)-60 f x \cos (5 e+4 f x)+200 \sin (e)-584 \sin (f x)+360 f x \cos (f x))}{30720 a^2 c^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^3),x]

[Out]

(Csc[e/2]*Csc[(e + f*x)/2]^5*Sec[e/2]*Sec[(e + f*x)/2]^3*(360*f*x*Cos[f*x] - 360*f*x*Cos[2*e + f*x] - 120*f*x*
Cos[e + 2*f*x] + 120*f*x*Cos[3*e + 2*f*x] - 120*f*x*Cos[2*e + 3*f*x] + 120*f*x*Cos[4*e + 3*f*x] + 60*f*x*Cos[3
*e + 4*f*x] - 60*f*x*Cos[5*e + 4*f*x] + 200*Sin[e] - 584*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)] - 520*Sin[2*e + f*x] + 248*Sin[e + 2*f*x] + 120*Sin[3*e + 2*f*x] +
248*Sin[2*e + 3*f*x] + 120*Sin[4*e + 3*f*x] - 184*Sin[3*e + 4*f*x]))/(30720*a^2*c^3*f)

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fricas [A]  time = 0.42, size = 154, normalized size = 1.57 \[ \frac {23 \, \cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{3} - 27 \, \cos \left (f x + e\right )^{2} + 15 \, {\left (f x \cos \left (f x + e\right )^{3} - f x \cos \left (f x + e\right )^{2} - f x \cos \left (f x + e\right ) + f x\right )} \sin \left (f x + e\right ) + 7 \, \cos \left (f x + e\right ) + 8}{15 \, {\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} - a^{2} c^{3} f \cos \left (f x + e\right )^{2} - a^{2} c^{3} f \cos \left (f x + e\right ) + a^{2} c^{3} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(23*cos(f*x + e)^4 - 8*cos(f*x + e)^3 - 27*cos(f*x + e)^2 + 15*(f*x*cos(f*x + e)^3 - f*x*cos(f*x + e)^2 -
f*x*cos(f*x + e) + f*x)*sin(f*x + e) + 7*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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giac [A]  time = 0.43, size = 116, normalized size = 1.18 \[ \frac {\frac {240 \, {\left (f x + e\right )}}{a^{2} c^{3}} + \frac {3 \, {\left (80 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 10 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}}{a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5}} + \frac {5 \, {\left (a^{4} c^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 18 \, a^{4} c^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{6} c^{9}}}{240 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/240*(240*(f*x + e)/(a^2*c^3) + 3*(80*tan(1/2*f*x + 1/2*e)^4 - 10*tan(1/2*f*x + 1/2*e)^2 + 1)/(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 - 18*a^4*c^6*tan(1/2*f*x + 1/2*e))/(a^6*c^9))/f

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maple [A]  time = 1.10, size = 130, normalized size = 1.33 \[ \frac {\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )}{48 f \,a^{2} c^{3}}-\frac {3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{8 f \,a^{2} c^{3}}+\frac {1}{80 f \,a^{2} c^{3} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{5}}-\frac {1}{8 f \,a^{2} c^{3} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{3}}+\frac {1}{f \,a^{2} c^{3} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}+\frac {2 \arctan \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{f \,a^{2} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^3,x)

[Out]

1/48/f/a^2/c^3*tan(1/2*e+1/2*f*x)^3-3/8/f/a^2/c^3*tan(1/2*e+1/2*f*x)+1/80/f/a^2/c^3/tan(1/2*e+1/2*f*x)^5-1/8/f
/a^2/c^3/tan(1/2*e+1/2*f*x)^3+1/f/a^2/c^3/tan(1/2*e+1/2*f*x)+2/f/a^2/c^3*arctan(tan(1/2*e+1/2*f*x))

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maxima [A]  time = 0.43, size = 147, normalized size = 1.50 \[ -\frac {\frac {5 \, {\left (\frac {18 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2} c^{3}} - \frac {480 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{2} c^{3}} + \frac {3 \, {\left (\frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {80 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{a^{2} c^{3} \sin \left (f x + e\right )^{5}}}{240 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

-1/240*(5*(18*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^2*c^3) - 480*arctan(si
n(f*x + e)/(cos(f*x + e) + 1))/(a^2*c^3) + 3*(10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 80*sin(f*x + e)^4/(cos(
f*x + e) + 1)^4 - 1)*(cos(f*x + e) + 1)^5/(a^2*c^3*sin(f*x + e)^5))/f

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mupad [B]  time = 1.54, size = 161, normalized size = 1.64 \[ \frac {3\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-90\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+240\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-30\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+240\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (e+f\,x\right )}{240\,a^2\,c^3\,f\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a/cos(e + f*x))^2*(c - c/cos(e + f*x))^3),x)

[Out]

(3*cos(e/2 + (f*x)/2)^8 + 5*sin(e/2 + (f*x)/2)^8 - 90*cos(e/2 + (f*x)/2)^2*sin(e/2 + (f*x)/2)^6 + 240*cos(e/2
+ (f*x)/2)^4*sin(e/2 + (f*x)/2)^4 - 30*cos(e/2 + (f*x)/2)^6*sin(e/2 + (f*x)/2)^2 + 240*cos(e/2 + (f*x)/2)^3*si
n(e/2 + (f*x)/2)^5*(e + f*x))/(240*a^2*c^3*f*cos(e/2 + (f*x)/2)^3*sin(e/2 + (f*x)/2)^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(1/(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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