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

Optimal. Leaf size=100 \[ \frac {\cot ^5(e+f x) (1-\sec (e+f x))}{5 a^3 c^2 f}-\frac {\cot ^3(e+f x) (5-4 \sec (e+f x))}{15 a^3 c^2 f}+\frac {\cot (e+f x) (15-8 \sec (e+f x))}{15 a^3 c^2 f}+\frac {x}{a^3 c^2} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 100, 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} \[ \frac {\cot ^5(e+f x) (1-\sec (e+f x))}{5 a^3 c^2 f}-\frac {\cot ^3(e+f x) (5-4 \sec (e+f x))}{15 a^3 c^2 f}+\frac {\cot (e+f x) (15-8 \sec (e+f x))}{15 a^3 c^2 f}+\frac {x}{a^3 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [B]  time = 1.01, size = 257, normalized size = 2.57 \[ \frac {\csc \left (\frac {e}{2}\right ) \sec \left (\frac {e}{2}\right ) \csc ^3\left (\frac {1}{2} (e+f x)\right ) \sec ^5\left (\frac {1}{2} (e+f x)\right ) (534 \sin (e+f x)+178 \sin (2 (e+f x))-178 \sin (3 (e+f x))-89 \sin (4 (e+f x))-520 \sin (2 e+f x)-248 \sin (e+2 f x)-120 \sin (3 e+2 f x)+248 \sin (2 e+3 f x)+120 \sin (4 e+3 f x)+184 \sin (3 e+4 f x)-360 f x \cos (2 e+f x)+120 f x \cos (e+2 f x)-120 f x \cos (3 e+2 f x)-120 f x \cos (2 e+3 f x)+120 f x \cos (4 e+3 f x)-60 f x \cos (3 e+4 f x)+60 f x \cos (5 e+4 f x)-200 \sin (e)-584 \sin (f x)+360 f x \cos (f x))}{30720 a^3 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[e/2]*Csc[(e + f*x)/2]^3*Sec[e/2]*Sec[(e + f*x)/2]^5*(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^3*c^2*f)

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fricas [A]  time = 0.43, size = 154, normalized size = 1.54 \[ \frac {23 \, \cos \left (f x + e\right )^{4} + 8 \, \cos \left (f x + e\right )^{3} - 27 \, \cos \left (f x + e\right )^{2} + 15 \, {\left (f x \cos \left (f x + e\right )^{3} + f x \cos \left (f x + e\right )^{2} - f x \cos \left (f x + e\right ) - f x\right )} \sin \left (f x + e\right ) - 7 \, \cos \left (f x + e\right ) + 8}{15 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} + a^{3} c^{2} f \cos \left (f x + e\right )^{2} - a^{3} c^{2} f \cos \left (f x + e\right ) - a^{3} c^{2} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [A]  time = 0.35, size = 122, normalized size = 1.22 \[ \frac {\frac {240 \, {\left (f x + e\right )}}{a^{3} c^{2}} + \frac {5 \, {\left (18 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}}{a^{3} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}} - \frac {3 \, {\left (a^{12} c^{8} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 10 \, a^{12} c^{8} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 80 \, a^{12} c^{8} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{15} c^{10}}}{240 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(240*(f*x + e)/(a^3*c^2) + 5*(18*tan(1/2*f*x + 1/2*e)^2 - 1)/(a^3*c^2*tan(1/2*f*x + 1/2*e)^3) - 3*(a^12*
c^8*tan(1/2*f*x + 1/2*e)^5 - 10*a^12*c^8*tan(1/2*f*x + 1/2*e)^3 + 80*a^12*c^8*tan(1/2*f*x + 1/2*e))/(a^15*c^10
))/f

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maple [A]  time = 1.08, size = 131, normalized size = 1.31 \[ -\frac {\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )}{80 f \,a^{3} c^{2}}+\frac {\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )}{8 f \,a^{3} c^{2}}-\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{f \,a^{3} c^{2}}-\frac {1}{48 f \,a^{3} c^{2} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{3}}+\frac {3}{8 f \,a^{3} c^{2} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}+\frac {2 \arctan \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{f \,a^{3} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 146, normalized size = 1.46 \[ -\frac {\frac {3 \, {\left (\frac {80 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3} c^{2}} - \frac {480 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{3} c^{2}} - \frac {5 \, {\left (\frac {18 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{a^{3} c^{2} \sin \left (f x + e\right )^{3}}}{240 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.50, size = 161, normalized size = 1.61 \[ -\frac {5\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+3\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-30\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+240\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-90\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-240\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (e+f\,x\right )}{240\,a^3\,c^2\,f\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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