3.1.26 \(\int \frac {1}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))} \, dx\) [26]

Optimal. Leaf size=69 \[ \frac {x}{a^2 c}+\frac {\cot (e+f x) (3-2 \sec (e+f x))}{3 a^2 c f}-\frac {\cot ^3(e+f x) (1-\sec (e+f x))}{3 a^2 c f} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3989, 3967, 8} \begin {gather*} -\frac {\cot ^3(e+f x) (1-\sec (e+f x))}{3 a^2 c f}+\frac {\cot (e+f x) (3-2 \sec (e+f x))}{3 a^2 c f}+\frac {x}{a^2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 3967

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 3989

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

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Mathematica [A]
time = 0.58, size = 135, normalized size = 1.96 \begin {gather*} \frac {\csc \left (\frac {e}{2}\right ) \csc \left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {e}{2}\right ) \sec ^3\left (\frac {1}{2} (e+f x)\right ) (6 f x \cos (f x)-6 f x \cos (2 e+f x)+3 f x \cos (e+2 f x)-3 f x \cos (3 e+2 f x)-10 \sin (f x)+10 \sin (e+f x)+5 \sin (2 (e+f x))-6 \sin (2 e+f x)-8 \sin (e+2 f x))}{96 a^2 c f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[e/2]*Csc[(e + f*x)/2]*Sec[e/2]*Sec[(e + f*x)/2]^3*(6*f*x*Cos[f*x] - 6*f*x*Cos[2*e + f*x] + 3*f*x*Cos[e +
2*f*x] - 3*f*x*Cos[3*e + 2*f*x] - 10*Sin[f*x] + 10*Sin[e + f*x] + 5*Sin[2*(e + f*x)] - 6*Sin[2*e + f*x] - 8*Si
n[e + 2*f*x]))/(96*a^2*c*f)

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Maple [A]
time = 0.10, size = 60, normalized size = 0.87

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}-4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+8 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f \,a^{2} c}\)	\(60\)
default	\(\frac {\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}-4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+8 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f \,a^{2} c}\)	\(60\)
risch	\(\frac {x}{a^{2} c}-\frac {2 i \left (3 \,{\mathrm e}^{3 i \left (f x +e \right )}-5 \,{\mathrm e}^{i \left (f x +e \right )}-4\right )}{3 f \,a^{2} c \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}\)	\(72\)
norman	\(\frac {\frac {x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c a}+\frac {1}{4 a c f}-\frac {\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}{a c f}+\frac {\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )}{12 a c f}}{a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}\)	\(89\)

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)),x,method=_RETURNVERBOSE)

[Out]

1/4/f/a^2/c*(1/3*tan(1/2*f*x+1/2*e)^3-4*tan(1/2*f*x+1/2*e)+1/tan(1/2*f*x+1/2*e)+8*arctan(tan(1/2*f*x+1/2*e)))

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Maxima [A]
time = 0.50, size = 110, normalized size = 1.59 \begin {gather*} -\frac {\frac {\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}}}{a^{2} c} - \frac {24 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{2} c} - \frac {3 \, {\left (\cos \left (f x + e\right ) + 1\right )}}{a^{2} c \sin \left (f x + e\right )}}{12 \, f} \end {gather*}

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)),x, algorithm="maxima")

[Out]

-1/12*((12*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^2*c) - 24*arctan(sin(f*x
+ e)/(cos(f*x + e) + 1))/(a^2*c) - 3*(cos(f*x + e) + 1)/(a^2*c*sin(f*x + e)))/f

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Fricas [A]
time = 2.40, size = 76, normalized size = 1.10 \begin {gather*} \frac {4 \, \cos \left (f x + e\right )^{2} + 3 \, {\left (f x \cos \left (f x + e\right ) + f x\right )} \sin \left (f x + e\right ) + \cos \left (f x + e\right ) - 2}{3 \, {\left (a^{2} c f \cos \left (f x + e\right ) + a^{2} c f\right )} \sin \left (f x + e\right )} \end {gather*}

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)),x, algorithm="fricas")

[Out]

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

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)),x)

[Out]

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

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Giac [A]
time = 0.49, size = 81, normalized size = 1.17 \begin {gather*} \frac {\frac {12 \, {\left (f x + e\right )}}{a^{2} c} + \frac {3}{a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \frac {a^{4} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 12 \, a^{4} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{6} c^{3}}}{12 \, f} \end {gather*}

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)),x, algorithm="giac")

[Out]

1/12*(12*(f*x + e)/(a^2*c) + 3/(a^2*c*tan(1/2*f*x + 1/2*e)) + (a^4*c^2*tan(1/2*f*x + 1/2*e)^3 - 12*a^4*c^2*tan
(1/2*f*x + 1/2*e))/(a^6*c^3))/f

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Mupad [B]
time = 1.42, size = 69, normalized size = 1.00 \begin {gather*} \frac {x}{a^2\,c}+\frac {\frac {4\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{3}-\frac {7\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{6}+\frac {1}{12}}{a^2\,c\,f\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )} \end {gather*}

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))),x)

[Out]

x/(a^2*c) + ((4*cos(e/2 + (f*x)/2)^4)/3 - (7*cos(e/2 + (f*x)/2)^2)/6 + 1/12)/(a^2*c*f*cos(e/2 + (f*x)/2)^3*sin
(e/2 + (f*x)/2))

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