3.1.24 \(\int \frac {(c-c \sec (e+f x))^2}{(a+a \sec (e+f x))^2} \, dx\) [24]

Optimal. Leaf size=67 \[ \frac {c^2 x}{a^2}-\frac {4 c^2 \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))^2}-\frac {4 c^2 \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3988, 3862, 4004, 3879, 3881, 3882} \begin {gather*} -\frac {4 c^2 \tan (e+f x)}{3 a^2 f (\sec (e+f x)+1)}-\frac {4 c^2 \tan (e+f x)}{3 a^2 f (\sec (e+f x)+1)^2}+\frac {c^2 x}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*x)/a^2 - (4*c^2*Tan[e + f*x])/(3*a^2*f*(1 + Sec[e + f*x])^2) - (4*c^2*Tan[e + f*x])/(3*a^2*f*(1 + Sec[e +
f*x]))

Rule 3862

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-Cot[c + d*x])*((a + b*Csc[c + d*x])^n/(d*
(2*n + 1))), x] + Dist[1/(a^2*(2*n + 1)), Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*
x]), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 3879

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[-Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3881

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[b*Cot[e + f*x]*((a
+ b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Dist[(m + 1)/(a*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(
m + 1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && IntegerQ[2*m]

Rule 3882

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*(
(a + b*Csc[e + f*x])^m/(f*(2*m + 1))), x] + Dist[m/(b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1
), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 3988

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Dis
t[c^n, Int[ExpandTrig[(1 + (d/c)*csc[e + f*x])^n, (a + b*csc[e + f*x])^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] && IGtQ[m, 0] && ILtQ[n, 0] && LtQ[m + n, 2]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a),
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 67, normalized size = 1.00 \begin {gather*} \frac {c^2 \left (\frac {2 \text {ArcTan}\left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{f}-\frac {2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{f}+\frac {2 \tan ^3\left (\frac {e}{2}+\frac {f x}{2}\right )}{3 f}\right )}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*((2*ArcTan[Tan[e/2 + (f*x)/2]])/f - (2*Tan[e/2 + (f*x)/2])/f + (2*Tan[e/2 + (f*x)/2]^3)/(3*f)))/a^2

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Maple [A]
time = 0.13, size = 47, normalized size = 0.70

method	result	size
derivativedivides	\(\frac {2 c^{2} \left (\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}-\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,a^{2}}\)	\(47\)
default	\(\frac {2 c^{2} \left (\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}-\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,a^{2}}\)	\(47\)
risch	\(\frac {c^{2} x}{a^{2}}-\frac {8 i c^{2} \left (3 \,{\mathrm e}^{2 i \left (f x +e \right )}+3 \,{\mathrm e}^{i \left (f x +e \right )}+2\right )}{3 f \,a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3}}\)	\(59\)
norman	\(\frac {\frac {c^{2} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {c^{2} x}{a}+\frac {2 c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}-\frac {8 c^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a f}+\frac {2 c^{2} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a f}}{a \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\)	\(113\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sec(f*x+e))^2/(a+a*sec(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (67) = 134\).
time = 0.53, size = 184, normalized size = 2.75 \begin {gather*} -\frac {c^{2} {\left (\frac {\frac {9 \, \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}} - \frac {12 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{2}}\right )} - \frac {c^{2} {\left (\frac {3 \, \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}} + \frac {2 \, c^{2} {\left (\frac {3 \, \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}}}{6 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(c^2*((9*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 12*arctan(sin(f*x +
e)/(cos(f*x + e) + 1))/a^2) - c^2*(3*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a
^2 + 2*c^2*(3*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2)/f

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Fricas [A]
time = 2.86, size = 100, normalized size = 1.49 \begin {gather*} \frac {3 \, c^{2} f x \cos \left (f x + e\right )^{2} + 6 \, c^{2} f x \cos \left (f x + e\right ) + 3 \, c^{2} f x - 4 \, {\left (2 \, c^{2} \cos \left (f x + e\right ) + c^{2}\right )} \sin \left (f x + e\right )}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.47, size = 60, normalized size = 0.90 \begin {gather*} \frac {\frac {3 \, {\left (f x + e\right )} c^{2}}{a^{2}} + \frac {2 \, {\left (a^{4} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a^{4} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{6}}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.38, size = 38, normalized size = 0.57 \begin {gather*} \frac {2\,c^2\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3-3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+\frac {3\,f\,x}{2}\right )}{3\,a^2\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*c^2*(tan(e/2 + (f*x)/2)^3 - 3*tan(e/2 + (f*x)/2) + (3*f*x)/2))/(3*a^2*f)

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