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

Optimal. Leaf size=56 \[ -\frac{a^2 \tanh ^{-1}(\sin (e+f x))}{c f}-\frac{4 a^2 \tan (e+f x)}{c f (1-\sec (e+f x))}+\frac{a^2 x}{c} \]

[Out]

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

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Rubi [A]  time = 0.163921, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3903, 3777, 8, 3794, 3789, 3770} \[ -\frac{a^2 \tanh ^{-1}(\sin (e+f x))}{c f}-\frac{4 a^2 \tan (e+f x)}{c f (1-\sec (e+f x))}+\frac{a^2 x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3903

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*csc[e + f*x])/c)^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 3777

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 8

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

Rule 3794

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 3789

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [B]  time = 0.285926, size = 169, normalized size = 3.02 \[ \frac{a^2 \csc \left (\frac{e}{2}\right ) \sin \left (\frac{1}{2} (e+f x)\right ) \left (-\cos \left (\frac{f x}{2}\right ) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+f x\right )+\cos \left (e+\frac{f x}{2}\right ) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+f x\right )+8 \sin \left (\frac{f x}{2}\right )\right )}{c f (\cos (e+f x)-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Csc[e/2]*(-(Cos[(f*x)/2]*(f*x + Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e
+ f*x)/2]])) + Cos[e + (f*x)/2]*(f*x + Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(
e + f*x)/2]]) + 8*Sin[(f*x)/2])*Sin[(e + f*x)/2])/(c*f*(-1 + Cos[e + f*x]))

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Maple [A]  time = 0.079, size = 90, normalized size = 1.6 \begin{align*} 2\,{\frac{{a}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{fc}}-{\frac{{a}^{2}}{fc}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }+{\frac{{a}^{2}}{fc}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) }+4\,{\frac{{a}^{2}}{fc\tan \left ( 1/2\,fx+e/2 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.59887, size = 207, normalized size = 3.7 \begin{align*} \frac{a^{2}{\left (\frac{2 \, \arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} + \frac{\cos \left (f x + e\right ) + 1}{c \sin \left (f x + e\right )}\right )} - a^{2}{\left (\frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c} - \frac{\cos \left (f x + e\right ) + 1}{c \sin \left (f x + e\right )}\right )} + \frac{2 \, a^{2}{\left (\cos \left (f x + e\right ) + 1\right )}}{c \sin \left (f x + e\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.09638, size = 217, normalized size = 3.88 \begin{align*} \frac{2 \, a^{2} f x \sin \left (f x + e\right ) - a^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + a^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + 8 \, a^{2} \cos \left (f x + e\right ) + 8 \, a^{2}}{2 \, c f \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^2*f*x*sin(f*x + e) - a^2*log(sin(f*x + e) + 1)*sin(f*x + e) + a^2*log(-sin(f*x + e) + 1)*sin(f*x + e)
+ 8*a^2*cos(f*x + e) + 8*a^2)/(c*f*sin(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.38402, size = 109, normalized size = 1.95 \begin{align*} \frac{\frac{{\left (f x + e\right )} a^{2}}{c} - \frac{a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c} + \frac{a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c} + \frac{4 \, a^{2}}{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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