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

Optimal. Leaf size=88 \[ \frac{c^2 \tanh ^{-1}(\sin (e+f x))}{a^2 f}-\frac{2 c^2 \tan (e+f x)}{f \left (a^2 \sec (e+f x)+a^2\right )}+\frac{2 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.131277, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {3957, 3770} \[ \frac{c^2 \tanh ^{-1}(\sin (e+f x))}{a^2 f}-\frac{2 c^2 \tan (e+f x)}{f \left (a^2 \sec (e+f x)+a^2\right )}+\frac{2 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )}{3 f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3957

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

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

Mathematica [A]  time = 0.0938139, size = 109, normalized size = 1.24 \[ \frac{c^2 \left (-\frac{4 \tan \left (\frac{1}{2} (e+f x)\right )}{3 f}-\frac{2 \tan \left (\frac{1}{2} (e+f x)\right ) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{3 f}-\frac{\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{f}+\frac{\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}{f}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.07, size = 89, normalized size = 1. \begin{align*} -{\frac{2\,{c}^{2}}{3\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-2\,{\frac{{c}^{2}\tan \left ( 1/2\,fx+e/2 \right ) }{f{a}^{2}}}-{\frac{{c}^{2}}{f{a}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) }+{\frac{{c}^{2}}{f{a}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.99547, size = 265, normalized size = 3.01 \begin{align*} -\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{6 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac{6 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} + \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}} - \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}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(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 - 6*log(sin(f*x + e)/
(cos(f*x + e) + 1) + 1)/a^2 + 6*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/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 - 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 = 0.477391, size = 340, normalized size = 3.86 \begin{align*} \frac{3 \,{\left (c^{2} \cos \left (f x + e\right )^{2} + 2 \, c^{2} \cos \left (f x + e\right ) + c^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \,{\left (c^{2} \cos \left (f x + e\right )^{2} + 2 \, c^{2} \cos \left (f x + e\right ) + c^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 8 \,{\left (c^{2} \cos \left (f x + e\right ) + 2 \, c^{2}\right )} \sin \left (f x + e\right )}{6 \,{\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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(c^2*cos(f*x + e)^2 + 2*c^2*cos(f*x + e) + c^2)*log(sin(f*x + e) + 1) - 3*(c^2*cos(f*x + e)^2 + 2*c^2*c
os(f*x + e) + c^2)*log(-sin(f*x + e) + 1) - 8*(c^2*cos(f*x + e) + 2*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., size = 0, normalized size = 0. \begin{align*} \frac{c^{2} \left (\int \frac{\sec{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int - \frac{2 \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{\sec ^{3}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.313, size = 126, normalized size = 1.43 \begin{align*} \frac{\frac{3 \, c^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} - \frac{3 \, c^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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