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

Optimal. Leaf size=89 \[ \frac{a^2 \tanh ^{-1}(\sin (e+f x))}{c^2 f}+\frac{2 a^2 \tan (e+f x)}{f \left (c^2-c^2 \sec (e+f x)\right )}-\frac{2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{3 f (c-c \sec (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.127974, antiderivative size = 89, 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} \[ \frac{a^2 \tanh ^{-1}(\sin (e+f x))}{c^2 f}+\frac{2 a^2 \tan (e+f x)}{f \left (c^2-c^2 \sec (e+f x)\right )}-\frac{2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{3 f (c-c \sec (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0846671, size = 109, normalized size = 1.22 \[ \frac{a^2 \left (-\frac{4 \cot \left (\frac{1}{2} (e+f x)\right )}{3 f}-\frac{2 \cot \left (\frac{1}{2} (e+f x)\right ) \csc ^2\left (\frac{1}{2} (e+f x)\right )}{3 f}-\frac{\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{f}+\frac{\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}{f}\right )}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.077, size = 91, normalized size = 1. \begin{align*}{\frac{{a}^{2}}{f{c}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }-{\frac{{a}^{2}}{f{c}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) }-{\frac{2\,{a}^{2}}{3\,f{c}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-3}}-2\,{\frac{{a}^{2}}{f{c}^{2}\tan \left ( 1/2\,fx+e/2 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.983811, size = 271, normalized size = 3.04 \begin{align*} \frac{a^{2}{\left (\frac{6 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c^{2}} - \frac{6 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{2}} - \frac{{\left (\frac{9 \, \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}}{c^{2} \sin \left (f x + e\right )^{3}}\right )} - \frac{2 \, a^{2}{\left (\frac{3 \, \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}}{c^{2} \sin \left (f x + e\right )^{3}} + \frac{a^{2}{\left (\frac{3 \, \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}}{c^{2} \sin \left (f x + e\right )^{3}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.478872, size = 313, normalized size = 3.52 \begin{align*} -\frac{8 \, a^{2} \cos \left (f x + e\right )^{2} - 8 \, a^{2} \cos \left (f x + e\right ) - 3 \,{\left (a^{2} \cos \left (f x + e\right ) - a^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + 3 \,{\left (a^{2} \cos \left (f x + e\right ) - a^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 16 \, a^{2}}{6 \,{\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(8*a^2*cos(f*x + e)^2 - 8*a^2*cos(f*x + e) - 3*(a^2*cos(f*x + e) - a^2)*log(sin(f*x + e) + 1)*sin(f*x + e
) + 3*(a^2*cos(f*x + e) - a^2)*log(-sin(f*x + e) + 1)*sin(f*x + e) - 16*a^2)/((c^2*f*cos(f*x + e) - c^2*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{\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 )}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**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))/c**2

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Giac [A]  time = 1.26181, size = 119, normalized size = 1.34 \begin{align*} \frac{\frac{3 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c^{2}} - \frac{3 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c^{2}} - \frac{2 \,{\left (3 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a^{2}\right )}}{c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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