3.17 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=38 \[ -\frac{\tan (e+f x) (a \sec (e+f x)+a)^2}{5 f (c-c \sec (e+f x))^3} \]

[Out]

-((a + a*Sec[e + f*x])^2*Tan[e + f*x])/(5*f*(c - c*Sec[e + f*x])^3)

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Rubi [A]  time = 0.0751821, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.031, Rules used = {3950} \[ -\frac{\tan (e+f x) (a \sec (e+f x)+a)^2}{5 f (c-c \sec (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + a*Sec[e + f*x])^2*Tan[e + f*x])/(5*f*(c - c*Sec[e + f*x])^3)

Rule 3950

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

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^3} \, dx &=-\frac{(a+a \sec (e+f x))^2 \tan (e+f x)}{5 f (c-c \sec (e+f x))^3}\\ \end{align*}

Mathematica [A]  time = 0.115587, size = 25, normalized size = 0.66 \[ \frac{a^2 \cot ^5\left (\frac{1}{2} (e+f x)\right )}{5 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Cot[(e + f*x)/2]^5)/(5*c^3*f)

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Maple [A]  time = 0.083, size = 23, normalized size = 0.6 \begin{align*}{\frac{{a}^{2}}{5\,f{c}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-5}} \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))^3,x)

[Out]

1/5/f*a^2/c^3/tan(1/2*f*x+1/2*e)^5

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Maxima [B]  time = 0.986597, size = 255, normalized size = 6.71 \begin{align*} \frac{\frac{a^{2}{\left (\frac{10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 3\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}} - \frac{a^{2}{\left (\frac{10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 3\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}} - \frac{6 \, a^{2}{\left (\frac{5 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 1\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}}}{60 \, 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))^3,x, algorithm="maxima")

[Out]

1/60*(a^2*(10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 3)*(cos(f*x + e)
+ 1)^5/(c^3*sin(f*x + e)^5) - a^2*(10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 15*sin(f*x + e)^4/(cos(f*x + e) +
1)^4 - 3)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x + e)^5) - 6*a^2*(5*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 1)*(cos(f
*x + e) + 1)^5/(c^3*sin(f*x + e)^5))/f

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Fricas [B]  time = 0.439728, size = 196, normalized size = 5.16 \begin{align*} \frac{a^{2} \cos \left (f x + e\right )^{3} + 3 \, a^{2} \cos \left (f x + e\right )^{2} + 3 \, a^{2} \cos \left (f x + e\right ) + a^{2}}{5 \,{\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} 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))^3,x, algorithm="fricas")

[Out]

1/5*(a^2*cos(f*x + e)^3 + 3*a^2*cos(f*x + e)^2 + 3*a^2*cos(f*x + e) + a^2)/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*co
s(f*x + e) + c^3*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 ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} - 1}\, dx + \int \frac{2 \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} - 1}\, dx + \int \frac{\sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} - 1}\, dx\right )}{c^{3}} \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))**3,x)

[Out]

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

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Giac [A]  time = 1.34113, size = 31, normalized size = 0.82 \begin{align*} \frac{a^{2}}{5 \, c^{3} f \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5}} \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))^3,x, algorithm="giac")

[Out]

1/5*a^2/(c^3*f*tan(1/2*f*x + 1/2*e)^5)