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

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

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 38, normalized size of antiderivative = 1.00, 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} \[ \frac {\tan (e+f x) (c-c \sec (e+f x))^2}{5 f (a \sec (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c - c*Sec[e + f*x])^2*Tan[e + f*x])/(5*f*(a + a*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) (c-c \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx &=\frac {(c-c \sec (e+f x))^2 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 25, normalized size = 0.66 \[ \frac {c^2 \tan ^5\left (\frac {1}{2} (e+f x)\right )}{5 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*Tan[(e + f*x)/2]^5)/(5*a^3*f)

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fricas [B]  time = 0.42, size = 82, normalized size = 2.16 \[ \frac {{\left (c^{2} \cos \left (f x + e\right )^{2} - 2 \, c^{2} \cos \left (f x + e\right ) + c^{2}\right )} \sin \left (f x + e\right )}{5 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \]

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))^3,x, algorithm="fricas")

[Out]

1/5*(c^2*cos(f*x + e)^2 - 2*c^2*cos(f*x + e) + c^2)*sin(f*x + e)/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^
2 + 3*a^3*f*cos(f*x + e) + a^3*f)

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giac [A]  time = 1.80, size = 23, normalized size = 0.61 \[ \frac {c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5}}{5 \, a^{3} f} \]

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))^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.84, size = 23, normalized size = 0.61 \[ \frac {c^{2} \left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{5 f \,a^{3}} \]

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))^3,x)

[Out]

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

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maxima [B]  time = 0.34, size = 185, normalized size = 4.87 \[ \frac {\frac {c^{2} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {c^{2} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {6 \, c^{2} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \]

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))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.59, size = 22, normalized size = 0.58 \[ \frac {c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{5\,a^3\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c/cos(e + f*x))^2/(cos(e + f*x)*(a + a/cos(e + f*x))^3),x)

[Out]

(c^2*tan(e/2 + (f*x)/2)^5)/(5*a^3*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {c^{2} \left (\int \frac {\sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {2 \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {\sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx\right )}{a^{3}} \]

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))**3,x)

[Out]

c**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))/a**3

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