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

Optimal. Leaf size=76 \[ \frac{\tan (e+f x) (c-c \sec (e+f x))}{15 a f (a \sec (e+f x)+a)^2}+\frac{\tan (e+f x) (c-c \sec (e+f x))}{5 f (a \sec (e+f x)+a)^3} \]

[Out]

((c - c*Sec[e + f*x])*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) + ((c - c*Sec[e + f*x])*Tan[e + f*x])/(15*a*f
*(a + a*Sec[e + f*x])^2)

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Rubi [A]  time = 0.0987711, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3951, 3950} \[ \frac{\tan (e+f x) (c-c \sec (e+f x))}{15 a f (a \sec (e+f x)+a)^2}+\frac{\tan (e+f x) (c-c \sec (e+f x))}{5 f (a \sec (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c - c*Sec[e + f*x])*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) + ((c - c*Sec[e + f*x])*Tan[e + f*x])/(15*a*f
*(a + a*Sec[e + f*x])^2)

Rule 3951

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] +
Dist[(m + n + 1)/(a*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n, x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[m + n + 1, 0] && NeQ[2
*m + 1, 0] &&  !LtQ[n, 0] &&  !(IGtQ[n + 1/2, 0] && LtQ[n + 1/2, -(m + n)])

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))}{(a+a \sec (e+f x))^3} \, dx &=\frac{(c-c \sec (e+f x)) \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{\int \frac{\sec (e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx}{5 a}\\ &=\frac{(c-c \sec (e+f x)) \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{(c-c \sec (e+f x)) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}\\ \end{align*}

Mathematica [A]  time = 0.328775, size = 87, normalized size = 1.14 \[ \frac{c \sec \left (\frac{e}{2}\right ) \left (-15 \sin \left (e+\frac{f x}{2}\right )+5 \sin \left (e+\frac{3 f x}{2}\right )-15 \sin \left (2 e+\frac{3 f x}{2}\right )+4 \sin \left (2 e+\frac{5 f x}{2}\right )+25 \sin \left (\frac{f x}{2}\right )\right ) \sec ^5\left (\frac{1}{2} (e+f x)\right )}{240 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sec[e/2]*Sec[(e + f*x)/2]^5*(25*Sin[(f*x)/2] - 15*Sin[e + (f*x)/2] + 5*Sin[e + (3*f*x)/2] - 15*Sin[2*e + (3
*f*x)/2] + 4*Sin[2*e + (5*f*x)/2]))/(240*a^3*f)

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Maple [A]  time = 0.069, size = 37, normalized size = 0.5 \begin{align*}{\frac{c}{2\,f{a}^{3}} \left ({\frac{1}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}-{\frac{1}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}} \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))/(a+a*sec(f*x+e))^3,x)

[Out]

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

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Maxima [A]  time = 1.0071, size = 155, normalized size = 2.04 \begin{align*} \frac{\frac{c{\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{3 \, c{\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))/(a+a*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

1/60*(c*(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 - 3*c*(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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Fricas [A]  time = 0.438331, size = 192, normalized size = 2.53 \begin{align*} \frac{{\left (4 \, c \cos \left (f x + e\right )^{2} - 3 \, c \cos \left (f x + e\right ) - c\right )} \sin \left (f x + e\right )}{15 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))/(a+a*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(4*c*cos(f*x + e)^2 - 3*c*cos(f*x + e) - c)*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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{c \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{\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\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))/(a+a*sec(f*x+e))**3,x)

[Out]

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

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Giac [A]  time = 1.27811, size = 53, normalized size = 0.7 \begin{align*} \frac{3 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 5 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3}}{30 \, a^{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))/(a+a*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/30*(3*c*tan(1/2*f*x + 1/2*e)^5 - 5*c*tan(1/2*f*x + 1/2*e)^3)/(a^3*f)