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

Optimal. Leaf size=86 \[ -\frac{4 c \tan (e+f x)}{15 f \left (a^3 \sec (e+f x)+a^3\right )}+\frac{11 c \tan (e+f x)}{15 a f (a \sec (e+f x)+a)^2}-\frac{2 c \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3} \]

[Out]

(-2*c*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) + (11*c*Tan[e + f*x])/(15*a*f*(a + a*Sec[e + f*x])^2) - (4*c*
Tan[e + f*x])/(15*f*(a^3 + a^3*Sec[e + f*x]))

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Rubi [A]  time = 0.163336, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {4008, 4000, 3794} \[ -\frac{4 c \tan (e+f x)}{15 f \left (a^3 \sec (e+f x)+a^3\right )}+\frac{11 c \tan (e+f x)}{15 a f (a \sec (e+f x)+a)^2}-\frac{2 c \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) + (11*c*Tan[e + f*x])/(15*a*f*(a + a*Sec[e + f*x])^2) - (4*c*
Tan[e + f*x])/(15*f*(a^3 + a^3*Sec[e + f*x]))

Rule 4008

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(b*f*(2*m + 1)), x] + Dist[1/(b^2*(2*
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b*B*(2*m + 1)*Csc[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 4000

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(a*B*m + A*b*
(m + 1))/(a*b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, A, B, e, f}, x
] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] && LtQ[m, -2^(-1)]

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\sec ^2(e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx &=-\frac{2 c \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{\int \frac{\sec (e+f x) (-6 a c+5 a c \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx}{5 a^2}\\ &=-\frac{2 c \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{11 c \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}-\frac{(4 c) \int \frac{\sec (e+f x)}{a+a \sec (e+f x)} \, dx}{15 a^2}\\ &=-\frac{2 c \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{11 c \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}-\frac{4 c \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.17772, size = 43, normalized size = 0.5 \[ -\frac{c (\cos (e+f x)+4) \tan ^3\left (\frac{1}{2} (e+f x)\right ) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{30 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*(4 + Cos[e + f*x])*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]^3)/(30*a^3*f)

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Maple [A]  time = 0.071, size = 37, normalized size = 0.4 \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)^2*(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 = 1.00829, size = 155, normalized size = 1.8 \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)^2*(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 = 0.43997, size = 192, normalized size = 2.23 \begin{align*} \frac{{\left (c \cos \left (f x + e\right )^{2} + 3 \, c \cos \left (f x + e\right ) - 4 \, 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)^2*(c-c*sec(f*x+e))/(a+a*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(c*cos(f*x + e)^2 + 3*c*cos(f*x + e) - 4*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 ^{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 )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*(Integral(-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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Giac [A]  time = 1.25364, size = 53, normalized size = 0.62 \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)^2*(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)