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

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

[Out]

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

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Rubi [A]  time = 0.0980012, 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} \[ -\frac{\tan (e+f x) (a \sec (e+f x)+a)}{15 c f (c-c \sec (e+f x))^2}-\frac{\tan (e+f x) (a \sec (e+f x)+a)}{5 f (c-c \sec (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.332773, size = 87, normalized size = 1.14 \[ -\frac{a \csc \left (\frac{e}{2}\right ) \left (15 \sin \left (e+\frac{f x}{2}\right )-5 \sin \left (e+\frac{3 f x}{2}\right )-15 \sin \left (2 e+\frac{3 f x}{2}\right )+4 \sin \left (2 e+\frac{5 f x}{2}\right )+25 \sin \left (\frac{f x}{2}\right )\right ) \csc ^5\left (\frac{1}{2} (e+f x)\right )}{240 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Csc[e/2]*Csc[(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*c^3*f)

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Maple [A]  time = 0.079, size = 37, normalized size = 0.5 \begin{align*}{\frac{a}{2\,f{c}^{3}} \left ( -{\frac{1}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-3}}+{\frac{1}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-5}} \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))/(c-c*sec(f*x+e))^3,x)

[Out]

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

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Maxima [A]  time = 0.9913, size = 158, normalized size = 2.08 \begin{align*} -\frac{\frac{a{\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{3 \, a{\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))/(c-c*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

-1/60*(a*(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) + 3*a*(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 [A]  time = 0.441625, size = 189, normalized size = 2.49 \begin{align*} \frac{4 \, a \cos \left (f x + e\right )^{3} + 7 \, a \cos \left (f x + e\right )^{2} + 2 \, a \cos \left (f x + e\right ) - a}{15 \,{\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))/(c-c*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(4*a*cos(f*x + e)^3 + 7*a*cos(f*x + e)^2 + 2*a*cos(f*x + e) - a)/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(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 \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 )}{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))/(c-c*sec(f*x+e))**3,x)

[Out]

-a*(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))/c**3

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Giac [A]  time = 1.29068, size = 53, normalized size = 0.7 \begin{align*} -\frac{5 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 3 \, a}{30 \, 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))/(c-c*sec(f*x+e))^3,x, algorithm="giac")

[Out]

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