3.9 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^5} \, dx\)

Optimal. Leaf size=158 \[ -\frac{2 \tan (e+f x) (a \sec (e+f x)+a)}{315 c f \left (c^2-c^2 \sec (e+f x)\right )^2}-\frac{2 \tan (e+f x) (a \sec (e+f x)+a)}{105 c^2 f (c-c \sec (e+f x))^3}-\frac{\tan (e+f x) (a \sec (e+f x)+a)}{21 c f (c-c \sec (e+f x))^4}-\frac{\tan (e+f x) (a \sec (e+f x)+a)}{9 f (c-c \sec (e+f x))^5} \]

[Out]

-((a + a*Sec[e + f*x])*Tan[e + f*x])/(9*f*(c - c*Sec[e + f*x])^5) - ((a + a*Sec[e + f*x])*Tan[e + f*x])/(21*c*
f*(c - c*Sec[e + f*x])^4) - (2*(a + a*Sec[e + f*x])*Tan[e + f*x])/(105*c^2*f*(c - c*Sec[e + f*x])^3) - (2*(a +
a*Sec[e + f*x])*Tan[e + f*x])/(315*c*f*(c^2 - c^2*Sec[e + f*x])^2)

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Rubi [A]  time = 0.209773, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3951, 3950} \[ -\frac{2 \tan (e+f x) (a \sec (e+f x)+a)}{315 c f \left (c^2-c^2 \sec (e+f x)\right )^2}-\frac{2 \tan (e+f x) (a \sec (e+f x)+a)}{105 c^2 f (c-c \sec (e+f x))^3}-\frac{\tan (e+f x) (a \sec (e+f x)+a)}{21 c f (c-c \sec (e+f x))^4}-\frac{\tan (e+f x) (a \sec (e+f x)+a)}{9 f (c-c \sec (e+f x))^5} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x]))/(c - c*Sec[e + f*x])^5,x]

[Out]

-((a + a*Sec[e + f*x])*Tan[e + f*x])/(9*f*(c - c*Sec[e + f*x])^5) - ((a + a*Sec[e + f*x])*Tan[e + f*x])/(21*c*
f*(c - c*Sec[e + f*x])^4) - (2*(a + a*Sec[e + f*x])*Tan[e + f*x])/(105*c^2*f*(c - c*Sec[e + f*x])^3) - (2*(a +
a*Sec[e + f*x])*Tan[e + f*x])/(315*c*f*(c^2 - c^2*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))^5} \, dx &=-\frac{(a+a \sec (e+f x)) \tan (e+f x)}{9 f (c-c \sec (e+f x))^5}+\frac{\int \frac{\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^4} \, dx}{3 c}\\ &=-\frac{(a+a \sec (e+f x)) \tan (e+f x)}{9 f (c-c \sec (e+f x))^5}-\frac{(a+a \sec (e+f x)) \tan (e+f x)}{21 c f (c-c \sec (e+f x))^4}+\frac{2 \int \frac{\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^3} \, dx}{21 c^2}\\ &=-\frac{(a+a \sec (e+f x)) \tan (e+f x)}{9 f (c-c \sec (e+f x))^5}-\frac{(a+a \sec (e+f x)) \tan (e+f x)}{21 c f (c-c \sec (e+f x))^4}-\frac{2 (a+a \sec (e+f x)) \tan (e+f x)}{105 c^2 f (c-c \sec (e+f x))^3}+\frac{2 \int \frac{\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^2} \, dx}{105 c^3}\\ &=-\frac{(a+a \sec (e+f x)) \tan (e+f x)}{9 f (c-c \sec (e+f x))^5}-\frac{(a+a \sec (e+f x)) \tan (e+f x)}{21 c f (c-c \sec (e+f x))^4}-\frac{2 (a+a \sec (e+f x)) \tan (e+f x)}{105 c^2 f (c-c \sec (e+f x))^3}-\frac{2 (a+a \sec (e+f x)) \tan (e+f x)}{315 c^3 f (c-c \sec (e+f x))^2}\\ \end{align*}

Mathematica [A]  time = 0.356075, size = 139, normalized size = 0.88 \[ -\frac{a \csc \left (\frac{e}{2}\right ) \left (3465 \sin \left (e+\frac{f x}{2}\right )-2247 \sin \left (e+\frac{3 f x}{2}\right )-2625 \sin \left (2 e+\frac{3 f x}{2}\right )+1143 \sin \left (2 e+\frac{5 f x}{2}\right )+945 \sin \left (3 e+\frac{5 f x}{2}\right )-207 \sin \left (3 e+\frac{7 f x}{2}\right )-315 \sin \left (4 e+\frac{7 f x}{2}\right )+58 \sin \left (4 e+\frac{9 f x}{2}\right )+3843 \sin \left (\frac{f x}{2}\right )\right ) \csc ^9\left (\frac{1}{2} (e+f x)\right )}{80640 c^5 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x]))/(c - c*Sec[e + f*x])^5,x]

[Out]

-(a*Csc[e/2]*Csc[(e + f*x)/2]^9*(3843*Sin[(f*x)/2] + 3465*Sin[e + (f*x)/2] - 2247*Sin[e + (3*f*x)/2] - 2625*Si
n[2*e + (3*f*x)/2] + 1143*Sin[2*e + (5*f*x)/2] + 945*Sin[3*e + (5*f*x)/2] - 207*Sin[3*e + (7*f*x)/2] - 315*Sin
[4*e + (7*f*x)/2] + 58*Sin[4*e + (9*f*x)/2]))/(80640*c^5*f)

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Maple [A]  time = 0.093, size = 63, normalized size = 0.4 \begin{align*}{\frac{a}{8\,f{c}^{5}} \left ( -{\frac{1}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-3}}+{\frac{3}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-5}}-{\frac{3}{7} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-7}}+{\frac{1}{9} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-9}} \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))^5,x)

[Out]

1/8/f*a/c^5*(-1/3/tan(1/2*f*x+1/2*e)^3+3/5/tan(1/2*f*x+1/2*e)^5-3/7/tan(1/2*f*x+1/2*e)^7+1/9/tan(1/2*f*x+1/2*e
)^9)

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Maxima [A]  time = 1.02831, size = 266, normalized size = 1.68 \begin{align*} -\frac{\frac{a{\left (\frac{180 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{378 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{420 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac{315 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - 35\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{c^{5} \sin \left (f x + e\right )^{9}} + \frac{5 \, a{\left (\frac{18 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{42 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac{63 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - 7\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{c^{5} \sin \left (f x + e\right )^{9}}}{5040 \, 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))^5,x, algorithm="maxima")

[Out]

-1/5040*(a*(180*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 378*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 420*sin(f*x +
e)^6/(cos(f*x + e) + 1)^6 - 315*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - 35)*(cos(f*x + e) + 1)^9/(c^5*sin(f*x +
e)^9) + 5*a*(18*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 42*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 63*sin(f*x + e)
^8/(cos(f*x + e) + 1)^8 - 7)*(cos(f*x + e) + 1)^9/(c^5*sin(f*x + e)^9))/f

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Fricas [A]  time = 0.453771, size = 321, normalized size = 2.03 \begin{align*} \frac{58 \, a \cos \left (f x + e\right )^{5} + 83 \, a \cos \left (f x + e\right )^{4} + 4 \, a \cos \left (f x + e\right )^{3} - 11 \, a \cos \left (f x + e\right )^{2} + 8 \, a \cos \left (f x + e\right ) - 2 \, a}{315 \,{\left (c^{5} f \cos \left (f x + e\right )^{4} - 4 \, c^{5} f \cos \left (f x + e\right )^{3} + 6 \, c^{5} f \cos \left (f x + e\right )^{2} - 4 \, c^{5} f \cos \left (f x + e\right ) + c^{5} 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))^5,x, algorithm="fricas")

[Out]

1/315*(58*a*cos(f*x + e)^5 + 83*a*cos(f*x + e)^4 + 4*a*cos(f*x + e)^3 - 11*a*cos(f*x + e)^2 + 8*a*cos(f*x + e)
- 2*a)/((c^5*f*cos(f*x + e)^4 - 4*c^5*f*cos(f*x + e)^3 + 6*c^5*f*cos(f*x + e)^2 - 4*c^5*f*cos(f*x + e) + c^5*
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 ^{5}{\left (e + f x \right )} - 5 \sec ^{4}{\left (e + f x \right )} + 10 \sec ^{3}{\left (e + f x \right )} - 10 \sec ^{2}{\left (e + f x \right )} + 5 \sec{\left (e + f x \right )} - 1}\, dx + \int \frac{\sec ^{2}{\left (e + f x \right )}}{\sec ^{5}{\left (e + f x \right )} - 5 \sec ^{4}{\left (e + f x \right )} + 10 \sec ^{3}{\left (e + f x \right )} - 10 \sec ^{2}{\left (e + f x \right )} + 5 \sec{\left (e + f x \right )} - 1}\, dx\right )}{c^{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))**5,x)

[Out]

-a*(Integral(sec(e + f*x)/(sec(e + f*x)**5 - 5*sec(e + f*x)**4 + 10*sec(e + f*x)**3 - 10*sec(e + f*x)**2 + 5*s
ec(e + f*x) - 1), x) + Integral(sec(e + f*x)**2/(sec(e + f*x)**5 - 5*sec(e + f*x)**4 + 10*sec(e + f*x)**3 - 10
*sec(e + f*x)**2 + 5*sec(e + f*x) - 1), x))/c**5

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Giac [A]  time = 1.22664, size = 93, normalized size = 0.59 \begin{align*} -\frac{105 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 189 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 135 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 35 \, a}{2520 \, c^{5} f \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9}} \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))^5,x, algorithm="giac")

[Out]

-1/2520*(105*a*tan(1/2*f*x + 1/2*e)^6 - 189*a*tan(1/2*f*x + 1/2*e)^4 + 135*a*tan(1/2*f*x + 1/2*e)^2 - 35*a)/(c
^5*f*tan(1/2*f*x + 1/2*e)^9)