3.18 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^4} \, dx\)

Optimal. Leaf size=80 \[ -\frac{\tan (e+f x) (a \sec (e+f x)+a)^2}{35 c f (c-c \sec (e+f x))^3}-\frac{\tan (e+f x) (a \sec (e+f x)+a)^2}{7 f (c-c \sec (e+f x))^4} \]

[Out]

-((a + a*Sec[e + f*x])^2*Tan[e + f*x])/(7*f*(c - c*Sec[e + f*x])^4) - ((a + a*Sec[e + f*x])^2*Tan[e + f*x])/(3
5*c*f*(c - c*Sec[e + f*x])^3)

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Rubi [A]  time = 0.150209, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {3951, 3950} \[ -\frac{\tan (e+f x) (a \sec (e+f x)+a)^2}{35 c f (c-c \sec (e+f x))^3}-\frac{\tan (e+f x) (a \sec (e+f x)+a)^2}{7 f (c-c \sec (e+f x))^4} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^2)/(c - c*Sec[e + f*x])^4,x]

[Out]

-((a + a*Sec[e + f*x])^2*Tan[e + f*x])/(7*f*(c - c*Sec[e + f*x])^4) - ((a + a*Sec[e + f*x])^2*Tan[e + f*x])/(3
5*c*f*(c - c*Sec[e + f*x])^3)

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))^2}{(c-c \sec (e+f x))^4} \, dx &=-\frac{(a+a \sec (e+f x))^2 \tan (e+f x)}{7 f (c-c \sec (e+f x))^4}+\frac{\int \frac{\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^3} \, dx}{7 c}\\ &=-\frac{(a+a \sec (e+f x))^2 \tan (e+f x)}{7 f (c-c \sec (e+f x))^4}-\frac{(a+a \sec (e+f x))^2 \tan (e+f x)}{35 c f (c-c \sec (e+f x))^3}\\ \end{align*}

Mathematica [A]  time = 0.43033, size = 115, normalized size = 1.44 \[ -\frac{a^2 \csc \left (\frac{e}{2}\right ) \left (140 \sin \left (e+\frac{f x}{2}\right )-91 \sin \left (e+\frac{3 f x}{2}\right )-35 \sin \left (2 e+\frac{3 f x}{2}\right )+7 \sin \left (2 e+\frac{5 f x}{2}\right )+35 \sin \left (3 e+\frac{5 f x}{2}\right )-6 \sin \left (3 e+\frac{7 f x}{2}\right )+70 \sin \left (\frac{f x}{2}\right )\right ) \csc ^7\left (\frac{1}{2} (e+f x)\right )}{2240 c^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^2)/(c - c*Sec[e + f*x])^4,x]

[Out]

-(a^2*Csc[e/2]*Csc[(e + f*x)/2]^7*(70*Sin[(f*x)/2] + 140*Sin[e + (f*x)/2] - 91*Sin[e + (3*f*x)/2] - 35*Sin[2*e
+ (3*f*x)/2] + 7*Sin[2*e + (5*f*x)/2] + 35*Sin[3*e + (5*f*x)/2] - 6*Sin[3*e + (7*f*x)/2]))/(2240*c^4*f)

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Maple [A]  time = 0.097, size = 39, normalized size = 0.5 \begin{align*}{\frac{{a}^{2}}{2\,f{c}^{4}} \left ({\frac{1}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-5}}-{\frac{1}{7} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-7}} \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))^2/(c-c*sec(f*x+e))^4,x)

[Out]

1/2/f*a^2/c^4*(1/5/tan(1/2*f*x+1/2*e)^5-1/7/tan(1/2*f*x+1/2*e)^7)

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Maxima [B]  time = 1.05842, size = 365, normalized size = 4.56 \begin{align*} \frac{\frac{2 \, a^{2}{\left (\frac{21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{35 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{105 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 15\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}} + \frac{3 \, a^{2}{\left (\frac{21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{35 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{35 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 5\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}} - \frac{a^{2}{\left (\frac{21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{35 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{105 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + 15\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}}}{840 \, 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))^2/(c-c*sec(f*x+e))^4,x, algorithm="maxima")

[Out]

1/840*(2*a^2*(21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 35*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 105*sin(f*x +
e)^6/(cos(f*x + e) + 1)^6 - 15)*(cos(f*x + e) + 1)^7/(c^4*sin(f*x + e)^7) + 3*a^2*(21*sin(f*x + e)^2/(cos(f*x
+ e) + 1)^2 - 35*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 35*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 5)*(cos(f*x +
e) + 1)^7/(c^4*sin(f*x + e)^7) - a^2*(21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35*sin(f*x + e)^4/(cos(f*x + e)
+ 1)^4 - 105*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 15)*(cos(f*x + e) + 1)^7/(c^4*sin(f*x + e)^7))/f

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Fricas [A]  time = 0.443729, size = 267, normalized size = 3.34 \begin{align*} \frac{6 \, a^{2} \cos \left (f x + e\right )^{4} + 17 \, a^{2} \cos \left (f x + e\right )^{3} + 15 \, a^{2} \cos \left (f x + e\right )^{2} + 3 \, a^{2} \cos \left (f x + e\right ) - a^{2}}{35 \,{\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} 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))^2/(c-c*sec(f*x+e))^4,x, algorithm="fricas")

[Out]

1/35*(6*a^2*cos(f*x + e)^4 + 17*a^2*cos(f*x + e)^3 + 15*a^2*cos(f*x + e)^2 + 3*a^2*cos(f*x + e) - a^2)/((c^4*f
*cos(f*x + e)^3 - 3*c^4*f*cos(f*x + e)^2 + 3*c^4*f*cos(f*x + e) - c^4*f)*sin(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{2} \left (\int \frac{\sec{\left (e + f x \right )}}{\sec ^{4}{\left (e + f x \right )} - 4 \sec ^{3}{\left (e + f x \right )} + 6 \sec ^{2}{\left (e + f x \right )} - 4 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{2 \sec ^{2}{\left (e + f x \right )}}{\sec ^{4}{\left (e + f x \right )} - 4 \sec ^{3}{\left (e + f x \right )} + 6 \sec ^{2}{\left (e + f x \right )} - 4 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{\sec ^{3}{\left (e + f x \right )}}{\sec ^{4}{\left (e + f x \right )} - 4 \sec ^{3}{\left (e + f x \right )} + 6 \sec ^{2}{\left (e + f x \right )} - 4 \sec{\left (e + f x \right )} + 1}\, dx\right )}{c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**2/(c-c*sec(f*x+e))**4,x)

[Out]

a**2*(Integral(sec(e + f*x)/(sec(e + f*x)**4 - 4*sec(e + f*x)**3 + 6*sec(e + f*x)**2 - 4*sec(e + f*x) + 1), x)
+ Integral(2*sec(e + f*x)**2/(sec(e + f*x)**4 - 4*sec(e + f*x)**3 + 6*sec(e + f*x)**2 - 4*sec(e + f*x) + 1),
x) + Integral(sec(e + f*x)**3/(sec(e + f*x)**4 - 4*sec(e + f*x)**3 + 6*sec(e + f*x)**2 - 4*sec(e + f*x) + 1),
x))/c**4

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Giac [A]  time = 1.33813, size = 58, normalized size = 0.72 \begin{align*} \frac{7 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 5 \, a^{2}}{70 \, c^{4} f \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^4,x, algorithm="giac")

[Out]

1/70*(7*a^2*tan(1/2*f*x + 1/2*e)^2 - 5*a^2)/(c^4*f*tan(1/2*f*x + 1/2*e)^7)