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

Optimal. Leaf size=116 \[ -\frac{2 \tan (e+f x) (a \sec (e+f x)+a)}{105 f \left (c^2-c^2 \sec (e+f x)\right )^2}-\frac{2 \tan (e+f x) (a \sec (e+f x)+a)}{35 c f (c-c \sec (e+f x))^3}-\frac{\tan (e+f x) (a \sec (e+f x)+a)}{7 f (c-c \sec (e+f x))^4} \]

[Out]

-((a + a*Sec[e + f*x])*Tan[e + f*x])/(7*f*(c - c*Sec[e + f*x])^4) - (2*(a + a*Sec[e + f*x])*Tan[e + f*x])/(35*
c*f*(c - c*Sec[e + f*x])^3) - (2*(a + a*Sec[e + f*x])*Tan[e + f*x])/(105*f*(c^2 - c^2*Sec[e + f*x])^2)

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Rubi [A]  time = 0.150324, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3951, 3950} \[ -\frac{2 \tan (e+f x) (a \sec (e+f x)+a)}{105 f \left (c^2-c^2 \sec (e+f x)\right )^2}-\frac{2 \tan (e+f x) (a \sec (e+f x)+a)}{35 c f (c-c \sec (e+f x))^3}-\frac{\tan (e+f x) (a \sec (e+f x)+a)}{7 f (c-c \sec (e+f x))^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + a*Sec[e + f*x])*Tan[e + f*x])/(7*f*(c - c*Sec[e + f*x])^4) - (2*(a + a*Sec[e + f*x])*Tan[e + f*x])/(35*
c*f*(c - c*Sec[e + f*x])^3) - (2*(a + a*Sec[e + f*x])*Tan[e + f*x])/(105*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))^4} \, dx &=-\frac{(a+a \sec (e+f x)) \tan (e+f x)}{7 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}{7 c}\\ &=-\frac{(a+a \sec (e+f x)) \tan (e+f x)}{7 f (c-c \sec (e+f x))^4}-\frac{2 (a+a \sec (e+f x)) \tan (e+f x)}{35 c 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}{35 c^2}\\ &=-\frac{(a+a \sec (e+f x)) \tan (e+f x)}{7 f (c-c \sec (e+f x))^4}-\frac{2 (a+a \sec (e+f x)) \tan (e+f x)}{35 c f (c-c \sec (e+f x))^3}-\frac{2 (a+a \sec (e+f x)) \tan (e+f x)}{105 f \left (c^2-c^2 \sec (e+f x)\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.39332, size = 113, normalized size = 0.97 \[ -\frac{a \csc \left (\frac{e}{2}\right ) \left (455 \sin \left (e+\frac{f x}{2}\right )-273 \sin \left (e+\frac{3 f x}{2}\right )-210 \sin \left (2 e+\frac{3 f x}{2}\right )+56 \sin \left (2 e+\frac{5 f x}{2}\right )+105 \sin \left (3 e+\frac{5 f x}{2}\right )-23 \sin \left (3 e+\frac{7 f x}{2}\right )+350 \sin \left (\frac{f x}{2}\right )\right ) \csc ^7\left (\frac{1}{2} (e+f x)\right )}{6720 c^4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Csc[e/2]*Csc[(e + f*x)/2]^7*(350*Sin[(f*x)/2] + 455*Sin[e + (f*x)/2] - 273*Sin[e + (3*f*x)/2] - 210*Sin[2*
e + (3*f*x)/2] + 56*Sin[2*e + (5*f*x)/2] + 105*Sin[3*e + (5*f*x)/2] - 23*Sin[3*e + (7*f*x)/2]))/(6720*c^4*f)

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Maple [A]  time = 0.084, size = 50, normalized size = 0.4 \begin{align*}{\frac{a}{4\,f{c}^{4}} \left ( -{\frac{1}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-3}}+{\frac{2}{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))/(c-c*sec(f*x+e))^4,x)

[Out]

1/4/f*a/c^4*(-1/3/tan(1/2*f*x+1/2*e)^3+2/5/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 [A]  time = 1.02754, size = 239, normalized size = 2.06 \begin{align*} \frac{\frac{a{\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{\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}}}{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))/(c-c*sec(f*x+e))^4,x, algorithm="maxima")

[Out]

1/840*(a*(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*(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))/f

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Fricas [A]  time = 0.446291, size = 258, normalized size = 2.22 \begin{align*} \frac{23 \, a \cos \left (f x + e\right )^{4} + 36 \, a \cos \left (f x + e\right )^{3} + 5 \, a \cos \left (f x + e\right )^{2} - 6 \, a \cos \left (f x + e\right ) + 2 \, a}{105 \,{\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))/(c-c*sec(f*x+e))^4,x, algorithm="fricas")

[Out]

1/105*(23*a*cos(f*x + e)^4 + 36*a*cos(f*x + e)^3 + 5*a*cos(f*x + e)^2 - 6*a*cos(f*x + e) + 2*a)/((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 \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{\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\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))/(c-c*sec(f*x+e))**4,x)

[Out]

a*(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(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))/c
**4

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Giac [A]  time = 1.31515, size = 73, normalized size = 0.63 \begin{align*} -\frac{35 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 42 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 15 \, a}{420 \, 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))/(c-c*sec(f*x+e))^4,x, algorithm="giac")

[Out]

-1/420*(35*a*tan(1/2*f*x + 1/2*e)^4 - 42*a*tan(1/2*f*x + 1/2*e)^2 + 15*a)/(c^4*f*tan(1/2*f*x + 1/2*e)^7)