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

Optimal. Leaf size=88 \[ -\frac{8 c \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)}-\frac{3 c \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)^2}-\frac{2 c \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)^3}+\frac{c x}{a^3} \]

[Out]

(c*x)/a^3 - (2*c*Tan[e + f*x])/(5*a^3*f*(1 + Sec[e + f*x])^3) - (3*c*Tan[e + f*x])/(5*a^3*f*(1 + Sec[e + f*x])
^2) - (8*c*Tan[e + f*x])/(5*a^3*f*(1 + Sec[e + f*x]))

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Rubi [A]  time = 0.202123, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3903, 3777, 3922, 3919, 3794, 3796} \[ -\frac{8 c \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)}-\frac{3 c \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)^2}-\frac{2 c \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)^3}+\frac{c x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x)/a^3 - (2*c*Tan[e + f*x])/(5*a^3*f*(1 + Sec[e + f*x])^3) - (3*c*Tan[e + f*x])/(5*a^3*f*(1 + Sec[e + f*x])
^2) - (8*c*Tan[e + f*x])/(5*a^3*f*(1 + Sec[e + f*x]))

Rule 3903

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Dis
t[c^n, Int[ExpandTrig[(1 + (d*csc[e + f*x])/c)^n, (a + b*csc[e + f*x])^m, x], x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0] && ILtQ[n, 0] && LtQ[m + n, 2]

Rule 3777

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(Cot[c + d*x]*(a + b*Csc[c + d*x])^n)/(d*(
2*n + 1)), x] + Dist[1/(a^2*(2*n + 1)), Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*x]
), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 3922

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

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

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]

Rule 3796

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

Rubi steps

\begin{align*} \int \frac{c-c \sec (e+f x)}{(a+a \sec (e+f x))^3} \, dx &=\frac{\int \left (\frac{c}{(1+\sec (e+f x))^3}-\frac{c \sec (e+f x)}{(1+\sec (e+f x))^3}\right ) \, dx}{a^3}\\ &=\frac{c \int \frac{1}{(1+\sec (e+f x))^3} \, dx}{a^3}-\frac{c \int \frac{\sec (e+f x)}{(1+\sec (e+f x))^3} \, dx}{a^3}\\ &=-\frac{2 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac{c \int \frac{-5+2 \sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}-\frac{(2 c) \int \frac{\sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}\\ &=-\frac{2 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac{3 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^2}+\frac{c \int \frac{15-7 \sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}-\frac{(2 c) \int \frac{\sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}\\ &=\frac{c x}{a^3}-\frac{2 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac{3 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^2}-\frac{2 c \tan (e+f x)}{15 a^3 f (1+\sec (e+f x))}-\frac{(22 c) \int \frac{\sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}\\ &=\frac{c x}{a^3}-\frac{2 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac{3 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^2}-\frac{8 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.456782, size = 169, normalized size = 1.92 \[ \frac{c \sec \left (\frac{e}{2}\right ) \sec ^5\left (\frac{1}{2} (e+f x)\right ) \left (110 \sin \left (e+\frac{f x}{2}\right )-90 \sin \left (e+\frac{3 f x}{2}\right )+40 \sin \left (2 e+\frac{3 f x}{2}\right )-26 \sin \left (2 e+\frac{5 f x}{2}\right )+50 f x \cos \left (e+\frac{f x}{2}\right )+25 f x \cos \left (e+\frac{3 f x}{2}\right )+25 f x \cos \left (2 e+\frac{3 f x}{2}\right )+5 f x \cos \left (2 e+\frac{5 f x}{2}\right )+5 f x \cos \left (3 e+\frac{5 f x}{2}\right )-150 \sin \left (\frac{f x}{2}\right )+50 f x \cos \left (\frac{f x}{2}\right )\right )}{160 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sec[e/2]*Sec[(e + f*x)/2]^5*(50*f*x*Cos[(f*x)/2] + 50*f*x*Cos[e + (f*x)/2] + 25*f*x*Cos[e + (3*f*x)/2] + 25
*f*x*Cos[2*e + (3*f*x)/2] + 5*f*x*Cos[2*e + (5*f*x)/2] + 5*f*x*Cos[3*e + (5*f*x)/2] - 150*Sin[(f*x)/2] + 110*S
in[e + (f*x)/2] - 90*Sin[e + (3*f*x)/2] + 40*Sin[2*e + (3*f*x)/2] - 26*Sin[2*e + (5*f*x)/2]))/(160*a^3*f)

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Maple [A]  time = 0.086, size = 79, normalized size = 0.9 \begin{align*} -{\frac{c}{10\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}+{\frac{c}{2\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-2\,{\frac{c\tan \left ( 1/2\,fx+e/2 \right ) }{f{a}^{3}}}+2\,{\frac{c\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sec(f*x+e))/(a+a*sec(f*x+e))^3,x)

[Out]

-1/10/f*c/a^3*tan(1/2*f*x+1/2*e)^5+1/2/f*c/a^3*tan(1/2*f*x+1/2*e)^3-2/f*c/a^3*tan(1/2*f*x+1/2*e)+2/f*c/a^3*arc
tan(tan(1/2*f*x+1/2*e))

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Maxima [A]  time = 1.5744, size = 215, normalized size = 2.44 \begin{align*} -\frac{c{\left (\frac{\frac{105 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{20 \, \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}}}{a^{3}} - \frac{120 \, \arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{3}}\right )} + \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}}}{60 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))/(a+a*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

-1/60*(c*((105*sin(f*x + e)/(cos(f*x + e) + 1) - 20*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(co
s(f*x + e) + 1)^5)/a^3 - 120*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^3) + 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)/f

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Fricas [A]  time = 0.99497, size = 313, normalized size = 3.56 \begin{align*} \frac{5 \, c f x \cos \left (f x + e\right )^{3} + 15 \, c f x \cos \left (f x + e\right )^{2} + 15 \, c f x \cos \left (f x + e\right ) + 5 \, c f x -{\left (13 \, c \cos \left (f x + e\right )^{2} + 19 \, c \cos \left (f x + e\right ) + 8 \, c\right )} \sin \left (f x + e\right )}{5 \,{\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((c-c*sec(f*x+e))/(a+a*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/5*(5*c*f*x*cos(f*x + e)^3 + 15*c*f*x*cos(f*x + e)^2 + 15*c*f*x*cos(f*x + e) + 5*c*f*x - (13*c*cos(f*x + e)^2
+ 19*c*cos(f*x + e) + 8*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{\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{1}{\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((c-c*sec(f*x+e))/(a+a*sec(f*x+e))**3,x)

[Out]

-c*(Integral(sec(e + f*x)/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(-1/(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 = 1.37685, size = 101, normalized size = 1.15 \begin{align*} \frac{\frac{10 \,{\left (f x + e\right )} c}{a^{3}} - \frac{a^{12} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 5 \, a^{12} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 20 \, a^{12} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{15}}}{10 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))/(a+a*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/10*(10*(f*x + e)*c/a^3 - (a^12*c*tan(1/2*f*x + 1/2*e)^5 - 5*a^12*c*tan(1/2*f*x + 1/2*e)^3 + 20*a^12*c*tan(1/
2*f*x + 1/2*e))/a^15)/f