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

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Rubi [A]  time = 0.20, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.46, 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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fricas [A]  time = 0.43, size = 124, normalized size = 1.41 \[ \frac {5 \, c f x \cos \left (f x + e\right )^{3} + 15 \, c f x \cos \left (f x + e\right )^{2} + 15 \, c f x \cos \left (f x + e\right ) + 5 \, c f x - {\left (13 \, c \cos \left (f x + e\right )^{2} + 19 \, c \cos \left (f x + e\right ) + 8 \, c\right )} \sin \left (f x + e\right )}{5 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \]

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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giac [A]  time = 1.19, size = 75, normalized size = 0.85 \[ \frac {\frac {10 \, {\left (f x + e\right )} c}{a^{3}} - \frac {a^{12} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 5 \, a^{12} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 20 \, a^{12} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{15}}}{10 \, f} \]

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

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maple [A]  time = 0.75, size = 79, normalized size = 0.90 \[ -\frac {c \left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{10 f \,a^{3}}+\frac {c \left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{2 f \,a^{3}}-\frac {2 c \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{f \,a^{3}}+\frac {2 c \arctan \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{f \,a^{3}} \]

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

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maxima [A]  time = 0.43, size = 159, normalized size = 1.81 \[ -\frac {c {\left (\frac {\frac {105 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{3}}\right )} + \frac {c {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \]

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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mupad [B]  time = 1.38, size = 85, normalized size = 0.97 \[ \frac {c\,x}{a^3}-\frac {\frac {13\,c\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{5}-\frac {7\,c\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{10}+\frac {c\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{10}}{a^3\,f\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*x)/a^3 - ((c*sin(e/2 + (f*x)/2))/10 - (7*c*cos(e/2 + (f*x)/2)^2*sin(e/2 + (f*x)/2))/10 + (13*c*cos(e/2 + (f
*x)/2)^4*sin(e/2 + (f*x)/2))/5)/(a^3*f*cos(e/2 + (f*x)/2)^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {c \left (\int \frac {\sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {1}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\right )\, dx\right )}{a^{3}} \]

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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