3.34 \(\int \frac {(c-c \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=96 \[ -\frac {23 c^2 \tan (e+f x)}{15 a^3 f (\sec (e+f x)+1)}-\frac {8 c^2 \tan (e+f x)}{15 a^3 f (\sec (e+f x)+1)^2}-\frac {4 c^2 \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)^3}+\frac {c^2 x}{a^3} \]

[Out]

c^2*x/a^3-4/5*c^2*tan(f*x+e)/a^3/f/(1+sec(f*x+e))^3-8/15*c^2*tan(f*x+e)/a^3/f/(1+sec(f*x+e))^2-23/15*c^2*tan(f
*x+e)/a^3/f/(1+sec(f*x+e))

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Rubi [A]  time = 0.30, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3903, 3777, 3922, 3919, 3794, 3796, 3797} \[ -\frac {23 c^2 \tan (e+f x)}{15 a^3 f (\sec (e+f x)+1)}-\frac {8 c^2 \tan (e+f x)}{15 a^3 f (\sec (e+f x)+1)^2}-\frac {4 c^2 \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)^3}+\frac {c^2 x}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[(c - c*Sec[e + f*x])^2/(a + a*Sec[e + f*x])^3,x]

[Out]

(c^2*x)/a^3 - (4*c^2*Tan[e + f*x])/(5*a^3*f*(1 + Sec[e + f*x])^3) - (8*c^2*Tan[e + f*x])/(15*a^3*f*(1 + Sec[e
+ f*x])^2) - (23*c^2*Tan[e + f*x])/(15*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 3797

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^m)/(f*(2*m + 1)), x] + Dist[m/(b*(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)]

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))^2}{(a+a \sec (e+f x))^3} \, dx &=\frac {\int \left (\frac {c^2}{(1+\sec (e+f x))^3}-\frac {2 c^2 \sec (e+f x)}{(1+\sec (e+f x))^3}+\frac {c^2 \sec ^2(e+f x)}{(1+\sec (e+f x))^3}\right ) \, dx}{a^3}\\ &=\frac {c^2 \int \frac {1}{(1+\sec (e+f x))^3} \, dx}{a^3}+\frac {c^2 \int \frac {\sec ^2(e+f x)}{(1+\sec (e+f x))^3} \, dx}{a^3}-\frac {\left (2 c^2\right ) \int \frac {\sec (e+f x)}{(1+\sec (e+f x))^3} \, dx}{a^3}\\ &=-\frac {4 c^2 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac {c^2 \int \frac {-5+2 \sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}+\frac {\left (3 c^2\right ) \int \frac {\sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}-\frac {\left (4 c^2\right ) \int \frac {\sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}\\ &=-\frac {4 c^2 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac {8 c^2 \tan (e+f x)}{15 a^3 f (1+\sec (e+f x))^2}+\frac {c^2 \int \frac {15-7 \sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}+\frac {c^2 \int \frac {\sec (e+f x)}{1+\sec (e+f x)} \, dx}{5 a^3}-\frac {\left (4 c^2\right ) \int \frac {\sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}\\ &=\frac {c^2 x}{a^3}-\frac {4 c^2 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac {8 c^2 \tan (e+f x)}{15 a^3 f (1+\sec (e+f x))^2}-\frac {c^2 \tan (e+f x)}{15 a^3 f (1+\sec (e+f x))}-\frac {\left (22 c^2\right ) \int \frac {\sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}\\ &=\frac {c^2 x}{a^3}-\frac {4 c^2 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac {8 c^2 \tan (e+f x)}{15 a^3 f (1+\sec (e+f x))^2}-\frac {23 c^2 \tan (e+f x)}{15 a^3 f (1+\sec (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 0.47, size = 171, normalized size = 1.78 \[ \frac {c^2 \sec \left (\frac {e}{2}\right ) \sec ^5\left (\frac {1}{2} (e+f x)\right ) \left (360 \sin \left (e+\frac {f x}{2}\right )-280 \sin \left (e+\frac {3 f x}{2}\right )+150 \sin \left (2 e+\frac {3 f x}{2}\right )-86 \sin \left (2 e+\frac {5 f x}{2}\right )+150 f x \cos \left (e+\frac {f x}{2}\right )+75 f x \cos \left (e+\frac {3 f x}{2}\right )+75 f x \cos \left (2 e+\frac {3 f x}{2}\right )+15 f x \cos \left (2 e+\frac {5 f x}{2}\right )+15 f x \cos \left (3 e+\frac {5 f x}{2}\right )-500 \sin \left (\frac {f x}{2}\right )+150 f x \cos \left (\frac {f x}{2}\right )\right )}{480 a^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sec[e + f*x])^2/(a + a*Sec[e + f*x])^3,x]

[Out]

(c^2*Sec[e/2]*Sec[(e + f*x)/2]^5*(150*f*x*Cos[(f*x)/2] + 150*f*x*Cos[e + (f*x)/2] + 75*f*x*Cos[e + (3*f*x)/2]
+ 75*f*x*Cos[2*e + (3*f*x)/2] + 15*f*x*Cos[2*e + (5*f*x)/2] + 15*f*x*Cos[3*e + (5*f*x)/2] - 500*Sin[(f*x)/2] +
360*Sin[e + (f*x)/2] - 280*Sin[e + (3*f*x)/2] + 150*Sin[2*e + (3*f*x)/2] - 86*Sin[2*e + (5*f*x)/2]))/(480*a^3
*f)

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fricas [A]  time = 0.44, size = 138, normalized size = 1.44 \[ \frac {15 \, c^{2} f x \cos \left (f x + e\right )^{3} + 45 \, c^{2} f x \cos \left (f x + e\right )^{2} + 45 \, c^{2} f x \cos \left (f x + e\right ) + 15 \, c^{2} f x - {\left (43 \, c^{2} \cos \left (f x + e\right )^{2} + 54 \, c^{2} \cos \left (f x + e\right ) + 23 \, c^{2}\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^2/(a+a*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(15*c^2*f*x*cos(f*x + e)^3 + 45*c^2*f*x*cos(f*x + e)^2 + 45*c^2*f*x*cos(f*x + e) + 15*c^2*f*x - (43*c^2*c
os(f*x + e)^2 + 54*c^2*cos(f*x + e) + 23*c^2)*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 = 0.33, size = 84, normalized size = 0.88 \[ \frac {\frac {15 \, {\left (f x + e\right )} c^{2}}{a^{3}} - \frac {3 \, a^{12} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 10 \, a^{12} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 \, a^{12} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{15}}}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^2/(a+a*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/15*(15*(f*x + e)*c^2/a^3 - (3*a^12*c^2*tan(1/2*f*x + 1/2*e)^5 - 10*a^12*c^2*tan(1/2*f*x + 1/2*e)^3 + 30*a^12
*c^2*tan(1/2*f*x + 1/2*e))/a^15)/f

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maple [A]  time = 0.77, size = 87, normalized size = 0.91 \[ -\frac {c^{2} \left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{5 f \,a^{3}}+\frac {2 c^{2} \left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{3 f \,a^{3}}-\frac {2 c^{2} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{f \,a^{3}}+\frac {2 c^{2} \arctan \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{f \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sec(f*x+e))^2/(a+a*sec(f*x+e))^3,x)

[Out]

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

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maxima [B]  time = 0.43, size = 211, normalized size = 2.20 \[ -\frac {c^{2} {\left (\frac {\frac {105 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{3}}\right )} + \frac {2 \, c^{2} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {3 \, c^{2} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^2/(a+a*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

-1/60*(c^2*((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/(
cos(f*x + e) + 1)^5)/a^3 - 120*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^3) + 2*c^2*(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 - 3*c^2*(5*sin
(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3)/f

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mupad [B]  time = 1.40, size = 93, normalized size = 0.97 \[ \frac {c^2\,x}{a^3}-\frac {\frac {43\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c^2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{15}-\frac {16\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c^2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{15}+\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c^2}{5}}{a^3\,f\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c/cos(e + f*x))^2/(a + a/cos(e + f*x))^3,x)

[Out]

(c^2*x)/a^3 - ((c^2*sin(e/2 + (f*x)/2))/5 - (16*c^2*cos(e/2 + (f*x)/2)^2*sin(e/2 + (f*x)/2))/15 + (43*c^2*cos(
e/2 + (f*x)/2)^4*sin(e/2 + (f*x)/2))/15)/(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 \[ \frac {c^{2} \left (\int \left (- \frac {2 \sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {\sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {1}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx\right )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))**2/(a+a*sec(f*x+e))**3,x)

[Out]

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

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