3.3 \(\int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^2 \, dx\)

Optimal. Leaf size=61 \[ \frac {a c^2 \tan ^3(e+f x)}{3 f}+\frac {a c^2 \tanh ^{-1}(\sin (e+f x))}{2 f}-\frac {a c^2 \tan (e+f x) \sec (e+f x)}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3958, 2611, 3770, 2607, 30} \[ \frac {a c^2 \tan ^3(e+f x)}{3 f}+\frac {a c^2 \tanh ^{-1}(\sin (e+f x))}{2 f}-\frac {a c^2 \tan (e+f x) \sec (e+f x)}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c^2*ArcTanh[Sin[e + f*x]])/(2*f) - (a*c^2*Sec[e + f*x]*Tan[e + f*x])/(2*f) + (a*c^2*Tan[e + f*x]^3)/(3*f)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3958

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)
)^(n_.), x_Symbol] :> Dist[(-(a*c))^m, Int[ExpandTrig[csc[e + f*x]*cot[e + f*x]^(2*m), (c + d*csc[e + f*x])^(n
- 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] && IntegersQ[m,
n] && GeQ[n - m, 0] && GtQ[m*n, 0]

Rubi steps

\begin {align*} \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^2 \, dx &=-\left ((a c) \int \left (c \sec (e+f x) \tan ^2(e+f x)-c \sec ^2(e+f x) \tan ^2(e+f x)\right ) \, dx\right )\\ &=-\left (\left (a c^2\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx\right )+\left (a c^2\right ) \int \sec ^2(e+f x) \tan ^2(e+f x) \, dx\\ &=-\frac {a c^2 \sec (e+f x) \tan (e+f x)}{2 f}+\frac {1}{2} \left (a c^2\right ) \int \sec (e+f x) \, dx+\frac {\left (a c^2\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a c^2 \tanh ^{-1}(\sin (e+f x))}{2 f}-\frac {a c^2 \sec (e+f x) \tan (e+f x)}{2 f}+\frac {a c^2 \tan ^3(e+f x)}{3 f}\\ \end {align*}

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Mathematica [B]  time = 0.72, size = 313, normalized size = 5.13 \[ -\frac {a c^2 \sec (e) \sec ^3(e+f x) \left (-12 \sin (2 e+f x)+6 \sin (e+2 f x)+6 \sin (3 e+2 f x)+4 \sin (2 e+3 f x)+3 \cos (2 e+3 f x) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+3 \cos (4 e+3 f x) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+9 \cos (f x) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )+9 \cos (2 e+f x) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-3 \cos (2 e+3 f x) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )-3 \cos (4 e+3 f x) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )}{48 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/48*(a*c^2*Sec[e]*Sec[e + f*x]^3*(3*Cos[2*e + 3*f*x]*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 3*Cos[4*e +
3*f*x]*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 9*Cos[f*x]*(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[C
os[(e + f*x)/2] + Sin[(e + f*x)/2]]) + 9*Cos[2*e + f*x]*(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e
+ f*x)/2] + Sin[(e + f*x)/2]]) - 3*Cos[2*e + 3*f*x]*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] - 3*Cos[4*e + 3*
f*x]*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] - 12*Sin[2*e + f*x] + 6*Sin[e + 2*f*x] + 6*Sin[3*e + 2*f*x] + 4*
Sin[2*e + 3*f*x]))/f

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fricas [A]  time = 0.45, size = 103, normalized size = 1.69 \[ \frac {3 \, a c^{2} \cos \left (f x + e\right )^{3} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, a c^{2} \cos \left (f x + e\right )^{3} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (2 \, a c^{2} \cos \left (f x + e\right )^{2} + 3 \, a c^{2} \cos \left (f x + e\right ) - 2 \, a c^{2}\right )} \sin \left (f x + e\right )}{12 \, f \cos \left (f x + e\right )^{3}} \]

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))^2,x, algorithm="fricas")

[Out]

1/12*(3*a*c^2*cos(f*x + e)^3*log(sin(f*x + e) + 1) - 3*a*c^2*cos(f*x + e)^3*log(-sin(f*x + e) + 1) - 2*(2*a*c^
2*cos(f*x + e)^2 + 3*a*c^2*cos(f*x + e) - 2*a*c^2)*sin(f*x + e))/(f*cos(f*x + e)^3)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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))^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f*(-a*c^2/4*ln(abs(tan((f*x+exp(1))/2)-1))+a*c^2/4*ln(abs(ta
n((f*x+exp(1))/2)+1))+(-3*tan((f*x+exp(1))/2)^5*a*c^2-8*tan((f*x+exp(1))/2)^3*a*c^2+3*tan((f*x+exp(1))/2)*a*c^
2)*1/6/(tan((f*x+exp(1))/2)^2-1)^3)

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maple [A]  time = 1.17, size = 84, normalized size = 1.38 \[ -\frac {a \,c^{2} \sec \left (f x +e \right ) \tan \left (f x +e \right )}{2 f}+\frac {a \,c^{2} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2 f}-\frac {a \,c^{2} \tan \left (f x +e \right )}{3 f}+\frac {a \,c^{2} \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{3 f} \]

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))^2,x)

[Out]

-1/2*a*c^2*sec(f*x+e)*tan(f*x+e)/f+1/2/f*a*c^2*ln(sec(f*x+e)+tan(f*x+e))-1/3*a*c^2*tan(f*x+e)/f+1/3/f*a*c^2*ta
n(f*x+e)*sec(f*x+e)^2

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maxima [A]  time = 0.59, size = 108, normalized size = 1.77 \[ \frac {4 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a c^{2} + 3 \, a c^{2} {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 12 \, a c^{2} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) - 12 \, a c^{2} \tan \left (f x + e\right )}{12 \, f} \]

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))^2,x, algorithm="maxima")

[Out]

1/12*(4*(tan(f*x + e)^3 + 3*tan(f*x + e))*a*c^2 + 3*a*c^2*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x +
e) + 1) + log(sin(f*x + e) - 1)) + 12*a*c^2*log(sec(f*x + e) + tan(f*x + e)) - 12*a*c^2*tan(f*x + e))/f

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mupad [B]  time = 3.81, size = 114, normalized size = 1.87 \[ \frac {a\,c^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{f}-\frac {a\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+\frac {8\,a\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3}-a\,c^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*c^2*atanh(tan(e/2 + (f*x)/2)))/f - ((8*a*c^2*tan(e/2 + (f*x)/2)^3)/3 - a*c^2*tan(e/2 + (f*x)/2) + a*c^2*tan
(e/2 + (f*x)/2)^5)/(f*(3*tan(e/2 + (f*x)/2)^2 - 3*tan(e/2 + (f*x)/2)^4 + tan(e/2 + (f*x)/2)^6 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a c^{2} \left (\int \sec {\left (e + f x \right )}\, dx + \int \left (- \sec ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- \sec ^{3}{\left (e + f x \right )}\right )\, dx + \int \sec ^{4}{\left (e + f x \right )}\, dx\right ) \]

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))**2,x)

[Out]

a*c**2*(Integral(sec(e + f*x), x) + Integral(-sec(e + f*x)**2, x) + Integral(-sec(e + f*x)**3, x) + Integral(s
ec(e + f*x)**4, x))

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