3.2 \(\int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^3 \, dx\)

Optimal. Leaf size=86 \[ \frac {2 a c^3 \tan ^3(e+f x)}{3 f}+\frac {5 a c^3 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {a c^3 \tan (e+f x) \sec ^3(e+f x)}{4 f}-\frac {3 a c^3 \tan (e+f x) \sec (e+f x)}{8 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3958, 2611, 3770, 2607, 30, 3768} \[ \frac {2 a c^3 \tan ^3(e+f x)}{3 f}+\frac {5 a c^3 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {a c^3 \tan (e+f x) \sec ^3(e+f x)}{4 f}-\frac {3 a c^3 \tan (e+f x) \sec (e+f x)}{8 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a*c^3*ArcTanh[Sin[e + f*x]])/(8*f) - (3*a*c^3*Sec[e + f*x]*Tan[e + f*x])/(8*f) - (a*c^3*Sec[e + f*x]^3*Tan[
e + f*x])/(4*f) + (2*a*c^3*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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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))^3 \, dx &=-\left ((a c) \int \left (c^2 \sec (e+f x) \tan ^2(e+f x)-2 c^2 \sec ^2(e+f x) \tan ^2(e+f x)+c^2 \sec ^3(e+f x) \tan ^2(e+f x)\right ) \, dx\right )\\ &=-\left (\left (a c^3\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx\right )-\left (a c^3\right ) \int \sec ^3(e+f x) \tan ^2(e+f x) \, dx+\left (2 a c^3\right ) \int \sec ^2(e+f x) \tan ^2(e+f x) \, dx\\ &=-\frac {a c^3 \sec (e+f x) \tan (e+f x)}{2 f}-\frac {a c^3 \sec ^3(e+f x) \tan (e+f x)}{4 f}+\frac {1}{4} \left (a c^3\right ) \int \sec ^3(e+f x) \, dx+\frac {1}{2} \left (a c^3\right ) \int \sec (e+f x) \, dx+\frac {\left (2 a c^3\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a c^3 \tanh ^{-1}(\sin (e+f x))}{2 f}-\frac {3 a c^3 \sec (e+f x) \tan (e+f x)}{8 f}-\frac {a c^3 \sec ^3(e+f x) \tan (e+f x)}{4 f}+\frac {2 a c^3 \tan ^3(e+f x)}{3 f}+\frac {1}{8} \left (a c^3\right ) \int \sec (e+f x) \, dx\\ &=\frac {5 a c^3 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {3 a c^3 \sec (e+f x) \tan (e+f x)}{8 f}-\frac {a c^3 \sec ^3(e+f x) \tan (e+f x)}{4 f}+\frac {2 a c^3 \tan ^3(e+f x)}{3 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 6.48, size = 887, normalized size = 10.31 \[ a \left (\frac {5 \cos ^3(e+f x) \log \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )-\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right ) (c-c \sec (e+f x))^3 \csc ^6\left (\frac {e}{2}+\frac {f x}{2}\right )}{64 f}-\frac {5 \cos ^3(e+f x) \log \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right ) (c-c \sec (e+f x))^3 \csc ^6\left (\frac {e}{2}+\frac {f x}{2}\right )}{64 f}+\frac {\cos ^3(e+f x) (c-c \sec (e+f x))^3 \sin \left (\frac {f x}{2}\right ) \csc ^6\left (\frac {e}{2}+\frac {f x}{2}\right )}{12 f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )-\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}+\frac {\cos ^3(e+f x) (c-c \sec (e+f x))^3 \sin \left (\frac {f x}{2}\right ) \csc ^6\left (\frac {e}{2}+\frac {f x}{2}\right )}{12 f \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}+\frac {\cos ^3(e+f x) (c-c \sec (e+f x))^3 \left (\cos \left (\frac {e}{2}\right )-17 \sin \left (\frac {e}{2}\right )\right ) \csc ^6\left (\frac {e}{2}+\frac {f x}{2}\right )}{384 f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )-\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )^2}+\frac {\cos ^3(e+f x) (c-c \sec (e+f x))^3 \left (-\cos \left (\frac {e}{2}\right )-17 \sin \left (\frac {e}{2}\right )\right ) \csc ^6\left (\frac {e}{2}+\frac {f x}{2}\right )}{384 f \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )^2}-\frac {\cos ^3(e+f x) (c-c \sec (e+f x))^3 \sin \left (\frac {f x}{2}\right ) \csc ^6\left (\frac {e}{2}+\frac {f x}{2}\right )}{24 f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )-\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )^3}-\frac {\cos ^3(e+f x) (c-c \sec (e+f x))^3 \sin \left (\frac {f x}{2}\right ) \csc ^6\left (\frac {e}{2}+\frac {f x}{2}\right )}{24 f \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )^3}+\frac {\cos ^3(e+f x) (c-c \sec (e+f x))^3 \csc ^6\left (\frac {e}{2}+\frac {f x}{2}\right )}{128 f \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )-\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )^4}-\frac {\cos ^3(e+f x) (c-c \sec (e+f x))^3 \csc ^6\left (\frac {e}{2}+\frac {f x}{2}\right )}{128 f \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )^4}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*((5*Cos[e + f*x]^3*Csc[e/2 + (f*x)/2]^6*Log[Cos[e/2 + (f*x)/2] - Sin[e/2 + (f*x)/2]]*(c - c*Sec[e + f*x])^3)
/(64*f) - (5*Cos[e + f*x]^3*Csc[e/2 + (f*x)/2]^6*Log[Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2]]*(c - c*Sec[e + f
*x])^3)/(64*f) + (Cos[e + f*x]^3*Csc[e/2 + (f*x)/2]^6*(c - c*Sec[e + f*x])^3)/(128*f*(Cos[e/2 + (f*x)/2] - Sin
[e/2 + (f*x)/2])^4) - (Cos[e + f*x]^3*Csc[e/2 + (f*x)/2]^6*(c - c*Sec[e + f*x])^3*Sin[(f*x)/2])/(24*f*(Cos[e/2
] - Sin[e/2])*(Cos[e/2 + (f*x)/2] - Sin[e/2 + (f*x)/2])^3) + (Cos[e + f*x]^3*Csc[e/2 + (f*x)/2]^6*(c - c*Sec[e
+ f*x])^3*(Cos[e/2] - 17*Sin[e/2]))/(384*f*(Cos[e/2] - Sin[e/2])*(Cos[e/2 + (f*x)/2] - Sin[e/2 + (f*x)/2])^2)
+ (Cos[e + f*x]^3*Csc[e/2 + (f*x)/2]^6*(c - c*Sec[e + f*x])^3*Sin[(f*x)/2])/(12*f*(Cos[e/2] - Sin[e/2])*(Cos[
e/2 + (f*x)/2] - Sin[e/2 + (f*x)/2])) - (Cos[e + f*x]^3*Csc[e/2 + (f*x)/2]^6*(c - c*Sec[e + f*x])^3)/(128*f*(C
os[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^4) - (Cos[e + f*x]^3*Csc[e/2 + (f*x)/2]^6*(c - c*Sec[e + f*x])^3*Sin[(
f*x)/2])/(24*f*(Cos[e/2] + Sin[e/2])*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^3) + (Cos[e + f*x]^3*Csc[e/2 +
(f*x)/2]^6*(c - c*Sec[e + f*x])^3*(-Cos[e/2] - 17*Sin[e/2]))/(384*f*(Cos[e/2] + Sin[e/2])*(Cos[e/2 + (f*x)/2]
+ Sin[e/2 + (f*x)/2])^2) + (Cos[e + f*x]^3*Csc[e/2 + (f*x)/2]^6*(c - c*Sec[e + f*x])^3*Sin[(f*x)/2])/(12*f*(Co
s[e/2] + Sin[e/2])*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])))

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 117, normalized size = 1.36 \[ \frac {15 \, a c^{3} \cos \left (f x + e\right )^{4} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, a c^{3} \cos \left (f x + e\right )^{4} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (16 \, a c^{3} \cos \left (f x + e\right )^{3} + 9 \, a c^{3} \cos \left (f x + e\right )^{2} - 16 \, a c^{3} \cos \left (f x + e\right ) + 6 \, a c^{3}\right )} \sin \left (f x + e\right )}{48 \, f \cos \left (f x + e\right )^{4}} \]

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))^3,x, algorithm="fricas")

[Out]

1/48*(15*a*c^3*cos(f*x + e)^4*log(sin(f*x + e) + 1) - 15*a*c^3*cos(f*x + e)^4*log(-sin(f*x + e) + 1) - 2*(16*a
*c^3*cos(f*x + e)^3 + 9*a*c^3*cos(f*x + e)^2 - 16*a*c^3*cos(f*x + e) + 6*a*c^3)*sin(f*x + e))/(f*cos(f*x + e)^
4)

________________________________________________________________________________________

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))^3,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*(5*a*c^3/16*ln(abs(tan((f*x+exp(1))/2)-1))-5*a*c^3/16*ln(
abs(tan((f*x+exp(1))/2)+1))+(15*tan((f*x+exp(1))/2)^7*a*c^3+73*tan((f*x+exp(1))/2)^5*a*c^3-55*tan((f*x+exp(1))
/2)^3*a*c^3+15*tan((f*x+exp(1))/2)*a*c^3)*1/24/(tan((f*x+exp(1))/2)^2-1)^4)

________________________________________________________________________________________

maple [A]  time = 1.31, size = 107, normalized size = 1.24 \[ -\frac {2 a \,c^{3} \tan \left (f x +e \right )}{3 f}+\frac {2 a \,c^{3} \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{3 f}+\frac {5 a \,c^{3} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8 f}-\frac {a \,c^{3} \left (\sec ^{3}\left (f x +e \right )\right ) \tan \left (f x +e \right )}{4 f}-\frac {3 a \,c^{3} \sec \left (f x +e \right ) \tan \left (f x +e \right )}{8 f} \]

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))^3,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 133, normalized size = 1.55 \[ \frac {32 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a c^{3} + 3 \, a c^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 48 \, a c^{3} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) - 96 \, a c^{3} \tan \left (f x + e\right )}{48 \, f} \]

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))^3,x, algorithm="maxima")

[Out]

1/48*(32*(tan(f*x + e)^3 + 3*tan(f*x + e))*a*c^3 + 3*a*c^3*(2*(3*sin(f*x + e)^3 - 5*sin(f*x + e))/(sin(f*x + e
)^4 - 2*sin(f*x + e)^2 + 1) - 3*log(sin(f*x + e) + 1) + 3*log(sin(f*x + e) - 1)) + 48*a*c^3*log(sec(f*x + e) +
tan(f*x + e)) - 96*a*c^3*tan(f*x + e))/f

________________________________________________________________________________________

mupad [B]  time = 5.20, size = 146, normalized size = 1.70 \[ \frac {5\,a\,c^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{4\,f}-\frac {\frac {5\,a\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{4}+\frac {73\,a\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{12}-\frac {55\,a\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{12}+\frac {5\,a\,c^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{4}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a*c^3*atanh(tan(e/2 + (f*x)/2)))/(4*f) - ((5*a*c^3*tan(e/2 + (f*x)/2))/4 - (55*a*c^3*tan(e/2 + (f*x)/2)^3)/
12 + (73*a*c^3*tan(e/2 + (f*x)/2)^5)/12 + (5*a*c^3*tan(e/2 + (f*x)/2)^7)/4)/(f*(6*tan(e/2 + (f*x)/2)^4 - 4*tan
(e/2 + (f*x)/2)^2 - 4*tan(e/2 + (f*x)/2)^6 + tan(e/2 + (f*x)/2)^8 + 1))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a c^{3} \left (\int \left (- \sec {\left (e + f x \right )}\right )\, dx + \int 2 \sec ^{2}{\left (e + f x \right )}\, dx + \int \left (- 2 \sec ^{4}{\left (e + f x \right )}\right )\, dx + \int \sec ^{5}{\left (e + f x \right )}\, dx\right ) \]

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))**3,x)

[Out]

-a*c**3*(Integral(-sec(e + f*x), x) + Integral(2*sec(e + f*x)**2, x) + Integral(-2*sec(e + f*x)**4, x) + Integ
ral(sec(e + f*x)**5, x))

________________________________________________________________________________________