3.168 \(\int \sec ^2(e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx\)

Optimal. Leaf size=105 \[ -\frac {a^3 c \tan ^5(e+f x)}{5 f}-\frac {2 a^3 c \tan ^3(e+f x)}{3 f}+\frac {a^3 c \tanh ^{-1}(\sin (e+f x))}{4 f}-\frac {a^3 c \tan (e+f x) \sec ^3(e+f x)}{2 f}+\frac {a^3 c \tan (e+f x) \sec (e+f x)}{4 f} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3962, 2607, 30, 2611, 3768, 3770, 14} \[ -\frac {a^3 c \tan ^5(e+f x)}{5 f}-\frac {2 a^3 c \tan ^3(e+f x)}{3 f}+\frac {a^3 c \tanh ^{-1}(\sin (e+f x))}{4 f}-\frac {a^3 c \tan (e+f x) \sec ^3(e+f x)}{2 f}+\frac {a^3 c \tan (e+f x) \sec (e+f x)}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 3962

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)
]*(d_.) + (c_))^(n_.), x_Symbol] :> Dist[(-(a*c))^m, Int[ExpandTrig[(g*csc[e + f*x])^p*cot[e + f*x]^(2*m), (c
+ d*csc[e + f*x])^(n - m), x], x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, 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 ^2(e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx &=-\left ((a c) \int \left (a^2 \sec ^2(e+f x) \tan ^2(e+f x)+2 a^2 \sec ^3(e+f x) \tan ^2(e+f x)+a^2 \sec ^4(e+f x) \tan ^2(e+f x)\right ) \, dx\right )\\ &=-\left (\left (a^3 c\right ) \int \sec ^2(e+f x) \tan ^2(e+f x) \, dx\right )-\left (a^3 c\right ) \int \sec ^4(e+f x) \tan ^2(e+f x) \, dx-\left (2 a^3 c\right ) \int \sec ^3(e+f x) \tan ^2(e+f x) \, dx\\ &=-\frac {a^3 c \sec ^3(e+f x) \tan (e+f x)}{2 f}+\frac {1}{2} \left (a^3 c\right ) \int \sec ^3(e+f x) \, dx-\frac {\left (a^3 c\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,\tan (e+f x)\right )}{f}-\frac {\left (a^3 c\right ) \operatorname {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a^3 c \sec (e+f x) \tan (e+f x)}{4 f}-\frac {a^3 c \sec ^3(e+f x) \tan (e+f x)}{2 f}-\frac {a^3 c \tan ^3(e+f x)}{3 f}+\frac {1}{4} \left (a^3 c\right ) \int \sec (e+f x) \, dx-\frac {\left (a^3 c\right ) \operatorname {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a^3 c \tanh ^{-1}(\sin (e+f x))}{4 f}+\frac {a^3 c \sec (e+f x) \tan (e+f x)}{4 f}-\frac {a^3 c \sec ^3(e+f x) \tan (e+f x)}{2 f}-\frac {2 a^3 c \tan ^3(e+f x)}{3 f}-\frac {a^3 c \tan ^5(e+f x)}{5 f}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 68, normalized size = 0.65 \[ \frac {a^3 c \left (15 \tanh ^{-1}(\sin (e+f x))-\tan (e+f x) \left (12 \tan ^4(e+f x)+40 \tan ^2(e+f x)+30 \sec ^3(e+f x)-15 \sec (e+f x)\right )\right )}{60 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*c*(15*ArcTanh[Sin[e + f*x]] - Tan[e + f*x]*(-15*Sec[e + f*x] + 30*Sec[e + f*x]^3 + 40*Tan[e + f*x]^2 + 12
*Tan[e + f*x]^4)))/(60*f)

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fricas [A]  time = 0.43, size = 131, normalized size = 1.25 \[ \frac {15 \, a^{3} c \cos \left (f x + e\right )^{5} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, a^{3} c \cos \left (f x + e\right )^{5} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (28 \, a^{3} c \cos \left (f x + e\right )^{4} + 15 \, a^{3} c \cos \left (f x + e\right )^{3} - 16 \, a^{3} c \cos \left (f x + e\right )^{2} - 30 \, a^{3} c \cos \left (f x + e\right ) - 12 \, a^{3} c\right )} \sin \left (f x + e\right )}{120 \, f \cos \left (f x + e\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*a^3*c*cos(f*x + e)^5*log(sin(f*x + e) + 1) - 15*a^3*c*cos(f*x + e)^5*log(-sin(f*x + e) + 1) + 2*(28*
a^3*c*cos(f*x + e)^4 + 15*a^3*c*cos(f*x + e)^3 - 16*a^3*c*cos(f*x + e)^2 - 30*a^3*c*cos(f*x + e) - 12*a^3*c)*s
in(f*x + e))/(f*cos(f*x + e)^5)

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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)^2*(a+a*sec(f*x+e))^3*(c-c*sec(f*x+e)),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^3*c/8*ln(abs(tan((f*x+exp(1))/2)-1))-a^3*c/8*ln(abs(ta
n((f*x+exp(1))/2)+1))-(-15*tan((f*x+exp(1))/2)^9*a^3*c+70*tan((f*x+exp(1))/2)^7*a^3*c-128*tan((f*x+exp(1))/2)^
5*a^3*c+250*tan((f*x+exp(1))/2)^3*a^3*c+15*tan((f*x+exp(1))/2)*a^3*c)*1/60/(tan((f*x+exp(1))/2)^2-1)^5)

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maple [A]  time = 1.40, size = 130, normalized size = 1.24 \[ \frac {a^{3} c \sec \left (f x +e \right ) \tan \left (f x +e \right )}{4 f}+\frac {a^{3} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{4 f}+\frac {7 a^{3} c \tan \left (f x +e \right )}{15 f}-\frac {a^{3} c \left (\sec ^{3}\left (f x +e \right )\right ) \tan \left (f x +e \right )}{2 f}-\frac {a^{3} c \tan \left (f x +e \right ) \left (\sec ^{4}\left (f x +e \right )\right )}{5 f}-\frac {4 a^{3} c \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{15 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^3*c*sec(f*x+e)*tan(f*x+e)/f+1/4/f*a^3*c*ln(sec(f*x+e)+tan(f*x+e))+7/15*a^3*c*tan(f*x+e)/f-1/2*a^3*c*sec(
f*x+e)^3*tan(f*x+e)/f-1/5/f*a^3*c*tan(f*x+e)*sec(f*x+e)^4-4/15/f*a^3*c*tan(f*x+e)*sec(f*x+e)^2

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maxima [A]  time = 0.46, size = 172, normalized size = 1.64 \[ -\frac {8 \, {\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} a^{3} c - 15 \, a^{3} c {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 60 \, a^{3} c {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 120 \, a^{3} c \tan \left (f x + e\right )}{120 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(8*(3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^3*c - 15*a^3*c*(2*(3*sin(f*x + e)^3 - 5*s
in(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)) + 60
*a^3*c*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1) + log(sin(f*x + e) - 1)) - 120*a^3*c*tan(f
*x + e))/f

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mupad [B]  time = 6.32, size = 175, normalized size = 1.67 \[ \frac {-\frac {c\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{2}+\frac {7\,c\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{3}-\frac {64\,c\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{15}+\frac {25\,c\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3}+\frac {c\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{2}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}+\frac {a^3\,c\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{2\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^3*c*tan(e/2 + (f*x)/2))/2 + (25*a^3*c*tan(e/2 + (f*x)/2)^3)/3 - (64*a^3*c*tan(e/2 + (f*x)/2)^5)/15 + (7*a^
3*c*tan(e/2 + (f*x)/2)^7)/3 - (a^3*c*tan(e/2 + (f*x)/2)^9)/2)/(f*(5*tan(e/2 + (f*x)/2)^2 - 10*tan(e/2 + (f*x)/
2)^4 + 10*tan(e/2 + (f*x)/2)^6 - 5*tan(e/2 + (f*x)/2)^8 + tan(e/2 + (f*x)/2)^10 - 1)) + (a^3*c*atanh(tan(e/2 +
(f*x)/2)))/(2*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{3} c \left (\int \left (- \sec ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- 2 \sec ^{3}{\left (e + f x \right )}\right )\, dx + \int 2 \sec ^{5}{\left (e + f x \right )}\, dx + \int \sec ^{6}{\left (e + f x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**3*c*(Integral(-sec(e + f*x)**2, x) + Integral(-2*sec(e + f*x)**3, x) + Integral(2*sec(e + f*x)**5, x) + In
tegral(sec(e + f*x)**6, x))

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