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]

(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)

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Rubi [A]  time = 0.193214, antiderivative size = 105, normalized size of antiderivative = 1., 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 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]

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 30

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

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*}

Mathematica [A]  time = 0.309133, 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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Maple [A]  time = 0.027, size = 130, normalized size = 1.2 \begin{align*}{\frac{{a}^{3}c\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{4\,f}}+{\frac{{a}^{3}c\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{4\,f}}+{\frac{7\,{a}^{3}c\tan \left ( fx+e \right ) }{15\,f}}-{\frac{{a}^{3}c \left ( \sec \left ( fx+e \right ) \right ) ^{3}\tan \left ( fx+e \right ) }{2\,f}}-{\frac{{a}^{3}c\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{4}}{5\,f}}-{\frac{4\,{a}^{3}c\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{15\,f}} \end{align*}

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/f*a^3*c*tan(f*x+e)-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 = 1.03722, size = 232, normalized size = 2.21 \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 0.493847, size = 342, normalized size = 3.26 \begin{align*} \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}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - a^{3} c \left (\int - \sec ^{2}{\left (e + f x \right )}\, dx + \int - 2 \sec ^{3}{\left (e + f x \right )}\, dx + \int 2 \sec ^{5}{\left (e + f x \right )}\, dx + \int \sec ^{6}{\left (e + f x \right )}\, dx\right ) \end{align*}

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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Giac [A]  time = 1.22543, size = 207, normalized size = 1.97 \begin{align*} \frac{15 \, a^{3} c \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right ) - 15 \, a^{3} c \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right ) - \frac{2 \,{\left (15 \, a^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9} - 70 \, a^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 128 \, a^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 250 \, a^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 15 \, a^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}^{5}}}{60 \, f} \end{align*}

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]

1/60*(15*a^3*c*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 15*a^3*c*log(abs(tan(1/2*f*x + 1/2*e) - 1)) - 2*(15*a^3*c*
tan(1/2*f*x + 1/2*e)^9 - 70*a^3*c*tan(1/2*f*x + 1/2*e)^7 + 128*a^3*c*tan(1/2*f*x + 1/2*e)^5 - 250*a^3*c*tan(1/
2*f*x + 1/2*e)^3 - 15*a^3*c*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^5)/f