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

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

[Out]

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

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Rubi [A]  time = 0.15574, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3958, 2611, 3770, 2607, 30, 3768} \[ -\frac{2 a^3 c \tan ^3(e+f x)}{3 f}+\frac{5 a^3 c \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac{a^3 c \tan (e+f x) \sec ^3(e+f x)}{4 f}-\frac{3 a^3 c \tan (e+f x) \sec (e+f x)}{8 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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]

Rubi steps

\begin{align*} \int \sec (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 (e+f x) \tan ^2(e+f x)+2 a^2 \sec ^2(e+f x) \tan ^2(e+f x)+a^2 \sec ^3(e+f x) \tan ^2(e+f x)\right ) \, dx\right )\\ &=-\left (\left (a^3 c\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx\right )-\left (a^3 c\right ) \int \sec ^3(e+f x) \tan ^2(e+f x) \, dx-\left (2 a^3 c\right ) \int \sec ^2(e+f x) \tan ^2(e+f x) \, dx\\ &=-\frac{a^3 c \sec (e+f x) \tan (e+f x)}{2 f}-\frac{a^3 c \sec ^3(e+f x) \tan (e+f x)}{4 f}+\frac{1}{4} \left (a^3 c\right ) \int \sec ^3(e+f x) \, dx+\frac{1}{2} \left (a^3 c\right ) \int \sec (e+f x) \, dx-\frac{\left (2 a^3 c\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{a^3 c \tanh ^{-1}(\sin (e+f x))}{2 f}-\frac{3 a^3 c \sec (e+f x) \tan (e+f x)}{8 f}-\frac{a^3 c \sec ^3(e+f x) \tan (e+f x)}{4 f}-\frac{2 a^3 c \tan ^3(e+f x)}{3 f}+\frac{1}{8} \left (a^3 c\right ) \int \sec (e+f x) \, dx\\ &=\frac{5 a^3 c \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac{3 a^3 c \sec (e+f x) \tan (e+f x)}{8 f}-\frac{a^3 c \sec ^3(e+f x) \tan (e+f x)}{4 f}-\frac{2 a^3 c \tan ^3(e+f x)}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.616023, size = 70, normalized size = 0.81 \[ \frac{a^3 c \left (60 \tanh ^{-1}(\sin (e+f x))-(33 \sin (e+f x)+16 \sin (2 (e+f x))+9 \sin (3 (e+f x))-8 \sin (4 (e+f x))) \sec ^4(e+f x)\right )}{96 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*c*(60*ArcTanh[Sin[e + f*x]] - Sec[e + f*x]^4*(33*Sin[e + f*x] + 16*Sin[2*(e + f*x)] + 9*Sin[3*(e + f*x)]
- 8*Sin[4*(e + f*x)])))/(96*f)

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Maple [A]  time = 0.024, size = 107, normalized size = 1.2 \begin{align*}{\frac{2\,{a}^{3}c\tan \left ( fx+e \right ) }{3\,f}}+{\frac{5\,{a}^{3}c\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{8\,f}}-{\frac{2\,{a}^{3}c\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{3\,f}}-{\frac{{a}^{3}c \left ( \sec \left ( fx+e \right ) \right ) ^{3}\tan \left ( fx+e \right ) }{4\,f}}-{\frac{3\,{a}^{3}c\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{8\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01203, size = 180, normalized size = 2.09 \begin{align*} -\frac{32 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{3} c - 3 \, 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 )} - 48 \, a^{3} c \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) - 96 \, a^{3} c \tan \left (f x + e\right )}{48 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.487776, size = 302, normalized size = 3.51 \begin{align*} \frac{15 \, a^{3} c \cos \left (f x + e\right )^{4} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, a^{3} c \cos \left (f x + e\right )^{4} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (16 \, a^{3} c \cos \left (f x + e\right )^{3} - 9 \, a^{3} c \cos \left (f x + e\right )^{2} - 16 \, a^{3} c \cos \left (f x + e\right ) - 6 \, a^{3} c\right )} \sin \left (f x + e\right )}{48 \, f \cos \left (f x + e\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

-a**3*c*(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))

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Giac [A]  time = 1.31749, size = 182, normalized size = 2.12 \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 )^{7} - 55 \, a^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 73 \, 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 )}^{4}}}{24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(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)^7 - 55*a^3*c*tan(1/2*f*x + 1/2*e)^5 + 73*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)^4)/f