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

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

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3962, 2607, 30, 2611, 3768, 3770} \[ -\frac {a^2 c \tan ^3(e+f x)}{3 f}+\frac {a^2 c \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {a^2 c \tan (e+f x) \sec ^3(e+f x)}{4 f}+\frac {a^2 c \tan (e+f x) \sec (e+f x)}{8 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c*ArcTanh[Sin[e + f*x]])/(8*f) + (a^2*c*Sec[e + f*x]*Tan[e + f*x])/(8*f) - (a^2*c*Sec[e + f*x]^3*Tan[e +
f*x])/(4*f) - (a^2*c*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 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))^2 (c-c \sec (e+f x)) \, dx &=-\left ((a c) \int \left (a \sec ^2(e+f x) \tan ^2(e+f x)+a \sec ^3(e+f x) \tan ^2(e+f x)\right ) \, dx\right )\\ &=-\left (\left (a^2 c\right ) \int \sec ^2(e+f x) \tan ^2(e+f x) \, dx\right )-\left (a^2 c\right ) \int \sec ^3(e+f x) \tan ^2(e+f x) \, dx\\ &=-\frac {a^2 c \sec ^3(e+f x) \tan (e+f x)}{4 f}+\frac {1}{4} \left (a^2 c\right ) \int \sec ^3(e+f x) \, dx-\frac {\left (a^2 c\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a^2 c \sec (e+f x) \tan (e+f x)}{8 f}-\frac {a^2 c \sec ^3(e+f x) \tan (e+f x)}{4 f}-\frac {a^2 c \tan ^3(e+f x)}{3 f}+\frac {1}{8} \left (a^2 c\right ) \int \sec (e+f x) \, dx\\ &=\frac {a^2 c \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac {a^2 c \sec (e+f x) \tan (e+f x)}{8 f}-\frac {a^2 c \sec ^3(e+f x) \tan (e+f x)}{4 f}-\frac {a^2 c \tan ^3(e+f x)}{3 f}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 57, normalized size = 0.66 \[ \frac {a^2 c \left (3 \tanh ^{-1}(\sin (e+f x))+\tan (e+f x) \left (-8 \tan ^2(e+f x)-6 \sec ^3(e+f x)+3 \sec (e+f x)\right )\right )}{24 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c*(3*ArcTanh[Sin[e + f*x]] + Tan[e + f*x]*(3*Sec[e + f*x] - 6*Sec[e + f*x]^3 - 8*Tan[e + f*x]^2)))/(24*f)

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fricas [A]  time = 0.46, size = 117, normalized size = 1.36 \[ \frac {3 \, a^{2} c \cos \left (f x + e\right )^{4} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, a^{2} c \cos \left (f x + e\right )^{4} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (8 \, a^{2} c \cos \left (f x + e\right )^{3} + 3 \, a^{2} c \cos \left (f x + e\right )^{2} - 8 \, a^{2} c \cos \left (f x + e\right ) - 6 \, a^{2} c\right )} \sin \left (f x + e\right )}{48 \, f \cos \left (f x + e\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*a^2*c*cos(f*x + e)^4*log(sin(f*x + e) + 1) - 3*a^2*c*cos(f*x + e)^4*log(-sin(f*x + e) + 1) + 2*(8*a^2*
c*cos(f*x + e)^3 + 3*a^2*c*cos(f*x + e)^2 - 8*a^2*c*cos(f*x + e) - 6*a^2*c)*sin(f*x + e))/(f*cos(f*x + e)^4)

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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))^2*(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^2*c/16*ln(abs(tan((f*x+exp(1))/2)-1))-a^2*c/16*ln(abs(
tan((f*x+exp(1))/2)+1))+(3*tan((f*x+exp(1))/2)^7*a^2*c-11*tan((f*x+exp(1))/2)^5*a^2*c+53*tan((f*x+exp(1))/2)^3
*a^2*c+3*tan((f*x+exp(1))/2)*a^2*c)*1/24/(tan((f*x+exp(1))/2)^2-1)^4)

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maple [A]  time = 1.37, size = 107, normalized size = 1.24 \[ \frac {a^{2} c \sec \left (f x +e \right ) \tan \left (f x +e \right )}{8 f}+\frac {a^{2} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8 f}+\frac {a^{2} c \tan \left (f x +e \right )}{3 f}-\frac {a^{2} c \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{3 f}-\frac {a^{2} c \left (\sec ^{3}\left (f x +e \right )\right ) \tan \left (f x +e \right )}{4 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.48, size = 160, normalized size = 1.86 \[ -\frac {16 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c - 3 \, a^{2} c {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 12 \, a^{2} c {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 48 \, a^{2} c \tan \left (f x + e\right )}{48 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(16*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*c - 3*a^2*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)) + 12*a^2*c*(2*sin(f*x + e)/(
sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1) + log(sin(f*x + e) - 1)) - 48*a^2*c*tan(f*x + e))/f

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mupad [B]  time = 4.92, size = 146, normalized size = 1.70 \[ \frac {a^2\,c\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{4\,f}-\frac {\frac {c\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{4}-\frac {11\,c\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{12}+\frac {53\,c\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{12}+\frac {c\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{4}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*c*atanh(tan(e/2 + (f*x)/2)))/(4*f) - ((a^2*c*tan(e/2 + (f*x)/2))/4 + (53*a^2*c*tan(e/2 + (f*x)/2)^3)/12 -
(11*a^2*c*tan(e/2 + (f*x)/2)^5)/12 + (a^2*c*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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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