3.1.14 \(\int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x)) \, dx\) [14]

Optimal. Leaf size=61 \[ \frac {a^2 c \tanh ^{-1}(\sin (e+f x))}{2 f}-\frac {a^2 c \sec (e+f x) \tan (e+f x)}{2 f}-\frac {a^2 c \tan ^3(e+f x)}{3 f} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4043, 2691, 3855, 2687, 30} \begin {gather*} -\frac {a^2 c \tan ^3(e+f x)}{3 f}+\frac {a^2 c \tanh ^{-1}(\sin (e+f x))}{2 f}-\frac {a^2 c \tan (e+f x) \sec (e+f x)}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c*ArcTanh[Sin[e + f*x]])/(2*f) - (a^2*c*Sec[e + f*x]*Tan[e + f*x])/(2*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 2687

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 2691

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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4043

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

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Mathematica [A]
time = 0.12, size = 45, normalized size = 0.74 \begin {gather*} \frac {a^2 c \left (3 \tanh ^{-1}(\sin (e+f x))-3 \sec (e+f x) \tan (e+f x)-2 \tan ^3(e+f x)\right )}{6 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.19, size = 96, normalized size = 1.57

method	result	size
derivativedivides	\(\frac {a^{2} c \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )-a^{2} c \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )+a^{2} c \tan \left (f x +e \right )+a^{2} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}\)	\(96\)
default	\(\frac {a^{2} c \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )-a^{2} c \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )+a^{2} c \tan \left (f x +e \right )+a^{2} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}\)	\(96\)
risch	\(\frac {i a^{2} c \left (3 \,{\mathrm e}^{5 i \left (f x +e \right )}+6 \,{\mathrm e}^{4 i \left (f x +e \right )}-3 \,{\mathrm e}^{i \left (f x +e \right )}+2\right )}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}+\frac {a^{2} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{2 f}-\frac {a^{2} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{2 f}\)	\(104\)
norman	\(\frac {\frac {a^{2} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {8 a^{2} c \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}-\frac {a^{2} c \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {a^{2} c \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 f}+\frac {a^{2} c \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 f}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 117, normalized size = 1.92 \begin {gather*} -\frac {4 \, {\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 \, \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 )} - 12 \, a^{2} c \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) - 12 \, a^{2} c \tan \left (f x + e\right )}{12 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(4*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*c - 3*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)) - 12*a^2*c*log(sec(f*x + e) + tan(f*x + e)) - 12*a^2*c*tan(f*x + e))/f

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Fricas [A]
time = 2.50, size = 111, normalized size = 1.82 \begin {gather*} \frac {3 \, a^{2} c \cos \left (f x + e\right )^{3} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, a^{2} c \cos \left (f x + e\right )^{3} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (2 \, a^{2} c \cos \left (f x + e\right )^{2} - 3 \, a^{2} c \cos \left (f x + e\right ) - 2 \, a^{2} c\right )} \sin \left (f x + e\right )}{12 \, f \cos \left (f x + e\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (55) = 110\).
time = 0.55, size = 111, normalized size = 1.82 \begin {gather*} \frac {3 \, a^{2} c \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right ) - 3 \, a^{2} c \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 8 \, a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a^{2} 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 )}^{3}}}{6 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*a^2*c*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 3*a^2*c*log(abs(tan(1/2*f*x + 1/2*e) - 1)) - 2*(3*a^2*c*tan(
1/2*f*x + 1/2*e)^5 - 8*a^2*c*tan(1/2*f*x + 1/2*e)^3 - 3*a^2*c*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 -
1)^3)/f

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Mupad [B]
time = 3.79, size = 113, normalized size = 1.85 \begin {gather*} \frac {-c\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+\frac {8\,c\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3}+c\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}+\frac {a^2\,c\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{f} \end {gather*}

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),x)

[Out]

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

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