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

Optimal. Leaf size=55 \[ \frac{a^2 c \tanh ^{-1}(\sin (e+f x))}{2 f}-\frac{c \tan (e+f x) \left (a^2 \sec (e+f x)+2 a^2\right )}{2 f}+a^2 c x \]

[Out]

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

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Rubi [A]  time = 0.0626252, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3904, 3881, 3770} \[ \frac{a^2 c \tanh ^{-1}(\sin (e+f x))}{2 f}-\frac{c \tan (e+f x) \left (a^2 \sec (e+f x)+2 a^2\right )}{2 f}+a^2 c x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3904

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[(-(a*c))^m, Int[Cot[e + f*x]^(2*m)*(c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] &&  !(IntegerQ[n] && GtQ[m - n, 0])

Rule 3881

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(e*(e*Cot[
c + d*x])^(m - 1)*(a*m + b*(m - 1)*Csc[c + d*x]))/(d*m*(m - 1)), x] - Dist[e^2/m, Int[(e*Cot[c + d*x])^(m - 2)
*(a*m + b*(m - 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m, 1]

Rule 3770

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

Rubi steps

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

Mathematica [A]  time = 0.291376, size = 72, normalized size = 1.31 \[ \frac{a^2 c \sec ^2(e+f x) \left (-\sin (e+f x)-\sin (2 (e+f x))+(e+f x) \cos (2 (e+f x))+\cos ^2(e+f x) \tanh ^{-1}(\sin (e+f x))+e+f x\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.02, size = 76, normalized size = 1.4 \begin{align*}{\frac{{a}^{2}c\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{2\,f}}+{a}^{2}cx+{\frac{{a}^{2}ce}{f}}-{\frac{{a}^{2}c\tan \left ( fx+e \right ) }{f}}-{\frac{{a}^{2}c\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{2\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.07022, size = 128, normalized size = 2.33 \begin{align*} \frac{4 \,{\left (f x + e\right )} a^{2} c + 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 )} + 4 \, a^{2} c \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) - 4 \, a^{2} c \tan \left (f x + e\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.07603, size = 259, normalized size = 4.71 \begin{align*} \frac{4 \, a^{2} c f x \cos \left (f x + e\right )^{2} + a^{2} c \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) - a^{2} c \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (2 \, a^{2} c \cos \left (f x + e\right ) + a^{2} c\right )} \sin \left (f x + e\right )}{4 \, f \cos \left (f x + e\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - a^{2} c \left (\int \left (-1\right )\, dx + \int - \sec{\left (e + f x \right )}\, dx + \int \sec ^{2}{\left (e + f x \right )}\, dx + \int \sec ^{3}{\left (e + f x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*c*(Integral(-1, x) + Integral(-sec(e + f*x), x) + Integral(sec(e + f*x)**2, x) + Integral(sec(e + f*x)**
3, x))

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Giac [B]  time = 1.36062, size = 147, normalized size = 2.67 \begin{align*} \frac{2 \,{\left (f x + e\right )} a^{2} c + a^{2} c \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right ) - a^{2} c \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right ) + \frac{2 \,{\left (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 )}^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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