3.15 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))^2}{c-c \sec (e+f x)} \, dx\)

Optimal. Leaf size=74 \[ -\frac{3 a^2 \tan (e+f x)}{c f}-\frac{3 a^2 \tanh ^{-1}(\sin (e+f x))}{c f}-\frac{2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))} \]

[Out]

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

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Rubi [A]  time = 0.0993604, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {3957, 3787, 3770, 3767, 8} \[ -\frac{3 a^2 \tan (e+f x)}{c f}-\frac{3 a^2 \tanh ^{-1}(\sin (e+f x))}{c f}-\frac{2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))} \]

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]

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

Rule 3957

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
^(n_.), x_Symbol] :> Simp[(2*a*c*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(c + d*Csc[e + f*x])^(n - 1))/(b*f*(2*m +
1)), x] - Dist[(d*(2*n - 1))/(b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x]
)^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0] && L
tQ[m, -2^(-1)] && IntegerQ[2*m]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3770

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [B]  time = 1.71673, size = 220, normalized size = 2.97 \[ \frac{2 a^2 \sin \left (\frac{1}{2} (e+f x)\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \sec (e+f x) \left (4 \csc \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \sec \left (\frac{1}{2} (e+f x)\right )+\tan \left (\frac{1}{2} (e+f x)\right ) \left (\frac{\sin (f x)}{\left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}-3 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+3 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )\right )}{f (c-c \sec (e+f x))} \]

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]

(2*a^2*Cos[(e + f*x)/2]*Sec[e + f*x]*Sin[(e + f*x)/2]*(4*Csc[e/2]*Sec[(e + f*x)/2]*Sin[(f*x)/2] + (-3*Log[Cos[
(e + f*x)/2] - Sin[(e + f*x)/2]] + 3*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + Sin[f*x]/((Cos[e/2] - Sin[e/2]
)*(Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])))*Tan[(e +
f*x)/2]))/(f*(c - c*Sec[e + f*x]))

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Maple [A]  time = 0.063, size = 116, normalized size = 1.6 \begin{align*}{\frac{{a}^{2}}{fc} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}-3\,{\frac{{a}^{2}\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }{fc}}+{\frac{{a}^{2}}{fc} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}+3\,{\frac{{a}^{2}\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }{fc}}+4\,{\frac{{a}^{2}}{fc\tan \left ( 1/2\,fx+e/2 \right ) }} \end{align*}

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)

[Out]

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

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Maxima [B]  time = 1.03162, size = 304, normalized size = 4.11 \begin{align*} -\frac{a^{2}{\left (\frac{\frac{3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1}{\frac{c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c}\right )} + 2 \, a^{2}{\left (\frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c} - \frac{\cos \left (f x + e\right ) + 1}{c \sin \left (f x + e\right )}\right )} - \frac{a^{2}{\left (\cos \left (f x + e\right ) + 1\right )}}{c \sin \left (f x + e\right )}}{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))^2/(c-c*sec(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.476103, size = 275, normalized size = 3.72 \begin{align*} -\frac{3 \, a^{2} \cos \left (f x + e\right ) \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 3 \, a^{2} \cos \left (f x + e\right ) \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 10 \, a^{2} \cos \left (f x + e\right )^{2} - 8 \, a^{2} \cos \left (f x + e\right ) + 2 \, a^{2}}{2 \, c f \cos \left (f x + e\right ) \sin \left (f x + e\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))^2/(c-c*sec(f*x+e)),x, algorithm="fricas")

[Out]

-1/2*(3*a^2*cos(f*x + e)*log(sin(f*x + e) + 1)*sin(f*x + e) - 3*a^2*cos(f*x + e)*log(-sin(f*x + e) + 1)*sin(f*
x + e) - 10*a^2*cos(f*x + e)^2 - 8*a^2*cos(f*x + e) + 2*a^2)/(c*f*cos(f*x + e)*sin(f*x + e))

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

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*(Integral(sec(e + f*x)/(sec(e + f*x) - 1), x) + Integral(2*sec(e + f*x)**2/(sec(e + f*x) - 1), x) + Inte
gral(sec(e + f*x)**3/(sec(e + f*x) - 1), x))/c

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Giac [A]  time = 1.3107, size = 142, normalized size = 1.92 \begin{align*} -\frac{\frac{3 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c} - \frac{3 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c} - \frac{2 \,{\left (3 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 \, a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )} c}}{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))^2/(c-c*sec(f*x+e)),x, algorithm="giac")

[Out]

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