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

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

[Out]

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

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Rubi [A]  time = 0.102935, 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} \[ \frac{3 c^2 \tan (e+f x)}{a f}-\frac{3 c^2 \tanh ^{-1}(\sin (e+f x))}{a f}+\frac{2 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )}{f (a \sec (e+f x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 1.57997, size = 220, normalized size = 2.97 \[ \frac{2 c^2 \sin \left (\frac{1}{2} (e+f x)\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \sec (e+f x) \left (4 \sec \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \csc \left (\frac{1}{2} (e+f x)\right )+\cot \left (\frac{1}{2} (e+f x)\right ) \left (\frac{\sin (f x)}{\left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}+3 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-3 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )\right )}{a f (\sec (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^2*Cos[(e + f*x)/2]*Sec[e + f*x]*Sin[(e + f*x)/2]*(4*Csc[(e + f*x)/2]*Sec[e/2]*Sin[(f*x)/2] + Cot[(e + 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])))))/(a*f*(1 + Sec[e + f*x]))

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Maple [A]  time = 0.059, size = 116, normalized size = 1.6 \begin{align*} 4\,{\frac{{c}^{2}\tan \left ( 1/2\,fx+e/2 \right ) }{fa}}-{\frac{{c}^{2}}{fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}-3\,{\frac{{c}^{2}\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }{fa}}-{\frac{{c}^{2}}{fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}+3\,{\frac{{c}^{2}\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }{fa}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.961418, size = 302, normalized size = 4.08 \begin{align*} -\frac{c^{2}{\left (\frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac{2 \, \sin \left (f x + e\right )}{{\left (a - \frac{a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (f x + e\right ) + 1\right )}} - \frac{\sin \left (f x + e\right )}{a{\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + 2 \, c^{2}{\left (\frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac{\sin \left (f x + e\right )}{a{\left (\cos \left (f x + e\right ) + 1\right )}}\right )} - \frac{c^{2} \sin \left (f x + e\right )}{a{\left (\cos \left (f x + e\right ) + 1\right )}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.478713, size = 301, normalized size = 4.07 \begin{align*} -\frac{3 \,{\left (c^{2} \cos \left (f x + e\right )^{2} + c^{2} \cos \left (f x + e\right )\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \,{\left (c^{2} \cos \left (f x + e\right )^{2} + c^{2} \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (5 \, c^{2} \cos \left (f x + e\right ) + c^{2}\right )} \sin \left (f x + e\right )}{2 \,{\left (a f \cos \left (f x + e\right )^{2} + a f \cos \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*(c^2*cos(f*x + e)^2 + c^2*cos(f*x + e))*log(sin(f*x + e) + 1) - 3*(c^2*cos(f*x + e)^2 + c^2*cos(f*x +
e))*log(-sin(f*x + e) + 1) - 2*(5*c^2*cos(f*x + e) + c^2)*sin(f*x + e))/(a*f*cos(f*x + e)^2 + a*f*cos(f*x + e)
)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{c^{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 )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**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))/a

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Giac [A]  time = 1.32756, size = 138, normalized size = 1.86 \begin{align*} -\frac{\frac{3 \, c^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a} - \frac{3 \, c^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a} - \frac{4 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a} + \frac{2 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} a}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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