3.213 \(\int \frac{\sec (e+f x) (c+d \sec (e+f x))}{a+a \sec (e+f x)} \, dx\)

Optimal. Leaf size=43 \[ \frac{(c-d) \tan (e+f x)}{f (a \sec (e+f x)+a)}+\frac{d \tanh ^{-1}(\sin (e+f x))}{a f} \]

[Out]

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

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Rubi [A]  time = 0.084563, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {3998, 3770, 3794} \[ \frac{(c-d) \tan (e+f x)}{f (a \sec (e+f x)+a)}+\frac{d \tanh ^{-1}(\sin (e+f x))}{a f} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c + d*Sec[e + f*x]))/(a + a*Sec[e + f*x]),x]

[Out]

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

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 3770

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

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

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

Mathematica [B]  time = 0.262645, size = 109, normalized size = 2.53 \[ \frac{2 \cos \left (\frac{1}{2} (e+f x)\right ) \left ((c-d) \sec \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right )+d \cos \left (\frac{1}{2} (e+f x)\right ) \left (\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )\right )\right )}{a f (\cos (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c + d*Sec[e + f*x]))/(a + a*Sec[e + f*x]),x]

[Out]

(2*Cos[(e + f*x)/2]*(d*Cos[(e + f*x)/2]*(-Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + Log[Cos[(e + f*x)/2] + Si
n[(e + f*x)/2]]) + (c - d)*Sec[e/2]*Sin[(f*x)/2]))/(a*f*(1 + Cos[e + f*x]))

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Maple [A]  time = 0.043, size = 78, normalized size = 1.8 \begin{align*}{\frac{c}{fa}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }-{\frac{d}{fa}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }-{\frac{d}{fa}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) }+{\frac{d}{fa}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c+d*sec(f*x+e))/(a+a*sec(f*x+e)),x)

[Out]

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

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Maxima [B]  time = 0.986664, size = 134, normalized size = 3.12 \begin{align*} \frac{d{\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 \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+d*sec(f*x+e))/(a+a*sec(f*x+e)),x, algorithm="maxima")

[Out]

(d*(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*sin(f*x + e)/(a*(cos(f*x + e) + 1)))/f

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Fricas [A]  time = 0.476746, size = 197, normalized size = 4.58 \begin{align*} \frac{{\left (d \cos \left (f x + e\right ) + d\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (d \cos \left (f x + e\right ) + d\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (c - d\right )} \sin \left (f x + e\right )}{2 \,{\left (a f \cos \left (f x + e\right ) + a f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(c*sec(e + f*x)/(sec(e + f*x) + 1), x) + Integral(d*sec(e + f*x)**2/(sec(e + f*x) + 1), x))/a

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Giac [A]  time = 1.2826, size = 100, normalized size = 2.33 \begin{align*} \frac{\frac{d \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a} - \frac{d \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a} + \frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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