3.154 \(\int \sec (e+f x) (a+b \sec ^2(e+f x)) \, dx\)

Optimal. Leaf size=40 \[ \frac{(2 a+b) \tanh ^{-1}(\sin (e+f x))}{2 f}+\frac{b \tan (e+f x) \sec (e+f x)}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.0248988, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {4046, 3770} \[ \frac{(2 a+b) \tanh ^{-1}(\sin (e+f x))}{2 f}+\frac{b \tan (e+f x) \sec (e+f x)}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[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 \sec (e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx &=\frac{b \sec (e+f x) \tan (e+f x)}{2 f}+\frac{1}{2} (2 a+b) \int \sec (e+f x) \, dx\\ &=\frac{(2 a+b) \tanh ^{-1}(\sin (e+f x))}{2 f}+\frac{b \sec (e+f x) \tan (e+f x)}{2 f}\\ \end{align*}

Mathematica [A]  time = 0.0206032, size = 48, normalized size = 1.2 \[ \frac{a \tanh ^{-1}(\sin (e+f x))}{f}+\frac{b \tanh ^{-1}(\sin (e+f x))}{2 f}+\frac{b \tan (e+f x) \sec (e+f x)}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.027, size = 59, normalized size = 1.5 \begin{align*}{\frac{a\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{f}}+{\frac{b\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{2\,f}}+{\frac{b\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{2\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*a*ln(sec(f*x+e)+tan(f*x+e))+1/2*b*sec(f*x+e)*tan(f*x+e)/f+1/2/f*b*ln(sec(f*x+e)+tan(f*x+e))

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Maxima [A]  time = 0.986785, size = 78, normalized size = 1.95 \begin{align*} \frac{{\left (2 \, a + b\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (2 \, a + b\right )} \log \left (\sin \left (f x + e\right ) - 1\right ) - \frac{2 \, b \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1}}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.494837, size = 192, normalized size = 4.8 \begin{align*} \frac{{\left (2 \, a + b\right )} \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (2 \, a + b\right )} \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, b \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(sec(f*x+e)*(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sec(e + f*x)**2)*sec(e + f*x), x)

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Giac [A]  time = 1.29418, size = 86, normalized size = 2.15 \begin{align*} \frac{{\left (2 \, a + b\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (2 \, a + b\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - \frac{2 \, b \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1}}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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