3.1.5 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))}{c-c \sec (e+f x)} \, dx\) [5]

Optimal. Leaf size=42 \[ -\frac {a \tanh ^{-1}(\sin (e+f x))}{c f}-\frac {2 a \tan (e+f x)}{f (c-c \sec (e+f x))} \]

[Out]

-a*arctanh(sin(f*x+e))/c/f-2*a*tan(f*x+e)/f/(c-c*sec(f*x+e))

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Rubi [A]
time = 0.04, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {4042, 3855} \begin {gather*} -\frac {a \tanh ^{-1}(\sin (e+f x))}{c f}-\frac {2 a \tan (e+f x)}{f (c-c \sec (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3855

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

Rule 4042

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]

Rubi steps

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

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Mathematica [A]
time = 0.08, size = 77, normalized size = 1.83 \begin {gather*} -\frac {a \left (-\frac {2 \cot \left (\frac {1}{2} (e+f x)\right )}{f}-\frac {\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}{f}+\frac {\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{f}\right )}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*((-2*Cot[(e + f*x)/2])/f - Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]/f + Log[Cos[(e + f*x)/2] + Sin[(e + f
*x)/2]]/f))/c)

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Maple [A]
time = 0.13, size = 50, normalized size = 1.19

method	result	size
derivativedivides	\(\frac {2 a \left (-\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2}+\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2}+\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{f c}\)	\(50\)
default	\(\frac {2 a \left (-\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2}+\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2}+\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{f c}\)	\(50\)
risch	\(\frac {4 i a}{f c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}+\frac {a \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{c f}-\frac {a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{c f}\)	\(68\)
norman	\(\frac {-\frac {2 a}{c f}+\frac {2 a \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+\frac {a \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{c f}-\frac {a \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{c f}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))/(c-c*sec(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (46) = 92\).
time = 0.27, size = 109, normalized size = 2.60 \begin {gather*} -\frac {a {\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 {\left (\cos \left (f x + e\right ) + 1\right )}}{c \sin \left (f x + e\right )}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.50, size = 72, normalized size = 1.71 \begin {gather*} -\frac {a \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - a \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 4 \, a \cos \left (f x + e\right ) - 4 \, a}{2 \, c f \sin \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {a \left (\int \frac {\sec {\left (e + f x \right )}}{\sec {\left (e + f x \right )} - 1}\, dx + \int \frac {\sec ^{2}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} - 1}\, dx\right )}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.57, size = 60, normalized size = 1.43 \begin {gather*} -\frac {\frac {a \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{c} - \frac {a \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{c} - \frac {2 \, a}{c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.85, size = 31, normalized size = 0.74 \begin {gather*} -\frac {2\,a\,\left (\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{c\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))/(cos(e + f*x)*(c - c/cos(e + f*x))),x)

[Out]

-(2*a*(atanh(tan(e/2 + (f*x)/2)) - cot(e/2 + (f*x)/2)))/(c*f)

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