3.1.45 \(\int \frac {\sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx\) [45]

Optimal. Leaf size=38 \[ \frac {\tanh ^{-1}(\sin (c+d x))}{a d}-\frac {\tan (c+d x)}{d (a+a \sec (c+d x))} \]

[Out]

arctanh(sin(d*x+c))/a/d-tan(d*x+c)/d/(a+a*sec(d*x+c))

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Rubi [A]
time = 0.05, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3874, 3855, 3879} \begin {gather*} \frac {\tanh ^{-1}(\sin (c+d x))}{a d}-\frac {\tan (c+d x)}{d (a \sec (c+d x)+a)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a + a*Sec[c + d*x]),x]

[Out]

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

Rule 3855

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

Rule 3874

Int[csc[(e_.) + (f_.)*(x_)]^2/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[Csc[e + f*x],
x], x] - Dist[a/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x]

Rule 3879

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(109\) vs. \(2(38)=76\).
time = 0.19, size = 109, normalized size = 2.87 \begin {gather*} -\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (\cos \left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )\right )}{a d (1+\sec (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/(a + a*Sec[c + d*x]),x]

[Out]

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

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Maple [A]
time = 0.04, size = 46, normalized size = 1.21

method	result	size
derivativedivides	\(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\)	\(46\)
default	\(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\)	\(46\)
risch	\(-\frac {2 i}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a d}\)	\(65\)
norman	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}}{\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a d}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 75, normalized size = 1.97 \begin {gather*} \frac {\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a - sin(d*x + c)/(a*(co
s(d*x + c) + 1)))/d

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Fricas [A]
time = 2.71, size = 65, normalized size = 1.71 \begin {gather*} \frac {{\left (\cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (\cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2/(a+a*sec(d*x+c)),x)

[Out]

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

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Giac [A]
time = 0.44, size = 54, normalized size = 1.42 \begin {gather*} \frac {\frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.65, size = 31, normalized size = 0.82 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^2*(a + a/cos(c + d*x))),x)

[Out]

(2*atanh(tan(c/2 + (d*x)/2)) - tan(c/2 + (d*x)/2))/(a*d)

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