3.449 \(\int \sec (c+d x) (a+b \sec (c+d x)) \, dx\)

Optimal. Leaf size=24 \[ \frac {a \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3787, 3770, 3767, 8} \[ \frac {a \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + b*Sec[c + d*x]),x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 3770

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 24, normalized size = 1.00 \[ \frac {a \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + b*Sec[c + d*x]),x]

[Out]

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

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fricas [B]  time = 0.46, size = 60, normalized size = 2.50 \[ \frac {a \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - a \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, b \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.20, size = 63, normalized size = 2.62 \[ \frac {a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.56, size = 32, normalized size = 1.33 \[ \frac {b \tan \left (d x +c \right )}{d}+\frac {a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+b*sec(d*x+c)),x)

[Out]

b*tan(d*x+c)/d+1/d*a*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.54, size = 29, normalized size = 1.21 \[ \frac {a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + b \tan \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

(a*log(sec(d*x + c) + tan(d*x + c)) + b*tan(d*x + c))/d

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mupad [B]  time = 0.79, size = 47, normalized size = 1.96 \[ \frac {2\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))/cos(c + d*x),x)

[Out]

(2*a*atanh(tan(c/2 + (d*x)/2)))/d - (2*b*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d*x)/2)^2 - 1))

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sympy [A]  time = 4.27, size = 37, normalized size = 1.54 \[ \begin {cases} \frac {a \log {\left (\tan {\left (c + d x \right )} + \sec {\left (c + d x \right )} \right )} + b \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \sec {\relax (c )}\right ) \sec {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*sec(d*x+c)),x)

[Out]

Piecewise(((a*log(tan(c + d*x) + sec(c + d*x)) + b*tan(c + d*x))/d, Ne(d, 0)), (x*(a + b*sec(c))*sec(c), True)
)

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