3.20 \(\int \frac{x}{a+b \sec (c+d x^2)} \, dx\)

Optimal. Leaf size=66 \[ \frac{x^2}{2 a}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} \left (c+d x^2\right )\right )}{\sqrt{a+b}}\right )}{a d \sqrt{a-b} \sqrt{a+b}} \]

[Out]

x^2/(2*a) - (b*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x^2)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d)

________________________________________________________________________________________

Rubi [A]  time = 0.108982, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4204, 3783, 2659, 208} \[ \frac{x^2}{2 a}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} \left (c+d x^2\right )\right )}{\sqrt{a+b}}\right )}{a d \sqrt{a-b} \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Sec[c + d*x^2]),x]

[Out]

x^2/(2*a) - (b*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x^2)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d)

Rule 4204

Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sec[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[
(m + 1)/n], 0] && IntegerQ[p]

Rule 3783

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[1/a, Int[1/(1 + (a*Sin[c + d
*x])/b), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.153877, size = 67, normalized size = 1.02 \[ \frac{\frac{2 b \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} \left (c+d x^2\right )\right )}{\sqrt{a^2-b^2}}\right )}{d \sqrt{a^2-b^2}}+\frac{c}{d}+x^2}{2 a} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*Sec[c + d*x^2]),x]

[Out]

(c/d + x^2 + (2*b*ArcTanh[((-a + b)*Tan[(c + d*x^2)/2])/Sqrt[a^2 - b^2]])/(Sqrt[a^2 - b^2]*d))/(2*a)

________________________________________________________________________________________

Maple [A]  time = 0.046, size = 70, normalized size = 1.1 \begin{align*}{\frac{1}{da}\arctan \left ( \tan \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{b}{da}{\it Artanh} \left ({(a-b)\tan \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a+b*sec(d*x^2+c)),x)

[Out]

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

________________________________________________________________________________________

Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 1.76205, size = 535, normalized size = 8.11 \begin{align*} \left [\frac{2 \,{\left (a^{2} - b^{2}\right )} d x^{2} + \sqrt{a^{2} - b^{2}} b \log \left (\frac{2 \, a b \cos \left (d x^{2} + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x^{2} + c\right )^{2} - 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x^{2} + c\right ) + a\right )} \sin \left (d x^{2} + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x^{2} + c\right )^{2} + 2 \, a b \cos \left (d x^{2} + c\right ) + b^{2}}\right )}{4 \,{\left (a^{3} - a b^{2}\right )} d}, \frac{{\left (a^{2} - b^{2}\right )} d x^{2} - \sqrt{-a^{2} + b^{2}} b \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x^{2} + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x^{2} + c\right )}\right )}{2 \,{\left (a^{3} - a b^{2}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*(a^2 - b^2)*d*x^2 + sqrt(a^2 - b^2)*b*log((2*a*b*cos(d*x^2 + c) - (a^2 - 2*b^2)*cos(d*x^2 + c)^2 - 2*s
qrt(a^2 - b^2)*(b*cos(d*x^2 + c) + a)*sin(d*x^2 + c) + 2*a^2 - b^2)/(a^2*cos(d*x^2 + c)^2 + 2*a*b*cos(d*x^2 +
c) + b^2)))/((a^3 - a*b^2)*d), 1/2*((a^2 - b^2)*d*x^2 - sqrt(-a^2 + b^2)*b*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x
^2 + c) + a)/((a^2 - b^2)*sin(d*x^2 + c))))/((a^3 - a*b^2)*d)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.25145, size = 140, normalized size = 2.12 \begin{align*} -\frac{{\left (\pi \left \lfloor \frac{d x^{2} + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x^{2} + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x^{2} + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )} b}{\sqrt{-a^{2} + b^{2}} a d} + \frac{d x^{2} + c}{2 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(pi*floor(1/2*(d*x^2 + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x^2 + 1/2*c) - b*tan(1/2*d*x^2 + 1
/2*c))/sqrt(-a^2 + b^2)))*b/(sqrt(-a^2 + b^2)*a*d) + 1/2*(d*x^2 + c)/(a*d)