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

Optimal. Leaf size=57 \[ \frac{b \log \left (a \cos \left (c+d x^2\right )+b \sin \left (c+d x^2\right )\right )}{2 d \left (a^2+b^2\right )}+\frac{a x^2}{2 \left (a^2+b^2\right )} \]

[Out]

(a*x^2)/(2*(a^2 + b^2)) + (b*Log[a*Cos[c + d*x^2] + b*Sin[c + d*x^2]])/(2*(a^2 + b^2)*d)

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Rubi [A]  time = 0.0779404, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3747, 3484, 3530} \[ \frac{b \log \left (a \cos \left (c+d x^2\right )+b \sin \left (c+d x^2\right )\right )}{2 d \left (a^2+b^2\right )}+\frac{a x^2}{2 \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Tan[c + d*x^2]),x]

[Out]

(a*x^2)/(2*(a^2 + b^2)) + (b*Log[a*Cos[c + d*x^2] + b*Sin[c + d*x^2]])/(2*(a^2 + b^2)*d)

Rule 3747

Int[(x_)^(m_.)*((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Tan[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 3484

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

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rubi steps

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

Mathematica [C]  time = 0.135128, size = 82, normalized size = 1.44 \[ \frac{(-b-i a) \log \left (-\tan \left (c+d x^2\right )+i\right )+i (a+i b) \log \left (\tan \left (c+d x^2\right )+i\right )+2 b \log \left (a+b \tan \left (c+d x^2\right )\right )}{4 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*Tan[c + d*x^2]),x]

[Out]

(((-I)*a - b)*Log[I - Tan[c + d*x^2]] + I*(a + I*b)*Log[I + Tan[c + d*x^2]] + 2*b*Log[a + b*Tan[c + d*x^2]])/(
4*(a^2 + b^2)*d)

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Maple [A]  time = 0.028, size = 82, normalized size = 1.4 \begin{align*} -{\frac{b\ln \left ( 1+ \left ( \tan \left ( d{x}^{2}+c \right ) \right ) ^{2} \right ) }{4\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{a\arctan \left ( \tan \left ( d{x}^{2}+c \right ) \right ) }{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{b\ln \left ( a+b\tan \left ( d{x}^{2}+c \right ) \right ) }{2\,d \left ({a}^{2}+{b}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d/(a^2+b^2)*b*ln(1+tan(d*x^2+c)^2)+1/2/d/(a^2+b^2)*a*arctan(tan(d*x^2+c))+1/2/d*b/(a^2+b^2)*ln(a+b*tan(d*
x^2+c))

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Maxima [B]  time = 1.21489, size = 193, normalized size = 3.39 \begin{align*} \frac{2 \, a d x^{2} + b \log \left (\frac{{\left (a^{2} + b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + 4 \, a b \sin \left (2 \, d x^{2} + 2 \, c\right ) +{\left (a^{2} + b^{2}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + a^{2} + b^{2} + 2 \,{\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )}{{\left (a^{2} + b^{2}\right )} \cos \left (2 \, c\right )^{2} +{\left (a^{2} + b^{2}\right )} \sin \left (2 \, c\right )^{2}}\right )}{4 \,{\left (a^{2} + b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*d*x^2 + b*log(((a^2 + b^2)*cos(2*d*x^2 + 2*c)^2 + 4*a*b*sin(2*d*x^2 + 2*c) + (a^2 + b^2)*sin(2*d*x^2
+ 2*c)^2 + a^2 + b^2 + 2*(a^2 - b^2)*cos(2*d*x^2 + 2*c))/((a^2 + b^2)*cos(2*c)^2 + (a^2 + b^2)*sin(2*c)^2)))/(
(a^2 + b^2)*d)

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Fricas [A]  time = 1.60962, size = 158, normalized size = 2.77 \begin{align*} \frac{2 \, a d x^{2} + b \log \left (\frac{b^{2} \tan \left (d x^{2} + c\right )^{2} + 2 \, a b \tan \left (d x^{2} + c\right ) + a^{2}}{\tan \left (d x^{2} + c\right )^{2} + 1}\right )}{4 \,{\left (a^{2} + b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*d*x^2 + b*log((b^2*tan(d*x^2 + c)^2 + 2*a*b*tan(d*x^2 + c) + a^2)/(tan(d*x^2 + c)^2 + 1)))/((a^2 + b^
2)*d)

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Sympy [A]  time = 3.03411, size = 360, normalized size = 6.32 \begin{align*} \begin{cases} \frac{\tilde{\infty } x^{2}}{\tan{\left (c \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac{x^{2}}{2 a} & \text{for}\: b = 0 \\- \frac{i \left (\operatorname{atan}{\left (\tan{\left (c + d x^{2} \right )} \right )} + \pi \left \lfloor{\frac{c + d x^{2} - \frac{\pi }{2}}{\pi }}\right \rfloor \right ) \tan{\left (c + d x^{2} \right )}}{- 4 b d \tan{\left (c + d x^{2} \right )} + 4 i b d} - \frac{\operatorname{atan}{\left (\tan{\left (c + d x^{2} \right )} \right )} + \pi \left \lfloor{\frac{c + d x^{2} - \frac{\pi }{2}}{\pi }}\right \rfloor }{- 4 b d \tan{\left (c + d x^{2} \right )} + 4 i b d} - \frac{i}{- 4 b d \tan{\left (c + d x^{2} \right )} + 4 i b d} & \text{for}\: a = - i b \\- \frac{i \left (\operatorname{atan}{\left (\tan{\left (c + d x^{2} \right )} \right )} + \pi \left \lfloor{\frac{c + d x^{2} - \frac{\pi }{2}}{\pi }}\right \rfloor \right ) \tan{\left (c + d x^{2} \right )}}{4 b d \tan{\left (c + d x^{2} \right )} + 4 i b d} + \frac{\operatorname{atan}{\left (\tan{\left (c + d x^{2} \right )} \right )} + \pi \left \lfloor{\frac{c + d x^{2} - \frac{\pi }{2}}{\pi }}\right \rfloor }{4 b d \tan{\left (c + d x^{2} \right )} + 4 i b d} - \frac{i}{4 b d \tan{\left (c + d x^{2} \right )} + 4 i b d} & \text{for}\: a = i b \\\frac{x^{2}}{2 \left (a + b \tan{\left (c \right )}\right )} & \text{for}\: d = 0 \\\frac{2 a d x^{2}}{4 a^{2} d + 4 b^{2} d} + \frac{2 b \log{\left (\frac{a}{b} + \tan{\left (c + d x^{2} \right )} \right )}}{4 a^{2} d + 4 b^{2} d} - \frac{b \log{\left (\tan ^{2}{\left (c + d x^{2} \right )} + 1 \right )}}{4 a^{2} d + 4 b^{2} d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x**2/tan(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (x**2/(2*a), Eq(b, 0)), (-I*(atan(tan(c + d*x**2)
) + pi*floor((c + d*x**2 - pi/2)/pi))*tan(c + d*x**2)/(-4*b*d*tan(c + d*x**2) + 4*I*b*d) - (atan(tan(c + d*x**
2)) + pi*floor((c + d*x**2 - pi/2)/pi))/(-4*b*d*tan(c + d*x**2) + 4*I*b*d) - I/(-4*b*d*tan(c + d*x**2) + 4*I*b
*d), Eq(a, -I*b)), (-I*(atan(tan(c + d*x**2)) + pi*floor((c + d*x**2 - pi/2)/pi))*tan(c + d*x**2)/(4*b*d*tan(c
+ d*x**2) + 4*I*b*d) + (atan(tan(c + d*x**2)) + pi*floor((c + d*x**2 - pi/2)/pi))/(4*b*d*tan(c + d*x**2) + 4*
I*b*d) - I/(4*b*d*tan(c + d*x**2) + 4*I*b*d), Eq(a, I*b)), (x**2/(2*(a + b*tan(c))), Eq(d, 0)), (2*a*d*x**2/(4
*a**2*d + 4*b**2*d) + 2*b*log(a/b + tan(c + d*x**2))/(4*a**2*d + 4*b**2*d) - b*log(tan(c + d*x**2)**2 + 1)/(4*
a**2*d + 4*b**2*d), True))

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Giac [A]  time = 1.18423, size = 116, normalized size = 2.04 \begin{align*} \frac{b^{2} \log \left ({\left | b \tan \left (d x^{2} + c\right ) + a \right |}\right )}{2 \,{\left (a^{2} b d + b^{3} d\right )}} + \frac{{\left (d x^{2} + c\right )} a}{2 \,{\left (a^{2} d + b^{2} d\right )}} - \frac{b \log \left (\tan \left (d x^{2} + c\right )^{2} + 1\right )}{4 \,{\left (a^{2} d + b^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^2*log(abs(b*tan(d*x^2 + c) + a))/(a^2*b*d + b^3*d) + 1/2*(d*x^2 + c)*a/(a^2*d + b^2*d) - 1/4*b*log(tan(d
*x^2 + c)^2 + 1)/(a^2*d + b^2*d)