3.157 \(\int \frac{x \sec ^2(x)}{(a+b \tan (x))^2} \, dx\)

Optimal. Leaf size=50 \[ \frac{a x}{b \left (a^2+b^2\right )}+\frac{\log (a \cos (x)+b \sin (x))}{a^2+b^2}-\frac{x}{b (a+b \tan (x))} \]

[Out]

(a*x)/(b*(a^2 + b^2)) + Log[a*Cos[x] + b*Sin[x]]/(a^2 + b^2) - x/(b*(a + b*Tan[x]))

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Rubi [A]  time = 0.0826108, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4424, 3484, 3530} \[ \frac{a x}{b \left (a^2+b^2\right )}+\frac{\log (a \cos (x)+b \sin (x))}{a^2+b^2}-\frac{x}{b (a+b \tan (x))} \]

Antiderivative was successfully verified.

[In]

Int[(x*Sec[x]^2)/(a + b*Tan[x])^2,x]

[Out]

(a*x)/(b*(a^2 + b^2)) + Log[a*Cos[x] + b*Sin[x]]/(a^2 + b^2) - x/(b*(a + b*Tan[x]))

Rule 4424

Int[((e_.) + (f_.)*(x_))^(m_.)*Sec[(c_.) + (d_.)*(x_)]^2*((a_) + (b_.)*Tan[(c_.) + (d_.)*(x_)])^(n_.), x_Symbo
l] :> Simp[((e + f*x)^m*(a + b*Tan[c + d*x])^(n + 1))/(b*d*(n + 1)), x] - Dist[(f*m)/(b*d*(n + 1)), Int[(e + f
*x)^(m - 1)*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[m, 0] && NeQ[n, -1
]

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

Mathematica [A]  time = 0.176216, size = 48, normalized size = 0.96 \[ \frac{a \log (a \cos (x)+b \sin (x))-b x}{a^3+a b^2}+\frac{x \sin (x)}{a^2 \cos (x)+a b \sin (x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Sec[x]^2)/(a + b*Tan[x])^2,x]

[Out]

(-(b*x) + a*Log[a*Cos[x] + b*Sin[x]])/(a^3 + a*b^2) + (x*Sin[x])/(a^2*Cos[x] + a*b*Sin[x])

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Maple [C]  time = 0.164, size = 86, normalized size = 1.7 \begin{align*}{\frac{-2\,ix}{{a}^{2}+{b}^{2}}}+{\frac{2\,ix}{ \left ( -ib{{\rm e}^{2\,ix}}+a{{\rm e}^{2\,ix}}+ib+a \right ) \left ( -ib+a \right ) }}+{\frac{1}{{a}^{2}+{b}^{2}}\ln \left ({{\rm e}^{2\,ix}}-{\frac{ib+a}{ib-a}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I/(a^2+b^2)*x+2*I*x/(-I*b*exp(2*I*x)+a*exp(2*I*x)+I*b+a)/(-I*b+a)+1/(a^2+b^2)*ln(exp(2*I*x)-(I*b+a)/(I*b-a)
)

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Maxima [B]  time = 1.039, size = 338, normalized size = 6.76 \begin{align*} -\frac{8 \, a b x \cos \left (2 \, x\right ) - 4 \,{\left (a^{2} - b^{2}\right )} x \sin \left (2 \, x\right ) -{\left ({\left (a^{2} + b^{2}\right )} \cos \left (2 \, x\right )^{2} + 4 \, a b \sin \left (2 \, x\right ) +{\left (a^{2} + b^{2}\right )} \sin \left (2 \, x\right )^{2} + a^{2} + b^{2} + 2 \,{\left (a^{2} - b^{2}\right )} \cos \left (2 \, x\right )\right )} \log \left (\frac{{\left (a^{2} + b^{2}\right )} \cos \left (2 \, x\right )^{2} + 4 \, a b \sin \left (2 \, x\right ) +{\left (a^{2} + b^{2}\right )} \sin \left (2 \, x\right )^{2} + a^{2} + b^{2} + 2 \,{\left (a^{2} - b^{2}\right )} \cos \left (2 \, x\right )}{a^{2} + b^{2}}\right )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (2 \, x\right )^{2} +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (2 \, x\right )^{2} + 2 \,{\left (a^{4} - b^{4}\right )} \cos \left (2 \, x\right ) + 4 \,{\left (a^{3} b + a b^{3}\right )} \sin \left (2 \, x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(8*a*b*x*cos(2*x) - 4*(a^2 - b^2)*x*sin(2*x) - ((a^2 + b^2)*cos(2*x)^2 + 4*a*b*sin(2*x) + (a^2 + b^2)*sin
(2*x)^2 + a^2 + b^2 + 2*(a^2 - b^2)*cos(2*x))*log(((a^2 + b^2)*cos(2*x)^2 + 4*a*b*sin(2*x) + (a^2 + b^2)*sin(2
*x)^2 + a^2 + b^2 + 2*(a^2 - b^2)*cos(2*x))/(a^2 + b^2)))/(a^4 + 2*a^2*b^2 + b^4 + (a^4 + 2*a^2*b^2 + b^4)*cos
(2*x)^2 + (a^4 + 2*a^2*b^2 + b^4)*sin(2*x)^2 + 2*(a^4 - b^4)*cos(2*x) + 4*(a^3*b + a*b^3)*sin(2*x))

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Fricas [A]  time = 2.5764, size = 216, normalized size = 4.32 \begin{align*} -\frac{2 \, b x \cos \left (x\right ) - 2 \, a x \sin \left (x\right ) -{\left (a \cos \left (x\right ) + b \sin \left (x\right )\right )} \log \left (2 \, a b \cos \left (x\right ) \sin \left (x\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2}\right )}{2 \,{\left ({\left (a^{3} + a b^{2}\right )} \cos \left (x\right ) +{\left (a^{2} b + b^{3}\right )} \sin \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*x*cos(x) - 2*a*x*sin(x) - (a*cos(x) + b*sin(x))*log(2*a*b*cos(x)*sin(x) + (a^2 - b^2)*cos(x)^2 + b^2
))/((a^3 + a*b^2)*cos(x) + (a^2*b + b^3)*sin(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.362, size = 435, normalized size = 8.7 \begin{align*} -\frac{2 \, b x \tan \left (\frac{1}{2} \, x\right )^{2} - a \log \left (\frac{4 \,{\left (a^{2} \tan \left (\frac{1}{2} \, x\right )^{4} - 4 \, a b \tan \left (\frac{1}{2} \, x\right )^{3} - 2 \, a^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 4 \, b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 4 \, a b \tan \left (\frac{1}{2} \, x\right ) + a^{2}\right )}}{\tan \left (\frac{1}{2} \, x\right )^{4} + 2 \, \tan \left (\frac{1}{2} \, x\right )^{2} + 1}\right ) \tan \left (\frac{1}{2} \, x\right )^{2} + 4 \, a x \tan \left (\frac{1}{2} \, x\right ) + 2 \, b \log \left (\frac{4 \,{\left (a^{2} \tan \left (\frac{1}{2} \, x\right )^{4} - 4 \, a b \tan \left (\frac{1}{2} \, x\right )^{3} - 2 \, a^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 4 \, b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 4 \, a b \tan \left (\frac{1}{2} \, x\right ) + a^{2}\right )}}{\tan \left (\frac{1}{2} \, x\right )^{4} + 2 \, \tan \left (\frac{1}{2} \, x\right )^{2} + 1}\right ) \tan \left (\frac{1}{2} \, x\right ) - 2 \, b x + a \log \left (\frac{4 \,{\left (a^{2} \tan \left (\frac{1}{2} \, x\right )^{4} - 4 \, a b \tan \left (\frac{1}{2} \, x\right )^{3} - 2 \, a^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 4 \, b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 4 \, a b \tan \left (\frac{1}{2} \, x\right ) + a^{2}\right )}}{\tan \left (\frac{1}{2} \, x\right )^{4} + 2 \, \tan \left (\frac{1}{2} \, x\right )^{2} + 1}\right )}{2 \,{\left (a^{3} \tan \left (\frac{1}{2} \, x\right )^{2} + a b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} - 2 \, a^{2} b \tan \left (\frac{1}{2} \, x\right ) - 2 \, b^{3} \tan \left (\frac{1}{2} \, x\right ) - a^{3} - a b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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