3.293 \(\int \frac {\tan (x)}{b \cos (x)+a \sin (x)} \, dx\)

Optimal. Leaf size=47 \[ \frac {b \tanh ^{-1}\left (\frac {a \cos (x)-b \sin (x)}{\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2}}+\frac {\tanh ^{-1}(\sin (x))}{a} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3110, 3770, 3074, 206} \[ \frac {b \tanh ^{-1}\left (\frac {a \cos (x)-b \sin (x)}{\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2}}+\frac {\tanh ^{-1}(\sin (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]/(b*Cos[x] + a*Sin[x]),x]

[Out]

ArcTanh[Sin[x]]/a + (b*ArcTanh[(a*Cos[x] - b*Sin[x])/Sqrt[a^2 + b^2]])/(a*Sqrt[a^2 + b^2])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3074

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

Rule 3110

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(cos[c + d*x]^m*sin[c + d*x]^n)/(a*cos[c + d*x] + b*sin[c + d
*x]), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IntegersQ[m, n]

Rule 3770

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

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 76, normalized size = 1.62 \[ \frac {-\frac {2 b \tanh ^{-1}\left (\frac {b \tan \left (\frac {x}{2}\right )-a}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}-\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+\log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]/(b*Cos[x] + a*Sin[x]),x]

[Out]

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

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fricas [B]  time = 0.70, size = 140, normalized size = 2.98 \[ \frac {\sqrt {a^{2} + b^{2}} b \log \left (\frac {2 \, a b \cos \relax (x) \sin \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - a^{2} - 2 \, b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cos \relax (x) - b \sin \relax (x)\right )}}{2 \, a b \cos \relax (x) \sin \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + a^{2}}\right ) + {\left (a^{2} + b^{2}\right )} \log \left (\sin \relax (x) + 1\right ) - {\left (a^{2} + b^{2}\right )} \log \left (-\sin \relax (x) + 1\right )}{2 \, {\left (a^{3} + a b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(b*cos(x)+a*sin(x)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.78, size = 90, normalized size = 1.91 \[ \frac {b \log \left (\frac {{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{a} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(b*cos(x)+a*sin(x)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.11, size = 63, normalized size = 1.34 \[ -\frac {\ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{a}-\frac {2 b \arctanh \left (\frac {2 b \tan \left (\frac {x}{2}\right )-2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a \sqrt {a^{2}+b^{2}}}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(b*cos(x)+a*sin(x)),x)

[Out]

-1/a*ln(tan(1/2*x)-1)-2*b/a/(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*tan(1/2*x)-2*a)/(a^2+b^2)^(1/2))+1/a*ln(tan(1/2*x
)+1)

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maxima [B]  time = 0.45, size = 98, normalized size = 2.09 \[ \frac {b \log \left (\frac {a - \frac {b \sin \relax (x)}{\cos \relax (x) + 1} + \sqrt {a^{2} + b^{2}}}{a - \frac {b \sin \relax (x)}{\cos \relax (x) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a} + \frac {\log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1} - 1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(b*cos(x)+a*sin(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.48, size = 408, normalized size = 8.68 \[ \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a}-\frac {2\,b\,\mathrm {atanh}\left (\frac {64\,b^3}{\sqrt {a^2+b^2}\,\left (128\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}+\frac {64\,a\,b^3}{a^2+b^2}\right )}-\frac {64\,b^5}{{\left (a^2+b^2\right )}^{3/2}\,\left (128\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}+\frac {64\,a\,b^3}{a^2+b^2}\right )}+\frac {128\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}\,\left (\frac {64\,a^2\,b^3}{a^2+b^2}+128\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,a\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}\right )}-\frac {128\,b^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\left (a^2+b^2\right )}^{3/2}\,\left (\frac {64\,a^2\,b^3}{a^2+b^2}+128\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,a\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}\right )}+\frac {128\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}\,\left (128\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}+\frac {64\,a\,b^3}{a^2+b^2}\right )}-\frac {192\,a\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\left (a^2+b^2\right )}^{3/2}\,\left (128\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}+\frac {64\,a\,b^3}{a^2+b^2}\right )}\right )}{a\,\sqrt {a^2+b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(b*cos(x) + a*sin(x)),x)

[Out]

(2*atanh(tan(x/2)))/a - (2*b*atanh((64*b^3)/((a^2 + b^2)^(1/2)*(128*b^2*tan(x/2) - (128*b^4*tan(x/2))/(a^2 + b
^2) + (64*a*b^3)/(a^2 + b^2))) - (64*b^5)/((a^2 + b^2)^(3/2)*(128*b^2*tan(x/2) - (128*b^4*tan(x/2))/(a^2 + b^2
) + (64*a*b^3)/(a^2 + b^2))) + (128*b^4*tan(x/2))/((a^2 + b^2)^(1/2)*((64*a^2*b^3)/(a^2 + b^2) + 128*a*b^2*tan
(x/2) - (128*a*b^4*tan(x/2))/(a^2 + b^2))) - (128*b^6*tan(x/2))/((a^2 + b^2)^(3/2)*((64*a^2*b^3)/(a^2 + b^2) +
128*a*b^2*tan(x/2) - (128*a*b^4*tan(x/2))/(a^2 + b^2))) + (128*a*b^2*tan(x/2))/((a^2 + b^2)^(1/2)*(128*b^2*ta
n(x/2) - (128*b^4*tan(x/2))/(a^2 + b^2) + (64*a*b^3)/(a^2 + b^2))) - (192*a*b^4*tan(x/2))/((a^2 + b^2)^(3/2)*(
128*b^2*tan(x/2) - (128*b^4*tan(x/2))/(a^2 + b^2) + (64*a*b^3)/(a^2 + b^2)))))/(a*(a^2 + b^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan {\relax (x )}}{a \sin {\relax (x )} + b \cos {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(b*cos(x)+a*sin(x)),x)

[Out]

Integral(tan(x)/(a*sin(x) + b*cos(x)), x)

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