3.552 \(\int \frac{\cos (c+d x)}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=90 \[ \frac{a \sin (c+d x)}{d \left (a^2+b^2\right )}+\frac{b \cos (c+d x)}{d \left (a^2+b^2\right )}-\frac{b^2 \tanh ^{-1}\left (\frac{\cos (c+d x) (b-a \tan (c+d x))}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}} \]

[Out]

-((b^2*ArcTanh[(Cos[c + d*x]*(b - a*Tan[c + d*x]))/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(3/2)*d)) + (b*Cos[c + d*x])
/((a^2 + b^2)*d) + (a*Sin[c + d*x])/((a^2 + b^2)*d)

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Rubi [A]  time = 0.100533, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3511, 3486, 2637, 3509, 206} \[ \frac{a \sin (c+d x)}{d \left (a^2+b^2\right )}+\frac{b \cos (c+d x)}{d \left (a^2+b^2\right )}-\frac{b^2 \tanh ^{-1}\left (\frac{\cos (c+d x) (b-a \tan (c+d x))}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]/(a + b*Tan[c + d*x]),x]

[Out]

-((b^2*ArcTanh[(Cos[c + d*x]*(b - a*Tan[c + d*x]))/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(3/2)*d)) + (b*Cos[c + d*x])
/((a^2 + b^2)*d) + (a*Sin[c + d*x])/((a^2 + b^2)*d)

Rule 3511

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^
2), Int[(d*Sec[e + f*x])^m*(a - b*Tan[e + f*x]), x], x] + Dist[b^2/(d^2*(a^2 + b^2)), Int[(d*Sec[e + f*x])^(m
+ 2)/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 + b^2, 0] && ILtQ[m, 0]

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3509

Int[sec[(e_.) + (f_.)*(x_)]/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Dist[f^(-1), Subst[Int[1/(a^
2 + b^2 - x^2), x], x, (b - a*Tan[e + f*x])/Sec[e + f*x]], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 + b^2, 0]

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])

Rubi steps

\begin{align*} \int \frac{\cos (c+d x)}{a+b \tan (c+d x)} \, dx &=\frac{\int \cos (c+d x) (a-b \tan (c+d x)) \, dx}{a^2+b^2}+\frac{b^2 \int \frac{\sec (c+d x)}{a+b \tan (c+d x)} \, dx}{a^2+b^2}\\ &=\frac{b \cos (c+d x)}{\left (a^2+b^2\right ) d}+\frac{a \int \cos (c+d x) \, dx}{a^2+b^2}-\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{a^2+b^2-x^2} \, dx,x,\cos (c+d x) (b-a \tan (c+d x))\right )}{\left (a^2+b^2\right ) d}\\ &=-\frac{b^2 \tanh ^{-1}\left (\frac{\cos (c+d x) (b-a \tan (c+d x))}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d}+\frac{b \cos (c+d x)}{\left (a^2+b^2\right ) d}+\frac{a \sin (c+d x)}{\left (a^2+b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 0.305512, size = 79, normalized size = 0.88 \[ \frac{\sqrt{a^2+b^2} (a \sin (c+d x)+b \cos (c+d x))+2 b^2 \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^2*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]] + Sqrt[a^2 + b^2]*(b*Cos[c + d*x] + a*Sin[c + d*x]))
/((a^2 + b^2)^(3/2)*d)

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Maple [A]  time = 0.08, size = 90, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{-a\tan \left ( 1/2\,dx+c/2 \right ) -b}{ \left ({a}^{2}+{b}^{2} \right ) \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}+2\,{\frac{{b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-2/(a^2+b^2)*(-a*tan(1/2*d*x+1/2*c)-b)/(1+tan(1/2*d*x+1/2*c)^2)+2*b^2/(a^2+b^2)^(3/2)*arctanh(1/2*(2*a*ta
n(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.0526, size = 436, normalized size = 4.84 \begin{align*} \frac{\sqrt{a^{2} + b^{2}} b^{2} \log \left (-\frac{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 2 \,{\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right ) + 2 \,{\left (a^{3} + a b^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.52948, size = 159, normalized size = 1.77 \begin{align*} -\frac{\frac{b^{2} \log \left (\frac{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac{3}{2}}} - \frac{2 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b\right )}}{{\left (a^{2} + b^{2}\right )}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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