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

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

[Out]

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

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Rubi [A]  time = 0.105716, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {3518, 3109, 2637, 2638, 3074, 206} \[ \frac{b \sin (c+d x)}{d \left (a^2+b^2\right )}-\frac{a \cos (c+d x)}{d \left (a^2+b^2\right )}+\frac{a b \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3109

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

Rule 2637

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

Rule 2638

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

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

Mathematica [A]  time = 0.336073, size = 79, normalized size = 0.88 \[ \frac{\sqrt{a^2+b^2} (b \sin (c+d x)-a \cos (c+d x))-2 a b \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.05, size = 100, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{-\tan \left ( 1/2\,dx+c/2 \right ) b+a}{ \left ({a}^{2}+{b}^{2} \right ) \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-4\,{\frac{ab}{ \left ( 2\,{a}^{2}+2\,{b}^{2} \right ) \sqrt{{a}^{2}+{b}^{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(sin(d*x+c)/(a+b*tan(d*x+c)),x)

[Out]

1/d*(-2/(a^2+b^2)*(-tan(1/2*d*x+1/2*c)*b+a)/(1+tan(1/2*d*x+1/2*c)^2)-4*a*b/(2*a^2+2*b^2)/(a^2+b^2)^(1/2)*arcta
nh(1/2*(2*a*tan(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(sin(d*x+c)/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.0465, size = 435, normalized size = 4.83 \begin{align*} \frac{\sqrt{a^{2} + b^{2}} a b \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^{3} + a b^{2}\right )} \cos \left (d x + c\right ) + 2 \,{\left (a^{2} b + b^{3}\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(sin(d*x+c)/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(sqrt(a^2 + b^2)*a*b*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*s
qrt(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^3 + a*b^2)*cos(d*x + c) + 2*(a^2*b + b^3)*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{\sin{\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(sin(d*x+c)/(a+b*tan(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.33428, size = 159, normalized size = 1.77 \begin{align*} \frac{\frac{a b \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 (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\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(sin(d*x+c)/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

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