\(\int \frac {\sin (c+d x)}{a+b \tan (c+d x)} \, dx\) [55]

Optimal result

Integrand size = 19, antiderivative size = 90 \[ \int \frac {\sin (c+d x)}{a+b \tan (c+d x)} \, dx=\frac {a b \text {arctanh}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d}-\frac {a \cos (c+d x)}{\left (a^2+b^2\right ) d}+\frac {b \sin (c+d x)}{\left (a^2+b^2\right ) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3599, 3188, 2717, 2718, 3153, 212} \[ \int \frac {\sin (c+d x)}{a+b \tan (c+d x)} \, dx=\frac {a b \text {arctanh}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}}+\frac {b \sin (c+d x)}{d \left (a^2+b^2\right )}-\frac {a \cos (c+d x)}{d \left (a^2+b^2\right )} \]

[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 212

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

Rule 2717

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

Rule 2718

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

Rule 3153

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 3188

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 3599

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]))

Rubi steps \begin{align*} \text {integral}& = \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) \text {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 \text {arctanh}\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] (verified)

Time = 0.46 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {\sin (c+d x)}{a+b \tan (c+d x)} \, dx=\frac {-2 a b \text {arctanh}\left (\frac {-b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )+\sqrt {a^2+b^2} (-a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^{3/2} d} \]

[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)

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(\frac {\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 a}{\left (a^{2}+b^{2}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {4 a b \,\operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (2 a^{2}+2 b^{2}\right ) \sqrt {a^{2}+b^{2}}}}{d}\)	\(101\)
default	\(\frac {\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 a}{\left (a^{2}+b^{2}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {4 a b \,\operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (2 a^{2}+2 b^{2}\right ) \sqrt {a^{2}+b^{2}}}}{d}\)	\(101\)
risch	\(-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 \left (-i b +a \right ) d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 \left (i b +a \right ) d}+\frac {i b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i b +a}{\sqrt {-a^{2}-b^{2}}}\right )}{\sqrt {-a^{2}-b^{2}}\, \left (a^{2}+b^{2}\right ) d}-\frac {i b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i b +a}{\sqrt {-a^{2}-b^{2}}}\right )}{\sqrt {-a^{2}-b^{2}}\, \left (a^{2}+b^{2}\right ) d}\)	\(169\)

[In]

int(sin(d*x+c)/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (86) = 172\).

Time = 0.28 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.06 \[ \int \frac {\sin (c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\sqrt {a^{2} + b^{2}} a b \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) - 2 \, {\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right ) + 2 \, {\left (a^{2} b + b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} \]

[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)

Sympy [F]

\[ \int \frac {\sin (c+d x)}{a+b \tan (c+d x)} \, dx=\int \frac {\sin {\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.57 \[ \int \frac {\sin (c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {a b \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (a - \frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}}{a^{2} + b^{2} + \frac {{\left (a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.31 \[ \int \frac {\sin (c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {a b \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}{{\left (a^{2} + b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}}}{d} \]

[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

Mupad [B] (verification not implemented)

Time = 4.45 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.22 \[ \int \frac {\sin (c+d x)}{a+b \tan (c+d x)} \, dx=\frac {2\,a\,b\,\mathrm {atanh}\left (\frac {a^2\,b+b^3-a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+b^2\right )}{{\left (a^2+b^2\right )}^{3/2}}\right )}{d\,{\left (a^2+b^2\right )}^{3/2}}-\frac {\frac {2\,a}{a^2+b^2}-\frac {2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2+b^2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

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

[Out]

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