\(\int \frac {1}{3+5 \sin (c+d x)} \, dx\) [34]

Optimal result

Integrand size = 12, antiderivative size = 63 \[ \int \frac {1}{3+5 \sin (c+d x)} \, dx=-\frac {\log \left (3 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}+\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+3 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \]

[Out]

-1/4*ln(3*cos(1/2*d*x+1/2*c)+sin(1/2*d*x+1/2*c))/d+1/4*ln(cos(1/2*d*x+1/2*c)+3*sin(1/2*d*x+1/2*c))/d

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2739, 630, 31} \[ \int \frac {1}{3+5 \sin (c+d x)} \, dx=\frac {\log \left (3 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}-\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+3 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \]

[In]

Int[(3 + 5*Sin[c + d*x])^(-1),x]

[Out]

-1/4*Log[3*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]/d + Log[Cos[(c + d*x)/2] + 3*Sin[(c + d*x)/2]]/(4*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 630

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \text {Subst}\left (\int \frac {1}{3+10 x+3 x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{d} \\ & = \frac {3 \text {Subst}\left (\int \frac {1}{1+3 x} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}-\frac {3 \text {Subst}\left (\int \frac {1}{9+3 x} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \\ & = -\frac {\log \left (3+\tan \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}+\frac {\log \left (1+3 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int \frac {1}{3+5 \sin (c+d x)} \, dx=-\frac {\log \left (3 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}+\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+3 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \]

[In]

Integrate[(3 + 5*Sin[c + d*x])^(-1),x]

[Out]

-1/4*Log[3*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]/d + Log[Cos[(c + d*x)/2] + 3*Sin[(c + d*x)/2]]/(4*d)

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.57

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )}{4}+\frac {\ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4}}{d}\)	\(36\)
default	\(\frac {-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )}{4}+\frac {\ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4}}{d}\)	\(36\)
parallelrisch	\(\frac {-\ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+9\right )+\ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\)	\(37\)
norman	\(-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )}{4 d}+\frac {\ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\)	\(38\)
risch	\(\frac {\ln \left (-\frac {4}{5}+\frac {3 i}{5}+{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {4}{5}+\frac {3 i}{5}\right )}{4 d}\)	\(40\)

[In]

int(1/(3+5*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/4*ln(tan(1/2*d*x+1/2*c)+3)+1/4*ln(3*tan(1/2*d*x+1/2*c)+1))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.73 \[ \int \frac {1}{3+5 \sin (c+d x)} \, dx=-\frac {\log \left (4 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right ) + 5\right ) - \log \left (-4 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right ) + 5\right )}{8 \, d} \]

[In]

integrate(1/(3+5*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/8*(log(4*cos(d*x + c) + 3*sin(d*x + c) + 5) - log(-4*cos(d*x + c) + 3*sin(d*x + c) + 5))/d

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.67 \[ \int \frac {1}{3+5 \sin (c+d x)} \, dx=\begin {cases} - \frac {\log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 \right )}}{4 d} + \frac {\log {\left (3 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \right )}}{4 d} & \text {for}\: d \neq 0 \\\frac {x}{5 \sin {\left (c \right )} + 3} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(3+5*sin(d*x+c)),x)

[Out]

Piecewise((-log(tan(c/2 + d*x/2) + 3)/(4*d) + log(3*tan(c/2 + d*x/2) + 1)/(4*d), Ne(d, 0)), (x/(5*sin(c) + 3),
True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.78 \[ \int \frac {1}{3+5 \sin (c+d x)} \, dx=\frac {\log \left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) - \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 3\right )}{4 \, d} \]

[In]

integrate(1/(3+5*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/4*(log(3*sin(d*x + c)/(cos(d*x + c) + 1) + 1) - log(sin(d*x + c)/(cos(d*x + c) + 1) + 3))/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.57 \[ \int \frac {1}{3+5 \sin (c+d x)} \, dx=\frac {\log \left ({\left | 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \right |}\right )}{4 \, d} \]

[In]

integrate(1/(3+5*sin(d*x+c)),x, algorithm="giac")

[Out]

1/4*(log(abs(3*tan(1/2*d*x + 1/2*c) + 1)) - log(abs(tan(1/2*d*x + 1/2*c) + 3)))/d

Mupad [B] (verification not implemented)

Time = 6.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.30 \[ \int \frac {1}{3+5 \sin (c+d x)} \, dx=-\frac {\mathrm {atanh}\left (\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {5}{4}\right )}{2\,d} \]

[In]

int(1/(5*sin(c + d*x) + 3),x)

[Out]

-atanh((3*tan(c/2 + (d*x)/2))/4 + 5/4)/(2*d)