### 3.38 $$\int \frac{1}{3-5 \sin (c+d x)} \, dx$$

Optimal. Leaf size=65 $\frac{\log \left (3 \cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{4 d}-\frac{\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-3 \sin \left (\frac{1}{2} (c+d x)\right )\right )}{4 d}$

[Out]

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

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Rubi [A]  time = 0.0229075, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {2660, 616, 31} $\frac{\log \left (3 \cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{4 d}-\frac{\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-3 \sin \left (\frac{1}{2} (c+d x)\right )\right )}{4 d}$

Antiderivative was successfully veriﬁed.

[In]

Int[(3 - 5*Sin[c + d*x])^(-1),x]

[Out]

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

Rule 2660

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]

Rule 616

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 31

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

Rubi steps

\begin{align*} \int \frac{1}{3-5 \sin (c+d x)} \, dx &=\frac{2 \operatorname{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 \operatorname{Subst}\left (\int \frac{1}{-9+3 x} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{4 d}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{-1+3 x} \, dx,x,\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}+\frac{\log \left (3-\tan \left (\frac{1}{2} (c+d x)\right )\right )}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.0229769, size = 65, normalized size = 1. $\frac{\log \left (3 \cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{4 d}-\frac{\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-3 \sin \left (\frac{1}{2} (c+d x)\right )\right )}{4 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.017, size = 38, normalized size = 0.6 \begin{align*} -{\frac{1}{4\,d}\ln \left ( 3\,\tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }+{\frac{1}{4\,d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -3 \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.955084, size = 66, normalized size = 1.02 \begin{align*} -\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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.70096, size = 127, normalized size = 1.95 \begin{align*} \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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 0.760065, size = 42, normalized size = 0.65 \begin{align*} \begin{cases} \frac{\log{\left (\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 3 \right )}}{4 d} - \frac{\log{\left (\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} - \frac{1}{3} \right )}}{4 d} & \text{for}\: d \neq 0 \\\frac{x}{3 - 5 \sin{\left (c \right )}} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18952, size = 49, normalized size = 0.75 \begin{align*} -\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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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