3.35 \(\int \frac{1}{(3+5 \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=88 \[ -\frac{5 \cos (c+d x)}{16 d (5 \sin (c+d x)+3)}+\frac{3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+3 \cos \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}-\frac{3 \log \left (3 \sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{64 d} \]

[Out]

(3*Log[3*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(64*d) - (3*Log[Cos[(c + d*x)/2] + 3*Sin[(c + d*x)/2]])/(64*d)
- (5*Cos[c + d*x])/(16*d*(3 + 5*Sin[c + d*x]))

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Rubi [A]  time = 0.0482032, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {2664, 12, 2660, 616, 31} \[ -\frac{5 \cos (c+d x)}{16 d (5 \sin (c+d x)+3)}+\frac{3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+3 \cos \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}-\frac{3 \log \left (3 \sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{64 d} \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*Sin[c + d*x])^(-2),x]

[Out]

(3*Log[3*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(64*d) - (3*Log[Cos[(c + d*x)/2] + 3*Sin[(c + d*x)/2]])/(64*d)
- (5*Cos[c + d*x])/(16*d*(3 + 5*Sin[c + d*x]))

Rule 2664

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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))^2} \, dx &=-\frac{5 \cos (c+d x)}{16 d (3+5 \sin (c+d x))}+\frac{1}{16} \int -\frac{3}{3+5 \sin (c+d x)} \, dx\\ &=-\frac{5 \cos (c+d x)}{16 d (3+5 \sin (c+d x))}-\frac{3}{16} \int \frac{1}{3+5 \sin (c+d x)} \, dx\\ &=-\frac{5 \cos (c+d x)}{16 d (3+5 \sin (c+d x))}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{3+10 x+3 x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}\\ &=-\frac{5 \cos (c+d x)}{16 d (3+5 \sin (c+d x))}-\frac{9 \operatorname{Subst}\left (\int \frac{1}{1+3 x} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}+\frac{9 \operatorname{Subst}\left (\int \frac{1}{9+3 x} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}\\ &=\frac{3 \log \left (3+\tan \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}-\frac{3 \log \left (1+3 \tan \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}-\frac{5 \cos (c+d x)}{16 d (3+5 \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.186162, size = 126, normalized size = 1.43 \[ \frac{20 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\frac{3}{3 \sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{1}{\sin \left (\frac{1}{2} (c+d x)\right )+3 \cos \left (\frac{1}{2} (c+d x)\right )}\right )+9 \left (\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+3 \cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (3 \sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{192 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*Sin[c + d*x])^(-2),x]

[Out]

(9*(Log[3*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + 3*Sin[(c + d*x)/2]]) + 20*Sin[(c + d*x
)/2]*((3*Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^(-1) + 3/(Cos[(c + d*x)/2] + 3*Sin[(c + d*x)/2])))/(192*d)

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Maple [A]  time = 0.035, size = 76, normalized size = 0.9 \begin{align*} -{\frac{5}{48\,d} \left ( 3\,\tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-1}}-{\frac{3}{64\,d}\ln \left ( 3\,\tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }-{\frac{5}{16\,d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +3 \right ) ^{-1}}+{\frac{3}{64\,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))^2,x)

[Out]

-5/48/d/(3*tan(1/2*d*x+1/2*c)+1)-3/64/d*ln(3*tan(1/2*d*x+1/2*c)+1)-5/16/d/(tan(1/2*d*x+1/2*c)+3)+3/64/d*ln(tan
(1/2*d*x+1/2*c)+3)

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Maxima [A]  time = 0.969562, size = 155, normalized size = 1.76 \begin{align*} -\frac{\frac{40 \,{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 3\right )}}{\frac{10 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 3} + 9 \, \log \left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) - 9 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 3\right )}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/192*(40*(5*sin(d*x + c)/(cos(d*x + c) + 1) + 3)/(10*sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^2/(cos
(d*x + c) + 1)^2 + 3) + 9*log(3*sin(d*x + c)/(cos(d*x + c) + 1) + 1) - 9*log(sin(d*x + c)/(cos(d*x + c) + 1) +
3))/d

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Fricas [A]  time = 1.86458, size = 247, normalized size = 2.81 \begin{align*} \frac{3 \,{\left (5 \, \sin \left (d x + c\right ) + 3\right )} \log \left (4 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right ) + 5\right ) - 3 \,{\left (5 \, \sin \left (d x + c\right ) + 3\right )} \log \left (-4 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right ) + 5\right ) - 40 \, \cos \left (d x + c\right )}{128 \,{\left (5 \, d \sin \left (d x + c\right ) + 3 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/128*(3*(5*sin(d*x + c) + 3)*log(4*cos(d*x + c) + 3*sin(d*x + c) + 5) - 3*(5*sin(d*x + c) + 3)*log(-4*cos(d*x
+ c) + 3*sin(d*x + c) + 5) - 40*cos(d*x + c))/(5*d*sin(d*x + c) + 3*d)

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Sympy [A]  time = 2.17932, size = 466, normalized size = 5.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*sin(d*x+c))**2,x)

[Out]

Piecewise((x/(3 - 5*sin(2*atan(1/3)))**2, Eq(c, -d*x - 2*atan(1/3))), (x/(3 - 5*sin(2*atan(3)))**2, Eq(c, -d*x
- 2*atan(3))), (x/(5*sin(c) + 3)**2, Eq(d, 0)), (-27*log(tan(c/2 + d*x/2) + 1/3)*tan(c/2 + d*x/2)**2/(576*d*t
an(c/2 + d*x/2)**2 + 1920*d*tan(c/2 + d*x/2) + 576*d) - 90*log(tan(c/2 + d*x/2) + 1/3)*tan(c/2 + d*x/2)/(576*d
*tan(c/2 + d*x/2)**2 + 1920*d*tan(c/2 + d*x/2) + 576*d) - 27*log(tan(c/2 + d*x/2) + 1/3)/(576*d*tan(c/2 + d*x/
2)**2 + 1920*d*tan(c/2 + d*x/2) + 576*d) + 27*log(tan(c/2 + d*x/2) + 3)*tan(c/2 + d*x/2)**2/(576*d*tan(c/2 + d
*x/2)**2 + 1920*d*tan(c/2 + d*x/2) + 576*d) + 90*log(tan(c/2 + d*x/2) + 3)*tan(c/2 + d*x/2)/(576*d*tan(c/2 + d
*x/2)**2 + 1920*d*tan(c/2 + d*x/2) + 576*d) + 27*log(tan(c/2 + d*x/2) + 3)/(576*d*tan(c/2 + d*x/2)**2 + 1920*d
*tan(c/2 + d*x/2) + 576*d) - 200*tan(c/2 + d*x/2)/(576*d*tan(c/2 + d*x/2)**2 + 1920*d*tan(c/2 + d*x/2) + 576*d
) - 120/(576*d*tan(c/2 + d*x/2)**2 + 1920*d*tan(c/2 + d*x/2) + 576*d), True))

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Giac [A]  time = 1.17021, size = 109, normalized size = 1.24 \begin{align*} -\frac{\frac{40 \,{\left (5 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3\right )}}{3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 10 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3} + 9 \, \log \left ({\left | 3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 9 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \right |}\right )}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/192*(40*(5*tan(1/2*d*x + 1/2*c) + 3)/(3*tan(1/2*d*x + 1/2*c)^2 + 10*tan(1/2*d*x + 1/2*c) + 3) + 9*log(abs(3
*tan(1/2*d*x + 1/2*c) + 1)) - 9*log(abs(tan(1/2*d*x + 1/2*c) + 3)))/d