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

Optimal. Leaf size=90 \[ \frac {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}+\frac {3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}-\frac {3 \log \left (3 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 90, normalized size of antiderivative = 1.00, 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} \[ \frac {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}+\frac {3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}-\frac {3 \log \left (3 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d} \]

Antiderivative was successfully verified.

[In]

Int[(-3 + 5*Sin[c + d*x])^(-2),x]

[Out]

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

Rule 12

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

Rule 31

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

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 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 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]

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 (1-3 \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 {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 130, normalized size = 1.44 \[ \frac {20 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\frac {1}{3 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {3}{\cos \left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )}\right )+9 \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (3 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{192 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 88, normalized size = 0.98 \[ -\frac {3 \, {\left (5 \, \sin \left (d x + c\right ) - 3\right )} \log \left (4 \, \cos \left (d x + c\right ) - 3 \, \sin \left (d x + c\right ) + 5\right ) - 3 \, {\left (5 \, \sin \left (d x + c\right ) - 3\right )} \log \left (-4 \, \cos \left (d x + c\right ) - 3 \, \sin \left (d x + c\right ) + 5\right ) + 40 \, \cos \left (d x + c\right )}{128 \, {\left (5 \, d \sin \left (d x + c\right ) - 3 \, d\right )}} \]

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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giac [A]  time = 0.43, size = 81, normalized size = 0.90 \[ -\frac {\frac {40 \, {\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3\right )}}{3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3} - 9 \, \log \left ({\left | 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 9 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \right |}\right )}{192 \, d} \]

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

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maple [A]  time = 0.09, size = 76, normalized size = 0.84 \[ -\frac {5}{48 d \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64 d}-\frac {5}{16 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}{64 d} \]

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 = 0.37, size = 115, normalized size = 1.28 \[ \frac {\frac {40 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 3\right )}}{\frac {10 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 3} + 9 \, \log \left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right ) - 9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 3\right )}{192 \, d} \]

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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mupad [B]  time = 0.00, size = 64, normalized size = 0.71 \[ \frac {3\,\mathrm {atanh}\left (\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {5}{4}\right )}{32\,d}-\frac {\frac {25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{72}-\frac {5}{24}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {10\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*sin(c + d*x) - 3)^2,x)

[Out]

(3*atanh((3*tan(c/2 + (d*x)/2))/4 - 5/4))/(32*d) - ((25*tan(c/2 + (d*x)/2))/72 - 5/24)/(d*(tan(c/2 + (d*x)/2)^
2 - (10*tan(c/2 + (d*x)/2))/3 + 1))

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sympy [A]  time = 1.62, size = 462, normalized size = 5.13 \[ \begin {cases} \frac {x}{\left (-3 + 5 \sin {\left (2 \operatorname {atan}{\left (\frac {1}{3} \right )} \right )}\right )^{2}} & \text {for}\: c = - d x + 2 \operatorname {atan}{\left (\frac {1}{3} \right )} \\\frac {x}{\left (-3 + 5 \sin {\left (2 \operatorname {atan}{\relax (3 )} \right )}\right )^{2}} & \text {for}\: c = - d x + 2 \operatorname {atan}{\relax (3 )} \\\frac {x}{\left (5 \sin {\relax (c )} - 3\right )^{2}} & \text {for}\: d = 0 \\- \frac {27 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 \right )} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} + \frac {90 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 \right )} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} - \frac {27 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} + \frac {27 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - \frac {1}{3} \right )} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} - \frac {90 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - \frac {1}{3} \right )} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} + \frac {27 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - \frac {1}{3} \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} - \frac {200 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} + \frac {120}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} & \text {otherwise} \end {cases} \]

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) - 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) - 3)*tan(c/2 + d*x/2)/(576*d*t
an(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) + 27*log(tan(c/2 + d*x/2) - 1/3)*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) - 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) - 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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