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

Optimal. Leaf size=113 \[ \frac {45 \cos (c+d x)}{512 d (5 \sin (c+d x)+3)}-\frac {5 \cos (c+d x)}{32 d (5 \sin (c+d x)+3)^2}-\frac {43 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+3 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}+\frac {43 \log \left (3 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d} \]

[Out]

-43/2048*ln(3*cos(1/2*d*x+1/2*c)+sin(1/2*d*x+1/2*c))/d+43/2048*ln(cos(1/2*d*x+1/2*c)+3*sin(1/2*d*x+1/2*c))/d-5
/32*cos(d*x+c)/d/(3+5*sin(d*x+c))^2+45/512*cos(d*x+c)/d/(3+5*sin(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2664, 2754, 12, 2660, 616, 31} \[ \frac {45 \cos (c+d x)}{512 d (5 \sin (c+d x)+3)}-\frac {5 \cos (c+d x)}{32 d (5 \sin (c+d x)+3)^2}-\frac {43 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+3 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}+\frac {43 \log \left (3 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d} \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*Sin[c + d*x])^(-3),x]

[Out]

(-43*Log[3*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(2048*d) + (43*Log[Cos[(c + d*x)/2] + 3*Sin[(c + d*x)/2]])/(2
048*d) - (5*Cos[c + d*x])/(32*d*(3 + 5*Sin[c + d*x])^2) + (45*Cos[c + d*x])/(512*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]

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {1}{(3+5 \sin (c+d x))^3} \, dx &=-\frac {5 \cos (c+d x)}{32 d (3+5 \sin (c+d x))^2}+\frac {1}{32} \int \frac {-6+5 \sin (c+d x)}{(3+5 \sin (c+d x))^2} \, dx\\ &=-\frac {5 \cos (c+d x)}{32 d (3+5 \sin (c+d x))^2}+\frac {45 \cos (c+d x)}{512 d (3+5 \sin (c+d x))}+\frac {1}{512} \int \frac {43}{3+5 \sin (c+d x)} \, dx\\ &=-\frac {5 \cos (c+d x)}{32 d (3+5 \sin (c+d x))^2}+\frac {45 \cos (c+d x)}{512 d (3+5 \sin (c+d x))}+\frac {43}{512} \int \frac {1}{3+5 \sin (c+d x)} \, dx\\ &=-\frac {5 \cos (c+d x)}{32 d (3+5 \sin (c+d x))^2}+\frac {45 \cos (c+d x)}{512 d (3+5 \sin (c+d x))}+\frac {43 \operatorname {Subst}\left (\int \frac {1}{3+10 x+3 x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{256 d}\\ &=-\frac {5 \cos (c+d x)}{32 d (3+5 \sin (c+d x))^2}+\frac {45 \cos (c+d x)}{512 d (3+5 \sin (c+d x))}+\frac {129 \operatorname {Subst}\left (\int \frac {1}{1+3 x} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}-\frac {129 \operatorname {Subst}\left (\int \frac {1}{9+3 x} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}\\ &=-\frac {43 \log \left (3+\tan \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}+\frac {43 \log \left (1+3 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}-\frac {5 \cos (c+d x)}{32 d (3+5 \sin (c+d x))^2}+\frac {45 \cos (c+d x)}{512 d (3+5 \sin (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.50, size = 180, normalized size = 1.59 \[ \frac {\sin \left (\frac {1}{2} (c+d x)\right ) \left (-\frac {180}{3 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}-\frac {60}{\sin \left (\frac {1}{2} (c+d x)\right )+3 \cos \left (\frac {1}{2} (c+d x)\right )}\right )+\frac {40}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+3 \cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {40}{\left (3 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-43 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+3 \cos \left (\frac {1}{2} (c+d x)\right )\right )+43 \log \left (3 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-43*Log[3*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 43*Log[Cos[(c + d*x)/2] + 3*Sin[(c + d*x)/2]] + 40/(3*Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2])^2 - 40/(Cos[(c + d*x)/2] + 3*Sin[(c + d*x)/2])^2 + Sin[(c + d*x)/2]*(-60/(3*Cos
[(c + d*x)/2] + Sin[(c + d*x)/2]) - 180/(Cos[(c + d*x)/2] + 3*Sin[(c + d*x)/2])))/(2048*d)

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 133, normalized size = 1.18 \[ -\frac {43 \, {\left (25 \, \cos \left (d x + c\right )^{2} - 30 \, \sin \left (d x + c\right ) - 34\right )} \log \left (4 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right ) + 5\right ) - 43 \, {\left (25 \, \cos \left (d x + c\right )^{2} - 30 \, \sin \left (d x + c\right ) - 34\right )} \log \left (-4 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right ) + 5\right ) + 1800 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 440 \, \cos \left (d x + c\right )}{4096 \, {\left (25 \, d \cos \left (d x + c\right )^{2} - 30 \, d \sin \left (d x + c\right ) - 34 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4096*(43*(25*cos(d*x + c)^2 - 30*sin(d*x + c) - 34)*log(4*cos(d*x + c) + 3*sin(d*x + c) + 5) - 43*(25*cos(d
*x + c)^2 - 30*sin(d*x + c) - 34)*log(-4*cos(d*x + c) + 3*sin(d*x + c) + 5) + 1800*cos(d*x + c)*sin(d*x + c) +
440*cos(d*x + c))/(25*d*cos(d*x + c)^2 - 30*d*sin(d*x + c) - 34*d)

________________________________________________________________________________________

giac [A]  time = 0.55, size = 107, normalized size = 0.95 \[ -\frac {\frac {40 \, {\left (75 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 649 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 735 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 99\right )}}{{\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3\right )}^{2}} - 387 \, \log \left ({\left | 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 387 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \right |}\right )}{18432 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18432*(40*(75*tan(1/2*d*x + 1/2*c)^3 - 649*tan(1/2*d*x + 1/2*c)^2 - 735*tan(1/2*d*x + 1/2*c) - 99)/(3*tan(1
/2*d*x + 1/2*c)^2 + 10*tan(1/2*d*x + 1/2*c) + 3)^2 - 387*log(abs(3*tan(1/2*d*x + 1/2*c) + 1)) + 387*log(abs(ta
n(1/2*d*x + 1/2*c) + 3)))/d

________________________________________________________________________________________

maple [A]  time = 0.11, size = 114, normalized size = 1.01 \[ \frac {25}{128 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )^{2}}-\frac {15}{512 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )}-\frac {43 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )}{2048 d}-\frac {25}{1152 d \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {155}{4608 d \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {43 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2048 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/128/d/(tan(1/2*d*x+1/2*c)+3)^2-15/512/d/(tan(1/2*d*x+1/2*c)+3)-43/2048/d*ln(tan(1/2*d*x+1/2*c)+3)-25/1152/d
/(3*tan(1/2*d*x+1/2*c)+1)^2+155/4608/d/(3*tan(1/2*d*x+1/2*c)+1)+43/2048/d*ln(3*tan(1/2*d*x+1/2*c)+1)

________________________________________________________________________________________

maxima [A]  time = 0.31, size = 195, normalized size = 1.73 \[ \frac {\frac {40 \, {\left (\frac {735 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {649 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {75 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 99\right )}}{\frac {60 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {118 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {60 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {9 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 9} + 387 \, \log \left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) - 387 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 3\right )}{18432 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18432*(40*(735*sin(d*x + c)/(cos(d*x + c) + 1) + 649*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 75*sin(d*x + c)^3
/(cos(d*x + c) + 1)^3 + 99)/(60*sin(d*x + c)/(cos(d*x + c) + 1) + 118*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 60
*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 9*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 9) + 387*log(3*sin(d*x + c)/(co
s(d*x + c) + 1) + 1) - 387*log(sin(d*x + c)/(cos(d*x + c) + 1) + 3))/d

________________________________________________________________________________________

mupad [B]  time = 5.73, size = 115, normalized size = 1.02 \[ \frac {-\frac {125\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6912}+\frac {3245\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{20736}+\frac {1225\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{6912}+\frac {55}{2304}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {118\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{9}+\frac {20\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+1\right )}-\frac {43\,\mathrm {atanh}\left (\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {5}{4}\right )}{1024\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*sin(c + d*x) + 3)^3,x)

[Out]

((1225*tan(c/2 + (d*x)/2))/6912 + (3245*tan(c/2 + (d*x)/2)^2)/20736 - (125*tan(c/2 + (d*x)/2)^3)/6912 + 55/230
4)/(d*((20*tan(c/2 + (d*x)/2))/3 + (118*tan(c/2 + (d*x)/2)^2)/9 + (20*tan(c/2 + (d*x)/2)^3)/3 + tan(c/2 + (d*x
)/2)^4 + 1)) - (43*atanh((3*tan(c/2 + (d*x)/2))/4 + 5/4))/(1024*d)

________________________________________________________________________________________

sympy [A]  time = 3.54, size = 1227, normalized size = 10.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/(3 - 5*sin(2*atan(1/3)))**3, Eq(c, -d*x - 2*atan(1/3))), (x/(3 - 5*sin(2*atan(3)))**3, Eq(c, -d*x
- 2*atan(3))), (x/(5*sin(c) + 3)**3, Eq(d, 0)), (3483*log(tan(c/2 + d*x/2) + 1/3)*tan(c/2 + d*x/2)**4/(165888
*d*tan(c/2 + d*x/2)**4 + 1105920*d*tan(c/2 + d*x/2)**3 + 2174976*d*tan(c/2 + d*x/2)**2 + 1105920*d*tan(c/2 + d
*x/2) + 165888*d) + 23220*log(tan(c/2 + d*x/2) + 1/3)*tan(c/2 + d*x/2)**3/(165888*d*tan(c/2 + d*x/2)**4 + 1105
920*d*tan(c/2 + d*x/2)**3 + 2174976*d*tan(c/2 + d*x/2)**2 + 1105920*d*tan(c/2 + d*x/2) + 165888*d) + 45666*log
(tan(c/2 + d*x/2) + 1/3)*tan(c/2 + d*x/2)**2/(165888*d*tan(c/2 + d*x/2)**4 + 1105920*d*tan(c/2 + d*x/2)**3 + 2
174976*d*tan(c/2 + d*x/2)**2 + 1105920*d*tan(c/2 + d*x/2) + 165888*d) + 23220*log(tan(c/2 + d*x/2) + 1/3)*tan(
c/2 + d*x/2)/(165888*d*tan(c/2 + d*x/2)**4 + 1105920*d*tan(c/2 + d*x/2)**3 + 2174976*d*tan(c/2 + d*x/2)**2 + 1
105920*d*tan(c/2 + d*x/2) + 165888*d) + 3483*log(tan(c/2 + d*x/2) + 1/3)/(165888*d*tan(c/2 + d*x/2)**4 + 11059
20*d*tan(c/2 + d*x/2)**3 + 2174976*d*tan(c/2 + d*x/2)**2 + 1105920*d*tan(c/2 + d*x/2) + 165888*d) - 3483*log(t
an(c/2 + d*x/2) + 3)*tan(c/2 + d*x/2)**4/(165888*d*tan(c/2 + d*x/2)**4 + 1105920*d*tan(c/2 + d*x/2)**3 + 21749
76*d*tan(c/2 + d*x/2)**2 + 1105920*d*tan(c/2 + d*x/2) + 165888*d) - 23220*log(tan(c/2 + d*x/2) + 3)*tan(c/2 +
d*x/2)**3/(165888*d*tan(c/2 + d*x/2)**4 + 1105920*d*tan(c/2 + d*x/2)**3 + 2174976*d*tan(c/2 + d*x/2)**2 + 1105
920*d*tan(c/2 + d*x/2) + 165888*d) - 45666*log(tan(c/2 + d*x/2) + 3)*tan(c/2 + d*x/2)**2/(165888*d*tan(c/2 + d
*x/2)**4 + 1105920*d*tan(c/2 + d*x/2)**3 + 2174976*d*tan(c/2 + d*x/2)**2 + 1105920*d*tan(c/2 + d*x/2) + 165888
*d) - 23220*log(tan(c/2 + d*x/2) + 3)*tan(c/2 + d*x/2)/(165888*d*tan(c/2 + d*x/2)**4 + 1105920*d*tan(c/2 + d*x
/2)**3 + 2174976*d*tan(c/2 + d*x/2)**2 + 1105920*d*tan(c/2 + d*x/2) + 165888*d) - 3483*log(tan(c/2 + d*x/2) +
3)/(165888*d*tan(c/2 + d*x/2)**4 + 1105920*d*tan(c/2 + d*x/2)**3 + 2174976*d*tan(c/2 + d*x/2)**2 + 1105920*d*t
an(c/2 + d*x/2) + 165888*d) - 3000*tan(c/2 + d*x/2)**3/(165888*d*tan(c/2 + d*x/2)**4 + 1105920*d*tan(c/2 + d*x
/2)**3 + 2174976*d*tan(c/2 + d*x/2)**2 + 1105920*d*tan(c/2 + d*x/2) + 165888*d) + 25960*tan(c/2 + d*x/2)**2/(1
65888*d*tan(c/2 + d*x/2)**4 + 1105920*d*tan(c/2 + d*x/2)**3 + 2174976*d*tan(c/2 + d*x/2)**2 + 1105920*d*tan(c/
2 + d*x/2) + 165888*d) + 29400*tan(c/2 + d*x/2)/(165888*d*tan(c/2 + d*x/2)**4 + 1105920*d*tan(c/2 + d*x/2)**3
+ 2174976*d*tan(c/2 + d*x/2)**2 + 1105920*d*tan(c/2 + d*x/2) + 165888*d) + 3960/(165888*d*tan(c/2 + d*x/2)**4
+ 1105920*d*tan(c/2 + d*x/2)**3 + 2174976*d*tan(c/2 + d*x/2)**2 + 1105920*d*tan(c/2 + d*x/2) + 165888*d), True
))

________________________________________________________________________________________