Integrand size = 12, antiderivative size = 88 \[ \int \frac {1}{(3+5 \sin (c+d x))^2} \, dx=\frac {3 \log \left (3 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}-\frac {3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+3 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}-\frac {5 \cos (c+d x)}{16 d (3+5 \sin (c+d x))} \] Output:
3/64*ln(3*cos(1/2*d*x+1/2*c)+sin(1/2*d*x+1/2*c))/d-3/64*ln(cos(1/2*d*x+1/2 *c)+3*sin(1/2*d*x+1/2*c))/d-5/16*cos(d*x+c)/d/(3+5*sin(d*x+c))
Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.43 \[ \int \frac {1}{(3+5 \sin (c+d x))^2} \, dx=\frac {9 \left (\log \left (3 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+3 \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+20 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\frac {1}{3 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {3}{\cos \left (\frac {1}{2} (c+d x)\right )+3 \sin \left (\frac {1}{2} (c+d x)\right )}\right )}{192 d} \] Input:
Integrate[(3 + 5*Sin[c + d*x])^(-2),x]
Output:
(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)
Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {3042, 3143, 27, 3042, 3139, 1081, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{(5 \sin (c+d x)+3)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(5 \sin (c+d x)+3)^2}dx\) |
\(\Big \downarrow \) 3143 |
\(\displaystyle \frac {1}{16} \int -\frac {3}{5 \sin (c+d x)+3}dx-\frac {5 \cos (c+d x)}{16 d (5 \sin (c+d x)+3)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {3}{16} \int \frac {1}{5 \sin (c+d x)+3}dx-\frac {5 \cos (c+d x)}{16 d (5 \sin (c+d x)+3)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3}{16} \int \frac {1}{5 \sin (c+d x)+3}dx-\frac {5 \cos (c+d x)}{16 d (5 \sin (c+d x)+3)}\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle -\frac {3 \int \frac {1}{3 \tan ^2\left (\frac {1}{2} (c+d x)\right )+10 \tan \left (\frac {1}{2} (c+d x)\right )+3}d\tan \left (\frac {1}{2} (c+d x)\right )}{8 d}-\frac {5 \cos (c+d x)}{16 d (5 \sin (c+d x)+3)}\) |
\(\Big \downarrow \) 1081 |
\(\displaystyle -\frac {9 \int \left (\frac {1}{8 \left (3 \tan \left (\frac {1}{2} (c+d x)\right )+1\right )}-\frac {1}{24 \left (\tan \left (\frac {1}{2} (c+d x)\right )+3\right )}\right )d\tan \left (\frac {1}{2} (c+d x)\right )}{8 d}-\frac {5 \cos (c+d x)}{16 d (5 \sin (c+d x)+3)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {9 \left (\frac {1}{24} \log \left (3 \tan \left (\frac {1}{2} (c+d x)\right )+1\right )-\frac {1}{24} \log \left (\tan \left (\frac {1}{2} (c+d x)\right )+3\right )\right )}{8 d}-\frac {5 \cos (c+d x)}{16 d (5 \sin (c+d x)+3)}\) |
Input:
Int[(3 + 5*Sin[c + d*x])^(-2),x]
Output:
(-9*(-1/24*Log[3 + Tan[(c + d*x)/2]] + Log[1 + 3*Tan[(c + d*x)/2]]/24))/(8 *d) - (5*Cos[c + d*x])/(16*d*(3 + 5*Sin[c + d*x]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[c Int[ExpandIntegrand[1/((b/2 - q/2 + c*x)*(b/2 + q/2 + c*x)), x], x], x]] /; FreeQ[{a, b, c}, x] && NiceSqrtQ[b^2 - 4*a*c]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[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]
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] + Simp [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] && IntegerQ[2*n]
Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.77
method | result | size |
derivativedivides | \(\frac {-\frac {5}{48 \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64}-\frac {5}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )}{64}}{d}\) | \(68\) |
default | \(\frac {-\frac {5}{48 \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64}-\frac {5}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )}{64}}{d}\) | \(68\) |
risch | \(-\frac {3 \,{\mathrm e}^{i \left (d x +c \right )}+5 i}{8 d \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}-5+6 i {\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {3 \ln \left (\frac {4}{5}+\frac {3 i}{5}+{\mathrm e}^{i \left (d x +c \right )}\right )}{64 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {4}{5}+\frac {3 i}{5}\right )}{64 d}\) | \(86\) |
norman | \(\frac {-\frac {5}{8 d}-\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d}}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )}{64 d}-\frac {3 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64 d}\) | \(87\) |
parallelrisch | \(\frac {45 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right ) \sin \left (d x +c \right )-45 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {1}{3}\right ) \sin \left (d x +c \right )-100 \sin \left (d x +c \right )-60 \cos \left (d x +c \right )+27 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )-27 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {1}{3}\right )-60}{192 d \left (3+5 \sin \left (d x +c \right )\right )}\) | \(104\) |
Input:
int(1/(3+5*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-5/48/(3*tan(1/2*d*x+1/2*c)+1)-3/64*ln(3*tan(1/2*d*x+1/2*c)+1)-5/16/( tan(1/2*d*x+1/2*c)+3)+3/64*ln(tan(1/2*d*x+1/2*c)+3))
Time = 0.11 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(3+5 \sin (c+d x))^2} \, dx=\frac {3 \, {\left (5 \, \sin \left (d x + c\right ) + 3\right )} \log \left (4 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right ) + 5\right ) - 3 \, {\left (5 \, \sin \left (d x + c\right ) + 3\right )} \log \left (-4 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right ) + 5\right ) - 40 \, \cos \left (d x + c\right )}{128 \, {\left (5 \, d \sin \left (d x + c\right ) + 3 \, d\right )}} \] Input:
integrate(1/(3+5*sin(d*x+c))^2,x, algorithm="fricas")
Output:
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)
Leaf count of result is larger than twice the leaf count of optimal. 466 vs. \(2 (78) = 156\).
Time = 0.72 (sec) , antiderivative size = 466, normalized size of antiderivative = 5.30 \[ \int \frac {1}{(3+5 \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:
integrate(1/(3+5*sin(d*x+c))**2,x)
Output:
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*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 + 192 0*d*tan(c/2 + d*x/2) + 576*d) - 27*log(3*tan(c/2 + d*x/2) + 1)*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) - 9 0*log(3*tan(c/2 + d*x/2) + 1)*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(3*tan(c/2 + d*x/2) + 1)/(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))
Time = 0.03 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.31 \[ \int \frac {1}{(3+5 \sin (c+d x))^2} \, dx=-\frac {\frac {40 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 3\right )}}{\frac {10 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 3} + 9 \, \log \left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) - 9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 3\right )}{192 \, d} \] Input:
integrate(1/(3+5*sin(d*x+c))^2,x, algorithm="maxima")
Output:
-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
Time = 0.13 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(3+5 \sin (c+d x))^2} \, dx=-\frac {\frac {40 \, {\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3\right )}}{3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3} + 9 \, \log \left ({\left | 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 9 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \right |}\right )}{192 \, d} \] Input:
integrate(1/(3+5*sin(d*x+c))^2,x, algorithm="giac")
Output:
-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(ab s(tan(1/2*d*x + 1/2*c) + 3)))/d
Time = 25.97 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(3+5 \sin (c+d x))^2} \, dx=\frac {3\,\mathrm {atanh}\left (\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {5}{4}\right )}{32\,d}-\frac {\frac {25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{72}+\frac {5}{24}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {10\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+1\right )} \] Input:
int(1/(5*sin(c + d*x) + 3)^2,x)
Output:
(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*((10*tan(c/2 + (d*x)/2))/3 + tan(c/2 + (d*x)/2)^2 + 1))
Time = 0.18 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.11 \[ \int \frac {1}{(3+5 \sin (c+d x))^2} \, dx=\frac {-20 \cos \left (d x +c \right )+15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right ) \sin \left (d x +c \right )+9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )-15 \,\mathrm {log}\left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )-9 \,\mathrm {log}\left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64 d \left (5 \sin \left (d x +c \right )+3\right )} \] Input:
int(1/(3+5*sin(d*x+c))^2,x)
Output:
( - 20*cos(c + d*x) + 15*log(tan((c + d*x)/2) + 3)*sin(c + d*x) + 9*log(ta n((c + d*x)/2) + 3) - 15*log(3*tan((c + d*x)/2) + 1)*sin(c + d*x) - 9*log( 3*tan((c + d*x)/2) + 1))/(64*d*(5*sin(c + d*x) + 3))