3.3.35 \(\int \cot ^3(c+b x) \sin (a+b x) \, dx\) [235]
Optimal. Leaf size=74 \[ -\frac {\cos (a-c) \csc (c+b x)}{b}+\frac {3 \tanh ^{-1}(\cos (c+b x)) \sin (a-c)}{2 b}-\frac {\cot (c+b x) \csc (c+b x) \sin (a-c)}{2 b}-\frac {\sin (a+b x)}{b} \]
[Out]
-cos(a-c)*csc(b*x+c)/b+3/2*arctanh(cos(b*x+c))*sin(a-c)/b-1/2*cot(b*x+c)*csc(b*x+c)*sin(a-c)/b-sin(b*x+a)/b
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Rubi [A]
time = 0.06, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {4674, 4673,
2717, 3855, 2686, 8, 2691} \begin {gather*} -\frac {\cos (a-c) \csc (b x+c)}{b}+\frac {3 \sin (a-c) \tanh ^{-1}(\cos (b x+c))}{2 b}-\frac {\sin (a-c) \cot (b x+c) \csc (b x+c)}{2 b}-\frac {\sin (a+b x)}{b} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Cot[c + b*x]^3*Sin[a + b*x],x]
[Out]
-((Cos[a - c]*Csc[c + b*x])/b) + (3*ArcTanh[Cos[c + b*x]]*Sin[a - c])/(2*b) - (Cot[c + b*x]*Csc[c + b*x]*Sin[a
- c])/(2*b) - Sin[a + b*x]/b
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rule 2686
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Rule 2691
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]
Rule 2717
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Rule 3855
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rule 4673
Int[Cos[v_]*Cot[w_]^(n_.), x_Symbol] :> -Int[Sin[v]*Cot[w]^(n - 1), x] + Dist[Cos[v - w], Int[Csc[w]*Cot[w]^(n
- 1), x], x] /; GtQ[n, 0] && FreeQ[v - w, x] && NeQ[w, v]
Rule 4674
Int[Cot[w_]^(n_.)*Sin[v_], x_Symbol] :> Int[Cos[v]*Cot[w]^(n - 1), x] + Dist[Sin[v - w], Int[Csc[w]*Cot[w]^(n
- 1), x], x] /; GtQ[n, 0] && FreeQ[v - w, x] && NeQ[w, v]
Rubi steps
\begin {align*} \int \cot ^3(c+b x) \sin (a+b x) \, dx &=\sin (a-c) \int \cot ^2(c+b x) \csc (c+b x) \, dx+\int \cos (a+b x) \cot ^2(c+b x) \, dx\\ &=-\frac {\cot (c+b x) \csc (c+b x) \sin (a-c)}{2 b}+\cos (a-c) \int \cot (c+b x) \csc (c+b x) \, dx-\frac {1}{2} \sin (a-c) \int \csc (c+b x) \, dx-\int \cot (c+b x) \sin (a+b x) \, dx\\ &=\frac {\tanh ^{-1}(\cos (c+b x)) \sin (a-c)}{2 b}-\frac {\cot (c+b x) \csc (c+b x) \sin (a-c)}{2 b}-\frac {\cos (a-c) \text {Subst}(\int 1 \, dx,x,\csc (c+b x))}{b}-\sin (a-c) \int \csc (c+b x) \, dx-\int \cos (a+b x) \, dx\\ &=-\frac {\cos (a-c) \csc (c+b x)}{b}+\frac {3 \tanh ^{-1}(\cos (c+b x)) \sin (a-c)}{2 b}-\frac {\cot (c+b x) \csc (c+b x) \sin (a-c)}{2 b}-\frac {\sin (a+b x)}{b}\\ \end {align*}
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Mathematica [A]
time = 0.41, size = 71, normalized size = 0.96 \begin {gather*} \frac {12 \tanh ^{-1}\left (\cos (c)-\sin (c) \tan \left (\frac {b x}{2}\right )\right ) \sin (a-c)+\csc ^2(c+b x) (2 \sin (a-2 c-b x)-5 \sin (a+b x)+\sin (a+2 c+3 b x))}{4 b} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Cot[c + b*x]^3*Sin[a + b*x],x]
[Out]
(12*ArcTanh[Cos[c] - Sin[c]*Tan[(b*x)/2]]*Sin[a - c] + Csc[c + b*x]^2*(2*Sin[a - 2*c - b*x] - 5*Sin[a + b*x] +
Sin[a + 2*c + 3*b*x]))/(4*b)
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Maple [C] Result contains complex when optimal does not.
time = 0.16, size = 184, normalized size = 2.49
| | |
| method |
result |
size |
| | |
| risch |
\(\frac {i {\mathrm e}^{i \left (b x +a \right )}}{2 b}-\frac {i {\mathrm e}^{-i \left (b x +a \right )}}{2 b}+\frac {i \left (-3 \,{\mathrm e}^{i \left (3 b x +5 a +2 c \right )}-{\mathrm e}^{i \left (3 b x +3 a +4 c \right )}+{\mathrm e}^{i \left (b x +5 a \right )}+3 \,{\mathrm e}^{i \left (b x +3 a +2 c \right )}\right )}{2 b \left (-{\mathrm e}^{2 i \left (b x +a +c \right )}+{\mathrm e}^{2 i a}\right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+{\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{2 b}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-{\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{2 b}\) |
\(184\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(b*x+c)^3*sin(b*x+a),x,method=_RETURNVERBOSE)
[Out]
1/2*I*exp(I*(b*x+a))/b-1/2*I/b*exp(-I*(b*x+a))+1/2*I/b/(-exp(2*I*(b*x+a+c))+exp(2*I*a))^2*(-3*exp(I*(3*b*x+5*a
+2*c))-exp(I*(3*b*x+3*a+4*c))+exp(I*(b*x+5*a))+3*exp(I*(b*x+3*a+2*c)))+3/2*ln(exp(I*(b*x+a))+exp(I*(a-c)))/b*s
in(a-c)-3/2*ln(exp(I*(b*x+a))-exp(I*(a-c)))/b*sin(a-c)
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 1254 vs.
\(2 (70) = 140\).
time = 0.32, size = 1254, normalized size = 16.95 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(b*x+c)^3*sin(b*x+a),x, algorithm="maxima")
[Out]
1/4*(2*(sin(5*b*x + a + 4*c) - 2*sin(3*b*x + a + 2*c) + sin(b*x + a))*cos(6*b*x + 2*a + 4*c) + 2*(5*sin(4*b*x
+ 2*a + 2*c) + 2*sin(4*b*x + 4*c) - 2*sin(2*b*x + 2*a) - 5*sin(2*b*x + 2*c))*cos(5*b*x + a + 4*c) + 10*(2*sin(
3*b*x + a + 2*c) - sin(b*x + a))*cos(4*b*x + 2*a + 2*c) + 4*(2*sin(3*b*x + a + 2*c) - sin(b*x + a))*cos(4*b*x
+ 4*c) + 4*(2*sin(2*b*x + 2*a) + 5*sin(2*b*x + 2*c))*cos(3*b*x + a + 2*c) - 3*(cos(5*b*x + a + 4*c)^2*sin(-a +
c) + 4*cos(3*b*x + a + 2*c)^2*sin(-a + c) - 4*cos(3*b*x + a + 2*c)*cos(b*x + a)*sin(-a + c) + cos(b*x + a)^2*
sin(-a + c) + sin(5*b*x + a + 4*c)^2*sin(-a + c) + 4*sin(3*b*x + a + 2*c)^2*sin(-a + c) - 4*sin(3*b*x + a + 2*
c)*sin(b*x + a)*sin(-a + c) + sin(b*x + a)^2*sin(-a + c) - 2*(2*cos(3*b*x + a + 2*c)*sin(-a + c) - cos(b*x + a
)*sin(-a + c))*cos(5*b*x + a + 4*c) - 2*(2*sin(3*b*x + a + 2*c)*sin(-a + c) - sin(b*x + a)*sin(-a + c))*sin(5*
b*x + a + 4*c))*log(cos(b*x)^2 + 2*cos(b*x)*cos(c) + cos(c)^2 + sin(b*x)^2 - 2*sin(b*x)*sin(c) + sin(c)^2) + 3
*(cos(5*b*x + a + 4*c)^2*sin(-a + c) + 4*cos(3*b*x + a + 2*c)^2*sin(-a + c) - 4*cos(3*b*x + a + 2*c)*cos(b*x +
a)*sin(-a + c) + cos(b*x + a)^2*sin(-a + c) + sin(5*b*x + a + 4*c)^2*sin(-a + c) + 4*sin(3*b*x + a + 2*c)^2*s
in(-a + c) - 4*sin(3*b*x + a + 2*c)*sin(b*x + a)*sin(-a + c) + sin(b*x + a)^2*sin(-a + c) - 2*(2*cos(3*b*x + a
+ 2*c)*sin(-a + c) - cos(b*x + a)*sin(-a + c))*cos(5*b*x + a + 4*c) - 2*(2*sin(3*b*x + a + 2*c)*sin(-a + c) -
sin(b*x + a)*sin(-a + c))*sin(5*b*x + a + 4*c))*log(cos(b*x)^2 - 2*cos(b*x)*cos(c) + cos(c)^2 + sin(b*x)^2 +
2*sin(b*x)*sin(c) + sin(c)^2) - 2*(cos(5*b*x + a + 4*c) - 2*cos(3*b*x + a + 2*c) + cos(b*x + a))*sin(6*b*x + 2
*a + 4*c) - 2*(5*cos(4*b*x + 2*a + 2*c) + 2*cos(4*b*x + 4*c) - 2*cos(2*b*x + 2*a) - 5*cos(2*b*x + 2*c) + 1)*si
n(5*b*x + a + 4*c) - 10*(2*cos(3*b*x + a + 2*c) - cos(b*x + a))*sin(4*b*x + 2*a + 2*c) - 4*(2*cos(3*b*x + a +
2*c) - cos(b*x + a))*sin(4*b*x + 4*c) - 4*(2*cos(2*b*x + 2*a) + 5*cos(2*b*x + 2*c) - 1)*sin(3*b*x + a + 2*c) -
4*cos(b*x + a)*sin(2*b*x + 2*a) - 10*cos(b*x + a)*sin(2*b*x + 2*c) + 4*cos(2*b*x + 2*a)*sin(b*x + a) + 10*cos
(2*b*x + 2*c)*sin(b*x + a) - 2*sin(b*x + a))/(b*cos(5*b*x + a + 4*c)^2 + 4*b*cos(3*b*x + a + 2*c)^2 - 4*b*cos(
3*b*x + a + 2*c)*cos(b*x + a) + b*cos(b*x + a)^2 + b*sin(5*b*x + a + 4*c)^2 + 4*b*sin(3*b*x + a + 2*c)^2 - 4*b
*sin(3*b*x + a + 2*c)*sin(b*x + a) + b*sin(b*x + a)^2 - 2*(2*b*cos(3*b*x + a + 2*c) - b*cos(b*x + a))*cos(5*b*
x + a + 4*c) - 2*(2*b*sin(3*b*x + a + 2*c) - b*sin(b*x + a))*sin(5*b*x + a + 4*c))
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 372 vs.
\(2 (70) = 140\).
time = 2.30, size = 372, normalized size = 5.03 \begin {gather*} \frac {\frac {3 \, \sqrt {2} {\left (2 \, {\left (\cos \left (-2 \, a + 2 \, c\right )^{2} - 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + {\left (2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) - 1\right )} \sin \left (-2 \, a + 2 \, c\right )\right )} \log \left (-\frac {2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \frac {2 \, \sqrt {2} {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right )\right )}}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - \cos \left (-2 \, a + 2 \, c\right ) + 3}{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) - 1}\right )}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - 4 \, {\left (4 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 3 \, \cos \left (-2 \, a + 2 \, c\right ) - 5\right )} \sin \left (b x + a\right ) - 4 \, {\left (4 \, \cos \left (b x + a\right )^{3} - 5 \, \cos \left (b x + a\right )\right )} \sin \left (-2 \, a + 2 \, c\right )}{8 \, {\left (2 \, b \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, b \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - b \cos \left (-2 \, a + 2 \, c\right ) - b\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(b*x+c)^3*sin(b*x+a),x, algorithm="fricas")
[Out]
1/8*(3*sqrt(2)*(2*(cos(-2*a + 2*c)^2 - 1)*cos(b*x + a)*sin(b*x + a) + (2*cos(b*x + a)^2*cos(-2*a + 2*c) - cos(
-2*a + 2*c) - 1)*sin(-2*a + 2*c))*log(-(2*cos(b*x + a)^2*cos(-2*a + 2*c) - 2*cos(b*x + a)*sin(b*x + a)*sin(-2*
a + 2*c) - 2*sqrt(2)*((cos(-2*a + 2*c) + 1)*cos(b*x + a) - sin(b*x + a)*sin(-2*a + 2*c))/sqrt(cos(-2*a + 2*c)
+ 1) - cos(-2*a + 2*c) + 3)/(2*cos(b*x + a)^2*cos(-2*a + 2*c) - 2*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) -
cos(-2*a + 2*c) - 1))/sqrt(cos(-2*a + 2*c) + 1) - 4*(4*cos(b*x + a)^2*cos(-2*a + 2*c) - 3*cos(-2*a + 2*c) - 5)
*sin(b*x + a) - 4*(4*cos(b*x + a)^3 - 5*cos(b*x + a))*sin(-2*a + 2*c))/(2*b*cos(b*x + a)^2*cos(-2*a + 2*c) - 2
*b*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) - b*cos(-2*a + 2*c) - b)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin {\left (a + b x \right )} \cot ^{3}{\left (b x + c \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(b*x+c)**3*sin(b*x+a),x)
[Out]
Integral(sin(a + b*x)*cot(b*x + c)**3, x)
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 870 vs.
\(2 (70) = 140\).
time = 0.45, size = 870, normalized size = 11.76 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(b*x+c)^3*sin(b*x+a),x, algorithm="giac")
[Out]
1/4*(12*(tan(1/2*a)^2*tan(1/2*c)^2 - tan(1/2*a)*tan(1/2*c)^3 + tan(1/2*a)*tan(1/2*c) - tan(1/2*c)^2)*log(abs(t
an(1/2*b*x)*tan(1/2*c) - 1))/(tan(1/2*a)^2*tan(1/2*c)^3 + tan(1/2*a)^2*tan(1/2*c) + tan(1/2*c)^3 + tan(1/2*c))
- 12*(tan(1/2*a)^2*tan(1/2*c) - tan(1/2*a)*tan(1/2*c)^2 + tan(1/2*a) - tan(1/2*c))*log(abs(tan(1/2*b*x) + tan
(1/2*c)))/(tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1) + 8*(tan(1/2*b*x)*tan(1/2*a)^2 - tan(1
/2*b*x) - 2*tan(1/2*a))/((tan(1/2*b*x)^2 + 1)*(tan(1/2*a)^2 + 1)) - (2*tan(1/2*b*x)^3*tan(1/2*a)*tan(1/2*c)^7
+ tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*c)^7 + tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*c)^8 - 4*tan(1/2*b*x)^3*tan(1/2
*a)^2*tan(1/2*c)^4 + 6*tan(1/2*b*x)^3*tan(1/2*a)*tan(1/2*c)^5 - 5*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*c)^5 + 2
*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*c)^6 - 4*tan(1/2*b*x)*tan(1/2*a)^2*tan(1/2*c)^6 - tan(1/2*b*x)^2*tan(1/2*c)
^7 - 2*tan(1/2*b*x)*tan(1/2*a)*tan(1/2*c)^7 - 6*tan(1/2*b*x)^3*tan(1/2*a)*tan(1/2*c)^3 + 5*tan(1/2*b*x)^2*tan(
1/2*a)^2*tan(1/2*c)^3 + 4*tan(1/2*b*x)^3*tan(1/2*c)^4 - 22*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*c)^4 + 4*tan(1/2*
b*x)*tan(1/2*a)^2*tan(1/2*c)^4 + 5*tan(1/2*b*x)^2*tan(1/2*c)^5 - 14*tan(1/2*b*x)*tan(1/2*a)*tan(1/2*c)^5 + 2*t
an(1/2*a)^2*tan(1/2*c)^5 + 4*tan(1/2*b*x)*tan(1/2*c)^6 + 2*tan(1/2*a)*tan(1/2*c)^6 - 2*tan(1/2*b*x)^3*tan(1/2*
a)*tan(1/2*c) - tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*c) + 2*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*c)^2 - 4*tan(1/2*
b*x)*tan(1/2*a)^2*tan(1/2*c)^2 - 5*tan(1/2*b*x)^2*tan(1/2*c)^3 + 14*tan(1/2*b*x)*tan(1/2*a)*tan(1/2*c)^3 - 2*t
an(1/2*a)^2*tan(1/2*c)^3 - 4*tan(1/2*b*x)*tan(1/2*c)^4 + 12*tan(1/2*a)*tan(1/2*c)^4 - 2*tan(1/2*c)^5 + tan(1/2
*b*x)^2*tan(1/2*a) + tan(1/2*b*x)^2*tan(1/2*c) + 2*tan(1/2*b*x)*tan(1/2*a)*tan(1/2*c) + 4*tan(1/2*b*x)*tan(1/2
*c)^2 + 2*tan(1/2*a)*tan(1/2*c)^2 + 2*tan(1/2*c)^3)/((tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*c)^2)*(tan(1/2*b*x)^
2*tan(1/2*c) + tan(1/2*b*x)*tan(1/2*c)^2 - tan(1/2*b*x) - tan(1/2*c))^2))/b
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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \text {Hanged} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(c + b*x)^3*sin(a + b*x),x)
[Out]
\text{Hanged}
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