3.163 \(\int \cot ^3(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=116 \[ -\frac {a^3 \csc ^2(c+d x)}{2 d}-\frac {b \left (3 a^2-b^2\right ) \sin (c+d x)}{d}-\frac {a \left (a^2-3 b^2\right ) \log (\sin (c+d x))}{d}-\frac {3 a^2 b \csc (c+d x)}{d}-\frac {3 a b^2 \sin ^2(c+d x)}{2 d}-\frac {b^3 \sin ^3(c+d x)}{3 d} \]

[Out]

-3*a^2*b*csc(d*x+c)/d-1/2*a^3*csc(d*x+c)^2/d-a*(a^2-3*b^2)*ln(sin(d*x+c))/d-b*(3*a^2-b^2)*sin(d*x+c)/d-3/2*a*b
^2*sin(d*x+c)^2/d-1/3*b^3*sin(d*x+c)^3/d

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Rubi [A]  time = 0.09, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2721, 894} \[ -\frac {b \left (3 a^2-b^2\right ) \sin (c+d x)}{d}-\frac {a \left (a^2-3 b^2\right ) \log (\sin (c+d x))}{d}-\frac {3 a^2 b \csc (c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d}-\frac {3 a b^2 \sin ^2(c+d x)}{2 d}-\frac {b^3 \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3*(a + b*Sin[c + d*x])^3,x]

[Out]

(-3*a^2*b*Csc[c + d*x])/d - (a^3*Csc[c + d*x]^2)/(2*d) - (a*(a^2 - 3*b^2)*Log[Sin[c + d*x]])/d - (b*(3*a^2 - b
^2)*Sin[c + d*x])/d - (3*a*b^2*Sin[c + d*x]^2)/(2*d) - (b^3*Sin[c + d*x]^3)/(3*d)

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

\begin {align*} \int \cot ^3(c+d x) (a+b \sin (c+d x))^3 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(a+x)^3 \left (b^2-x^2\right )}{x^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-3 a^2 \left (1-\frac {b^2}{3 a^2}\right )+\frac {a^3 b^2}{x^3}+\frac {3 a^2 b^2}{x^2}+\frac {-a^3+3 a b^2}{x}-3 a x-x^2\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {3 a^2 b \csc (c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d}-\frac {a \left (a^2-3 b^2\right ) \log (\sin (c+d x))}{d}-\frac {b \left (3 a^2-b^2\right ) \sin (c+d x)}{d}-\frac {3 a b^2 \sin ^2(c+d x)}{2 d}-\frac {b^3 \sin ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 97, normalized size = 0.84 \[ -\frac {3 a^3 \csc ^2(c+d x)-6 b \left (b^2-3 a^2\right ) \sin (c+d x)+6 a \left (a^2-3 b^2\right ) \log (\sin (c+d x))+18 a^2 b \csc (c+d x)+9 a b^2 \sin ^2(c+d x)+2 b^3 \sin ^3(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3*(a + b*Sin[c + d*x])^3,x]

[Out]

-1/6*(18*a^2*b*Csc[c + d*x] + 3*a^3*Csc[c + d*x]^2 + 6*a*(a^2 - 3*b^2)*Log[Sin[c + d*x]] - 6*b*(-3*a^2 + b^2)*
Sin[c + d*x] + 9*a*b^2*Sin[c + d*x]^2 + 2*b^3*Sin[c + d*x]^3)/d

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fricas [A]  time = 0.47, size = 153, normalized size = 1.32 \[ \frac {18 \, a b^{2} \cos \left (d x + c\right )^{4} - 27 \, a b^{2} \cos \left (d x + c\right )^{2} + 6 \, a^{3} + 9 \, a b^{2} + 12 \, {\left (a^{3} - 3 \, a b^{2} - {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (b^{3} \cos \left (d x + c\right )^{4} + 18 \, a^{2} b - 2 \, b^{3} - {\left (9 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/12*(18*a*b^2*cos(d*x + c)^4 - 27*a*b^2*cos(d*x + c)^2 + 6*a^3 + 9*a*b^2 + 12*(a^3 - 3*a*b^2 - (a^3 - 3*a*b^2
)*cos(d*x + c)^2)*log(1/2*sin(d*x + c)) + 4*(b^3*cos(d*x + c)^4 + 18*a^2*b - 2*b^3 - (9*a^2*b - b^3)*cos(d*x +
c)^2)*sin(d*x + c))/(d*cos(d*x + c)^2 - d)

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giac [A]  time = 0.94, size = 131, normalized size = 1.13 \[ -\frac {2 \, b^{3} \sin \left (d x + c\right )^{3} + 9 \, a b^{2} \sin \left (d x + c\right )^{2} + 18 \, a^{2} b \sin \left (d x + c\right ) - 6 \, b^{3} \sin \left (d x + c\right ) + 6 \, {\left (a^{3} - 3 \, a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {3 \, {\left (3 \, a^{3} \sin \left (d x + c\right )^{2} - 9 \, a b^{2} \sin \left (d x + c\right )^{2} - 6 \, a^{2} b \sin \left (d x + c\right ) - a^{3}\right )}}{\sin \left (d x + c\right )^{2}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/6*(2*b^3*sin(d*x + c)^3 + 9*a*b^2*sin(d*x + c)^2 + 18*a^2*b*sin(d*x + c) - 6*b^3*sin(d*x + c) + 6*(a^3 - 3*
a*b^2)*log(abs(sin(d*x + c))) - 3*(3*a^3*sin(d*x + c)^2 - 9*a*b^2*sin(d*x + c)^2 - 6*a^2*b*sin(d*x + c) - a^3)
/sin(d*x + c)^2)/d

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maple [A]  time = 0.26, size = 165, normalized size = 1.42 \[ -\frac {a^{3} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {3 a^{2} b \left (\cos ^{4}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {3 a^{2} b \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}-\frac {6 a^{2} b \sin \left (d x +c \right )}{d}+\frac {3 a \,b^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{2 d}+\frac {3 a \,b^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}+\frac {b^{3} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {2 b^{3} \sin \left (d x +c \right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(a+b*sin(d*x+c))^3,x)

[Out]

-1/2/d*a^3*cot(d*x+c)^2-a^3*ln(sin(d*x+c))/d-3/d*a^2*b/sin(d*x+c)*cos(d*x+c)^4-3/d*a^2*b*cos(d*x+c)^2*sin(d*x+
c)-6*a^2*b*sin(d*x+c)/d+3/2/d*a*b^2*cos(d*x+c)^2+3/d*a*b^2*ln(sin(d*x+c))+1/3/d*b^3*cos(d*x+c)^2*sin(d*x+c)+2/
3/d*b^3*sin(d*x+c)

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maxima [A]  time = 1.14, size = 98, normalized size = 0.84 \[ -\frac {2 \, b^{3} \sin \left (d x + c\right )^{3} + 9 \, a b^{2} \sin \left (d x + c\right )^{2} + 6 \, {\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\sin \left (d x + c\right )\right ) + 6 \, {\left (3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right ) + \frac {3 \, {\left (6 \, a^{2} b \sin \left (d x + c\right ) + a^{3}\right )}}{\sin \left (d x + c\right )^{2}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/6*(2*b^3*sin(d*x + c)^3 + 9*a*b^2*sin(d*x + c)^2 + 6*(a^3 - 3*a*b^2)*log(sin(d*x + c)) + 6*(3*a^2*b - b^3)*
sin(d*x + c) + 3*(6*a^2*b*sin(d*x + c) + a^3)/sin(d*x + c)^2)/d

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mupad [B]  time = 6.95, size = 312, normalized size = 2.69 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,a\,b^2-a^3\right )}{d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (3\,a\,b^2-a^3\right )}{d}-\frac {\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {3\,a^3}{2}+24\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {a^3}{2}+24\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (30\,a^2\,b-8\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (42\,a^2\,b-8\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (66\,a^2\,b-\frac {16\,b^3}{3}\right )+\frac {a^3}{2}+6\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {3\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^3*(a + b*sin(c + d*x))^3,x)

[Out]

(log(tan(c/2 + (d*x)/2))*(3*a*b^2 - a^3))/d - (log(tan(c/2 + (d*x)/2)^2 + 1)*(3*a*b^2 - a^3))/d - ((3*a^3*tan(
c/2 + (d*x)/2)^2)/2 + tan(c/2 + (d*x)/2)^4*(24*a*b^2 + (3*a^3)/2) + tan(c/2 + (d*x)/2)^6*(24*a*b^2 + a^3/2) +
tan(c/2 + (d*x)/2)^7*(30*a^2*b - 8*b^3) + tan(c/2 + (d*x)/2)^3*(42*a^2*b - 8*b^3) + tan(c/2 + (d*x)/2)^5*(66*a
^2*b - (16*b^3)/3) + a^3/2 + 6*a^2*b*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 + 12*tan(c/2 + (d*x)/2)^4
+ 12*tan(c/2 + (d*x)/2)^6 + 4*tan(c/2 + (d*x)/2)^8)) - (a^3*tan(c/2 + (d*x)/2)^2)/(8*d) - (3*a^2*b*tan(c/2 + (
d*x)/2))/(2*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \cot ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3*(a+b*sin(d*x+c))**3,x)

[Out]

Integral((a + b*sin(c + d*x))**3*cot(c + d*x)**3, x)

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