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

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

[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)

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Rubi [A]  time = 0.0943571, antiderivative size = 116, normalized size of antiderivative = 1., 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 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]

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]))

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*}

Mathematica [A]  time = 0.320514, size = 97, normalized size = 0.84 \[ -\frac{-6 b \left (b^2-3 a^2\right ) \sin (c+d x)+6 a \left (a^2-3 b^2\right ) \log (\sin (c+d x))+18 a^2 b \csc (c+d x)+3 a^3 \csc ^2(c+d x)+9 a b^2 \sin ^2(c+d x)+2 b^3 \sin ^3(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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)/(6*d)

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Maple [A]  time = 0.065, size = 165, normalized size = 1.4 \begin{align*} -{\frac{{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-3\,{\frac{{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{d\sin \left ( dx+c \right ) }}-3\,{\frac{{a}^{2}b\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{d}}-6\,{\frac{{a}^{2}b\sin \left ( dx+c \right ) }{d}}+{\frac{3\,a{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+3\,{\frac{a{b}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{2\,{b}^{3}\sin \left ( dx+c \right ) }{3\,d}} \end{align*}

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*sin(d*x+c)*cos(d*x+c)
^2-6*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 = 1.89972, size = 132, normalized size = 1.14 \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 1.56308, size = 358, normalized size = 3.09 \begin{align*} \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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sin{\left (c + d x \right )}\right )^{3} \cot ^{3}{\left (c + d x \right )}\, dx \end{align*}

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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Giac [A]  time = 2.0221, size = 177, normalized size = 1.53 \begin{align*} -\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} \end{align*}

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