3.153 \(\int \cot ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=84 \[ -\frac{\left (a^2-b^2\right ) \log (\sin (c+d x))}{d}-\frac{a^2 \csc ^2(c+d x)}{2 d}-\frac{2 a b \sin (c+d x)}{d}-\frac{2 a b \csc (c+d x)}{d}-\frac{b^2 \sin ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0730727, antiderivative size = 84, 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} \[ -\frac{\left (a^2-b^2\right ) \log (\sin (c+d x))}{d}-\frac{a^2 \csc ^2(c+d x)}{2 d}-\frac{2 a b \sin (c+d x)}{d}-\frac{2 a b \csc (c+d x)}{d}-\frac{b^2 \sin ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]

[Out]

(-2*a*b*Csc[c + d*x])/d - (a^2*Csc[c + d*x]^2)/(2*d) - ((a^2 - b^2)*Log[Sin[c + d*x]])/d - (2*a*b*Sin[c + d*x]
)/d - (b^2*Sin[c + d*x]^2)/(2*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))^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a+x)^2 \left (b^2-x^2\right )}{x^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-2 a+\frac{a^2 b^2}{x^3}+\frac{2 a b^2}{x^2}+\frac{-a^2+b^2}{x}-x\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac{2 a b \csc (c+d x)}{d}-\frac{a^2 \csc ^2(c+d x)}{2 d}-\frac{\left (a^2-b^2\right ) \log (\sin (c+d x))}{d}-\frac{2 a b \sin (c+d x)}{d}-\frac{b^2 \sin ^2(c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.23557, size = 70, normalized size = 0.83 \[ -\frac{2 \left (a^2-b^2\right ) \log (\sin (c+d x))+a^2 \csc ^2(c+d x)+4 a b \sin (c+d x)+4 a b \csc (c+d x)+b^2 \sin ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]

[Out]

-(4*a*b*Csc[c + d*x] + a^2*Csc[c + d*x]^2 + 2*(a^2 - b^2)*Log[Sin[c + d*x]] + 4*a*b*Sin[c + d*x] + b^2*Sin[c +
d*x]^2)/(2*d)

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Maple [A]  time = 0.053, size = 120, normalized size = 1.4 \begin{align*} -{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-2\,{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{d\sin \left ( dx+c \right ) }}-2\,{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) }{d}}-4\,{\frac{ab\sin \left ( dx+c \right ) }{d}}+{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{{b}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{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))^2,x)

[Out]

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

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Maxima [A]  time = 1.56796, size = 93, normalized size = 1.11 \begin{align*} -\frac{b^{2} \sin \left (d x + c\right )^{2} + 4 \, a b \sin \left (d x + c\right ) + 2 \,{\left (a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right )\right ) + \frac{4 \, a b \sin \left (d x + c\right ) + a^{2}}{\sin \left (d x + c\right )^{2}}}{2 \, 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))^2,x, algorithm="maxima")

[Out]

-1/2*(b^2*sin(d*x + c)^2 + 4*a*b*sin(d*x + c) + 2*(a^2 - b^2)*log(sin(d*x + c)) + (4*a*b*sin(d*x + c) + a^2)/s
in(d*x + c)^2)/d

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Fricas [A]  time = 1.5208, size = 271, normalized size = 3.23 \begin{align*} \frac{2 \, b^{2} \cos \left (d x + c\right )^{4} - 3 \, b^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + b^{2} - 4 \,{\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + b^{2}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 8 \,{\left (a b \cos \left (d x + c\right )^{2} - 2 \, a b\right )} \sin \left (d x + c\right )}{4 \,{\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))^2,x, algorithm="fricas")

[Out]

1/4*(2*b^2*cos(d*x + c)^4 - 3*b^2*cos(d*x + c)^2 + 2*a^2 + b^2 - 4*((a^2 - b^2)*cos(d*x + c)^2 - a^2 + b^2)*lo
g(1/2*sin(d*x + c)) - 8*(a*b*cos(d*x + c)^2 - 2*a*b)*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 )^{2} \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))**2,x)

[Out]

Integral((a + b*sin(c + d*x))**2*cot(c + d*x)**3, x)

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Giac [A]  time = 2.31265, size = 134, normalized size = 1.6 \begin{align*} -\frac{b^{2} \sin \left (d x + c\right )^{2} + 4 \, a b \sin \left (d x + c\right ) + 2 \,{\left (a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac{3 \, a^{2} \sin \left (d x + c\right )^{2} - 3 \, b^{2} \sin \left (d x + c\right )^{2} - 4 \, a b \sin \left (d x + c\right ) - a^{2}}{\sin \left (d x + c\right )^{2}}}{2 \, 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))^2,x, algorithm="giac")

[Out]

-1/2*(b^2*sin(d*x + c)^2 + 4*a*b*sin(d*x + c) + 2*(a^2 - b^2)*log(abs(sin(d*x + c))) - (3*a^2*sin(d*x + c)^2 -
3*b^2*sin(d*x + c)^2 - 4*a*b*sin(d*x + c) - a^2)/sin(d*x + c)^2)/d