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

Optimal. Leaf size=78 \[ -\frac{a^2 \cot (c+d x)}{d}+a^2 (-x)+\frac{2 a b \cos (c+d x)}{d}-\frac{2 a b \tanh ^{-1}(\cos (c+d x))}{d}+\frac{b^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{b^2 x}{2} \]

[Out]

-(a^2*x) + (b^2*x)/2 - (2*a*b*ArcTanh[Cos[c + d*x]])/d + (2*a*b*Cos[c + d*x])/d - (a^2*Cot[c + d*x])/d + (b^2*
Cos[c + d*x]*Sin[c + d*x])/(2*d)

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Rubi [A]  time = 0.0855624, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2722, 2635, 8, 2592, 321, 206, 3473} \[ -\frac{a^2 \cot (c+d x)}{d}+a^2 (-x)+\frac{2 a b \cos (c+d x)}{d}-\frac{2 a b \tanh ^{-1}(\cos (c+d x))}{d}+\frac{b^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{b^2 x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*x) + (b^2*x)/2 - (2*a*b*ArcTanh[Cos[c + d*x]])/d + (2*a*b*Cos[c + d*x])/d - (a^2*Cot[c + d*x])/d + (b^2*
Cos[c + d*x]*Sin[c + d*x])/(2*d)

Rule 2722

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Int[Expan
dIntegrand[(g*Tan[e + f*x])^p, (a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2
, 0] && IGtQ[m, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rubi steps

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

Mathematica [A]  time = 0.416328, size = 116, normalized size = 1.49 \[ \frac{2 a^2 \tan \left (\frac{1}{2} (c+d x)\right )-2 a^2 \cot \left (\frac{1}{2} (c+d x)\right )-4 a^2 c-4 a^2 d x+8 a b \cos (c+d x)+8 a b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-8 a b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+b^2 \sin (2 (c+d x))+2 b^2 c+2 b^2 d x}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^2*c + 2*b^2*c - 4*a^2*d*x + 2*b^2*d*x + 8*a*b*Cos[c + d*x] - 2*a^2*Cot[(c + d*x)/2] - 8*a*b*Log[Cos[(c +
d*x)/2]] + 8*a*b*Log[Sin[(c + d*x)/2]] + b^2*Sin[2*(c + d*x)] + 2*a^2*Tan[(c + d*x)/2])/(4*d)

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Maple [A]  time = 0.039, size = 102, normalized size = 1.3 \begin{align*} -{a}^{2}x-{\frac{{a}^{2}\cot \left ( dx+c \right ) }{d}}-{\frac{{a}^{2}c}{d}}+2\,{\frac{ab\cos \left ( dx+c \right ) }{d}}+2\,{\frac{ab\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{2}x}{2}}+{\frac{{b}^{2}c}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^2*(a+b*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 2.59493, size = 107, normalized size = 1.37 \begin{align*} -\frac{4 \,{\left (d x + c + \frac{1}{\tan \left (d x + c\right )}\right )} a^{2} -{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b^{2} - 4 \, a b{\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.64996, size = 308, normalized size = 3.95 \begin{align*} -\frac{b^{2} \cos \left (d x + c\right )^{3} + 2 \, a b \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 2 \, a b \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right ) +{\left ({\left (2 \, a^{2} - b^{2}\right )} d x - 4 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, d \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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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 ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.23738, size = 200, normalized size = 2.56 \begin{align*} \frac{4 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) -{\left (2 \, a^{2} - b^{2}\right )}{\left (d x + c\right )} - \frac{4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} - \frac{2 \,{\left (b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a b\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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