3.3 \(\int (a+b \csc ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=41 \[ a^2 x-\frac{b (2 a+b) \cot (c+d x)}{d}-\frac{b^2 \cot ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0301582, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4128, 390, 203} \[ a^2 x-\frac{b (2 a+b) \cot (c+d x)}{d}-\frac{b^2 \cot ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csc[c + d*x]^2)^2,x]

[Out]

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

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 203

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

Rubi steps

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

Mathematica [B]  time = 0.630074, size = 83, normalized size = 2.02 \[ -\frac{4 \sin ^4(c+d x) \left (a+b \csc ^2(c+d x)\right )^2 \left (b \cot (c+d x) \left (6 a+b \csc ^2(c+d x)+2 b\right )-3 a^2 (c+d x)\right )}{3 d (a (-\cos (2 (c+d x)))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Csc[c + d*x]^2)^2,x]

[Out]

(-4*(a + b*Csc[c + d*x]^2)^2*(-3*a^2*(c + d*x) + b*Cot[c + d*x]*(6*a + 2*b + b*Csc[c + d*x]^2))*Sin[c + d*x]^4
)/(3*d*(a + 2*b - a*Cos[2*(c + d*x)])^2)

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Maple [A]  time = 0.032, size = 47, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( dx+c \right ) -2\,\cot \left ( dx+c \right ) ab+{b}^{2} \left ( -{\frac{2}{3}}-{\frac{ \left ( \csc \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \cot \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.987929, size = 66, normalized size = 1.61 \begin{align*} a^{2} x - \frac{2 \, a b}{d \tan \left (d x + c\right )} - \frac{{\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} b^{2}}{3 \, d \tan \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 0.477516, size = 216, normalized size = 5.27 \begin{align*} -\frac{2 \,{\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (2 \, a b + b^{2}\right )} \cos \left (d x + c\right ) - 3 \,{\left (a^{2} d x \cos \left (d x + c\right )^{2} - a^{2} d x\right )} \sin \left (d x + c\right )}{3 \,{\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \csc ^{2}{\left (c + d x \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.61886, size = 143, normalized size = 3.49 \begin{align*} \frac{b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \,{\left (d x + c\right )} a^{2} + 24 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{24 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 9 \, b^{2} \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 )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(b^2*tan(1/2*d*x + 1/2*c)^3 + 24*(d*x + c)*a^2 + 24*a*b*tan(1/2*d*x + 1/2*c) + 9*b^2*tan(1/2*d*x + 1/2*c)
- (24*a*b*tan(1/2*d*x + 1/2*c)^2 + 9*b^2*tan(1/2*d*x + 1/2*c)^2 + b^2)/tan(1/2*d*x + 1/2*c)^3)/d