3.37 \(\int (a+b \csc (c+d x))^3 \, dx\)

Optimal. Leaf size=73 \[ -\frac{b \left (6 a^2+b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}+a^3 x-\frac{5 a b^2 \cot (c+d x)}{2 d}-\frac{b^2 \cot (c+d x) (a+b \csc (c+d x))}{2 d} \]

[Out]

a^3*x - (b*(6*a^2 + b^2)*ArcTanh[Cos[c + d*x]])/(2*d) - (5*a*b^2*Cot[c + d*x])/(2*d) - (b^2*Cot[c + d*x]*(a +
b*Csc[c + d*x]))/(2*d)

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Rubi [A]  time = 0.0469065, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3782, 3770, 3767, 8} \[ -\frac{b \left (6 a^2+b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}+a^3 x-\frac{5 a b^2 \cot (c+d x)}{2 d}-\frac{b^2 \cot (c+d x) (a+b \csc (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csc[c + d*x])^3,x]

[Out]

a^3*x - (b*(6*a^2 + b^2)*ArcTanh[Cos[c + d*x]])/(2*d) - (5*a*b^2*Cot[c + d*x])/(2*d) - (b^2*Cot[c + d*x]*(a +
b*Csc[c + d*x]))/(2*d)

Rule 3782

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[1/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 3)*Simp[a^3*(n - 1) + (b*(b^2*(n - 2) +
3*a^2*(n - 1)))*Csc[c + d*x] + (a*b^2*(3*n - 4))*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rubi steps

\begin{align*} \int (a+b \csc (c+d x))^3 \, dx &=-\frac{b^2 \cot (c+d x) (a+b \csc (c+d x))}{2 d}+\frac{1}{2} \int \left (2 a^3+b \left (6 a^2+b^2\right ) \csc (c+d x)+5 a b^2 \csc ^2(c+d x)\right ) \, dx\\ &=a^3 x-\frac{b^2 \cot (c+d x) (a+b \csc (c+d x))}{2 d}+\frac{1}{2} \left (5 a b^2\right ) \int \csc ^2(c+d x) \, dx+\frac{1}{2} \left (b \left (6 a^2+b^2\right )\right ) \int \csc (c+d x) \, dx\\ &=a^3 x-\frac{b \left (6 a^2+b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{b^2 \cot (c+d x) (a+b \csc (c+d x))}{2 d}-\frac{\left (5 a b^2\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{2 d}\\ &=a^3 x-\frac{b \left (6 a^2+b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{5 a b^2 \cot (c+d x)}{2 d}-\frac{b^2 \cot (c+d x) (a+b \csc (c+d x))}{2 d}\\ \end{align*}

Mathematica [B]  time = 0.648397, size = 152, normalized size = 2.08 \[ \frac{24 a^2 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-24 a^2 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+8 a^3 c+8 a^3 d x+12 a b^2 \tan \left (\frac{1}{2} (c+d x)\right )-12 a b^2 \cot \left (\frac{1}{2} (c+d x)\right )-b^3 \csc ^2\left (\frac{1}{2} (c+d x)\right )+b^3 \sec ^2\left (\frac{1}{2} (c+d x)\right )+4 b^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-4 b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Csc[c + d*x])^3,x]

[Out]

(8*a^3*c + 8*a^3*d*x - 12*a*b^2*Cot[(c + d*x)/2] - b^3*Csc[(c + d*x)/2]^2 - 24*a^2*b*Log[Cos[(c + d*x)/2]] - 4
*b^3*Log[Cos[(c + d*x)/2]] + 24*a^2*b*Log[Sin[(c + d*x)/2]] + 4*b^3*Log[Sin[(c + d*x)/2]] + b^3*Sec[(c + d*x)/
2]^2 + 12*a*b^2*Tan[(c + d*x)/2])/(8*d)

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Maple [A]  time = 0.044, size = 99, normalized size = 1.4 \begin{align*}{a}^{3}x+{\frac{{a}^{3}c}{d}}+3\,{\frac{{a}^{2}b\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}-3\,{\frac{a{b}^{2}\cot \left ( dx+c \right ) }{d}}-{\frac{{b}^{3}\csc \left ( dx+c \right ) \cot \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00415, size = 128, normalized size = 1.75 \begin{align*} a^{3} x + \frac{b^{3}{\left (\frac{2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} - \frac{3 \, a^{2} b \log \left (\cot \left (d x + c\right ) + \csc \left (d x + c\right )\right )}{d} - \frac{3 \, a b^{2}}{d \tan \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))^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 0.510696, size = 383, normalized size = 5.25 \begin{align*} \frac{4 \, a^{3} d x \cos \left (d x + c\right )^{2} - 4 \, a^{3} d x + 12 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \, b^{3} \cos \left (d x + c\right ) +{\left (6 \, a^{2} b + b^{3} -{\left (6 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left (6 \, a^{2} b + b^{3} -{\left (6 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\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((a+b*csc(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(4*a^3*d*x*cos(d*x + c)^2 - 4*a^3*d*x + 12*a*b^2*cos(d*x + c)*sin(d*x + c) + 2*b^3*cos(d*x + c) + (6*a^2*b
+ b^3 - (6*a^2*b + b^3)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) - (6*a^2*b + b^3 - (6*a^2*b + b^3)*cos(d*
x + c)^2)*log(-1/2*cos(d*x + c) + 1/2))/(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 \csc{\left (c + d x \right )}\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*csc(c + d*x))**3, x)

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Giac [B]  time = 1.47343, size = 181, normalized size = 2.48 \begin{align*} \frac{b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 8 \,{\left (d x + c\right )} a^{3} + 12 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \,{\left (6 \, a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{36 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 6 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b^3*tan(1/2*d*x + 1/2*c)^2 + 8*(d*x + c)*a^3 + 12*a*b^2*tan(1/2*d*x + 1/2*c) + 4*(6*a^2*b + b^3)*log(abs(
tan(1/2*d*x + 1/2*c))) - (36*a^2*b*tan(1/2*d*x + 1/2*c)^2 + 6*b^3*tan(1/2*d*x + 1/2*c)^2 + 12*a*b^2*tan(1/2*d*
x + 1/2*c) + b^3)/tan(1/2*d*x + 1/2*c)^2)/d