3.2 \(\int (a+b \csc ^2(c+d x))^3 \, dx\)

Optimal. Leaf size=74 \[ a^3 x-\frac {b \left (3 a^2+3 a b+b^2\right ) \cot (c+d x)}{d}-\frac {b^2 (3 a+2 b) \cot ^3(c+d x)}{3 d}-\frac {b^3 \cot ^5(c+d x)}{5 d} \]

[Out]

a^3*x-b*(3*a^2+3*a*b+b^2)*cot(d*x+c)/d-1/3*b^2*(3*a+2*b)*cot(d*x+c)^3/d-1/5*b^3*cot(d*x+c)^5/d

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Rubi [A]  time = 0.05, antiderivative size = 74, normalized size of antiderivative = 1.00, 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} \[ -\frac {b \left (3 a^2+3 a b+b^2\right ) \cot (c+d x)}{d}+a^3 x-\frac {b^2 (3 a+2 b) \cot ^3(c+d x)}{3 d}-\frac {b^3 \cot ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csc[c + d*x]^2)^3,x]

[Out]

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

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])

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 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]

Rubi steps

\begin {align*} \int \left (a+b \csc ^2(c+d x)\right )^3 \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (a+b+b x^2\right )^3}{1+x^2} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (b \left (3 a^2+3 a b+b^2\right )+b^2 (3 a+2 b) x^2+b^3 x^4+\frac {a^3}{1+x^2}\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {b \left (3 a^2+3 a b+b^2\right ) \cot (c+d x)}{d}-\frac {b^2 (3 a+2 b) \cot ^3(c+d x)}{3 d}-\frac {b^3 \cot ^5(c+d x)}{5 d}-\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{d}\\ &=a^3 x-\frac {b \left (3 a^2+3 a b+b^2\right ) \cot (c+d x)}{d}-\frac {b^2 (3 a+2 b) \cot ^3(c+d x)}{3 d}-\frac {b^3 \cot ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 3.67, size = 112, normalized size = 1.51 \[ \frac {8 \sin ^6(c+d x) \left (a+b \csc ^2(c+d x)\right )^3 \left (b \cot (c+d x) \left (45 a^2+b (15 a+4 b) \csc ^2(c+d x)+30 a b+3 b^2 \csc ^4(c+d x)+8 b^2\right )-15 a^3 (c+d x)\right )}{15 d (a \cos (2 (c+d x))-a-2 b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Csc[c + d*x]^2)^3,x]

[Out]

(8*(a + b*Csc[c + d*x]^2)^3*(-15*a^3*(c + d*x) + b*Cot[c + d*x]*(45*a^2 + 30*a*b + 8*b^2 + b*(15*a + 4*b)*Csc[
c + d*x]^2 + 3*b^2*Csc[c + d*x]^4))*Sin[c + d*x]^6)/(15*d*(-a - 2*b + a*Cos[2*(c + d*x)])^3)

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fricas [B]  time = 0.47, size = 159, normalized size = 2.15 \[ -\frac {{\left (45 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )} \cos \left (d x + c\right )^{5} - 5 \, {\left (18 \, a^{2} b + 15 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right ) - 15 \, {\left (a^{3} d x \cos \left (d x + c\right )^{4} - 2 \, a^{3} d x \cos \left (d x + c\right )^{2} + a^{3} d x\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csc(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

-1/15*((45*a^2*b + 30*a*b^2 + 8*b^3)*cos(d*x + c)^5 - 5*(18*a^2*b + 15*a*b^2 + 4*b^3)*cos(d*x + c)^3 + 15*(3*a
^2*b + 3*a*b^2 + b^3)*cos(d*x + c) - 15*(a^3*d*x*cos(d*x + c)^4 - 2*a^3*d*x*cos(d*x + c)^2 + a^3*d*x)*sin(d*x
+ c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x + c))

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giac [B]  time = 0.36, size = 211, normalized size = 2.85 \[ \frac {3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 60 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 25 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 480 \, {\left (d x + c\right )} a^{3} + 720 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 540 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 150 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {720 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 540 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 150 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 60 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 25 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, b^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csc(d*x+c)^2)^3,x, algorithm="giac")

[Out]

1/480*(3*b^3*tan(1/2*d*x + 1/2*c)^5 + 60*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 25*b^3*tan(1/2*d*x + 1/2*c)^3 + 480*(d
*x + c)*a^3 + 720*a^2*b*tan(1/2*d*x + 1/2*c) + 540*a*b^2*tan(1/2*d*x + 1/2*c) + 150*b^3*tan(1/2*d*x + 1/2*c) -
(720*a^2*b*tan(1/2*d*x + 1/2*c)^4 + 540*a*b^2*tan(1/2*d*x + 1/2*c)^4 + 150*b^3*tan(1/2*d*x + 1/2*c)^4 + 60*a*
b^2*tan(1/2*d*x + 1/2*c)^2 + 25*b^3*tan(1/2*d*x + 1/2*c)^2 + 3*b^3)/tan(1/2*d*x + 1/2*c)^5)/d

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maple [A]  time = 1.36, size = 83, normalized size = 1.12 \[ \frac {a^{3} \left (d x +c \right )-3 a^{2} b \cot \left (d x +c \right )+3 b^{2} a \left (-\frac {2}{3}-\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{3}\right ) \cot \left (d x +c \right )+b^{3} \left (-\frac {8}{15}-\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\csc ^{2}\left (d x +c \right )\right )}{15}\right ) \cot \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*csc(d*x+c)^2)^3,x)

[Out]

1/d*(a^3*(d*x+c)-3*a^2*b*cot(d*x+c)+3*b^2*a*(-2/3-1/3*csc(d*x+c)^2)*cot(d*x+c)+b^3*(-8/15-1/5*csc(d*x+c)^4-4/1
5*csc(d*x+c)^2)*cot(d*x+c))

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maxima [A]  time = 0.32, size = 90, normalized size = 1.22 \[ a^{3} x - \frac {3 \, a^{2} b}{d \tan \left (d x + c\right )} - \frac {{\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} a b^{2}}{d \tan \left (d x + c\right )^{3}} - \frac {{\left (15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} + 3\right )} b^{3}}{15 \, d \tan \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csc(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

a^3*x - 3*a^2*b/(d*tan(d*x + c)) - (3*tan(d*x + c)^2 + 1)*a*b^2/(d*tan(d*x + c)^3) - 1/15*(15*tan(d*x + c)^4 +
10*tan(d*x + c)^2 + 3)*b^3/(d*tan(d*x + c)^5)

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mupad [B]  time = 0.43, size = 70, normalized size = 0.95 \[ a^3\,x-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {2\,b^3}{3}+a\,b^2\right )+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )+\frac {b^3}{5}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/sin(c + d*x)^2)^3,x)

[Out]

a^3*x - (tan(c + d*x)^2*(a*b^2 + (2*b^3)/3) + tan(c + d*x)^4*(3*a*b^2 + 3*a^2*b + b^3) + b^3/5)/(d*tan(c + d*x
)^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \csc ^{2}{\left (c + d x \right )}\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csc(d*x+c)**2)**3,x)

[Out]

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

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