3.55 \(\int (c+d x^2)^2 \cot ^{-1}(a x) \, dx\)

Optimal. Leaf size=109 \[ \frac {d x^2 \left (10 a^2 c-3 d\right )}{30 a^3}+\frac {\left (15 a^4 c^2-10 a^2 c d+3 d^2\right ) \log \left (a^2 x^2+1\right )}{30 a^5}+c^2 x \cot ^{-1}(a x)+\frac {2}{3} c d x^3 \cot ^{-1}(a x)+\frac {1}{5} d^2 x^5 \cot ^{-1}(a x)+\frac {d^2 x^4}{20 a} \]

[Out]

1/30*(10*a^2*c-3*d)*d*x^2/a^3+1/20*d^2*x^4/a+c^2*x*arccot(a*x)+2/3*c*d*x^3*arccot(a*x)+1/5*d^2*x^5*arccot(a*x)
+1/30*(15*a^4*c^2-10*a^2*c*d+3*d^2)*ln(a^2*x^2+1)/a^5

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Rubi [A]  time = 0.13, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {194, 4913, 1594, 1247, 698} \[ \frac {\left (15 a^4 c^2-10 a^2 c d+3 d^2\right ) \log \left (a^2 x^2+1\right )}{30 a^5}+\frac {d x^2 \left (10 a^2 c-3 d\right )}{30 a^3}+c^2 x \cot ^{-1}(a x)+\frac {2}{3} c d x^3 \cot ^{-1}(a x)+\frac {d^2 x^4}{20 a}+\frac {1}{5} d^2 x^5 \cot ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^2*ArcCot[a*x],x]

[Out]

((10*a^2*c - 3*d)*d*x^2)/(30*a^3) + (d^2*x^4)/(20*a) + c^2*x*ArcCot[a*x] + (2*c*d*x^3*ArcCot[a*x])/3 + (d^2*x^
5*ArcCot[a*x])/5 + ((15*a^4*c^2 - 10*a^2*c*d + 3*d^2)*Log[1 + a^2*x^2])/(30*a^5)

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 4913

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^q, x]}, Dist[a + b*ArcCot[c*x], u, x] + Dist[b*c, Int[u/(1 + c^2*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x]
&& (IntegerQ[q] || ILtQ[q + 1/2, 0])

Rubi steps

\begin {align*} \int \left (c+d x^2\right )^2 \cot ^{-1}(a x) \, dx &=c^2 x \cot ^{-1}(a x)+\frac {2}{3} c d x^3 \cot ^{-1}(a x)+\frac {1}{5} d^2 x^5 \cot ^{-1}(a x)+a \int \frac {c^2 x+\frac {2}{3} c d x^3+\frac {d^2 x^5}{5}}{1+a^2 x^2} \, dx\\ &=c^2 x \cot ^{-1}(a x)+\frac {2}{3} c d x^3 \cot ^{-1}(a x)+\frac {1}{5} d^2 x^5 \cot ^{-1}(a x)+a \int \frac {x \left (c^2+\frac {2}{3} c d x^2+\frac {d^2 x^4}{5}\right )}{1+a^2 x^2} \, dx\\ &=c^2 x \cot ^{-1}(a x)+\frac {2}{3} c d x^3 \cot ^{-1}(a x)+\frac {1}{5} d^2 x^5 \cot ^{-1}(a x)+\frac {1}{2} a \operatorname {Subst}\left (\int \frac {c^2+\frac {2 c d x}{3}+\frac {d^2 x^2}{5}}{1+a^2 x} \, dx,x,x^2\right )\\ &=c^2 x \cot ^{-1}(a x)+\frac {2}{3} c d x^3 \cot ^{-1}(a x)+\frac {1}{5} d^2 x^5 \cot ^{-1}(a x)+\frac {1}{2} a \operatorname {Subst}\left (\int \left (\frac {\left (10 a^2 c-3 d\right ) d}{15 a^4}+\frac {d^2 x}{5 a^2}+\frac {15 a^4 c^2-10 a^2 c d+3 d^2}{15 a^4 \left (1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac {\left (10 a^2 c-3 d\right ) d x^2}{30 a^3}+\frac {d^2 x^4}{20 a}+c^2 x \cot ^{-1}(a x)+\frac {2}{3} c d x^3 \cot ^{-1}(a x)+\frac {1}{5} d^2 x^5 \cot ^{-1}(a x)+\frac {\left (15 a^4 c^2-10 a^2 c d+3 d^2\right ) \log \left (1+a^2 x^2\right )}{30 a^5}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 97, normalized size = 0.89 \[ \frac {4 a^5 x \cot ^{-1}(a x) \left (15 c^2+10 c d x^2+3 d^2 x^4\right )+a^2 d x^2 \left (a^2 \left (20 c+3 d x^2\right )-6 d\right )+\left (30 a^4 c^2-20 a^2 c d+6 d^2\right ) \log \left (a^2 x^2+1\right )}{60 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^2*ArcCot[a*x],x]

[Out]

(a^2*d*x^2*(-6*d + a^2*(20*c + 3*d*x^2)) + 4*a^5*x*(15*c^2 + 10*c*d*x^2 + 3*d^2*x^4)*ArcCot[a*x] + (30*a^4*c^2
- 20*a^2*c*d + 6*d^2)*Log[1 + a^2*x^2])/(60*a^5)

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fricas [A]  time = 1.52, size = 108, normalized size = 0.99 \[ \frac {3 \, a^{4} d^{2} x^{4} + 2 \, {\left (10 \, a^{4} c d - 3 \, a^{2} d^{2}\right )} x^{2} + 4 \, {\left (3 \, a^{5} d^{2} x^{5} + 10 \, a^{5} c d x^{3} + 15 \, a^{5} c^{2} x\right )} \operatorname {arccot}\left (a x\right ) + 2 \, {\left (15 \, a^{4} c^{2} - 10 \, a^{2} c d + 3 \, d^{2}\right )} \log \left (a^{2} x^{2} + 1\right )}{60 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(3*a^4*d^2*x^4 + 2*(10*a^4*c*d - 3*a^2*d^2)*x^2 + 4*(3*a^5*d^2*x^5 + 10*a^5*c*d*x^3 + 15*a^5*c^2*x)*arcco
t(a*x) + 2*(15*a^4*c^2 - 10*a^2*c*d + 3*d^2)*log(a^2*x^2 + 1))/a^5

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giac [A]  time = 0.14, size = 171, normalized size = 1.57 \[ \frac {1}{60} \, {\left (\frac {4 \, {\left (3 \, d^{2} + \frac {10 \, c d}{x^{2}} + \frac {15 \, c^{2}}{x^{4}}\right )} x^{5} \arctan \left (\frac {1}{a x}\right )}{a} + \frac {{\left (3 \, d^{2} + \frac {20 \, c d}{x^{2}} + \frac {45 \, c^{2}}{x^{4}} - \frac {6 \, d^{2}}{a^{2} x^{2}} - \frac {30 \, c d}{a^{2} x^{4}} + \frac {9 \, d^{2}}{a^{4} x^{4}}\right )} x^{4}}{a^{2}} + \frac {2 \, {\left (15 \, a^{4} c^{2} - 10 \, a^{2} c d + 3 \, d^{2}\right )} \log \left (\frac {1}{a^{2} x^{2}} + 1\right )}{a^{6}} - \frac {2 \, {\left (15 \, a^{4} c^{2} - 10 \, a^{2} c d + 3 \, d^{2}\right )} \log \left (\frac {1}{a^{2} x^{2}}\right )}{a^{6}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(4*(3*d^2 + 10*c*d/x^2 + 15*c^2/x^4)*x^5*arctan(1/(a*x))/a + (3*d^2 + 20*c*d/x^2 + 45*c^2/x^4 - 6*d^2/(a^
2*x^2) - 30*c*d/(a^2*x^4) + 9*d^2/(a^4*x^4))*x^4/a^2 + 2*(15*a^4*c^2 - 10*a^2*c*d + 3*d^2)*log(1/(a^2*x^2) + 1
)/a^6 - 2*(15*a^4*c^2 - 10*a^2*c*d + 3*d^2)*log(1/(a^2*x^2))/a^6)*a

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maple [A]  time = 0.04, size = 119, normalized size = 1.09 \[ \frac {d^{2} x^{5} \mathrm {arccot}\left (a x \right )}{5}+\frac {2 c d \,x^{3} \mathrm {arccot}\left (a x \right )}{3}+c^{2} x \,\mathrm {arccot}\left (a x \right )+\frac {d c \,x^{2}}{3 a}+\frac {d^{2} x^{4}}{20 a}-\frac {x^{2} d^{2}}{10 a^{3}}+\frac {c^{2} \ln \left (a^{2} x^{2}+1\right )}{2 a}-\frac {\ln \left (a^{2} x^{2}+1\right ) c d}{3 a^{3}}+\frac {\ln \left (a^{2} x^{2}+1\right ) d^{2}}{10 a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*d^2*x^5*arccot(a*x)+2/3*c*d*x^3*arccot(a*x)+c^2*x*arccot(a*x)+1/3/a*d*c*x^2+1/20*d^2*x^4/a-1/10/a^3*x^2*d^
2+1/2*c^2*ln(a^2*x^2+1)/a-1/3/a^3*ln(a^2*x^2+1)*c*d+1/10/a^5*ln(a^2*x^2+1)*d^2

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maxima [A]  time = 0.31, size = 103, normalized size = 0.94 \[ \frac {1}{60} \, a {\left (\frac {3 \, a^{2} d^{2} x^{4} + 2 \, {\left (10 \, a^{2} c d - 3 \, d^{2}\right )} x^{2}}{a^{4}} + \frac {2 \, {\left (15 \, a^{4} c^{2} - 10 \, a^{2} c d + 3 \, d^{2}\right )} \log \left (a^{2} x^{2} + 1\right )}{a^{6}}\right )} + \frac {1}{15} \, {\left (3 \, d^{2} x^{5} + 10 \, c d x^{3} + 15 \, c^{2} x\right )} \operatorname {arccot}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*a*((3*a^2*d^2*x^4 + 2*(10*a^2*c*d - 3*d^2)*x^2)/a^4 + 2*(15*a^4*c^2 - 10*a^2*c*d + 3*d^2)*log(a^2*x^2 + 1
)/a^6) + 1/15*(3*d^2*x^5 + 10*c*d*x^3 + 15*c^2*x)*arccot(a*x)

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mupad [B]  time = 0.84, size = 116, normalized size = 1.06 \[ \frac {a^4\,\left (\frac {c^2\,\ln \left (a^2\,x^2+1\right )}{2}+\frac {d^2\,x^4}{20}+\frac {c\,d\,x^2}{3}\right )-a^2\,\left (\frac {d^2\,x^2}{10}+\frac {c\,d\,\ln \left (a^2\,x^2+1\right )}{3}\right )+\frac {d^2\,\ln \left (a^2\,x^2+1\right )}{10}}{a^5}+c^2\,x\,\mathrm {acot}\left (a\,x\right )+\frac {d^2\,x^5\,\mathrm {acot}\left (a\,x\right )}{5}+\frac {2\,c\,d\,x^3\,\mathrm {acot}\left (a\,x\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(a*x)*(c + d*x^2)^2,x)

[Out]

(a^4*((c^2*log(a^2*x^2 + 1))/2 + (d^2*x^4)/20 + (c*d*x^2)/3) - a^2*((d^2*x^2)/10 + (c*d*log(a^2*x^2 + 1))/3) +
(d^2*log(a^2*x^2 + 1))/10)/a^5 + c^2*x*acot(a*x) + (d^2*x^5*acot(a*x))/5 + (2*c*d*x^3*acot(a*x))/3

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sympy [A]  time = 1.36, size = 151, normalized size = 1.39 \[ \begin {cases} c^{2} x \operatorname {acot}{\left (a x \right )} + \frac {2 c d x^{3} \operatorname {acot}{\left (a x \right )}}{3} + \frac {d^{2} x^{5} \operatorname {acot}{\left (a x \right )}}{5} + \frac {c^{2} \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{2 a} + \frac {c d x^{2}}{3 a} + \frac {d^{2} x^{4}}{20 a} - \frac {c d \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{3 a^{3}} - \frac {d^{2} x^{2}}{10 a^{3}} + \frac {d^{2} \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{10 a^{5}} & \text {for}\: a \neq 0 \\\frac {\pi \left (c^{2} x + \frac {2 c d x^{3}}{3} + \frac {d^{2} x^{5}}{5}\right )}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**2*acot(a*x),x)

[Out]

Piecewise((c**2*x*acot(a*x) + 2*c*d*x**3*acot(a*x)/3 + d**2*x**5*acot(a*x)/5 + c**2*log(x**2 + a**(-2))/(2*a)
+ c*d*x**2/(3*a) + d**2*x**4/(20*a) - c*d*log(x**2 + a**(-2))/(3*a**3) - d**2*x**2/(10*a**3) + d**2*log(x**2 +
a**(-2))/(10*a**5), Ne(a, 0)), (pi*(c**2*x + 2*c*d*x**3/3 + d**2*x**5/5)/2, True))

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