3.2.60 \(\int (c+a^2 c x^2)^2 \text {ArcTan}(a x) \, dx\) [160]

Optimal. Leaf size=117 \[ -\frac {2 c^2 \left (1+a^2 x^2\right )}{15 a}-\frac {c^2 \left (1+a^2 x^2\right )^2}{20 a}+\frac {8}{15} c^2 x \text {ArcTan}(a x)+\frac {4}{15} c^2 x \left (1+a^2 x^2\right ) \text {ArcTan}(a x)+\frac {1}{5} c^2 x \left (1+a^2 x^2\right )^2 \text {ArcTan}(a x)-\frac {4 c^2 \log \left (1+a^2 x^2\right )}{15 a} \]

[Out]

-2/15*c^2*(a^2*x^2+1)/a-1/20*c^2*(a^2*x^2+1)^2/a+8/15*c^2*x*arctan(a*x)+4/15*c^2*x*(a^2*x^2+1)*arctan(a*x)+1/5
*c^2*x*(a^2*x^2+1)^2*arctan(a*x)-4/15*c^2*ln(a^2*x^2+1)/a

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Rubi [A]
time = 0.03, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4998, 4930, 266} \begin {gather*} \frac {1}{5} c^2 x \left (a^2 x^2+1\right )^2 \text {ArcTan}(a x)+\frac {4}{15} c^2 x \left (a^2 x^2+1\right ) \text {ArcTan}(a x)-\frac {c^2 \left (a^2 x^2+1\right )^2}{20 a}-\frac {2 c^2 \left (a^2 x^2+1\right )}{15 a}-\frac {4 c^2 \log \left (a^2 x^2+1\right )}{15 a}+\frac {8}{15} c^2 x \text {ArcTan}(a x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + a^2*c*x^2)^2*ArcTan[a*x],x]

[Out]

(-2*c^2*(1 + a^2*x^2))/(15*a) - (c^2*(1 + a^2*x^2)^2)/(20*a) + (8*c^2*x*ArcTan[a*x])/15 + (4*c^2*x*(1 + a^2*x^
2)*ArcTan[a*x])/15 + (c^2*x*(1 + a^2*x^2)^2*ArcTan[a*x])/5 - (4*c^2*Log[1 + a^2*x^2])/(15*a)

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Dist[b*c
*n*p, Int[x^n*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0
] && (EqQ[n, 1] || EqQ[p, 1])

Rule 4998

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(-b)*((d + e*x^2)^q/(2*c
*q*(2*q + 1))), x] + (Dist[2*d*(q/(2*q + 1)), Int[(d + e*x^2)^(q - 1)*(a + b*ArcTan[c*x]), x], x] + Simp[x*(d
+ e*x^2)^q*((a + b*ArcTan[c*x])/(2*q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[q, 0]

Rubi steps

\begin {align*} \int \left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x) \, dx &=-\frac {c^2 \left (1+a^2 x^2\right )^2}{20 a}+\frac {1}{5} c^2 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)+\frac {1}{5} (4 c) \int \left (c+a^2 c x^2\right ) \tan ^{-1}(a x) \, dx\\ &=-\frac {2 c^2 \left (1+a^2 x^2\right )}{15 a}-\frac {c^2 \left (1+a^2 x^2\right )^2}{20 a}+\frac {4}{15} c^2 x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)+\frac {1}{5} c^2 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)+\frac {1}{15} \left (8 c^2\right ) \int \tan ^{-1}(a x) \, dx\\ &=-\frac {2 c^2 \left (1+a^2 x^2\right )}{15 a}-\frac {c^2 \left (1+a^2 x^2\right )^2}{20 a}+\frac {8}{15} c^2 x \tan ^{-1}(a x)+\frac {4}{15} c^2 x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)+\frac {1}{5} c^2 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)-\frac {1}{15} \left (8 a c^2\right ) \int \frac {x}{1+a^2 x^2} \, dx\\ &=-\frac {2 c^2 \left (1+a^2 x^2\right )}{15 a}-\frac {c^2 \left (1+a^2 x^2\right )^2}{20 a}+\frac {8}{15} c^2 x \tan ^{-1}(a x)+\frac {4}{15} c^2 x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)+\frac {1}{5} c^2 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)-\frac {4 c^2 \log \left (1+a^2 x^2\right )}{15 a}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 65, normalized size = 0.56 \begin {gather*} \frac {c^2 \left (-14 a^2 x^2-3 a^4 x^4+4 a x \left (15+10 a^2 x^2+3 a^4 x^4\right ) \text {ArcTan}(a x)-16 \log \left (1+a^2 x^2\right )\right )}{60 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(-14*a^2*x^2 - 3*a^4*x^4 + 4*a*x*(15 + 10*a^2*x^2 + 3*a^4*x^4)*ArcTan[a*x] - 16*Log[1 + a^2*x^2]))/(60*a)

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Maple [A]
time = 0.10, size = 80, normalized size = 0.68

method	result	size
derivativedivides	\(\frac {\frac {c^{2} \arctan \left (a x \right ) a^{5} x^{5}}{5}+\frac {2 a^{3} c^{2} x^{3} \arctan \left (a x \right )}{3}+a \,c^{2} x \arctan \left (a x \right )-\frac {c^{2} \left (\frac {3 a^{4} x^{4}}{4}+\frac {7 a^{2} x^{2}}{2}+4 \ln \left (a^{2} x^{2}+1\right )\right )}{15}}{a}\)	\(80\)
default	\(\frac {\frac {c^{2} \arctan \left (a x \right ) a^{5} x^{5}}{5}+\frac {2 a^{3} c^{2} x^{3} \arctan \left (a x \right )}{3}+a \,c^{2} x \arctan \left (a x \right )-\frac {c^{2} \left (\frac {3 a^{4} x^{4}}{4}+\frac {7 a^{2} x^{2}}{2}+4 \ln \left (a^{2} x^{2}+1\right )\right )}{15}}{a}\)	\(80\)
risch	\(-\frac {i c^{2} x \left (3 a^{4} x^{4}+10 a^{2} x^{2}+15\right ) \ln \left (i a x +1\right )}{30}+\frac {i c^{2} a^{4} x^{5} \ln \left (-i a x +1\right )}{10}-\frac {c^{2} a^{3} x^{4}}{20}+\frac {i c^{2} a^{2} x^{3} \ln \left (-i a x +1\right )}{3}-\frac {7 a \,c^{2} x^{2}}{30}+\frac {i c^{2} x \ln \left (-i a x +1\right )}{2}-\frac {4 c^{2} \ln \left (-a^{2} x^{2}-1\right )}{15 a}-\frac {49 c^{2}}{180 a}\)	\(137\)
meijerg	\(\frac {c^{2} \left (\frac {a^{2} x^{2} \left (-3 a^{2} x^{2}+6\right )}{15}+\frac {4 a^{6} x^{6} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{5 \sqrt {a^{2} x^{2}}}-\frac {2 \ln \left (a^{2} x^{2}+1\right )}{5}\right )}{4 a}+\frac {c^{2} \left (-\frac {2 a^{2} x^{2}}{3}+\frac {4 a^{4} x^{4} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{3 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (a^{2} x^{2}+1\right )}{3}\right )}{2 a}+\frac {c^{2} \left (\frac {4 a^{2} x^{2} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (a^{2} x^{2}+1\right )\right )}{4 a}\)	\(172\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c*x^2+c)^2*arctan(a*x),x,method=_RETURNVERBOSE)

[Out]

1/a*(1/5*c^2*arctan(a*x)*a^5*x^5+2/3*a^3*c^2*x^3*arctan(a*x)+a*c^2*x*arctan(a*x)-1/15*c^2*(3/4*a^4*x^4+7/2*a^2
*x^2+4*ln(a^2*x^2+1)))

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Maxima [A]
time = 0.25, size = 77, normalized size = 0.66 \begin {gather*} -\frac {1}{60} \, {\left (3 \, a^{2} c^{2} x^{4} + 14 \, c^{2} x^{2} + \frac {16 \, c^{2} \log \left (a^{2} x^{2} + 1\right )}{a^{2}}\right )} a + \frac {1}{15} \, {\left (3 \, a^{4} c^{2} x^{5} + 10 \, a^{2} c^{2} x^{3} + 15 \, c^{2} x\right )} \arctan \left (a x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(3*a^2*c^2*x^4 + 14*c^2*x^2 + 16*c^2*log(a^2*x^2 + 1)/a^2)*a + 1/15*(3*a^4*c^2*x^5 + 10*a^2*c^2*x^3 + 15
*c^2*x)*arctan(a*x)

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Fricas [A]
time = 2.31, size = 79, normalized size = 0.68 \begin {gather*} -\frac {3 \, a^{4} c^{2} x^{4} + 14 \, a^{2} c^{2} x^{2} + 16 \, c^{2} \log \left (a^{2} x^{2} + 1\right ) - 4 \, {\left (3 \, a^{5} c^{2} x^{5} + 10 \, a^{3} c^{2} x^{3} + 15 \, a c^{2} x\right )} \arctan \left (a x\right )}{60 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(3*a^4*c^2*x^4 + 14*a^2*c^2*x^2 + 16*c^2*log(a^2*x^2 + 1) - 4*(3*a^5*c^2*x^5 + 10*a^3*c^2*x^3 + 15*a*c^2
*x)*arctan(a*x))/a

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Sympy [A]
time = 0.25, size = 88, normalized size = 0.75 \begin {gather*} \begin {cases} \frac {a^{4} c^{2} x^{5} \operatorname {atan}{\left (a x \right )}}{5} - \frac {a^{3} c^{2} x^{4}}{20} + \frac {2 a^{2} c^{2} x^{3} \operatorname {atan}{\left (a x \right )}}{3} - \frac {7 a c^{2} x^{2}}{30} + c^{2} x \operatorname {atan}{\left (a x \right )} - \frac {4 c^{2} \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{15 a} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2*c*x**2+c)**2*atan(a*x),x)

[Out]

Piecewise((a**4*c**2*x**5*atan(a*x)/5 - a**3*c**2*x**4/20 + 2*a**2*c**2*x**3*atan(a*x)/3 - 7*a*c**2*x**2/30 +
c**2*x*atan(a*x) - 4*c**2*log(x**2 + a**(-2))/(15*a), Ne(a, 0)), (0, True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [B]
time = 0.20, size = 69, normalized size = 0.59 \begin {gather*} -\frac {c^2\,\left (16\,\ln \left (a^2\,x^2+1\right )+14\,a^2\,x^2+3\,a^4\,x^4-40\,a^3\,x^3\,\mathrm {atan}\left (a\,x\right )-12\,a^5\,x^5\,\mathrm {atan}\left (a\,x\right )-60\,a\,x\,\mathrm {atan}\left (a\,x\right )\right )}{60\,a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a*x)*(c + a^2*c*x^2)^2,x)

[Out]

-(c^2*(16*log(a^2*x^2 + 1) + 14*a^2*x^2 + 3*a^4*x^4 - 40*a^3*x^3*atan(a*x) - 12*a^5*x^5*atan(a*x) - 60*a*x*ata
n(a*x)))/(60*a)

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