3.249 \(\int x^m (c+a^2 c x^2)^2 \tan ^{-1}(a x) \, dx\)

Optimal. Leaf size=201 \[ \frac {a^4 c^2 x^{m+5} \tan ^{-1}(a x)}{m+5}-\frac {a c^2 x^{m+2} \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-a^2 x^2\right )}{m^2+3 m+2}+\frac {2 a^2 c^2 x^{m+3} \tan ^{-1}(a x)}{m+3}-\frac {a^5 c^2 x^{m+6} \, _2F_1\left (1,\frac {m+6}{2};\frac {m+8}{2};-a^2 x^2\right )}{(m+5) (m+6)}-\frac {2 a^3 c^2 x^{m+4} \, _2F_1\left (1,\frac {m+4}{2};\frac {m+6}{2};-a^2 x^2\right )}{m^2+7 m+12}+\frac {c^2 x^{m+1} \tan ^{-1}(a x)}{m+1} \]

[Out]

c^2*x^(1+m)*arctan(a*x)/(1+m)+2*a^2*c^2*x^(3+m)*arctan(a*x)/(3+m)+a^4*c^2*x^(5+m)*arctan(a*x)/(5+m)-a*c^2*x^(2
+m)*hypergeom([1, 1+1/2*m],[2+1/2*m],-a^2*x^2)/(m^2+3*m+2)-2*a^3*c^2*x^(4+m)*hypergeom([1, 2+1/2*m],[3+1/2*m],
-a^2*x^2)/(m^2+7*m+12)-a^5*c^2*x^(6+m)*hypergeom([1, 3+1/2*m],[4+1/2*m],-a^2*x^2)/(5+m)/(6+m)

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Rubi [A]  time = 0.16, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4948, 4852, 364} \[ -\frac {a c^2 x^{m+2} \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-a^2 x^2\right )}{m^2+3 m+2}-\frac {2 a^3 c^2 x^{m+4} \, _2F_1\left (1,\frac {m+4}{2};\frac {m+6}{2};-a^2 x^2\right )}{m^2+7 m+12}-\frac {a^5 c^2 x^{m+6} \, _2F_1\left (1,\frac {m+6}{2};\frac {m+8}{2};-a^2 x^2\right )}{(m+5) (m+6)}+\frac {2 a^2 c^2 x^{m+3} \tan ^{-1}(a x)}{m+3}+\frac {a^4 c^2 x^{m+5} \tan ^{-1}(a x)}{m+5}+\frac {c^2 x^{m+1} \tan ^{-1}(a x)}{m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*x^(1 + m)*ArcTan[a*x])/(1 + m) + (2*a^2*c^2*x^(3 + m)*ArcTan[a*x])/(3 + m) + (a^4*c^2*x^(5 + m)*ArcTan[a*
x])/(5 + m) - (a*c^2*x^(2 + m)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -(a^2*x^2)])/(2 + 3*m + m^2) - (2*a^
3*c^2*x^(4 + m)*Hypergeometric2F1[1, (4 + m)/2, (6 + m)/2, -(a^2*x^2)])/(12 + 7*m + m^2) - (a^5*c^2*x^(6 + m)*
Hypergeometric2F1[1, (6 + m)/2, (8 + m)/2, -(a^2*x^2)])/((5 + m)*(6 + m))

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 4852

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

Rule 4948

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Int[Ex
pandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e,
c^2*d] && IGtQ[p, 0] && IGtQ[q, 1] && (EqQ[p, 1] || IntegerQ[m])

Rubi steps

\begin {align*} \int x^m \left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x) \, dx &=\int \left (c^2 x^m \tan ^{-1}(a x)+2 a^2 c^2 x^{2+m} \tan ^{-1}(a x)+a^4 c^2 x^{4+m} \tan ^{-1}(a x)\right ) \, dx\\ &=c^2 \int x^m \tan ^{-1}(a x) \, dx+\left (2 a^2 c^2\right ) \int x^{2+m} \tan ^{-1}(a x) \, dx+\left (a^4 c^2\right ) \int x^{4+m} \tan ^{-1}(a x) \, dx\\ &=\frac {c^2 x^{1+m} \tan ^{-1}(a x)}{1+m}+\frac {2 a^2 c^2 x^{3+m} \tan ^{-1}(a x)}{3+m}+\frac {a^4 c^2 x^{5+m} \tan ^{-1}(a x)}{5+m}-\frac {\left (a c^2\right ) \int \frac {x^{1+m}}{1+a^2 x^2} \, dx}{1+m}-\frac {\left (2 a^3 c^2\right ) \int \frac {x^{3+m}}{1+a^2 x^2} \, dx}{3+m}-\frac {\left (a^5 c^2\right ) \int \frac {x^{5+m}}{1+a^2 x^2} \, dx}{5+m}\\ &=\frac {c^2 x^{1+m} \tan ^{-1}(a x)}{1+m}+\frac {2 a^2 c^2 x^{3+m} \tan ^{-1}(a x)}{3+m}+\frac {a^4 c^2 x^{5+m} \tan ^{-1}(a x)}{5+m}-\frac {a c^2 x^{2+m} \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};-a^2 x^2\right )}{2+3 m+m^2}-\frac {2 a^3 c^2 x^{4+m} \, _2F_1\left (1,\frac {4+m}{2};\frac {6+m}{2};-a^2 x^2\right )}{12+7 m+m^2}-\frac {a^5 c^2 x^{6+m} \, _2F_1\left (1,\frac {6+m}{2};\frac {8+m}{2};-a^2 x^2\right )}{(5+m) (6+m)}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 175, normalized size = 0.87 \[ c^2 x^{m+1} \left (\frac {a^4 x^4 \tan ^{-1}(a x)}{m+5}-\frac {a x \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-a^2 x^2\right )}{m^2+3 m+2}+\frac {2 a^2 x^2 \tan ^{-1}(a x)}{m+3}-\frac {a^5 x^5 \, _2F_1\left (1,\frac {m+6}{2};\frac {m+8}{2};-a^2 x^2\right )}{(m+5) (m+6)}-\frac {2 a^3 x^3 \, _2F_1\left (1,\frac {m+4}{2};\frac {m+6}{2};-a^2 x^2\right )}{m^2+7 m+12}+\frac {\tan ^{-1}(a x)}{m+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

c^2*x^(1 + m)*(ArcTan[a*x]/(1 + m) + (2*a^2*x^2*ArcTan[a*x])/(3 + m) + (a^4*x^4*ArcTan[a*x])/(5 + m) - (a*x*Hy
pergeometric2F1[1, (2 + m)/2, (4 + m)/2, -(a^2*x^2)])/(2 + 3*m + m^2) - (2*a^3*x^3*Hypergeometric2F1[1, (4 + m
)/2, (6 + m)/2, -(a^2*x^2)])/(12 + 7*m + m^2) - (a^5*x^5*Hypergeometric2F1[1, (6 + m)/2, (8 + m)/2, -(a^2*x^2)
])/((5 + m)*(6 + m)))

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}\right )} x^{m} \arctan \left (a x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2)*x^m*arctan(a*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [C]  time = 1.40, size = 376, normalized size = 1.87 \[ \frac {a^{-1-m} c^{2} \left (-\frac {4 x^{m} a^{m} \left (a^{4} m^{2} x^{4}+2 m \,x^{4} a^{4}-a^{2} m^{2} x^{2}-4 m \,x^{2} a^{2}+m^{2}+6 m +8\right )}{\left (5+m \right ) m \left (2+m \right ) \left (4+m \right )}+\frac {8 x^{6+m} a^{6+m} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\left (10+2 m \right ) \sqrt {a^{2} x^{2}}}+\frac {2 x^{m} a^{m} \Phi \left (-a^{2} x^{2}, 1, \frac {m}{2}\right )}{5+m}\right )}{4}+\frac {a^{-1-m} c^{2} \left (-\frac {4 x^{m} a^{m} \left (m \,x^{2} a^{2}-m -2\right )}{\left (3+m \right ) m \left (2+m \right )}+\frac {8 x^{4+m} a^{4+m} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\left (6+2 m \right ) \sqrt {a^{2} x^{2}}}+\frac {2 x^{m} a^{m} \left (-4-m \right ) \Phi \left (-a^{2} x^{2}, 1, \frac {m}{2}\right )}{\left (4+m \right ) \left (3+m \right )}\right )}{2}+\frac {a^{-1-m} c^{2} \left (\frac {4 x^{m} a^{m} \left (-m -2\right )}{\left (2+m \right ) \left (1+m \right ) m}+\frac {8 x^{2+m} a^{2+m} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\left (2+2 m \right ) \sqrt {a^{2} x^{2}}}+\frac {2 x^{m} a^{m} \Phi \left (-a^{2} x^{2}, 1, \frac {m}{2}\right )}{1+m}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(a^2*c*x^2+c)^2*arctan(a*x),x)

[Out]

1/4*a^(-1-m)*c^2*(-4*x^m*a^m*(a^4*m^2*x^4+2*a^4*m*x^4-a^2*m^2*x^2-4*a^2*m*x^2+m^2+6*m+8)/(5+m)/m/(2+m)/(4+m)+8
*x^(6+m)*a^(6+m)/(10+2*m)/(a^2*x^2)^(1/2)*arctan((a^2*x^2)^(1/2))+2*x^m*a^m/(5+m)*LerchPhi(-a^2*x^2,1,1/2*m))+
1/2*a^(-1-m)*c^2*(-4*x^m*a^m*(a^2*m*x^2-m-2)/(3+m)/m/(2+m)+8*x^(4+m)*a^(4+m)/(6+2*m)/(a^2*x^2)^(1/2)*arctan((a
^2*x^2)^(1/2))+2/(4+m)*x^m*a^m*(-4-m)/(3+m)*LerchPhi(-a^2*x^2,1,1/2*m))+1/4*a^(-1-m)*c^2*(4/(2+m)*x^m*a^m*(-m-
2)/(1+m)/m+8*x^(2+m)*a^(2+m)/(2+2*m)/(a^2*x^2)^(1/2)*arctan((a^2*x^2)^(1/2))+2*x^m*a^m/(1+m)*LerchPhi(-a^2*x^2
,1,1/2*m))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left ({\left (a^{4} c^{2} m^{2} + 4 \, a^{4} c^{2} m + 3 \, a^{4} c^{2}\right )} x^{5} + 2 \, {\left (a^{2} c^{2} m^{2} + 6 \, a^{2} c^{2} m + 5 \, a^{2} c^{2}\right )} x^{3} + {\left (c^{2} m^{2} + 8 \, c^{2} m + 15 \, c^{2}\right )} x\right )} x^{m} \arctan \left (a x\right ) - {\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} \int \frac {{\left ({\left (a^{5} c^{2} m^{2} + 4 \, a^{5} c^{2} m + 3 \, a^{5} c^{2}\right )} x^{5} + 2 \, {\left (a^{3} c^{2} m^{2} + 6 \, a^{3} c^{2} m + 5 \, a^{3} c^{2}\right )} x^{3} + {\left (a c^{2} m^{2} + 8 \, a c^{2} m + 15 \, a c^{2}\right )} x\right )} x^{m}}{m^{3} + {\left (a^{2} m^{3} + 9 \, a^{2} m^{2} + 23 \, a^{2} m + 15 \, a^{2}\right )} x^{2} + 9 \, m^{2} + 23 \, m + 15}\,{d x}}{m^{3} + 9 \, m^{2} + 23 \, m + 15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((a^4*c^2*m^2 + 4*a^4*c^2*m + 3*a^4*c^2)*x^5 + 2*(a^2*c^2*m^2 + 6*a^2*c^2*m + 5*a^2*c^2)*x^3 + (c^2*m^2 + 8*c
^2*m + 15*c^2)*x)*x^m*arctan(a*x) - (m^3 + 9*m^2 + 23*m + 15)*integrate(((a^5*c^2*m^2 + 4*a^5*c^2*m + 3*a^5*c^
2)*x^5 + 2*(a^3*c^2*m^2 + 6*a^3*c^2*m + 5*a^3*c^2)*x^3 + (a*c^2*m^2 + 8*a*c^2*m + 15*a*c^2)*x)*x^m/(m^3 + (a^2
*m^3 + 9*a^2*m^2 + 23*a^2*m + 15*a^2)*x^2 + 9*m^2 + 23*m + 15), x))/(m^3 + 9*m^2 + 23*m + 15)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^m\,\mathrm {atan}\left (a\,x\right )\,{\left (c\,a^2\,x^2+c\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*atan(a*x)*(c + a^2*c*x^2)^2,x)

[Out]

int(x^m*atan(a*x)*(c + a^2*c*x^2)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ c^{2} \left (\int x^{m} \operatorname {atan}{\left (a x \right )}\, dx + \int 2 a^{2} x^{2} x^{m} \operatorname {atan}{\left (a x \right )}\, dx + \int a^{4} x^{4} x^{m} \operatorname {atan}{\left (a x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**2*(Integral(x**m*atan(a*x), x) + Integral(2*a**2*x**2*x**m*atan(a*x), x) + Integral(a**4*x**4*x**m*atan(a*x
), x))

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