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

Optimal. Leaf size=124 \[ -\frac{a c x^{m+2} \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-a^2 x^2\right )}{m^2+3 m+2}-\frac{a^3 c x^{m+4} \text{Hypergeometric2F1}\left (1,\frac{m+4}{2},\frac{m+6}{2},-a^2 x^2\right )}{m^2+7 m+12}+\frac{a^2 c x^{m+3} \tan ^{-1}(a x)}{m+3}+\frac{c x^{m+1} \tan ^{-1}(a x)}{m+1} \]

[Out]

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

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Rubi [A]  time = 0.0758609, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4950, 4852, 364} \[ -\frac{a c x^{m+2} \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};-a^2 x^2\right )}{m^2+3 m+2}-\frac{a^3 c x^{m+4} \, _2F_1\left (1,\frac{m+4}{2};\frac{m+6}{2};-a^2 x^2\right )}{m^2+7 m+12}+\frac{a^2 c x^{m+3} \tan ^{-1}(a x)}{m+3}+\frac{c x^{m+1} \tan ^{-1}(a x)}{m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4950

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

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 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])

Rubi steps

\begin{align*} \int x^m \left (c+a^2 c x^2\right ) \tan ^{-1}(a x) \, dx &=c \int x^m \tan ^{-1}(a x) \, dx+\left (a^2 c\right ) \int x^{2+m} \tan ^{-1}(a x) \, dx\\ &=\frac{c x^{1+m} \tan ^{-1}(a x)}{1+m}+\frac{a^2 c x^{3+m} \tan ^{-1}(a x)}{3+m}-\frac{(a c) \int \frac{x^{1+m}}{1+a^2 x^2} \, dx}{1+m}-\frac{\left (a^3 c\right ) \int \frac{x^{3+m}}{1+a^2 x^2} \, dx}{3+m}\\ &=\frac{c x^{1+m} \tan ^{-1}(a x)}{1+m}+\frac{a^2 c x^{3+m} \tan ^{-1}(a x)}{3+m}-\frac{a c 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{a^3 c 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}\\ \end{align*}

Mathematica [A]  time = 0.110708, size = 111, normalized size = 0.9 \[ c x^{m+1} \left (-\frac{a^3 x^3 \text{Hypergeometric2F1}\left (1,\frac{m+4}{2},\frac{m+6}{2},-a^2 x^2\right )}{m^2+7 m+12}-\frac{a x \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-a^2 x^2\right )}{m^2+3 m+2}+\left (\frac{a^2 x^2}{m+3}+\frac{1}{m+1}\right ) \tan ^{-1}(a x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.454, size = 222, normalized size = 1.8 \begin{align*}{\frac{{a}^{-1-m}c}{4} \left ( -4\,{\frac{{x}^{m}{a}^{m} \left ( m{x}^{2}{a}^{2}-m-2 \right ) }{ \left ( 3+m \right ) m \left ( 2+m \right ) }}+8\,{\frac{{x}^{4+m}{a}^{4+m}\arctan \left ( \sqrt{{a}^{2}{x}^{2}} \right ) }{ \left ( 6+2\,m \right ) \sqrt{{a}^{2}{x}^{2}}}}+2\,{\frac{{x}^{m}{a}^{m} \left ( -4-m \right ){\it LerchPhi} \left ( -{a}^{2}{x}^{2},1,m/2 \right ) }{ \left ( 4+m \right ) \left ( 3+m \right ) }} \right ) }+{\frac{{a}^{-1-m}c}{4} \left ( 4\,{\frac{{x}^{m}{a}^{m} \left ( -m-2 \right ) }{ \left ( 2+m \right ) \left ( 1+m \right ) m}}+8\,{\frac{{x}^{2+m}{a}^{2+m}\arctan \left ( \sqrt{{a}^{2}{x}^{2}} \right ) }{ \left ( 2+2\,m \right ) \sqrt{{a}^{2}{x}^{2}}}}+2\,{\frac{{x}^{m}{a}^{m}{\it LerchPhi} \left ( -{a}^{2}{x}^{2},1,m/2 \right ) }{1+m}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^(-1-m)*c*(-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*(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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a^{2} c x^{2} + c\right )} x^{m} \arctan \left (a x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} c \left (\int x^{m} \operatorname{atan}{\left (a x \right )}\, dx + \int a^{2} x^{2} x^{m} \operatorname{atan}{\left (a x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a^{2} c x^{2} + c\right )} x^{m} \arctan \left (a x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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