3.218 \(\int x (c+a^2 c x^2)^{5/2} \tan ^{-1}(a x) \, dx\)

Optimal. Leaf size=134 \[ -\frac{5 c^2 x \sqrt{a^2 c x^2+c}}{112 a}-\frac{5 c^{5/2} \tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{a^2 c x^2+c}}\right )}{112 a^2}-\frac{x \left (a^2 c x^2+c\right )^{5/2}}{42 a}-\frac{5 c x \left (a^2 c x^2+c\right )^{3/2}}{168 a}+\frac{\left (a^2 c x^2+c\right )^{7/2} \tan ^{-1}(a x)}{7 a^2 c} \]

[Out]

(-5*c^2*x*Sqrt[c + a^2*c*x^2])/(112*a) - (5*c*x*(c + a^2*c*x^2)^(3/2))/(168*a) - (x*(c + a^2*c*x^2)^(5/2))/(42
*a) + ((c + a^2*c*x^2)^(7/2)*ArcTan[a*x])/(7*a^2*c) - (5*c^(5/2)*ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x^2]])/(
112*a^2)

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Rubi [A]  time = 0.0853006, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4930, 195, 217, 206} \[ -\frac{5 c^2 x \sqrt{a^2 c x^2+c}}{112 a}-\frac{5 c^{5/2} \tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{a^2 c x^2+c}}\right )}{112 a^2}-\frac{x \left (a^2 c x^2+c\right )^{5/2}}{42 a}-\frac{5 c x \left (a^2 c x^2+c\right )^{3/2}}{168 a}+\frac{\left (a^2 c x^2+c\right )^{7/2} \tan ^{-1}(a x)}{7 a^2 c} \]

Antiderivative was successfully verified.

[In]

Int[x*(c + a^2*c*x^2)^(5/2)*ArcTan[a*x],x]

[Out]

(-5*c^2*x*Sqrt[c + a^2*c*x^2])/(112*a) - (5*c*x*(c + a^2*c*x^2)^(3/2))/(168*a) - (x*(c + a^2*c*x^2)^(5/2))/(42
*a) + ((c + a^2*c*x^2)^(7/2)*ArcTan[a*x])/(7*a^2*c) - (5*c^(5/2)*ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x^2]])/(
112*a^2)

Rule 4930

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x) \, dx &=\frac{\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)}{7 a^2 c}-\frac{\int \left (c+a^2 c x^2\right )^{5/2} \, dx}{7 a}\\ &=-\frac{x \left (c+a^2 c x^2\right )^{5/2}}{42 a}+\frac{\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)}{7 a^2 c}-\frac{(5 c) \int \left (c+a^2 c x^2\right )^{3/2} \, dx}{42 a}\\ &=-\frac{5 c x \left (c+a^2 c x^2\right )^{3/2}}{168 a}-\frac{x \left (c+a^2 c x^2\right )^{5/2}}{42 a}+\frac{\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)}{7 a^2 c}-\frac{\left (5 c^2\right ) \int \sqrt{c+a^2 c x^2} \, dx}{56 a}\\ &=-\frac{5 c^2 x \sqrt{c+a^2 c x^2}}{112 a}-\frac{5 c x \left (c+a^2 c x^2\right )^{3/2}}{168 a}-\frac{x \left (c+a^2 c x^2\right )^{5/2}}{42 a}+\frac{\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)}{7 a^2 c}-\frac{\left (5 c^3\right ) \int \frac{1}{\sqrt{c+a^2 c x^2}} \, dx}{112 a}\\ &=-\frac{5 c^2 x \sqrt{c+a^2 c x^2}}{112 a}-\frac{5 c x \left (c+a^2 c x^2\right )^{3/2}}{168 a}-\frac{x \left (c+a^2 c x^2\right )^{5/2}}{42 a}+\frac{\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)}{7 a^2 c}-\frac{\left (5 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-a^2 c x^2} \, dx,x,\frac{x}{\sqrt{c+a^2 c x^2}}\right )}{112 a}\\ &=-\frac{5 c^2 x \sqrt{c+a^2 c x^2}}{112 a}-\frac{5 c x \left (c+a^2 c x^2\right )^{3/2}}{168 a}-\frac{x \left (c+a^2 c x^2\right )^{5/2}}{42 a}+\frac{\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)}{7 a^2 c}-\frac{5 c^{5/2} \tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c+a^2 c x^2}}\right )}{112 a^2}\\ \end{align*}

Mathematica [A]  time = 0.211469, size = 111, normalized size = 0.83 \[ \frac{c^2 \left (-a x \left (8 a^4 x^4+26 a^2 x^2+33\right ) \sqrt{a^2 c x^2+c}-15 \sqrt{c} \log \left (\sqrt{c} \sqrt{a^2 c x^2+c}+a c x\right )+48 \left (a^2 x^2+1\right )^3 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)\right )}{336 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(c + a^2*c*x^2)^(5/2)*ArcTan[a*x],x]

[Out]

(c^2*(-(a*x*Sqrt[c + a^2*c*x^2]*(33 + 26*a^2*x^2 + 8*a^4*x^4)) + 48*(1 + a^2*x^2)^3*Sqrt[c + a^2*c*x^2]*ArcTan
[a*x] - 15*Sqrt[c]*Log[a*c*x + Sqrt[c]*Sqrt[c + a^2*c*x^2]]))/(336*a^2)

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Maple [C]  time = 0.329, size = 205, normalized size = 1.5 \begin{align*}{\frac{{c}^{2} \left ( 48\,\arctan \left ( ax \right ){x}^{6}{a}^{6}-8\,{a}^{5}{x}^{5}+144\,\arctan \left ( ax \right ){x}^{4}{a}^{4}-26\,{a}^{3}{x}^{3}+144\,\arctan \left ( ax \right ){a}^{2}{x}^{2}-33\,ax+48\,\arctan \left ( ax \right ) \right ) }{336\,{a}^{2}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}+{\frac{5\,{c}^{2}}{112\,{a}^{2}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }\ln \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-i \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-{\frac{5\,{c}^{2}}{112\,{a}^{2}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }\ln \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+i \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a^2*c*x^2+c)^(5/2)*arctan(a*x),x)

[Out]

1/336*c^2/a^2*(c*(a*x-I)*(a*x+I))^(1/2)*(48*arctan(a*x)*x^6*a^6-8*a^5*x^5+144*arctan(a*x)*x^4*a^4-26*a^3*x^3+1
44*arctan(a*x)*a^2*x^2-33*a*x+48*arctan(a*x))+5/112*c^2/a^2*(c*(a*x-I)*(a*x+I))^(1/2)*ln((1+I*a*x)/(a^2*x^2+1)
^(1/2)-I)/(a^2*x^2+1)^(1/2)-5/112*c^2/a^2*(c*(a*x-I)*(a*x+I))^(1/2)*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)+I)/(a^2*x^2
+1)^(1/2)

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Maxima [B]  time = 2.35587, size = 933, normalized size = 6.96 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a^2*c*x^2+c)^(5/2)*arctan(a*x),x, algorithm="maxima")

[Out]

1/1680*(560*(a^2*c^2*x^2 + c^2)*sqrt(a^2*x^2 + 1)*sqrt(c)*arctan(a*x) - 280*(a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*(
a*c^2*x*cos(1/2*arctan2(4*a*x, -a^2*x^2 + 3)) + 2*c^2*sin(1/2*arctan2(4*a*x, -a^2*x^2 + 3)))*sqrt(c) - ((a*(5*
(8*(a^2*x^2 + 1)^(3/2)*x^3/a^2 - 6*(a^2*x^2 + 1)^(3/2)*x/a^4 + 3*sqrt(a^2*x^2 + 1)*x/a^4 + 3*arcsinh(a^2*x/sqr
t(a^2))/(sqrt(a^2)*a^4))/a^2 - 24*(2*(a^2*x^2 + 1)^(3/2)*x/a^2 - sqrt(a^2*x^2 + 1)*x/a^2 - arcsinh(a^2*x/sqrt(
a^2))/(sqrt(a^2)*a^2))/a^4 + 64*(sqrt(a^2*x^2 + 1)*x + arcsinh(a^2*x/sqrt(a^2))/sqrt(a^2))/a^6) - 16*(15*(a^2*
x^2 + 1)^(3/2)*x^4/a^2 - 12*(a^2*x^2 + 1)^(3/2)*x^2/a^4 + 8*(a^2*x^2 + 1)^(3/2)/a^6)*arctan(a*x))*a^6*c^2 + 28
*(a*(3*(2*(a^2*x^2 + 1)^(3/2)*x/a^2 - sqrt(a^2*x^2 + 1)*x/a^2 - arcsinh(a^2*x/sqrt(a^2))/(sqrt(a^2)*a^2))/a^2
- 8*(sqrt(a^2*x^2 + 1)*x + arcsinh(a^2*x/sqrt(a^2))/sqrt(a^2))/a^4) - 8*(3*(a^2*x^2 + 1)^(3/2)*x^2/a^2 - 2*(a^
2*x^2 + 1)^(3/2)/a^4)*arctan(a*x))*a^4*c^2 - 140*c^2*arctan2((a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*sin(1/2*arctan2(
4*a*x, a^2*x^2 - 3)) + 2, a*x + (a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*cos(1/2*arctan2(4*a*x, a^2*x^2 - 3))) - 140*c
^2*arctan2((a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*sin(1/2*arctan2(4*a*x, a^2*x^2 - 3)) - 2, -a*x + (a^4*x^4 + 10*a^2
*x^2 + 9)^(1/4)*cos(1/2*arctan2(4*a*x, a^2*x^2 - 3))))*sqrt(c))/a^2

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Fricas [A]  time = 2.48669, size = 298, normalized size = 2.22 \begin{align*} \frac{15 \, c^{\frac{5}{2}} \log \left (-2 \, a^{2} c x^{2} + 2 \, \sqrt{a^{2} c x^{2} + c} a \sqrt{c} x - c\right ) - 2 \,{\left (8 \, a^{5} c^{2} x^{5} + 26 \, a^{3} c^{2} x^{3} + 33 \, a c^{2} x - 48 \,{\left (a^{6} c^{2} x^{6} + 3 \, a^{4} c^{2} x^{4} + 3 \, a^{2} c^{2} x^{2} + c^{2}\right )} \arctan \left (a x\right )\right )} \sqrt{a^{2} c x^{2} + c}}{672 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a^2*c*x^2+c)^(5/2)*arctan(a*x),x, algorithm="fricas")

[Out]

1/672*(15*c^(5/2)*log(-2*a^2*c*x^2 + 2*sqrt(a^2*c*x^2 + c)*a*sqrt(c)*x - c) - 2*(8*a^5*c^2*x^5 + 26*a^3*c^2*x^
3 + 33*a*c^2*x - 48*(a^6*c^2*x^6 + 3*a^4*c^2*x^4 + 3*a^2*c^2*x^2 + c^2)*arctan(a*x))*sqrt(a^2*c*x^2 + c))/a^2

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a**2*c*x**2+c)**(5/2)*atan(a*x),x)

[Out]

Timed out

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Giac [A]  time = 1.19344, size = 238, normalized size = 1.78 \begin{align*} -\frac{\sqrt{a^{2} c x^{2} + c}{\left (2 \,{\left (4 \, a^{4} c^{2} x^{2} + 13 \, a^{2} c^{2}\right )} x^{2} + 33 \, c^{2}\right )} x - \frac{15 \, c^{\frac{5}{2}} \log \left ({\left | -\sqrt{a^{2} c} x + \sqrt{a^{2} c x^{2} + c} \right |}\right )}{{\left | a \right |}}}{336 \, a} + \frac{{\left (42 \,{\left (a^{2} c x^{2} + c\right )}^{\frac{5}{2}} - 35 \,{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}} c + \frac{15 \,{\left (a^{2} c x^{2} + c\right )}^{\frac{7}{2}} - 42 \,{\left (a^{2} c x^{2} + c\right )}^{\frac{5}{2}} c + 35 \,{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}} c^{2}}{c}\right )} \arctan \left (a x\right )}{105 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a^2*c*x^2+c)^(5/2)*arctan(a*x),x, algorithm="giac")

[Out]

-1/336*(sqrt(a^2*c*x^2 + c)*(2*(4*a^4*c^2*x^2 + 13*a^2*c^2)*x^2 + 33*c^2)*x - 15*c^(5/2)*log(abs(-sqrt(a^2*c)*
x + sqrt(a^2*c*x^2 + c)))/abs(a))/a + 1/105*(42*(a^2*c*x^2 + c)^(5/2) - 35*(a^2*c*x^2 + c)^(3/2)*c + (15*(a^2*
c*x^2 + c)^(7/2) - 42*(a^2*c*x^2 + c)^(5/2)*c + 35*(a^2*c*x^2 + c)^(3/2)*c^2)/c)*arctan(a*x)/a^2