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

Optimal. Leaf size=387 \[ -\frac{5 i c^3 \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,-\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{56 a^2 \sqrt{a^2 c x^2+c}}+\frac{5 i c^3 \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{56 a^2 \sqrt{a^2 c x^2+c}}+\frac{5 c^2 \sqrt{a^2 c x^2+c}}{56 a^2}-\frac{5 c^2 x \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{56 a}+\frac{5 i c^3 \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{28 a^2 \sqrt{a^2 c x^2+c}}+\frac{\left (a^2 c x^2+c\right )^{5/2}}{105 a^2}+\frac{5 c \left (a^2 c x^2+c\right )^{3/2}}{252 a^2}+\frac{\left (a^2 c x^2+c\right )^{7/2} \tan ^{-1}(a x)^2}{7 a^2 c}-\frac{x \left (a^2 c x^2+c\right )^{5/2} \tan ^{-1}(a x)}{21 a}-\frac{5 c x \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)}{84 a} \]

[Out]

(5*c^2*Sqrt[c + a^2*c*x^2])/(56*a^2) + (5*c*(c + a^2*c*x^2)^(3/2))/(252*a^2) + (c + a^2*c*x^2)^(5/2)/(105*a^2)
- (5*c^2*x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(56*a) - (5*c*x*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/(84*a) - (x*(c
+ a^2*c*x^2)^(5/2)*ArcTan[a*x])/(21*a) + ((c + a^2*c*x^2)^(7/2)*ArcTan[a*x]^2)/(7*a^2*c) + (((5*I)/28)*c^3*Sq
rt[1 + a^2*x^2]*ArcTan[a*x]*ArcTan[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/(a^2*Sqrt[c + a^2*c*x^2]) - (((5*I)/56)*c
^3*Sqrt[1 + a^2*x^2]*PolyLog[2, ((-I)*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a^2*Sqrt[c + a^2*c*x^2]) + (((5*I)/5
6)*c^3*Sqrt[1 + a^2*x^2]*PolyLog[2, (I*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a^2*Sqrt[c + a^2*c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.280812, antiderivative size = 387, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4930, 4878, 4890, 4886} \[ -\frac{5 i c^3 \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,-\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{56 a^2 \sqrt{a^2 c x^2+c}}+\frac{5 i c^3 \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{56 a^2 \sqrt{a^2 c x^2+c}}+\frac{5 c^2 \sqrt{a^2 c x^2+c}}{56 a^2}-\frac{5 c^2 x \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{56 a}+\frac{5 i c^3 \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{28 a^2 \sqrt{a^2 c x^2+c}}+\frac{\left (a^2 c x^2+c\right )^{5/2}}{105 a^2}+\frac{5 c \left (a^2 c x^2+c\right )^{3/2}}{252 a^2}+\frac{\left (a^2 c x^2+c\right )^{7/2} \tan ^{-1}(a x)^2}{7 a^2 c}-\frac{x \left (a^2 c x^2+c\right )^{5/2} \tan ^{-1}(a x)}{21 a}-\frac{5 c x \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)}{84 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*c^2*Sqrt[c + a^2*c*x^2])/(56*a^2) + (5*c*(c + a^2*c*x^2)^(3/2))/(252*a^2) + (c + a^2*c*x^2)^(5/2)/(105*a^2)
- (5*c^2*x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(56*a) - (5*c*x*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/(84*a) - (x*(c
+ a^2*c*x^2)^(5/2)*ArcTan[a*x])/(21*a) + ((c + a^2*c*x^2)^(7/2)*ArcTan[a*x]^2)/(7*a^2*c) + (((5*I)/28)*c^3*Sq
rt[1 + a^2*x^2]*ArcTan[a*x]*ArcTan[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/(a^2*Sqrt[c + a^2*c*x^2]) - (((5*I)/56)*c
^3*Sqrt[1 + a^2*x^2]*PolyLog[2, ((-I)*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a^2*Sqrt[c + a^2*c*x^2]) + (((5*I)/5
6)*c^3*Sqrt[1 + a^2*x^2]*PolyLog[2, (I*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a^2*Sqrt[c + a^2*c*x^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 4878

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]

Rule 4890

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

Rule 4886

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

Rubi steps

\begin{align*} \int x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^2 \, dx &=\frac{\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)^2}{7 a^2 c}-\frac{2 \int \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x) \, dx}{7 a}\\ &=\frac{\left (c+a^2 c x^2\right )^{5/2}}{105 a^2}-\frac{x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{21 a}+\frac{\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)^2}{7 a^2 c}-\frac{(5 c) \int \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x) \, dx}{21 a}\\ &=\frac{5 c \left (c+a^2 c x^2\right )^{3/2}}{252 a^2}+\frac{\left (c+a^2 c x^2\right )^{5/2}}{105 a^2}-\frac{5 c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{84 a}-\frac{x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{21 a}+\frac{\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)^2}{7 a^2 c}-\frac{\left (5 c^2\right ) \int \sqrt{c+a^2 c x^2} \tan ^{-1}(a x) \, dx}{28 a}\\ &=\frac{5 c^2 \sqrt{c+a^2 c x^2}}{56 a^2}+\frac{5 c \left (c+a^2 c x^2\right )^{3/2}}{252 a^2}+\frac{\left (c+a^2 c x^2\right )^{5/2}}{105 a^2}-\frac{5 c^2 x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{56 a}-\frac{5 c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{84 a}-\frac{x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{21 a}+\frac{\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)^2}{7 a^2 c}-\frac{\left (5 c^3\right ) \int \frac{\tan ^{-1}(a x)}{\sqrt{c+a^2 c x^2}} \, dx}{56 a}\\ &=\frac{5 c^2 \sqrt{c+a^2 c x^2}}{56 a^2}+\frac{5 c \left (c+a^2 c x^2\right )^{3/2}}{252 a^2}+\frac{\left (c+a^2 c x^2\right )^{5/2}}{105 a^2}-\frac{5 c^2 x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{56 a}-\frac{5 c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{84 a}-\frac{x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{21 a}+\frac{\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)^2}{7 a^2 c}-\frac{\left (5 c^3 \sqrt{1+a^2 x^2}\right ) \int \frac{\tan ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx}{56 a \sqrt{c+a^2 c x^2}}\\ &=\frac{5 c^2 \sqrt{c+a^2 c x^2}}{56 a^2}+\frac{5 c \left (c+a^2 c x^2\right )^{3/2}}{252 a^2}+\frac{\left (c+a^2 c x^2\right )^{5/2}}{105 a^2}-\frac{5 c^2 x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{56 a}-\frac{5 c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{84 a}-\frac{x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{21 a}+\frac{\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)^2}{7 a^2 c}+\frac{5 i c^3 \sqrt{1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{28 a^2 \sqrt{c+a^2 c x^2}}-\frac{5 i c^3 \sqrt{1+a^2 x^2} \text{Li}_2\left (-\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{56 a^2 \sqrt{c+a^2 c x^2}}+\frac{5 i c^3 \sqrt{1+a^2 x^2} \text{Li}_2\left (\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{56 a^2 \sqrt{c+a^2 c x^2}}\\ \end{align*}

Mathematica [B]  time = 7.78639, size = 1087, normalized size = 2.81 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^2*(1 + a^2*x^2)*Sqrt[c*(1 + a^2*x^2)]*(2 + 4*ArcTan[a*x]^2 + 2*Cos[2*ArcTan[a*x]] - (3*ArcTan[a*x]*Log[1 -
I*E^(I*ArcTan[a*x])])/Sqrt[1 + a^2*x^2] - ArcTan[a*x]*Cos[3*ArcTan[a*x]]*Log[1 - I*E^(I*ArcTan[a*x])] + (3*Arc
Tan[a*x]*Log[1 + I*E^(I*ArcTan[a*x])])/Sqrt[1 + a^2*x^2] + ArcTan[a*x]*Cos[3*ArcTan[a*x]]*Log[1 + I*E^(I*ArcTa
n[a*x])] - ((4*I)*PolyLog[2, (-I)*E^(I*ArcTan[a*x])])/(1 + a^2*x^2)^(3/2) + ((4*I)*PolyLog[2, I*E^(I*ArcTan[a*
x])])/(1 + a^2*x^2)^(3/2) - 2*ArcTan[a*x]*Sin[2*ArcTan[a*x]]))/(12*a^2) - (c^2*(1 + a^2*x^2)^2*Sqrt[c*(1 + a^2
*x^2)]*(50 - 32*ArcTan[a*x]^2 + 72*Cos[2*ArcTan[a*x]] + 160*ArcTan[a*x]^2*Cos[2*ArcTan[a*x]] + 22*Cos[4*ArcTan
[a*x]] - (110*ArcTan[a*x]*Log[1 - I*E^(I*ArcTan[a*x])])/Sqrt[1 + a^2*x^2] - 55*ArcTan[a*x]*Cos[3*ArcTan[a*x]]*
Log[1 - I*E^(I*ArcTan[a*x])] - 11*ArcTan[a*x]*Cos[5*ArcTan[a*x]]*Log[1 - I*E^(I*ArcTan[a*x])] + (110*ArcTan[a*
x]*Log[1 + I*E^(I*ArcTan[a*x])])/Sqrt[1 + a^2*x^2] + 55*ArcTan[a*x]*Cos[3*ArcTan[a*x]]*Log[1 + I*E^(I*ArcTan[a
*x])] + 11*ArcTan[a*x]*Cos[5*ArcTan[a*x]]*Log[1 + I*E^(I*ArcTan[a*x])] - ((176*I)*PolyLog[2, (-I)*E^(I*ArcTan[
a*x])])/(1 + a^2*x^2)^(5/2) + ((176*I)*PolyLog[2, I*E^(I*ArcTan[a*x])])/(1 + a^2*x^2)^(5/2) + 4*ArcTan[a*x]*Si
n[2*ArcTan[a*x]] - 22*ArcTan[a*x]*Sin[4*ArcTan[a*x]]))/(480*a^2) + (c^2*(1 + a^2*x^2)^3*Sqrt[c*(1 + a^2*x^2)]*
(4116 + 10944*ArcTan[a*x]^2 + 6262*Cos[2*ArcTan[a*x]] - 5376*ArcTan[a*x]^2*Cos[2*ArcTan[a*x]] + 2764*Cos[4*Arc
Tan[a*x]] + 6720*ArcTan[a*x]^2*Cos[4*ArcTan[a*x]] + 618*Cos[6*ArcTan[a*x]] - (10815*ArcTan[a*x]*Log[1 - I*E^(I
*ArcTan[a*x])])/Sqrt[1 + a^2*x^2] - 6489*ArcTan[a*x]*Cos[3*ArcTan[a*x]]*Log[1 - I*E^(I*ArcTan[a*x])] - 2163*Ar
cTan[a*x]*Cos[5*ArcTan[a*x]]*Log[1 - I*E^(I*ArcTan[a*x])] - 309*ArcTan[a*x]*Cos[7*ArcTan[a*x]]*Log[1 - I*E^(I*
ArcTan[a*x])] + (10815*ArcTan[a*x]*Log[1 + I*E^(I*ArcTan[a*x])])/Sqrt[1 + a^2*x^2] + 6489*ArcTan[a*x]*Cos[3*Ar
cTan[a*x]]*Log[1 + I*E^(I*ArcTan[a*x])] + 2163*ArcTan[a*x]*Cos[5*ArcTan[a*x]]*Log[1 + I*E^(I*ArcTan[a*x])] + 3
09*ArcTan[a*x]*Cos[7*ArcTan[a*x]]*Log[1 + I*E^(I*ArcTan[a*x])] - ((19776*I)*PolyLog[2, (-I)*E^(I*ArcTan[a*x])]
)/(1 + a^2*x^2)^(7/2) + ((19776*I)*PolyLog[2, I*E^(I*ArcTan[a*x])])/(1 + a^2*x^2)^(7/2) - 1266*ArcTan[a*x]*Sin
[2*ArcTan[a*x]] + 360*ArcTan[a*x]*Sin[4*ArcTan[a*x]] - 618*ArcTan[a*x]*Sin[6*ArcTan[a*x]]))/(161280*a^2)

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Maple [A]  time = 0.351, size = 275, normalized size = 0.7 \begin{align*}{\frac{{c}^{2} \left ( 360\, \left ( \arctan \left ( ax \right ) \right ) ^{2}{x}^{6}{a}^{6}-120\,\arctan \left ( ax \right ){x}^{5}{a}^{5}+1080\, \left ( \arctan \left ( ax \right ) \right ) ^{2}{x}^{4}{a}^{4}+24\,{a}^{4}{x}^{4}-390\,\arctan \left ( ax \right ){x}^{3}{a}^{3}+1080\, \left ( \arctan \left ( ax \right ) \right ) ^{2}{x}^{2}{a}^{2}+98\,{a}^{2}{x}^{2}-495\,\arctan \left ( ax \right ) xa+360\, \left ( \arctan \left ( ax \right ) \right ) ^{2}+299 \right ) }{2520\,{a}^{2}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}+{\frac{5\,{c}^{2}}{56\,{a}^{2}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) } \left ( \arctan \left ( ax \right ) \ln \left ( 1+{i \left ( 1+iax \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) -\arctan \left ( ax \right ) \ln \left ( 1-{i \left ( 1+iax \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) -i{\it dilog} \left ( 1+{i \left ( 1+iax \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) +i{\it dilog} \left ( 1-{i \left ( 1+iax \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) \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)^2,x)

[Out]

1/2520*c^2/a^2*(c*(a*x-I)*(a*x+I))^(1/2)*(360*arctan(a*x)^2*x^6*a^6-120*arctan(a*x)*x^5*a^5+1080*arctan(a*x)^2
*x^4*a^4+24*a^4*x^4-390*arctan(a*x)*x^3*a^3+1080*arctan(a*x)^2*x^2*a^2+98*a^2*x^2-495*arctan(a*x)*x*a+360*arct
an(a*x)^2+299)+5/56*c^2*(c*(a*x-I)*(a*x+I))^(1/2)*(arctan(a*x)*ln(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-arctan(a*x)
*ln(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-I*dilog(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+I*dilog(1-I*(1+I*a*x)/(a^2*x^2+1
)^(1/2)))/a^2/(a^2*x^2+1)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a^{2} c x^{2} + c\right )}^{\frac{5}{2}} x \arctan \left (a x\right )^{2}\,{d x} \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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a^{4} c^{2} x^{5} + 2 \, a^{2} c^{2} x^{3} + c^{2} x\right )} \sqrt{a^{2} c x^{2} + c} \arctan \left (a x\right )^{2}, x\right ) \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)^2,x, algorithm="fricas")

[Out]

integral((a^4*c^2*x^5 + 2*a^2*c^2*x^3 + c^2*x)*sqrt(a^2*c*x^2 + c)*arctan(a*x)^2, x)

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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)**2,x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError