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

Optimal. Leaf size=387 \[ -\frac {5 i c^3 \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{56 a^2 \sqrt {a^2 c x^2+c}}+\frac {5 i c^3 \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{56 a^2 \sqrt {a^2 c x^2+c}}+\frac {5 i c^3 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{28 a^2 \sqrt {a^2 c x^2+c}}+\frac {5 c^2 \sqrt {a^2 c x^2+c}}{56 a^2}-\frac {5 c^2 x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{56 a}+\frac {\left (a^2 c x^2+c\right )^{5/2}}{105 a^2}+\frac {5 c \left (a^2 c x^2+c\right )^{3/2}}{252 a^2}+\frac {\left (a^2 c x^2+c\right )^{7/2} \tan ^{-1}(a x)^2}{7 a^2 c}-\frac {x \left (a^2 c x^2+c\right )^{5/2} \tan ^{-1}(a x)}{21 a}-\frac {5 c x \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)}{84 a} \]

[Out]

5/252*c*(a^2*c*x^2+c)^(3/2)/a^2+1/105*(a^2*c*x^2+c)^(5/2)/a^2-5/84*c*x*(a^2*c*x^2+c)^(3/2)*arctan(a*x)/a-1/21*
x*(a^2*c*x^2+c)^(5/2)*arctan(a*x)/a+1/7*(a^2*c*x^2+c)^(7/2)*arctan(a*x)^2/a^2/c+5/28*I*c^3*arctan(a*x)*arctan(
(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/a^2/(a^2*c*x^2+c)^(1/2)-5/56*I*c^3*polylog(2,-I*(1+I*a*x)^(
1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/a^2/(a^2*c*x^2+c)^(1/2)+5/56*I*c^3*polylog(2,I*(1+I*a*x)^(1/2)/(1-I*a*
x)^(1/2))*(a^2*x^2+1)^(1/2)/a^2/(a^2*c*x^2+c)^(1/2)+5/56*c^2*(a^2*c*x^2+c)^(1/2)/a^2-5/56*c^2*x*arctan(a*x)*(a
^2*c*x^2+c)^(1/2)/a

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Rubi [A]  time = 0.28, antiderivative size = 387, normalized size of antiderivative = 1.00, 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 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 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]

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 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]

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*}

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Mathematica [B]  time = 7.88, 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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fricas [F]  time = 1.32, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{4} c^{2} x^{5} + 2 \, a^{2} c^{2} x^{3} + c^{2} x\right )} \sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )^{2}, x\right ) \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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 >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 1.19, size = 275, normalized size = 0.71 \[ \frac {c^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (360 \arctan \left (a x \right )^{2} x^{6} a^{6}-120 \arctan \left (a x \right ) x^{5} a^{5}+1080 \arctan \left (a x \right )^{2} x^{4} a^{4}+24 a^{4} x^{4}-390 \arctan \left (a x \right ) x^{3} a^{3}+1080 \arctan \left (a x \right )^{2} x^{2} a^{2}+98 a^{2} x^{2}-495 \arctan \left (a x \right ) x a +360 \arctan \left (a x \right )^{2}+299\right )}{2520 a^{2}}+\frac {5 c^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i \dilog \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+\arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-\arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-i \dilog \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{56 a^{2} \sqrt {a^{2} x^{2}+1}} \]

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)*(I+a*x))^(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)*(I+a*x))^(1/2)*(I*dilog(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))-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)))/a^2/(a^2*x^2+1)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x \arctan \left (a x\right )^{2}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x\,{\mathrm {atan}\left (a\,x\right )}^2\,{\left (c\,a^2\,x^2+c\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \operatorname {atan}^{2}{\left (a x \right )}\, dx \]

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]

Integral(x*(c*(a**2*x**2 + 1))**(5/2)*atan(a*x)**2, x)

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