3.241 \(\int \frac {x^4 \tan ^{-1}(a x)}{(c+a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=308 \[ -\frac {x^3 \tan ^{-1}(a x)}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {i \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a^5 c^2 \sqrt {a^2 c x^2+c}}-\frac {i \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a^5 c^2 \sqrt {a^2 c x^2+c}}-\frac {4}{3 a^5 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^5 c^2 \sqrt {a^2 c x^2+c}}+\frac {1}{9 a^5 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {x \tan ^{-1}(a x)}{a^4 c^2 \sqrt {a^2 c x^2+c}} \]

[Out]

1/9/a^5/c/(a^2*c*x^2+c)^(3/2)-1/3*x^3*arctan(a*x)/a^2/c/(a^2*c*x^2+c)^(3/2)-4/3/a^5/c^2/(a^2*c*x^2+c)^(1/2)-x*
arctan(a*x)/a^4/c^2/(a^2*c*x^2+c)^(1/2)-2*I*arctan(a*x)*arctan((1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1
/2)/a^5/c^2/(a^2*c*x^2+c)^(1/2)+I*polylog(2,-I*(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/a^5/c^2/(a^2
*c*x^2+c)^(1/2)-I*polylog(2,I*(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/a^5/c^2/(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.37, antiderivative size = 308, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {4964, 4934, 4890, 4886, 4944, 266, 43} \[ \frac {i \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^5 c^2 \sqrt {a^2 c x^2+c}}-\frac {i \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^5 c^2 \sqrt {a^2 c x^2+c}}-\frac {4}{3 a^5 c^2 \sqrt {a^2 c x^2+c}}-\frac {x \tan ^{-1}(a x)}{a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^5 c^2 \sqrt {a^2 c x^2+c}}+\frac {1}{9 a^5 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {x^3 \tan ^{-1}(a x)}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*ArcTan[a*x])/(c + a^2*c*x^2)^(5/2),x]

[Out]

1/(9*a^5*c*(c + a^2*c*x^2)^(3/2)) - 4/(3*a^5*c^2*Sqrt[c + a^2*c*x^2]) - (x^3*ArcTan[a*x])/(3*a^2*c*(c + a^2*c*
x^2)^(3/2)) - (x*ArcTan[a*x])/(a^4*c^2*Sqrt[c + a^2*c*x^2]) - ((2*I)*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*ArcTan[Sqrt
[1 + I*a*x]/Sqrt[1 - I*a*x]])/(a^5*c^2*Sqrt[c + a^2*c*x^2]) + (I*Sqrt[1 + a^2*x^2]*PolyLog[2, ((-I)*Sqrt[1 + I
*a*x])/Sqrt[1 - I*a*x]])/(a^5*c^2*Sqrt[c + a^2*c*x^2]) - (I*Sqrt[1 + a^2*x^2]*PolyLog[2, (I*Sqrt[1 + I*a*x])/S
qrt[1 - I*a*x]])/(a^5*c^2*Sqrt[c + a^2*c*x^2])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 4934

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*(d + e*x^2)^(q
+ 1))/(4*c^3*d*(q + 1)^2), x] + (-Dist[1/(2*c^2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]), x],
x] + Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]))/(2*c^2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && E
qQ[e, c^2*d] && LtQ[q, -1] && NeQ[q, -5/2]

Rule 4944

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p)/(d*f*(m + 1)), x] - Dist[(b*c*p)/(f*(m + 1)), Int[(
f*x)^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e,
c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 4964

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

Rubi steps

\begin {align*} \int \frac {x^4 \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx &=-\frac {\int \frac {x^2 \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx}{a^2}+\frac {\int \frac {x^2 \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{a^2 c}\\ &=-\frac {1}{a^5 c^2 \sqrt {c+a^2 c x^2}}-\frac {x^3 \tan ^{-1}(a x)}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {x \tan ^{-1}(a x)}{a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\int \frac {x^3}{\left (c+a^2 c x^2\right )^{5/2}} \, dx}{3 a}+\frac {\int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx}{a^4 c^2}\\ &=-\frac {1}{a^5 c^2 \sqrt {c+a^2 c x^2}}-\frac {x^3 \tan ^{-1}(a x)}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {x \tan ^{-1}(a x)}{a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\operatorname {Subst}\left (\int \frac {x}{\left (c+a^2 c x\right )^{5/2}} \, dx,x,x^2\right )}{6 a}+\frac {\sqrt {1+a^2 x^2} \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{a^4 c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {1}{a^5 c^2 \sqrt {c+a^2 c x^2}}-\frac {x^3 \tan ^{-1}(a x)}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {x \tan ^{-1}(a x)}{a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {2 i \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 )}{a^5 c^2 \sqrt {c+a^2 c x^2}}+\frac {i \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^5 c^2 \sqrt {c+a^2 c x^2}}-\frac {i \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^5 c^2 \sqrt {c+a^2 c x^2}}+\frac {\operatorname {Subst}\left (\int \left (-\frac {1}{a^2 \left (c+a^2 c x\right )^{5/2}}+\frac {1}{a^2 c \left (c+a^2 c x\right )^{3/2}}\right ) \, dx,x,x^2\right )}{6 a}\\ &=\frac {1}{9 a^5 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {4}{3 a^5 c^2 \sqrt {c+a^2 c x^2}}-\frac {x^3 \tan ^{-1}(a x)}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {x \tan ^{-1}(a x)}{a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {2 i \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 )}{a^5 c^2 \sqrt {c+a^2 c x^2}}+\frac {i \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^5 c^2 \sqrt {c+a^2 c x^2}}-\frac {i \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^5 c^2 \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 177, normalized size = 0.57 \[ \frac {\sqrt {c \left (a^2 x^2+1\right )} \left (-\frac {45}{\sqrt {a^2 x^2+1}}-\frac {45 a x \tan ^{-1}(a x)}{\sqrt {a^2 x^2+1}}+36 i \left (\text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )-\text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )\right )+36 \tan ^{-1}(a x) \left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right )+3 \tan ^{-1}(a x) \sin \left (3 \tan ^{-1}(a x)\right )+\cos \left (3 \tan ^{-1}(a x)\right )\right )}{36 a^5 c^3 \sqrt {a^2 x^2+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*ArcTan[a*x])/(c + a^2*c*x^2)^(5/2),x]

[Out]

(Sqrt[c*(1 + a^2*x^2)]*(-45/Sqrt[1 + a^2*x^2] - (45*a*x*ArcTan[a*x])/Sqrt[1 + a^2*x^2] + Cos[3*ArcTan[a*x]] +
36*ArcTan[a*x]*(Log[1 - I*E^(I*ArcTan[a*x])] - Log[1 + I*E^(I*ArcTan[a*x])]) + (36*I)*(PolyLog[2, (-I)*E^(I*Ar
cTan[a*x])] - PolyLog[2, I*E^(I*ArcTan[a*x])]) + 3*ArcTan[a*x]*Sin[3*ArcTan[a*x]]))/(36*a^5*c^3*Sqrt[1 + a^2*x
^2])

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} x^{4} \arctan \left (a x\right )}{a^{6} c^{3} x^{6} + 3 \, a^{4} c^{3} x^{4} + 3 \, a^{2} c^{3} x^{2} + c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctan(a*x)/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(a^2*c*x^2 + c)*x^4*arctan(a*x)/(a^6*c^3*x^6 + 3*a^4*c^3*x^4 + 3*a^2*c^3*x^2 + c^3), 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^4*arctan(a*x)/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 3.24, size = 389, normalized size = 1.26 \[ -\frac {\left (i+3 \arctan \left (a x \right )\right ) \left (a^{3} x^{3}-3 i x^{2} a^{2}-3 a x +i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{72 \left (a^{2} x^{2}+1\right )^{2} c^{3} a^{5}}-\frac {5 \left (i+\arctan \left (a x \right )\right ) \left (a x -i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 a^{5} c^{3} \left (a^{2} x^{2}+1\right )}-\frac {5 \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a x +i\right ) \left (\arctan \left (a x \right )-i\right )}{8 a^{5} c^{3} \left (a^{2} x^{2}+1\right )}-\frac {\left (-i+3 \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a^{3} x^{3}+3 i x^{2} a^{2}-3 a x -i\right )}{72 \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right ) c^{3} a^{5}}+\frac {i \left (i \arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-i \arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+\dilog \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-\dilog \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, a^{5} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arctan(a*x)/(a^2*c*x^2+c)^(5/2),x)

[Out]

-1/72*(I+3*arctan(a*x))*(a^3*x^3-3*I*x^2*a^2-3*a*x+I)*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*x^2+1)^2/c^3/a^5-5/8*(I+a
rctan(a*x))*(a*x-I)*(c*(a*x-I)*(I+a*x))^(1/2)/a^5/c^3/(a^2*x^2+1)-5/8*(c*(a*x-I)*(I+a*x))^(1/2)*(I+a*x)*(arcta
n(a*x)-I)/a^5/c^3/(a^2*x^2+1)-1/72*(-I+3*arctan(a*x))*(c*(a*x-I)*(I+a*x))^(1/2)*(a^3*x^3+3*I*x^2*a^2-3*a*x-I)/
(a^4*x^4+2*a^2*x^2+1)/c^3/a^5+I*(I*arctan(a*x)*ln(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-I*arctan(a*x)*ln(1-I*(1+I*a
*x)/(a^2*x^2+1)^(1/2))+dilog(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-dilog(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2)))*(c*(a*x-
I)*(I+a*x))^(1/2)/(a^2*x^2+1)^(1/2)/a^5/c^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctan(a*x)/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^4*arctan(a*x)/(a^2*c*x^2 + c)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,\mathrm {atan}\left (a\,x\right )}{{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*atan(a*x))/(c + a^2*c*x^2)^(5/2),x)

[Out]

int((x^4*atan(a*x))/(c + a^2*c*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \operatorname {atan}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*atan(a*x)/(a**2*c*x**2+c)**(5/2),x)

[Out]

Integral(x**4*atan(a*x)/(c*(a**2*x**2 + 1))**(5/2), x)

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