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

Optimal. Leaf size=170 \[ -\frac {\tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{a^6 c^{5/2}}+\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{a^6 c^3}+\frac {5 \tan ^{-1}(a x)}{3 a^6 c^2 \sqrt {a^2 c x^2+c}}-\frac {5 x}{3 a^5 c^2 \sqrt {a^2 c x^2+c}}+\frac {x^2 \tan ^{-1}(a x)}{3 a^4 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {x^3}{9 a^3 c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

-1/9*x^3/a^3/c/(a^2*c*x^2+c)^(3/2)+1/3*x^2*arctan(a*x)/a^4/c/(a^2*c*x^2+c)^(3/2)-arctanh(a*x*c^(1/2)/(a^2*c*x^
2+c)^(1/2))/a^6/c^(5/2)-5/3*x/a^5/c^2/(a^2*c*x^2+c)^(1/2)+5/3*arctan(a*x)/a^6/c^2/(a^2*c*x^2+c)^(1/2)+arctan(a
*x)*(a^2*c*x^2+c)^(1/2)/a^6/c^3

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Rubi [A]  time = 0.43, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4964, 4930, 217, 206, 191, 4938} \[ -\frac {5 x}{3 a^5 c^2 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{a^6 c^3}+\frac {5 \tan ^{-1}(a x)}{3 a^6 c^2 \sqrt {a^2 c x^2+c}}-\frac {\tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{a^6 c^{5/2}}-\frac {x^3}{9 a^3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {x^2 \tan ^{-1}(a x)}{3 a^4 c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^3/(9*a^3*c*(c + a^2*c*x^2)^(3/2)) - (5*x)/(3*a^5*c^2*Sqrt[c + a^2*c*x^2]) + (x^2*ArcTan[a*x])/(3*a^4*c*(c +
a^2*c*x^2)^(3/2)) + (5*ArcTan[a*x])/(3*a^6*c^2*Sqrt[c + a^2*c*x^2]) + (Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(a^6*
c^3) - ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x^2]]/(a^6*c^(5/2))

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 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])

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 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 4938

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(b*(f*x
)^m*(d + e*x^2)^(q + 1))/(c*d*m^2), x] + (Dist[(f^2*(m - 1))/(c^2*d*m), Int[(f*x)^(m - 2)*(d + e*x^2)^(q + 1)*
(a + b*ArcTan[c*x]), x], x] - Simp[(f*(f*x)^(m - 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]))/(c^2*d*m), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && EqQ[m + 2*q + 2, 0] && LtQ[q, -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^5 \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx &=-\frac {\int \frac {x^3 \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx}{a^2}+\frac {\int \frac {x^3 \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{a^2 c}\\ &=-\frac {x^3}{9 a^3 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {x^2 \tan ^{-1}(a x)}{3 a^4 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {\int \frac {x \tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx}{a^4 c^2}-\frac {2 \int \frac {x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 a^4 c}-\frac {\int \frac {x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{a^4 c}\\ &=-\frac {x^3}{9 a^3 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {x^2 \tan ^{-1}(a x)}{3 a^4 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {5 \tan ^{-1}(a x)}{3 a^6 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a^6 c^3}-\frac {\int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx}{a^5 c^2}-\frac {2 \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 a^5 c}-\frac {\int \frac {1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{a^5 c}\\ &=-\frac {x^3}{9 a^3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 x}{3 a^5 c^2 \sqrt {c+a^2 c x^2}}+\frac {x^2 \tan ^{-1}(a x)}{3 a^4 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {5 \tan ^{-1}(a x)}{3 a^6 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a^6 c^3}-\frac {\operatorname {Subst}\left (\int \frac {1}{1-a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c+a^2 c x^2}}\right )}{a^5 c^2}\\ &=-\frac {x^3}{9 a^3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 x}{3 a^5 c^2 \sqrt {c+a^2 c x^2}}+\frac {x^2 \tan ^{-1}(a x)}{3 a^4 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {5 \tan ^{-1}(a x)}{3 a^6 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a^6 c^3}-\frac {\tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a^6 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 131, normalized size = 0.77 \[ -\frac {a x \left (16 a^2 x^2+15\right ) \sqrt {a^2 c x^2+c}+9 \sqrt {c} \left (a^2 x^2+1\right )^2 \log \left (\sqrt {c} \sqrt {a^2 c x^2+c}+a c x\right )-3 \left (3 a^4 x^4+12 a^2 x^2+8\right ) \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{9 a^6 c^3 \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/9*(a*x*(15 + 16*a^2*x^2)*Sqrt[c + a^2*c*x^2] - 3*Sqrt[c + a^2*c*x^2]*(8 + 12*a^2*x^2 + 3*a^4*x^4)*ArcTan[a*
x] + 9*Sqrt[c]*(1 + a^2*x^2)^2*Log[a*c*x + Sqrt[c]*Sqrt[c + a^2*c*x^2]])/(a^6*c^3*(1 + a^2*x^2)^2)

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fricas [A]  time = 0.53, size = 140, normalized size = 0.82 \[ \frac {9 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \sqrt {c} \log \left (-2 \, a^{2} c x^{2} + 2 \, \sqrt {a^{2} c x^{2} + c} a \sqrt {c} x - c\right ) - 2 \, {\left (16 \, a^{3} x^{3} + 15 \, a x - 3 \, {\left (3 \, a^{4} x^{4} + 12 \, a^{2} x^{2} + 8\right )} \arctan \left (a x\right )\right )} \sqrt {a^{2} c x^{2} + c}}{18 \, {\left (a^{10} c^{3} x^{4} + 2 \, a^{8} c^{3} x^{2} + a^{6} c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(9*(a^4*x^4 + 2*a^2*x^2 + 1)*sqrt(c)*log(-2*a^2*c*x^2 + 2*sqrt(a^2*c*x^2 + c)*a*sqrt(c)*x - c) - 2*(16*a^
3*x^3 + 15*a*x - 3*(3*a^4*x^4 + 12*a^2*x^2 + 8)*arctan(a*x))*sqrt(a^2*c*x^2 + c))/(a^10*c^3*x^4 + 2*a^8*c^3*x^
2 + a^6*c^3)

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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^5*arctan(a*x)/(a^2*c*x^2+c)^(5/2),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 [C]  time = 5.82, size = 386, normalized size = 2.27 \[ \frac {\left (i+3 \arctan \left (a x \right )\right ) \left (i x^{3} a^{3}+3 a^{2} x^{2}-3 i a x -1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{72 \left (a^{2} x^{2}+1\right )^{2} c^{3} a^{6}}+\frac {7 \left (i+\arctan \left (a x \right )\right ) \left (i a x +1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 a^{6} c^{3} \left (a^{2} x^{2}+1\right )}-\frac {7 \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i a x -1\right ) \left (\arctan \left (a x \right )-i\right )}{8 a^{6} c^{3} \left (a^{2} x^{2}+1\right )}-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i x^{3} a^{3}-3 a^{2} x^{2}-3 i a x +1\right ) \left (-i+3 \arctan \left (a x \right )\right )}{72 a^{6} c^{3} \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{c^{3} a^{6}}-\frac {\ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, a^{6} c^{3}}+\frac {\ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, a^{6} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(I+3*arctan(a*x))*(I*x^3*a^3+3*a^2*x^2-3*I*a*x-1)*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*x^2+1)^2/c^3/a^6+7/8*(I+
arctan(a*x))*(1+I*a*x)*(c*(a*x-I)*(I+a*x))^(1/2)/a^6/c^3/(a^2*x^2+1)-7/8*(c*(a*x-I)*(I+a*x))^(1/2)*(-1+I*a*x)*
(arctan(a*x)-I)/a^6/c^3/(a^2*x^2+1)-1/72*(c*(a*x-I)*(I+a*x))^(1/2)*(I*x^3*a^3-3*a^2*x^2-3*I*a*x+1)*(-I+3*arcta
n(a*x))/a^6/c^3/(a^4*x^4+2*a^2*x^2+1)+arctan(a*x)*(c*(a*x-I)*(I+a*x))^(1/2)/c^3/a^6-ln((1+I*a*x)/(a^2*x^2+1)^(
1/2)+I)/(a^2*x^2+1)^(1/2)*(c*(a*x-I)*(I+a*x))^(1/2)/a^6/c^3+ln((1+I*a*x)/(a^2*x^2+1)^(1/2)-I)/(a^2*x^2+1)^(1/2
)*(c*(a*x-I)*(I+a*x))^(1/2)/a^6/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^5*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 = -0.01 \[ \int \frac {x^5\,\mathrm {atan}\left (a\,x\right )}{{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x^5*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 \[ \int \frac {x^{5} \operatorname {atan}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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