3.210 \(\int x (c+a^2 c x^2)^{3/2} \tan ^{-1}(a x) \, dx\)

Optimal. Leaf size=109 \[ -\frac {3 c^{3/2} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{40 a^2}-\frac {x \left (a^2 c x^2+c\right )^{3/2}}{20 a}-\frac {3 c x \sqrt {a^2 c x^2+c}}{40 a}+\frac {\left (a^2 c x^2+c\right )^{5/2} \tan ^{-1}(a x)}{5 a^2 c} \]

[Out]

-1/20*x*(a^2*c*x^2+c)^(3/2)/a+1/5*(a^2*c*x^2+c)^(5/2)*arctan(a*x)/a^2/c-3/40*c^(3/2)*arctanh(a*x*c^(1/2)/(a^2*
c*x^2+c)^(1/2))/a^2-3/40*c*x*(a^2*c*x^2+c)^(1/2)/a

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Rubi [A]  time = 0.07, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4930, 195, 217, 206} \[ -\frac {3 c^{3/2} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{40 a^2}-\frac {x \left (a^2 c x^2+c\right )^{3/2}}{20 a}-\frac {3 c x \sqrt {a^2 c x^2+c}}{40 a}+\frac {\left (a^2 c x^2+c\right )^{5/2} \tan ^{-1}(a x)}{5 a^2 c} \]

Antiderivative was successfully verified.

[In]

Int[x*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x],x]

[Out]

(-3*c*x*Sqrt[c + a^2*c*x^2])/(40*a) - (x*(c + a^2*c*x^2)^(3/2))/(20*a) + ((c + a^2*c*x^2)^(5/2)*ArcTan[a*x])/(
5*a^2*c) - (3*c^(3/2)*ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x^2]])/(40*a^2)

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 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]

Rubi steps

\begin {align*} \int x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x) \, dx &=\frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{5 a^2 c}-\frac {\int \left (c+a^2 c x^2\right )^{3/2} \, dx}{5 a}\\ &=-\frac {x \left (c+a^2 c x^2\right )^{3/2}}{20 a}+\frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{5 a^2 c}-\frac {(3 c) \int \sqrt {c+a^2 c x^2} \, dx}{20 a}\\ &=-\frac {3 c x \sqrt {c+a^2 c x^2}}{40 a}-\frac {x \left (c+a^2 c x^2\right )^{3/2}}{20 a}+\frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{5 a^2 c}-\frac {\left (3 c^2\right ) \int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx}{40 a}\\ &=-\frac {3 c x \sqrt {c+a^2 c x^2}}{40 a}-\frac {x \left (c+a^2 c x^2\right )^{3/2}}{20 a}+\frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{5 a^2 c}-\frac {\left (3 c^2\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 )}{40 a}\\ &=-\frac {3 c x \sqrt {c+a^2 c x^2}}{40 a}-\frac {x \left (c+a^2 c x^2\right )^{3/2}}{20 a}+\frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{5 a^2 c}-\frac {3 c^{3/2} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{40 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 101, normalized size = 0.93 \[ -\frac {3 c^{3/2} \log \left (\sqrt {c} \sqrt {a^2 c x^2+c}+a c x\right )+a c x \left (2 a^2 x^2+5\right ) \sqrt {a^2 c x^2+c}-8 c \left (a^2 x^2+1\right )^2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{40 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x],x]

[Out]

-1/40*(a*c*x*(5 + 2*a^2*x^2)*Sqrt[c + a^2*c*x^2] - 8*c*(1 + a^2*x^2)^2*Sqrt[c + a^2*c*x^2]*ArcTan[a*x] + 3*c^(
3/2)*Log[a*c*x + Sqrt[c]*Sqrt[c + a^2*c*x^2]])/a^2

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fricas [A]  time = 0.72, size = 98, normalized size = 0.90 \[ \frac {3 \, c^{\frac {3}{2}} \log \left (-2 \, a^{2} c x^{2} + 2 \, \sqrt {a^{2} c x^{2} + c} a \sqrt {c} x - c\right ) - 2 \, {\left (2 \, a^{3} c x^{3} + 5 \, a c x - 8 \, {\left (a^{4} c x^{4} + 2 \, a^{2} c x^{2} + c\right )} \arctan \left (a x\right )\right )} \sqrt {a^{2} c x^{2} + c}}{80 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a^2*c*x^2+c)^(3/2)*arctan(a*x),x, algorithm="fricas")

[Out]

1/80*(3*c^(3/2)*log(-2*a^2*c*x^2 + 2*sqrt(a^2*c*x^2 + c)*a*sqrt(c)*x - c) - 2*(2*a^3*c*x^3 + 5*a*c*x - 8*(a^4*
c*x^4 + 2*a^2*c*x^2 + c)*arctan(a*x))*sqrt(a^2*c*x^2 + c))/a^2

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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)^(3/2)*arctan(a*x),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 = 0.89, size = 179, normalized size = 1.64 \[ \frac {c \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (8 \arctan \left (a x \right ) x^{4} a^{4}-2 a^{3} x^{3}+16 \arctan \left (a x \right ) x^{2} a^{2}-5 a x +8 \arctan \left (a x \right )\right )}{40 a^{2}}+\frac {3 c \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-i\right )}{40 a^{2} \sqrt {a^{2} x^{2}+1}}-\frac {3 c \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+i\right )}{40 a^{2} \sqrt {a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a^2*c*x^2+c)^(3/2)*arctan(a*x),x)

[Out]

1/40*c/a^2*(c*(a*x-I)*(I+a*x))^(1/2)*(8*arctan(a*x)*x^4*a^4-2*a^3*x^3+16*arctan(a*x)*x^2*a^2-5*a*x+8*arctan(a*
x))+3/40*c/a^2*(c*(a*x-I)*(I+a*x))^(1/2)*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)-I)/(a^2*x^2+1)^(1/2)-3/40*c/a^2*(c*(a*
x-I)*(I+a*x))^(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 = 0.60, size = 406, normalized size = 3.72 \[ \frac {40 \, {\left (a^{2} c x^{2} + c\right )} \sqrt {a^{2} x^{2} + 1} \sqrt {c} \arctan \left (a x\right ) - 20 \, {\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} {\left (a c x \cos \left (\frac {1}{2} \, \arctan \left (4 \, a x, -a^{2} x^{2} + 3\right )\right ) + 2 \, c \sin \left (\frac {1}{2} \, \arctan \left (4 \, a x, -a^{2} x^{2} + 3\right )\right )\right )} \sqrt {c} - {\left ({\left (a {\left (\frac {3 \, {\left (\frac {2 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{a^{2}} - \frac {\sqrt {a^{2} x^{2} + 1} x}{a^{2}} - \frac {\operatorname {arsinh}\left (a x\right )}{a^{3}}\right )}}{a^{2}} - \frac {8 \, {\left (\sqrt {a^{2} x^{2} + 1} x + \frac {\operatorname {arsinh}\left (a x\right )}{a}\right )}}{a^{4}}\right )} - 8 \, {\left (\frac {3 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{a^{2}} - \frac {2 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{4}}\right )} \arctan \left (a x\right )\right )} a^{4} c - 10 \, c \arctan \left ({\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right ) + 2, a x + {\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right )\right ) - 10 \, c \arctan \left ({\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right ) - 2, -a x + {\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right )\right )\right )} \sqrt {c}}{120 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a^2*c*x^2+c)^(3/2)*arctan(a*x),x, algorithm="maxima")

[Out]

1/120*(40*(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*sqrt(c)*arctan(a*x) - 20*(a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*(a*c*x*c
os(1/2*arctan2(4*a*x, -a^2*x^2 + 3)) + 2*c*sin(1/2*arctan2(4*a*x, -a^2*x^2 + 3)))*sqrt(c) - ((a*(3*(2*(a^2*x^2
+ 1)^(3/2)*x/a^2 - sqrt(a^2*x^2 + 1)*x/a^2 - arcsinh(a*x)/a^3)/a^2 - 8*(sqrt(a^2*x^2 + 1)*x + arcsinh(a*x)/a)
/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 - 10*c*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)*co
s(1/2*arctan2(4*a*x, a^2*x^2 - 3))) - 10*c*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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\mathrm {atan}\left (a\,x\right )\,{\left (c\,a^2\,x^2+c\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atan(a*x)*(c + a^2*c*x^2)^(3/2),x)

[Out]

int(x*atan(a*x)*(c + a^2*c*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \operatorname {atan}{\left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a**2*c*x**2+c)**(3/2)*atan(a*x),x)

[Out]

Integral(x*(c*(a**2*x**2 + 1))**(3/2)*atan(a*x), x)

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