3.1184 \(\int x^2 (d+e x^2)^{3/2} (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=119 \[ b \text {Int}\left (x^2 \tan ^{-1}(c x) \left (d+e x^2\right )^{3/2},x\right )-\frac {a d^3 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{16 e^{3/2}}+\frac {a d^2 x \sqrt {d+e x^2}}{16 e}+\frac {1}{8} a d x^3 \sqrt {d+e x^2}+\frac {1}{6} a x^3 \left (d+e x^2\right )^{3/2} \]

[Out]

1/6*a*x^3*(e*x^2+d)^(3/2)-1/16*a*d^3*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/e^(3/2)+1/16*a*d^2*x*(e*x^2+d)^(1/2)/e
+1/8*a*d*x^3*(e*x^2+d)^(1/2)+b*Unintegrable(x^2*(e*x^2+d)^(3/2)*arctan(c*x),x)

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Rubi [A]  time = 0.19, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int x^2 \left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

(a*d^2*x*Sqrt[d + e*x^2])/(16*e) + (a*d*x^3*Sqrt[d + e*x^2])/8 + (a*x^3*(d + e*x^2)^(3/2))/6 - (a*d^3*ArcTanh[
(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(16*e^(3/2)) + b*Defer[Int][x^2*(d + e*x^2)^(3/2)*ArcTan[c*x], x]

Rubi steps

\begin {align*} \int x^2 \left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right ) \, dx &=a \int x^2 \left (d+e x^2\right )^{3/2} \, dx+b \int x^2 \left (d+e x^2\right )^{3/2} \tan ^{-1}(c x) \, dx\\ &=\frac {1}{6} a x^3 \left (d+e x^2\right )^{3/2}+b \int x^2 \left (d+e x^2\right )^{3/2} \tan ^{-1}(c x) \, dx+\frac {1}{2} (a d) \int x^2 \sqrt {d+e x^2} \, dx\\ &=\frac {1}{8} a d x^3 \sqrt {d+e x^2}+\frac {1}{6} a x^3 \left (d+e x^2\right )^{3/2}+b \int x^2 \left (d+e x^2\right )^{3/2} \tan ^{-1}(c x) \, dx+\frac {1}{8} \left (a d^2\right ) \int \frac {x^2}{\sqrt {d+e x^2}} \, dx\\ &=\frac {a d^2 x \sqrt {d+e x^2}}{16 e}+\frac {1}{8} a d x^3 \sqrt {d+e x^2}+\frac {1}{6} a x^3 \left (d+e x^2\right )^{3/2}+b \int x^2 \left (d+e x^2\right )^{3/2} \tan ^{-1}(c x) \, dx-\frac {\left (a d^3\right ) \int \frac {1}{\sqrt {d+e x^2}} \, dx}{16 e}\\ &=\frac {a d^2 x \sqrt {d+e x^2}}{16 e}+\frac {1}{8} a d x^3 \sqrt {d+e x^2}+\frac {1}{6} a x^3 \left (d+e x^2\right )^{3/2}+b \int x^2 \left (d+e x^2\right )^{3/2} \tan ^{-1}(c x) \, dx-\frac {\left (a d^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{16 e}\\ &=\frac {a d^2 x \sqrt {d+e x^2}}{16 e}+\frac {1}{8} a d x^3 \sqrt {d+e x^2}+\frac {1}{6} a x^3 \left (d+e x^2\right )^{3/2}-\frac {a d^3 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{16 e^{3/2}}+b \int x^2 \left (d+e x^2\right )^{3/2} \tan ^{-1}(c x) \, dx\\ \end {align*}

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Mathematica [A]  time = 11.31, size = 0, normalized size = 0.00 \[ \int x^2 \left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a e x^{4} + a d x^{2} + {\left (b e x^{4} + b d x^{2}\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*e*x^4 + a*d*x^2 + (b*e*x^4 + b*d*x^2)*arctan(c*x))*sqrt(e*x^2 + d), x)

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giac [A]  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^2*(e*x^2+d)^(3/2)*(a+b*arctan(c*x)),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 1.08, size = 0, normalized size = 0.00 \[ \int x^{2} \left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \arctan \left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e-c^2*d>0)', see `assume?` for
more details)Is e-c^2*d positive or negative?

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mupad [A]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(a + b*atan(c*x))*(d + e*x**2)**(3/2), x)

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