3.1194 \(\int x^2 (d+e x^2)^{5/2} (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=140 \[ b \text{Unintegrable}\left (x^2 \tan ^{-1}(c x) \left (d+e x^2\right )^{5/2},x\right )-\frac{5 a d^4 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{128 e^{3/2}}+\frac{5 a d^3 x \sqrt{d+e x^2}}{128 e}+\frac{5}{64} a d^2 x^3 \sqrt{d+e x^2}+\frac{5}{48} a d x^3 \left (d+e x^2\right )^{3/2}+\frac{1}{8} a x^3 \left (d+e x^2\right )^{5/2} \]

[Out]

(5*a*d^3*x*Sqrt[d + e*x^2])/(128*e) + (5*a*d^2*x^3*Sqrt[d + e*x^2])/64 + (5*a*d*x^3*(d + e*x^2)^(3/2))/48 + (a
*x^3*(d + e*x^2)^(5/2))/8 - (5*a*d^4*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(128*e^(3/2)) + b*Unintegrable[x^2*
(d + e*x^2)^(5/2)*ArcTan[c*x], x]

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Rubi [A]  time = 0.227219, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int x^2 \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

(5*a*d^3*x*Sqrt[d + e*x^2])/(128*e) + (5*a*d^2*x^3*Sqrt[d + e*x^2])/64 + (5*a*d*x^3*(d + e*x^2)^(3/2))/48 + (a
*x^3*(d + e*x^2)^(5/2))/8 - (5*a*d^4*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(128*e^(3/2)) + b*Defer[Int][x^2*(d
+ e*x^2)^(5/2)*ArcTan[c*x], x]

Rubi steps

\begin{align*} \int x^2 \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right ) \, dx &=a \int x^2 \left (d+e x^2\right )^{5/2} \, dx+b \int x^2 \left (d+e x^2\right )^{5/2} \tan ^{-1}(c x) \, dx\\ &=\frac{1}{8} a x^3 \left (d+e x^2\right )^{5/2}+b \int x^2 \left (d+e x^2\right )^{5/2} \tan ^{-1}(c x) \, dx+\frac{1}{8} (5 a d) \int x^2 \left (d+e x^2\right )^{3/2} \, dx\\ &=\frac{5}{48} a d x^3 \left (d+e x^2\right )^{3/2}+\frac{1}{8} a x^3 \left (d+e x^2\right )^{5/2}+b \int x^2 \left (d+e x^2\right )^{5/2} \tan ^{-1}(c x) \, dx+\frac{1}{16} \left (5 a d^2\right ) \int x^2 \sqrt{d+e x^2} \, dx\\ &=\frac{5}{64} a d^2 x^3 \sqrt{d+e x^2}+\frac{5}{48} a d x^3 \left (d+e x^2\right )^{3/2}+\frac{1}{8} a x^3 \left (d+e x^2\right )^{5/2}+b \int x^2 \left (d+e x^2\right )^{5/2} \tan ^{-1}(c x) \, dx+\frac{1}{64} \left (5 a d^3\right ) \int \frac{x^2}{\sqrt{d+e x^2}} \, dx\\ &=\frac{5 a d^3 x \sqrt{d+e x^2}}{128 e}+\frac{5}{64} a d^2 x^3 \sqrt{d+e x^2}+\frac{5}{48} a d x^3 \left (d+e x^2\right )^{3/2}+\frac{1}{8} a x^3 \left (d+e x^2\right )^{5/2}+b \int x^2 \left (d+e x^2\right )^{5/2} \tan ^{-1}(c x) \, dx-\frac{\left (5 a d^4\right ) \int \frac{1}{\sqrt{d+e x^2}} \, dx}{128 e}\\ &=\frac{5 a d^3 x \sqrt{d+e x^2}}{128 e}+\frac{5}{64} a d^2 x^3 \sqrt{d+e x^2}+\frac{5}{48} a d x^3 \left (d+e x^2\right )^{3/2}+\frac{1}{8} a x^3 \left (d+e x^2\right )^{5/2}+b \int x^2 \left (d+e x^2\right )^{5/2} \tan ^{-1}(c x) \, dx-\frac{\left (5 a d^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{128 e}\\ &=\frac{5 a d^3 x \sqrt{d+e x^2}}{128 e}+\frac{5}{64} a d^2 x^3 \sqrt{d+e x^2}+\frac{5}{48} a d x^3 \left (d+e x^2\right )^{3/2}+\frac{1}{8} a x^3 \left (d+e x^2\right )^{5/2}-\frac{5 a d^4 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{128 e^{3/2}}+b \int x^2 \left (d+e x^2\right )^{5/2} \tan ^{-1}(c x) \, dx\\ \end{align*}

Mathematica [A]  time = 11.443, size = 0, normalized size = 0. \[ \int x^2 \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [A]  time = 0.586, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( e{x}^{2}+d \right ) ^{{\frac{5}{2}}} \left ( a+b\arctan \left ( cx \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a e^{2} x^{6} + 2 \, a d e x^{4} + a d^{2} x^{2} +{\left (b e^{2} x^{6} + 2 \, b d e x^{4} + b d^{2} x^{2}\right )} \arctan \left (c x\right )\right )} \sqrt{e x^{2} + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*e^2*x^6 + 2*a*d*e*x^4 + a*d^2*x^2 + (b*e^2*x^6 + 2*b*d*e*x^4 + b*d^2*x^2)*arctan(c*x))*sqrt(e*x^2
+ d), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d\right )}^{\frac{5}{2}}{\left (b \arctan \left (c x\right ) + a\right )} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x^2+d)^(5/2)*(a+b*arctan(c*x)),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^(5/2)*(b*arctan(c*x) + a)*x^2, x)