3.1199 \(\int \frac {(d+e x^2)^{5/2} (a+b \tan ^{-1}(c x))}{x^3} \, dx\)

Optimal. Leaf size=108 \[ b \text {Int}\left (\frac {\tan ^{-1}(c x) \left (d+e x^2\right )^{5/2}}{x^3},x\right )-\frac {5}{2} a d^{3/2} e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )-\frac {a \left (d+e x^2\right )^{5/2}}{2 x^2}+\frac {5}{6} a e \left (d+e x^2\right )^{3/2}+\frac {5}{2} a d e \sqrt {d+e x^2} \]

[Out]

5/6*a*e*(e*x^2+d)^(3/2)-1/2*a*(e*x^2+d)^(5/2)/x^2-5/2*a*d^(3/2)*e*arctanh((e*x^2+d)^(1/2)/d^(1/2))+5/2*a*d*e*(
e*x^2+d)^(1/2)+b*Unintegrable((e*x^2+d)^(5/2)*arctan(c*x)/x^3,x)

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Rubi [A]  time = 0.21, 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 = {} \[ \int \frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{x^3} \, dx \]

Verification is Not applicable to the result.

[In]

Int[((d + e*x^2)^(5/2)*(a + b*ArcTan[c*x]))/x^3,x]

[Out]

(5*a*d*e*Sqrt[d + e*x^2])/2 + (5*a*e*(d + e*x^2)^(3/2))/6 - (a*(d + e*x^2)^(5/2))/(2*x^2) - (5*a*d^(3/2)*e*Arc
Tanh[Sqrt[d + e*x^2]/Sqrt[d]])/2 + b*Defer[Int][((d + e*x^2)^(5/2)*ArcTan[c*x])/x^3, x]

Rubi steps

\begin {align*} \int \frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{x^3} \, dx &=a \int \frac {\left (d+e x^2\right )^{5/2}}{x^3} \, dx+b \int \frac {\left (d+e x^2\right )^{5/2} \tan ^{-1}(c x)}{x^3} \, dx\\ &=\frac {1}{2} a \operatorname {Subst}\left (\int \frac {(d+e x)^{5/2}}{x^2} \, dx,x,x^2\right )+b \int \frac {\left (d+e x^2\right )^{5/2} \tan ^{-1}(c x)}{x^3} \, dx\\ &=-\frac {a \left (d+e x^2\right )^{5/2}}{2 x^2}+b \int \frac {\left (d+e x^2\right )^{5/2} \tan ^{-1}(c x)}{x^3} \, dx+\frac {1}{4} (5 a e) \operatorname {Subst}\left (\int \frac {(d+e x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac {5}{6} a e \left (d+e x^2\right )^{3/2}-\frac {a \left (d+e x^2\right )^{5/2}}{2 x^2}+b \int \frac {\left (d+e x^2\right )^{5/2} \tan ^{-1}(c x)}{x^3} \, dx+\frac {1}{4} (5 a d e) \operatorname {Subst}\left (\int \frac {\sqrt {d+e x}}{x} \, dx,x,x^2\right )\\ &=\frac {5}{2} a d e \sqrt {d+e x^2}+\frac {5}{6} a e \left (d+e x^2\right )^{3/2}-\frac {a \left (d+e x^2\right )^{5/2}}{2 x^2}+b \int \frac {\left (d+e x^2\right )^{5/2} \tan ^{-1}(c x)}{x^3} \, dx+\frac {1}{4} \left (5 a d^2 e\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )\\ &=\frac {5}{2} a d e \sqrt {d+e x^2}+\frac {5}{6} a e \left (d+e x^2\right )^{3/2}-\frac {a \left (d+e x^2\right )^{5/2}}{2 x^2}+b \int \frac {\left (d+e x^2\right )^{5/2} \tan ^{-1}(c x)}{x^3} \, dx+\frac {1}{2} \left (5 a d^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )\\ &=\frac {5}{2} a d e \sqrt {d+e x^2}+\frac {5}{6} a e \left (d+e x^2\right )^{3/2}-\frac {a \left (d+e x^2\right )^{5/2}}{2 x^2}-\frac {5}{2} a d^{3/2} e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )+b \int \frac {\left (d+e x^2\right )^{5/2} \tan ^{-1}(c x)}{x^3} \, dx\\ \end {align*}

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Mathematica [A]  time = 55.47, size = 0, normalized size = 0.00 \[ \int \frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{x^3} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[((d + e*x^2)^(5/2)*(a + b*ArcTan[c*x]))/x^3,x]

[Out]

Integrate[((d + e*x^2)^(5/2)*(a + b*ArcTan[c*x]))/x^3, x]

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fricas [A]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a e^{2} x^{4} + 2 \, a d e x^{2} + a d^{2} + {\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(5/2)*(a+b*arctan(c*x))/x^3,x, algorithm="fricas")

[Out]

integral((a*e^2*x^4 + 2*a*d*e*x^2 + a*d^2 + (b*e^2*x^4 + 2*b*d*e*x^2 + b*d^2)*arctan(c*x))*sqrt(e*x^2 + d)/x^3
, x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(5/2)*(a+b*arctan(c*x))/x^3,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 1.08, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \,x^{2}+d \right )^{\frac {5}{2}} \left (a +b \arctan \left (c x \right )\right )}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^(5/2)*(a+b*arctan(c*x))/x^3,x)

[Out]

int((e*x^2+d)^(5/2)*(a+b*arctan(c*x))/x^3,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((e*x^2+d)^(5/2)*(a+b*arctan(c*x))/x^3,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 zero or nonzero?

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mupad [A]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^{5/2}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + e*x^2)^(5/2))/x^3,x)

[Out]

int(((a + b*atan(c*x))*(d + e*x^2)^(5/2))/x^3, x)

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sympy [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{\frac {5}{2}}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**(5/2)*(a+b*atan(c*x))/x**3,x)

[Out]

Integral((a + b*atan(c*x))*(d + e*x**2)**(5/2)/x**3, x)

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