3.1198 \(\int \frac {(d+e x^2)^{5/2} (a+b \tan ^{-1}(c x))}{x^2} \, dx\)

Optimal. Leaf size=111 \[ b \text {Int}\left (\frac {\tan ^{-1}(c x) \left (d+e x^2\right )^{5/2}}{x^2},x\right )+\frac {15}{8} a d^2 \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {a \left (d+e x^2\right )^{5/2}}{x}+\frac {5}{4} a e x \left (d+e x^2\right )^{3/2}+\frac {15}{8} a d e x \sqrt {d+e x^2} \]

[Out]

5/4*a*e*x*(e*x^2+d)^(3/2)-a*(e*x^2+d)^(5/2)/x+15/8*a*d^2*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))*e^(1/2)+15/8*a*d*e
*x*(e*x^2+d)^(1/2)+b*Unintegrable((e*x^2+d)^(5/2)*arctan(c*x)/x^2,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 = {} \[ \int \frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

(15*a*d*e*x*Sqrt[d + e*x^2])/8 + (5*a*e*x*(d + e*x^2)^(3/2))/4 - (a*(d + e*x^2)^(5/2))/x + (15*a*d^2*Sqrt[e]*A
rcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/8 + b*Defer[Int][((d + e*x^2)^(5/2)*ArcTan[c*x])/x^2, x]

Rubi steps

\begin {align*} \int \frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{x^2} \, dx &=a \int \frac {\left (d+e x^2\right )^{5/2}}{x^2} \, dx+b \int \frac {\left (d+e x^2\right )^{5/2} \tan ^{-1}(c x)}{x^2} \, dx\\ &=-\frac {a \left (d+e x^2\right )^{5/2}}{x}+b \int \frac {\left (d+e x^2\right )^{5/2} \tan ^{-1}(c x)}{x^2} \, dx+(5 a e) \int \left (d+e x^2\right )^{3/2} \, dx\\ &=\frac {5}{4} a e x \left (d+e x^2\right )^{3/2}-\frac {a \left (d+e x^2\right )^{5/2}}{x}+b \int \frac {\left (d+e x^2\right )^{5/2} \tan ^{-1}(c x)}{x^2} \, dx+\frac {1}{4} (15 a d e) \int \sqrt {d+e x^2} \, dx\\ &=\frac {15}{8} a d e x \sqrt {d+e x^2}+\frac {5}{4} a e x \left (d+e x^2\right )^{3/2}-\frac {a \left (d+e x^2\right )^{5/2}}{x}+b \int \frac {\left (d+e x^2\right )^{5/2} \tan ^{-1}(c x)}{x^2} \, dx+\frac {1}{8} \left (15 a d^2 e\right ) \int \frac {1}{\sqrt {d+e x^2}} \, dx\\ &=\frac {15}{8} a d e x \sqrt {d+e x^2}+\frac {5}{4} a e x \left (d+e x^2\right )^{3/2}-\frac {a \left (d+e x^2\right )^{5/2}}{x}+b \int \frac {\left (d+e x^2\right )^{5/2} \tan ^{-1}(c x)}{x^2} \, dx+\frac {1}{8} \left (15 a d^2 e\right ) \operatorname {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )\\ &=\frac {15}{8} a d e x \sqrt {d+e x^2}+\frac {5}{4} a e x \left (d+e x^2\right )^{3/2}-\frac {a \left (d+e x^2\right )^{5/2}}{x}+\frac {15}{8} a d^2 \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+b \int \frac {\left (d+e x^2\right )^{5/2} \tan ^{-1}(c x)}{x^2} \, dx\\ \end {align*}

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Mathematica [A]  time = 9.20, size = 0, normalized size = 0.00 \[ \int \frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a e^{2} x^{4} + 2 \, a d e x^{2} + a d^{2} + {\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 1.11, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \,x^{2}+d \right )^{\frac {5}{2}} \left (a +b \arctan \left (c x \right )\right )}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e*x^2+d)^(5/2)*(a+b*arctan(c*x))/x^2,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^2,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 \[ \int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^{5/2}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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