\(\int \frac {(d+e x^2)^{5/2} (a+b \arctan (c x))}{x} \, dx\) [1197]

Optimal result

Integrand size = 23, antiderivative size = 23 \[ \int \frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{x} \, dx=a d^2 \sqrt {d+e x^2}+\frac {1}{3} a d \left (d+e x^2\right )^{3/2}+\frac {1}{5} a \left (d+e x^2\right )^{5/2}-a d^{5/2} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )+b \text {Int}\left (\frac {\left (d+e x^2\right )^{5/2} \arctan (c x)}{x},x\right ) \]

[Out]

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

Rubi [N/A]

Not integrable

Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{x} \, dx=\int \frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{x} \, dx \]

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = a \int \frac {\left (d+e x^2\right )^{5/2}}{x} \, dx+b \int \frac {\left (d+e x^2\right )^{5/2} \arctan (c x)}{x} \, dx \\ & = \frac {1}{2} a \text {Subst}\left (\int \frac {(d+e x)^{5/2}}{x} \, dx,x,x^2\right )+b \int \frac {\left (d+e x^2\right )^{5/2} \arctan (c x)}{x} \, dx \\ & = \frac {1}{5} a \left (d+e x^2\right )^{5/2}+b \int \frac {\left (d+e x^2\right )^{5/2} \arctan (c x)}{x} \, dx+\frac {1}{2} (a d) \text {Subst}\left (\int \frac {(d+e x)^{3/2}}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{3} a d \left (d+e x^2\right )^{3/2}+\frac {1}{5} a \left (d+e x^2\right )^{5/2}+b \int \frac {\left (d+e x^2\right )^{5/2} \arctan (c x)}{x} \, dx+\frac {1}{2} \left (a d^2\right ) \text {Subst}\left (\int \frac {\sqrt {d+e x}}{x} \, dx,x,x^2\right ) \\ & = a d^2 \sqrt {d+e x^2}+\frac {1}{3} a d \left (d+e x^2\right )^{3/2}+\frac {1}{5} a \left (d+e x^2\right )^{5/2}+b \int \frac {\left (d+e x^2\right )^{5/2} \arctan (c x)}{x} \, dx+\frac {1}{2} \left (a d^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right ) \\ & = a d^2 \sqrt {d+e x^2}+\frac {1}{3} a d \left (d+e x^2\right )^{3/2}+\frac {1}{5} a \left (d+e x^2\right )^{5/2}+b \int \frac {\left (d+e x^2\right )^{5/2} \arctan (c x)}{x} \, dx+\frac {\left (a d^3\right ) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{e} \\ & = a d^2 \sqrt {d+e x^2}+\frac {1}{3} a d \left (d+e x^2\right )^{3/2}+\frac {1}{5} a \left (d+e x^2\right )^{5/2}-a d^{5/2} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )+b \int \frac {\left (d+e x^2\right )^{5/2} \arctan (c x)}{x} \, dx \\ \end{align*}

Mathematica [N/A]

Not integrable

Time = 10.80 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{x} \, dx=\int \frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{x} \, dx \]

[In]

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

[Out]

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

Maple [N/A] (verified)

Not integrable

Time = 0.62 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91

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

[In]

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

[Out]

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

Fricas [N/A]

Not integrable

Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.78 \[ \int \frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{\frac {5}{2}} {\left (b \arctan \left (c x\right ) + a\right )}}{x} \,d x } \]

[In]

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

Sympy [N/A]

Not integrable

Time = 40.39 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{x} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{\frac {5}{2}}}{x}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{x} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((e*x^2+d)^(5/2)*(a+b*arctan(c*x))/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>0)', see `assume?` for more
details)Is e

Giac [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{x} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [N/A]

Not integrable

Time = 0.99 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{x} \, dx=\int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^{5/2}}{x} \,d x \]

[In]

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

[Out]

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