\(\int \frac {x^5}{\sqrt {d x^2} (a+b x^2)} \, dx\) [671]

Optimal result

Integrand size = 22, antiderivative size = 72 \[ \int \frac {x^5}{\sqrt {d x^2} \left (a+b x^2\right )} \, dx=-\frac {a x^2}{b^2 \sqrt {d x^2}}+\frac {x^4}{3 b \sqrt {d x^2}}+\frac {a^{3/2} x \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2} \sqrt {d x^2}} \]

[Out]

-a*x^2/b^2/(d*x^2)^(1/2)+1/3*x^4/b/(d*x^2)^(1/2)+a^(3/2)*x*arctan(x*b^(1/2)/a^(1/2))/b^(5/2)/(d*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {15, 308, 211} \[ \int \frac {x^5}{\sqrt {d x^2} \left (a+b x^2\right )} \, dx=\frac {a^{3/2} x \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2} \sqrt {d x^2}}-\frac {a x^2}{b^2 \sqrt {d x^2}}+\frac {x^4}{3 b \sqrt {d x^2}} \]

[In]

Int[x^5/(Sqrt[d*x^2]*(a + b*x^2)),x]

[Out]

-((a*x^2)/(b^2*Sqrt[d*x^2])) + x^4/(3*b*Sqrt[d*x^2]) + (a^(3/2)*x*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(b^(5/2)*Sqrt[d
*x^2])

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rubi steps \begin{align*} \text {integral}& = \frac {x \int \frac {x^4}{a+b x^2} \, dx}{\sqrt {d x^2}} \\ & = \frac {x \int \left (-\frac {a}{b^2}+\frac {x^2}{b}+\frac {a^2}{b^2 \left (a+b x^2\right )}\right ) \, dx}{\sqrt {d x^2}} \\ & = -\frac {a x^2}{b^2 \sqrt {d x^2}}+\frac {x^4}{3 b \sqrt {d x^2}}+\frac {\left (a^2 x\right ) \int \frac {1}{a+b x^2} \, dx}{b^2 \sqrt {d x^2}} \\ & = -\frac {a x^2}{b^2 \sqrt {d x^2}}+\frac {x^4}{3 b \sqrt {d x^2}}+\frac {a^{3/2} x \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2} \sqrt {d x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.99 \[ \int \frac {x^5}{\sqrt {d x^2} \left (a+b x^2\right )} \, dx=\frac {\sqrt {d x^2} \left (-3 a+b x^2\right )}{3 b^2 d}+\frac {a^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {d x^2}}{\sqrt {a} \sqrt {d}}\right )}{b^{5/2} \sqrt {d}} \]

[In]

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

[Out]

(Sqrt[d*x^2]*(-3*a + b*x^2))/(3*b^2*d) + (a^(3/2)*ArcTan[(Sqrt[b]*Sqrt[d*x^2])/(Sqrt[a]*Sqrt[d])])/(b^(5/2)*Sq
rt[d])

Maple [A] (verified)

Time = 3.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.75

method	result	size
default	\(-\frac {x \left (-\sqrt {a b}\, b \,x^{3}+3 \sqrt {a b}\, a x -3 a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )\right )}{3 \sqrt {d \,x^{2}}\, b^{2} \sqrt {a b}}\)	\(54\)
pseudoelliptic	\(\frac {\arctan \left (\frac {b \sqrt {d \,x^{2}}}{\sqrt {a b d}}\right ) a^{2} d -\sqrt {d \,x^{2}}\, \left (-\frac {b \,x^{2}}{3}+a \right ) \sqrt {a b d}}{\sqrt {a b d}\, d \,b^{2}}\)	\(59\)
risch	\(\frac {x \left (\frac {1}{3} b \,x^{3}-a x \right )}{\sqrt {d \,x^{2}}\, b^{2}}+\frac {x \sqrt {-a b}\, a \ln \left (-\sqrt {-a b}\, x +a \right )}{2 \sqrt {d \,x^{2}}\, b^{3}}-\frac {x \sqrt {-a b}\, a \ln \left (\sqrt {-a b}\, x +a \right )}{2 \sqrt {d \,x^{2}}\, b^{3}}\)	\(88\)

[In]

int(x^5/(b*x^2+a)/(d*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*x*(-(a*b)^(1/2)*b*x^3+3*(a*b)^(1/2)*a*x-3*a^2*arctan(b*x/(a*b)^(1/2)))/(d*x^2)^(1/2)/b^2/(a*b)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 147, normalized size of antiderivative = 2.04 \[ \int \frac {x^5}{\sqrt {d x^2} \left (a+b x^2\right )} \, dx=\left [\frac {3 \, a d \sqrt {-\frac {a}{b d}} \log \left (\frac {b x^{2} + 2 \, \sqrt {d x^{2}} b \sqrt {-\frac {a}{b d}} - a}{b x^{2} + a}\right ) + 2 \, {\left (b x^{2} - 3 \, a\right )} \sqrt {d x^{2}}}{6 \, b^{2} d}, \frac {3 \, a d \sqrt {\frac {a}{b d}} \arctan \left (\frac {\sqrt {d x^{2}} b \sqrt {\frac {a}{b d}}}{a}\right ) + {\left (b x^{2} - 3 \, a\right )} \sqrt {d x^{2}}}{3 \, b^{2} d}\right ] \]

[In]

integrate(x^5/(b*x^2+a)/(d*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/6*(3*a*d*sqrt(-a/(b*d))*log((b*x^2 + 2*sqrt(d*x^2)*b*sqrt(-a/(b*d)) - a)/(b*x^2 + a)) + 2*(b*x^2 - 3*a)*sqr
t(d*x^2))/(b^2*d), 1/3*(3*a*d*sqrt(a/(b*d))*arctan(sqrt(d*x^2)*b*sqrt(a/(b*d))/a) + (b*x^2 - 3*a)*sqrt(d*x^2))
/(b^2*d)]

Sympy [A] (verification not implemented)

Time = 1.42 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.04 \[ \int \frac {x^5}{\sqrt {d x^2} \left (a+b x^2\right )} \, dx=\begin {cases} \frac {2 \left (\frac {a^{2} d^{3} \operatorname {atan}{\left (\frac {\sqrt {d x^{2}}}{\sqrt {\frac {a d}{b}}} \right )}}{2 b^{3} \sqrt {\frac {a d}{b}}} - \frac {a d^{2} \sqrt {d x^{2}}}{2 b^{2}} + \frac {d \left (d x^{2}\right )^{\frac {3}{2}}}{6 b}\right )}{d^{3}} & \text {for}\: d \neq 0 \\\tilde {\infty } x^{6} & \text {otherwise} \end {cases} \]

[In]

integrate(x**5/(b*x**2+a)/(d*x**2)**(1/2),x)

[Out]

Piecewise((2*(a**2*d**3*atan(sqrt(d*x**2)/sqrt(a*d/b))/(2*b**3*sqrt(a*d/b)) - a*d**2*sqrt(d*x**2)/(2*b**2) + d
*(d*x**2)**(3/2)/(6*b))/d**3, Ne(d, 0)), (zoo*x**6, True))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.93 \[ \int \frac {x^5}{\sqrt {d x^2} \left (a+b x^2\right )} \, dx=\frac {\frac {3 \, a^{2} d^{3} \arctan \left (\frac {\sqrt {d x^{2}} b}{\sqrt {a b d}}\right )}{\sqrt {a b d} b^{2}} + \frac {\left (d x^{2}\right )^{\frac {3}{2}} b d - 3 \, \sqrt {d x^{2}} a d^{2}}{b^{2}}}{3 \, d^{3}} \]

[In]

integrate(x^5/(b*x^2+a)/(d*x^2)^(1/2),x, algorithm="maxima")

[Out]

1/3*(3*a^2*d^3*arctan(sqrt(d*x^2)*b/sqrt(a*b*d))/(sqrt(a*b*d)*b^2) + ((d*x^2)^(3/2)*b*d - 3*sqrt(d*x^2)*a*d^2)
/b^2)/d^3

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.78 \[ \int \frac {x^5}{\sqrt {d x^2} \left (a+b x^2\right )} \, dx=\frac {a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2} \sqrt {d} \mathrm {sgn}\left (x\right )} + \frac {b^{2} d x^{3} - 3 \, a b d x}{3 \, b^{3} d^{\frac {3}{2}} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(x^5/(b*x^2+a)/(d*x^2)^(1/2),x, algorithm="giac")

[Out]

a^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^2*sqrt(d)*sgn(x)) + 1/3*(b^2*d*x^3 - 3*a*b*d*x)/(b^3*d^(3/2)*sgn(x))

Mupad [B] (verification not implemented)

Time = 5.23 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.71 \[ \int \frac {x^5}{\sqrt {d x^2} \left (a+b x^2\right )} \, dx=\frac {{\left (x^2\right )}^{3/2}}{3\,b\,\sqrt {d}}+\frac {a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x^2}}{\sqrt {a}}\right )}{b^{5/2}\,\sqrt {d}}-\frac {a\,\sqrt {x^2}}{b^2\,\sqrt {d}} \]

[In]

int(x^5/((a + b*x^2)*(d*x^2)^(1/2)),x)

[Out]

(x^2)^(3/2)/(3*b*d^(1/2)) + (a^(3/2)*atan((b^(1/2)*(x^2)^(1/2))/a^(1/2)))/(b^(5/2)*d^(1/2)) - (a*(x^2)^(1/2))/
(b^2*d^(1/2))