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

Optimal. Leaf size=72 \[ \frac {a^{3/2} x \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2} \sqrt {d x^2}}-\frac {a x^2}{b^2 \sqrt {d x^2}}+\frac {x^4}{3 b \sqrt {d x^2}} \]

[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)

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Rubi [A]  time = 0.03, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {15, 302, 205} \[ \frac {a^{3/2} x \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2} \sqrt {d x^2}}-\frac {a x^2}{b^2 \sqrt {d x^2}}+\frac {x^4}{3 b \sqrt {d x^2}} \]

Antiderivative was successfully verified.

[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 205

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

Rule 302

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*} \int \frac {x^5}{\sqrt {d x^2} \left (a+b x^2\right )} \, dx &=\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*}

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Mathematica [A]  time = 0.07, size = 56, normalized size = 0.78 \[ \frac {x \left (3 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )+\sqrt {b} x \left (b x^2-3 a\right )\right )}{3 b^{5/2} \sqrt {d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 147, normalized size = 2.04 \[ \left [\frac {3 \, a d \sqrt {-\frac {a}{b d}} \log \left (\frac {b x^{2} + 2 \, \sqrt {d x^{2}} b \sqrt {-\frac {a}{b d}} - a}{b x^{2} + a}\right ) + 2 \, {\left (b x^{2} - 3 \, a\right )} \sqrt {d x^{2}}}{6 \, b^{2} d}, \frac {3 \, a d \sqrt {\frac {a}{b d}} \arctan \left (\frac {\sqrt {d x^{2}} b \sqrt {\frac {a}{b d}}}{a}\right ) + {\left (b x^{2} - 3 \, a\right )} \sqrt {d x^{2}}}{3 \, b^{2} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 0.41, size = 70, normalized size = 0.97 \[ \frac {a^{2} \arctan \left (\frac {\sqrt {d x^{2}} b}{\sqrt {a b d}}\right )}{\sqrt {a b d} b^{2}} + \frac {\sqrt {d x^{2}} b^{2} d^{5} x^{2} - 3 \, \sqrt {d x^{2}} a b d^{5}}{3 \, b^{3} d^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*arctan(sqrt(d*x^2)*b/sqrt(a*b*d))/(sqrt(a*b*d)*b^2) + 1/3*(sqrt(d*x^2)*b^2*d^5*x^2 - 3*sqrt(d*x^2)*a*b*d^5
)/(b^3*d^6)

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maple [A]  time = 0.03, size = 53, normalized size = 0.74 \[ \frac {\left (\sqrt {a b}\, b \,x^{3}+3 a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )-3 \sqrt {a b}\, a x \right ) x}{3 \sqrt {d \,x^{2}}\, \sqrt {a b}\, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.83, size = 67, normalized size = 0.93 \[ \frac {\frac {3 \, a^{2} d^{3} \arctan \left (\frac {\sqrt {d x^{2}} b}{\sqrt {a b d}}\right )}{\sqrt {a b d} b^{2}} + \frac {\left (d x^{2}\right )^{\frac {3}{2}} b d - 3 \, \sqrt {d x^{2}} a d^{2}}{b^{2}}}{3 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.65, size = 51, normalized size = 0.71 \[ \frac {{\left (x^2\right )}^{3/2}}{3\,b\,\sqrt {d}}+\frac {a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x^2}}{\sqrt {a}}\right )}{b^{5/2}\,\sqrt {d}}-\frac {a\,\sqrt {x^2}}{b^2\,\sqrt {d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**5/(sqrt(d*x**2)*(a + b*x**2)), x)

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