3.7.72 \(\int \frac {x^3}{\sqrt {d x^2} (a+b x^2)} \, dx\) [672]

Optimal. Leaf size=52 \[ \frac {x^2}{b \sqrt {d x^2}}-\frac {\sqrt {a} x \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2} \sqrt {d x^2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {15, 327, 211} \begin {gather*} \frac {x^2}{b \sqrt {d x^2}}-\frac {\sqrt {a} x \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2} \sqrt {d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/(Sqrt[d*x^2]*(a + b*x^2)),x]

[Out]

x^2/(b*Sqrt[d*x^2]) - (Sqrt[a]*x*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(b^(3/2)*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 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 44, normalized size = 0.85 \begin {gather*} \frac {x \left (\sqrt {b} x-\sqrt {a} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right )}{b^{3/2} \sqrt {d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/(Sqrt[d*x^2]*(a + b*x^2)),x]

[Out]

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

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Maple [A]
time = 0.09, size = 38, normalized size = 0.73

method	result	size
default	\(\frac {x \left (\sqrt {a b}\, x -a \arctan \left (\frac {b x}{\sqrt {a b}}\right )\right )}{\sqrt {d \,x^{2}}\, b \sqrt {a b}}\)	\(38\)
risch	\(\frac {x^{2}}{b \sqrt {d \,x^{2}}}+\frac {x \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x -a \right )}{2 \sqrt {d \,x^{2}}\, b^{2}}-\frac {x \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x -a \right )}{2 \sqrt {d \,x^{2}}\, b^{2}}\)	\(81\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b*x^2+a)/(d*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.52, size = 49, normalized size = 0.94 \begin {gather*} -\frac {\frac {a d^{2} \arctan \left (\frac {\sqrt {d x^{2}} b}{\sqrt {a b d}}\right )}{\sqrt {a b d} b} - \frac {\sqrt {d x^{2}} d}{b}}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)/(d*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.83, size = 126, normalized size = 2.42 \begin {gather*} \left [\frac {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 \, \sqrt {d x^{2}}}{2 \, b d}, -\frac {d \sqrt {\frac {a}{b d}} \arctan \left (\frac {\sqrt {d x^{2}} b \sqrt {\frac {a}{b d}}}{a}\right ) - \sqrt {d x^{2}}}{b d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)/(d*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/2*(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*sqrt(d*x^2))/(b*d), -
(d*sqrt(a/(b*d))*arctan(sqrt(d*x^2)*b*sqrt(a/(b*d))/a) - sqrt(d*x^2))/(b*d)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\sqrt {d x^{2}} \left (a + b x^{2}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(b*x**2+a)/(d*x**2)**(1/2),x)

[Out]

Integral(x**3/(sqrt(d*x**2)*(a + b*x**2)), x)

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Giac [A]
time = 0.78, size = 40, normalized size = 0.77 \begin {gather*} -\frac {a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b \sqrt {d} \mathrm {sgn}\left (x\right )} + \frac {x}{b \sqrt {d} \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)/(d*x^2)^(1/2),x, algorithm="giac")

[Out]

-a*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b*sqrt(d)*sgn(x)) + x/(b*sqrt(d)*sgn(x))

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Mupad [B]
time = 0.32, size = 37, normalized size = 0.71 \begin {gather*} \frac {\sqrt {x^2}}{b\,\sqrt {d}}-\frac {\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x^2}}{\sqrt {a}}\right )}{b^{3/2}\,\sqrt {d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/((a + b*x^2)*(d*x^2)^(1/2)),x)

[Out]

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

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