\(\int \frac {1}{x^3 \sqrt {d x^2} (a+b x^2)} \, dx\) [675]

Optimal result

Integrand size = 22, antiderivative size = 68 \[ \int \frac {1}{x^3 \sqrt {d x^2} \left (a+b x^2\right )} \, dx=\frac {b}{a^2 \sqrt {d x^2}}-\frac {1}{3 a x^2 \sqrt {d x^2}}+\frac {b^{3/2} x \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} \sqrt {d x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 68, 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, 331, 211} \[ \int \frac {1}{x^3 \sqrt {d x^2} \left (a+b x^2\right )} \, dx=\frac {b^{3/2} x \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} \sqrt {d x^2}}+\frac {b}{a^2 \sqrt {d x^2}}-\frac {1}{3 a x^2 \sqrt {d x^2}} \]

[In]

Int[1/(x^3*Sqrt[d*x^2]*(a + b*x^2)),x]

[Out]

b/(a^2*Sqrt[d*x^2]) - 1/(3*a*x^2*Sqrt[d*x^2]) + (b^(3/2)*x*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(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 331

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

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^3 \sqrt {d x^2} \left (a+b x^2\right )} \, dx=\frac {d \left (-a+3 b x^2\right )}{3 a^2 \left (d x^2\right )^{3/2}}+\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {d x^2}}{\sqrt {a} \sqrt {d}}\right )}{a^{5/2} \sqrt {d}} \]

[In]

Integrate[1/(x^3*Sqrt[d*x^2]*(a + b*x^2)),x]

[Out]

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

Maple [A] (verified)

Time = 2.97 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.84

method	result	size
default	\(-\frac {-3 b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right ) x^{3}-3 \sqrt {a b}\, b \,x^{2}+\sqrt {a b}\, a}{3 x^{2} \sqrt {d \,x^{2}}\, a^{2} \sqrt {a b}}\)	\(57\)
pseudoelliptic	\(\frac {\arctan \left (\frac {b \sqrt {d \,x^{2}}}{\sqrt {a b d}}\right ) b^{2} x^{2} \sqrt {d \,x^{2}}-\frac {\left (-3 b \,x^{2}+a \right ) \sqrt {a b d}}{3}}{\sqrt {d \,x^{2}}\, \sqrt {a b d}\, a^{2} x^{2}}\)	\(68\)
risch	\(\frac {\frac {b \,x^{2}}{a^{2}}-\frac {1}{3 a}}{\sqrt {d \,x^{2}}\, x^{2}}+\frac {x \sqrt {-a b}\, b \ln \left (-b x -\sqrt {-a b}\right )}{2 \sqrt {d \,x^{2}}\, a^{3}}-\frac {x \sqrt {-a b}\, b \ln \left (-b x +\sqrt {-a b}\right )}{2 \sqrt {d \,x^{2}}\, a^{3}}\)	\(93\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.31 \[ \int \frac {1}{x^3 \sqrt {d x^2} \left (a+b x^2\right )} \, dx=\left [\frac {3 \, b d x^{4} \sqrt {-\frac {b}{a d}} \log \left (\frac {b x^{2} + 2 \, \sqrt {d x^{2}} a \sqrt {-\frac {b}{a d}} - a}{b x^{2} + a}\right ) + 2 \, {\left (3 \, b x^{2} - a\right )} \sqrt {d x^{2}}}{6 \, a^{2} d x^{4}}, \frac {3 \, b d x^{4} \sqrt {\frac {b}{a d}} \arctan \left (\sqrt {d x^{2}} \sqrt {\frac {b}{a d}}\right ) + {\left (3 \, b x^{2} - a\right )} \sqrt {d x^{2}}}{3 \, a^{2} d x^{4}}\right ] \]

[In]

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

[Out]

[1/6*(3*b*d*x^4*sqrt(-b/(a*d))*log((b*x^2 + 2*sqrt(d*x^2)*a*sqrt(-b/(a*d)) - a)/(b*x^2 + a)) + 2*(3*b*x^2 - a)
*sqrt(d*x^2))/(a^2*d*x^4), 1/3*(3*b*d*x^4*sqrt(b/(a*d))*arctan(sqrt(d*x^2)*sqrt(b/(a*d))) + (3*b*x^2 - a)*sqrt
(d*x^2))/(a^2*d*x^4)]

Sympy [A] (verification not implemented)

Time = 1.14 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^3 \sqrt {d x^2} \left (a+b x^2\right )} \, dx=\begin {cases} \frac {2 \left (- \frac {d^{2}}{6 a \left (d x^{2}\right )^{\frac {3}{2}}} + \frac {b d \operatorname {atan}{\left (\frac {\sqrt {d x^{2}}}{\sqrt {\frac {a d}{b}}} \right )}}{2 a^{2} \sqrt {\frac {a d}{b}}} + \frac {b d}{2 a^{2} \sqrt {d x^{2}}}\right )}{d} & \text {for}\: d \neq 0 \\\tilde {\infty } x^{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^3 \sqrt {d x^2} \left (a+b x^2\right )} \, dx=\frac {b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2} \sqrt {d}} + \frac {3 \, b \sqrt {d} x^{2} - a \sqrt {d}}{3 \, a^{2} d x^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^3 \sqrt {d x^2} \left (a+b x^2\right )} \, dx=\frac {b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2} \sqrt {d} \mathrm {sgn}\left (x\right )} + \frac {3 \, b x^{2} - a}{3 \, a^{2} \sqrt {d} x^{3} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.34 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^3 \sqrt {d x^2} \left (a+b x^2\right )} \, dx=\frac {b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x^2}}{\sqrt {a}}\right )}{a^{5/2}\,\sqrt {d}}-\frac {1}{3\,a\,\sqrt {d}\,{\left (x^2\right )}^{3/2}}+\frac {b\,x^2}{a^2\,\sqrt {d}\,{\left (x^2\right )}^{3/2}} \]

[In]

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

[Out]

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