\(\int \frac {(a+b x^2)^2}{x (c+d x^2)^{3/2}} \, dx\) [653]

Optimal result

Integrand size = 24, antiderivative size = 75 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{3/2}} \, dx=\frac {(b c-a d)^2}{c d^2 \sqrt {c+d x^2}}+\frac {b^2 \sqrt {c+d x^2}}{d^2}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {457, 89, 65, 214} \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{3/2}} \, dx=-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{3/2}}+\frac {(b c-a d)^2}{c d^2 \sqrt {c+d x^2}}+\frac {b^2 \sqrt {c+d x^2}}{d^2} \]

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 89

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], (c + d*x)^n*((e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

Rule 214

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

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{3/2}} \, dx=\frac {-2 a b c d+a^2 d^2+b^2 c \left (2 c+d x^2\right )}{c d^2 \sqrt {c+d x^2}}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.88 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.15

method	result	size
pseudoelliptic	\(\frac {b^{2} c^{\frac {3}{2}} d \,x^{2}-\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right ) a^{2} d^{2} \sqrt {d \,x^{2}+c}+a^{2} d^{2} \sqrt {c}-2 a b \,c^{\frac {3}{2}} d +2 b^{2} c^{\frac {5}{2}}}{c^{\frac {3}{2}} d^{2} \sqrt {d \,x^{2}+c}}\)	\(86\)
default	\(b^{2} \left (\frac {x^{2}}{d \sqrt {d \,x^{2}+c}}+\frac {2 c}{d^{2} \sqrt {d \,x^{2}+c}}\right )+a^{2} \left (\frac {1}{c \sqrt {d \,x^{2}+c}}-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{c^{\frac {3}{2}}}\right )-\frac {2 a b}{d \sqrt {d \,x^{2}+c}}\)	\(100\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 232, normalized size of antiderivative = 3.09 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{3/2}} \, dx=\left [\frac {{\left (a^{2} d^{3} x^{2} + a^{2} c d^{2}\right )} \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (b^{2} c^{2} d x^{2} + 2 \, b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {d x^{2} + c}}{2 \, {\left (c^{2} d^{3} x^{2} + c^{3} d^{2}\right )}}, \frac {{\left (a^{2} d^{3} x^{2} + a^{2} c d^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (b^{2} c^{2} d x^{2} + 2 \, b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {d x^{2} + c}}{c^{2} d^{3} x^{2} + c^{3} d^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 6.75 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.36 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {a^{2} d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{2 c \sqrt {- c}} + \frac {b^{2} \sqrt {c + d x^{2}}}{2 d} + \frac {\left (a d - b c\right )^{2}}{2 c d \sqrt {c + d x^{2}}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {a^{2} \log {\left (x^{2} \right )} + 2 a b x^{2} + \frac {b^{2} x^{4}}{2}}{2 c^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(a**2*d*atan(sqrt(c + d*x**2)/sqrt(-c))/(2*c*sqrt(-c)) + b**2*sqrt(c + d*x**2)/(2*d) + (a*d - b*c
)**2/(2*c*d*sqrt(c + d*x**2)))/d, Ne(d, 0)), ((a**2*log(x**2) + 2*a*b*x**2 + b**2*x**4/2)/(2*c**(3/2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{3/2}} \, dx=\frac {b^{2} x^{2}}{\sqrt {d x^{2} + c} d} - \frac {a^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{c^{\frac {3}{2}}} + \frac {a^{2}}{\sqrt {d x^{2} + c} c} + \frac {2 \, b^{2} c}{\sqrt {d x^{2} + c} d^{2}} - \frac {2 \, a b}{\sqrt {d x^{2} + c} d} \]

[In]

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

[Out]

b^2*x^2/(sqrt(d*x^2 + c)*d) - a^2*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(3/2) + a^2/(sqrt(d*x^2 + c)*c) + 2*b^2*c/(s
qrt(d*x^2 + c)*d^2) - 2*a*b/(sqrt(d*x^2 + c)*d)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{3/2}} \, dx=\frac {a^{2} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c} + \frac {\sqrt {d x^{2} + c} b^{2}}{d^{2}} + \frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{\sqrt {d x^{2} + c} c d^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.60 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{3/2}} \, dx=\frac {b^2\,\sqrt {d\,x^2+c}}{d^2}-\frac {a^2\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )}{c^{3/2}}+\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{c\,d^2\,\sqrt {d\,x^2+c}} \]

[In]

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

[Out]

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