\(\int \frac {(a+b x^2)^2}{x (c+d x^2)^{5/2}} \, dx\) [665]

Optimal result

Integrand size = 24, antiderivative size = 88 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{5/2}} \, dx=\frac {(b c-a d)^2}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {\frac {a^2}{c^2}-\frac {b^2}{d^2}}{\sqrt {c+d x^2}}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 88, 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} \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{5/2}} \, dx=-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{5/2}}+\frac {\frac {a^2}{c^2}-\frac {b^2}{d^2}}{\sqrt {c+d x^2}}+\frac {(b c-a d)^2}{3 c d^2 \left (c+d x^2\right )^{3/2}} \]

[In]

Int[(a + b*x^2)^2/(x*(c + d*x^2)^(5/2)),x]

[Out]

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

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)^{5/2}} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {(b c-a d)^2}{c d (c+d x)^{5/2}}+\frac {b^2 c^2-a^2 d^2}{c^2 d (c+d x)^{3/2}}+\frac {a^2}{c^2 x \sqrt {c+d x}}\right ) \, dx,x,x^2\right ) \\ & = \frac {(b c-a d)^2}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {\frac {a^2}{c^2}-\frac {b^2}{d^2}}{\sqrt {c+d x^2}}+\frac {a^2 \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{2 c^2} \\ & = \frac {(b c-a d)^2}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {\frac {a^2}{c^2}-\frac {b^2}{d^2}}{\sqrt {c+d x^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^2 d} \\ & = \frac {(b c-a d)^2}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {\frac {a^2}{c^2}-\frac {b^2}{d^2}}{\sqrt {c+d x^2}}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{5/2}} \, dx=-\frac {(b c-a d) \left (2 b c^2+4 a c d+3 b c d x^2+3 a d^2 x^2\right )}{3 c^2 d^2 \left (c+d x^2\right )^{3/2}}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{5/2}} \]

[In]

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

[Out]

-1/3*((b*c - a*d)*(2*b*c^2 + 4*a*c*d + 3*b*c*d*x^2 + 3*a*d^2*x^2))/(c^2*d^2*(c + d*x^2)^(3/2)) - (a^2*ArcTanh[
Sqrt[c + d*x^2]/Sqrt[c]])/c^(5/2)

Maple [A] (verified)

Time = 2.91 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.10

method	result	size
pseudoelliptic	\(-\frac {a^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}} d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )+\frac {2 b \left (\frac {3 b \,x^{2}}{2}+a \right ) d \,c^{\frac {5}{2}}}{3}-\sqrt {c}\, a^{2} d^{3} x^{2}-\frac {4 c^{\frac {3}{2}} a^{2} d^{2}}{3}+\frac {2 b^{2} c^{\frac {7}{2}}}{3}}{\left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{\frac {5}{2}} d^{2}}\)	\(97\)
default	\(b^{2} \left (-\frac {x^{2}}{d \left (d \,x^{2}+c \right )^{\frac {3}{2}}}-\frac {2 c}{3 d^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}\right )+a^{2} \left (\frac {1}{3 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {\frac {1}{c \sqrt {d \,x^{2}+c}}-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{c^{\frac {3}{2}}}}{c}\right )-\frac {2 a b}{3 d \left (d \,x^{2}+c \right )^{\frac {3}{2}}}\)	\(120\)

[In]

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

[Out]

-1/(d*x^2+c)^(3/2)/c^(5/2)*(a^2*(d*x^2+c)^(3/2)*d^2*arctanh((d*x^2+c)^(1/2)/c^(1/2))+2/3*b*(3/2*b*x^2+a)*d*c^(
5/2)-c^(1/2)*a^2*d^3*x^2-4/3*c^(3/2)*a^2*d^2+2/3*b^2*c^(7/2))/d^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 316, normalized size of antiderivative = 3.59 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (a^{2} d^{4} x^{4} + 2 \, a^{2} c d^{3} x^{2} + a^{2} c^{2} d^{2}\right )} \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (2 \, b^{2} c^{4} + 2 \, a b c^{3} d - 4 \, a^{2} c^{2} d^{2} + 3 \, {\left (b^{2} c^{3} d - a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, {\left (c^{3} d^{4} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{5} d^{2}\right )}}, \frac {3 \, {\left (a^{2} d^{4} x^{4} + 2 \, a^{2} c d^{3} x^{2} + a^{2} c^{2} d^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - {\left (2 \, b^{2} c^{4} + 2 \, a b c^{3} d - 4 \, a^{2} c^{2} d^{2} + 3 \, {\left (b^{2} c^{3} d - a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{3 \, {\left (c^{3} d^{4} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{5} d^{2}\right )}}\right ] \]

[In]

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

[Out]

[1/6*(3*(a^2*d^4*x^4 + 2*a^2*c*d^3*x^2 + a^2*c^2*d^2)*sqrt(c)*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x
^2) - 2*(2*b^2*c^4 + 2*a*b*c^3*d - 4*a^2*c^2*d^2 + 3*(b^2*c^3*d - a^2*c*d^3)*x^2)*sqrt(d*x^2 + c))/(c^3*d^4*x^
4 + 2*c^4*d^3*x^2 + c^5*d^2), 1/3*(3*(a^2*d^4*x^4 + 2*a^2*c*d^3*x^2 + a^2*c^2*d^2)*sqrt(-c)*arctan(sqrt(-c)/sq
rt(d*x^2 + c)) - (2*b^2*c^4 + 2*a*b*c^3*d - 4*a^2*c^2*d^2 + 3*(b^2*c^3*d - a^2*c*d^3)*x^2)*sqrt(d*x^2 + c))/(c
^3*d^4*x^4 + 2*c^4*d^3*x^2 + c^5*d^2)]

Sympy [A] (verification not implemented)

Time = 9.13 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{5/2}} \, dx=\begin {cases} \frac {2 \left (\frac {a^{2} d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{2 c^{2} \sqrt {- c}} + \frac {\left (a d - b c\right )^{2}}{6 c d \left (c + d x^{2}\right )^{\frac {3}{2}}} + \frac {\left (a d - b c\right ) \left (a d + b c\right )}{2 c^{2} d \sqrt {c + d x^{2}}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {a^{2} \log {\left (x^{2} \right )} + 2 a b x^{2} + \frac {b^{2} x^{4}}{2}}{2 c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(a**2*d*atan(sqrt(c + d*x**2)/sqrt(-c))/(2*c**2*sqrt(-c)) + (a*d - b*c)**2/(6*c*d*(c + d*x**2)**(
3/2)) + (a*d - b*c)*(a*d + b*c)/(2*c**2*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**(5/2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.23 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{5/2}} \, dx=-\frac {b^{2} x^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d} - \frac {a^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{c^{\frac {5}{2}}} + \frac {a^{2}}{\sqrt {d x^{2} + c} c^{2}} + \frac {a^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c} - \frac {2 \, b^{2} c}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{2}} - \frac {2 \, a b}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.16 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{5/2}} \, dx=\frac {a^{2} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{2}} - \frac {3 \, {\left (d x^{2} + c\right )} b^{2} c^{2} - b^{2} c^{3} + 2 \, a b c^{2} d - 3 \, {\left (d x^{2} + c\right )} a^{2} d^{2} - a^{2} c d^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2} d^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.87 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{5/2}} \, dx=\frac {\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{3\,c}+\frac {\left (a^2\,d^2-b^2\,c^2\right )\,\left (d\,x^2+c\right )}{c^2}}{d^2\,{\left (d\,x^2+c\right )}^{3/2}}-\frac {a^2\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )}{c^{5/2}} \]

[In]

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

[Out]

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