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

Optimal. Leaf size=75 \[ -\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{3/2}}+\frac {(b c-a d)^2}{c d^2 \sqrt {c+d x^2}}+\frac {b^2 \sqrt {c+d x^2}}{d^2} \]

[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

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Rubi [A]  time = 0.08, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {446, 87, 63, 208} \[ -\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{3/2}}+\frac {(b c-a d)^2}{c d^2 \sqrt {c+d x^2}}+\frac {b^2 \sqrt {c+d x^2}}{d^2} \]

Antiderivative was successfully verified.

[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 63

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 87

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 208

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

Rule 446

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*} \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^2}{x (c+d x)^{3/2}} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {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 \operatorname {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 \operatorname {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*}

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Mathematica [C]  time = 0.04, size = 62, normalized size = 0.83 \[ \frac {a^2 d^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {d x^2}{c}+1\right )+b c \left (-2 a d+2 b c+b d x^2\right )}{c d^2 \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*(2*b*c - 2*a*d + b*d*x^2) + a^2*d^2*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (d*x^2)/c])/(c*d^2*Sqrt[c + d*x^2
])

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fricas [A]  time = 0.63, size = 232, normalized size = 3.09 \[ \left [\frac {{\left (a^{2} d^{3} x^{2} + a^{2} c d^{2}\right )} \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (b^{2} c^{2} d x^{2} + 2 \, b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {d x^{2} + c}}{2 \, {\left (c^{2} d^{3} x^{2} + c^{3} d^{2}\right )}}, \frac {{\left (a^{2} d^{3} x^{2} + a^{2} c d^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (b^{2} c^{2} d x^{2} + 2 \, b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {d x^{2} + c}}{c^{2} d^{3} x^{2} + c^{3} d^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 0.33, size = 82, normalized size = 1.09 \[ \frac {a^{2} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c} + \frac {\sqrt {d x^{2} + c} b^{2}}{d^{2}} + \frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{\sqrt {d x^{2} + c} c d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.01, size = 102, normalized size = 1.36 \[ \frac {b^{2} x^{2}}{\sqrt {d \,x^{2}+c}\, d}-\frac {a^{2} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{c^{\frac {3}{2}}}+\frac {a^{2}}{\sqrt {d \,x^{2}+c}\, c}-\frac {2 a b}{\sqrt {d \,x^{2}+c}\, d}+\frac {2 b^{2} c}{\sqrt {d \,x^{2}+c}\, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.91, size = 90, normalized size = 1.20 \[ \frac {b^{2} x^{2}}{\sqrt {d x^{2} + c} d} - \frac {a^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{c^{\frac {3}{2}}} + \frac {a^{2}}{\sqrt {d x^{2} + c} c} + \frac {2 \, b^{2} c}{\sqrt {d x^{2} + c} d^{2}} - \frac {2 \, a b}{\sqrt {d x^{2} + c} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.86, size = 76, normalized size = 1.01 \[ \frac {b^2\,\sqrt {d\,x^2+c}}{d^2}-\frac {a^2\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )}{c^{3/2}}+\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{c\,d^2\,\sqrt {d\,x^2+c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 36.05, size = 70, normalized size = 0.93 \[ \frac {a^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{c \sqrt {- c}} + \frac {b^{2} \sqrt {c + d x^{2}}}{d^{2}} + \frac {\left (a d - b c\right )^{2}}{c d^{2} \sqrt {c + d x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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