3.7.43 \(\int \frac {(a+b x^2)^2}{x^3 \sqrt {c+d x^2}} \, dx\) [643]

Optimal. Leaf size=80 \[ \frac {b^2 \sqrt {c+d x^2}}{d}-\frac {a^2 \sqrt {c+d x^2}}{2 c x^2}-\frac {a (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 c^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {457, 91, 81, 65, 214} \begin {gather*} -\frac {a^2 \sqrt {c+d x^2}}{2 c x^2}-\frac {a (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 c^{3/2}}+\frac {b^2 \sqrt {c+d x^2}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*Sqrt[c + d*x^2])/d - (a^2*Sqrt[c + d*x^2])/(2*c*x^2) - (a*(4*b*c - a*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])
/(2*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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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*} \int \frac {\left (a+b x^2\right )^2}{x^3 \sqrt {c+d x^2}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^2}{x^2 \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=-\frac {a^2 \sqrt {c+d x^2}}{2 c x^2}+\frac {\text {Subst}\left (\int \frac {\frac {1}{2} a (4 b c-a d)+b^2 c x}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{2 c}\\ &=\frac {b^2 \sqrt {c+d x^2}}{d}-\frac {a^2 \sqrt {c+d x^2}}{2 c x^2}+\frac {(a (4 b c-a d)) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{4 c}\\ &=\frac {b^2 \sqrt {c+d x^2}}{d}-\frac {a^2 \sqrt {c+d x^2}}{2 c x^2}+\frac {(a (4 b c-a d)) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 c d}\\ &=\frac {b^2 \sqrt {c+d x^2}}{d}-\frac {a^2 \sqrt {c+d x^2}}{2 c x^2}-\frac {a (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 c^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 77, normalized size = 0.96 \begin {gather*} \frac {\left (-a^2 d+2 b^2 c x^2\right ) \sqrt {c+d x^2}}{2 c d x^2}+\frac {a (-4 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.11, size = 99, normalized size = 1.24

method	result	size
default	\(\frac {b^{2} \sqrt {d \,x^{2}+c}}{d}+a^{2} \left (-\frac {\sqrt {d \,x^{2}+c}}{2 c \,x^{2}}+\frac {d \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{2 c^{\frac {3}{2}}}\right )-\frac {2 a b \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{\sqrt {c}}\)	\(99\)
risch	\(-\frac {a^{2} \sqrt {d \,x^{2}+c}}{2 c \,x^{2}}+\frac {b^{2} \sqrt {d \,x^{2}+c}}{d}+\frac {a^{2} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) d}{2 c^{\frac {3}{2}}}-\frac {2 a b \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{\sqrt {c}}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 77, normalized size = 0.96 \begin {gather*} -\frac {2 \, a b \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{\sqrt {c}} + \frac {a^{2} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{2 \, c^{\frac {3}{2}}} + \frac {\sqrt {d x^{2} + c} b^{2}}{d} - \frac {\sqrt {d x^{2} + c} a^{2}}{2 \, c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.63, size = 175, normalized size = 2.19 \begin {gather*} \left [-\frac {{\left (4 \, a b c d - a^{2} d^{2}\right )} \sqrt {c} x^{2} \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^{2} x^{2} - a^{2} c d\right )} \sqrt {d x^{2} + c}}{4 \, c^{2} d x^{2}}, \frac {{\left (4 \, a b c d - a^{2} d^{2}\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (2 \, b^{2} c^{2} x^{2} - a^{2} c d\right )} \sqrt {d x^{2} + c}}{2 \, c^{2} d x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 39.45, size = 99, normalized size = 1.24 \begin {gather*} - \frac {a^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{2 c x} + \frac {a^{2} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{2 c^{\frac {3}{2}}} - \frac {2 a b \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{\sqrt {c}} + b^{2} \left (\begin {cases} \frac {x^{2}}{2 \sqrt {c}} & \text {for}\: d = 0 \\\frac {\sqrt {c + d x^{2}}}{d} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(2*c*x) + a**2*d*asinh(sqrt(c)/(sqrt(d)*x))/(2*c**(3/2)) - 2*a*b*asinh(sqrt
(c)/(sqrt(d)*x))/sqrt(c) + b**2*Piecewise((x**2/(2*sqrt(c)), Eq(d, 0)), (sqrt(c + d*x**2)/d, True))

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Giac [A]
time = 0.92, size = 81, normalized size = 1.01 \begin {gather*} \frac {2 \, \sqrt {d x^{2} + c} b^{2} - \frac {\sqrt {d x^{2} + c} a^{2} d}{c x^{2}} + \frac {{\left (4 \, a b c d - a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*sqrt(d*x^2 + c)*b^2 - sqrt(d*x^2 + c)*a^2*d/(c*x^2) + (4*a*b*c*d - a^2*d^2)*arctan(sqrt(d*x^2 + c)/sqrt
(-c))/(sqrt(-c)*c))/d

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Mupad [B]
time = 0.45, size = 65, normalized size = 0.81 \begin {gather*} \frac {b^2\,\sqrt {d\,x^2+c}}{d}+\frac {a\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )\,\left (a\,d-4\,b\,c\right )}{2\,c^{3/2}}-\frac {a^2\,\sqrt {d\,x^2+c}}{2\,c\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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