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

Optimal. Leaf size=103 \[ \frac {-\frac {3 a^2 d}{c}+4 a b-\frac {2 b^2 c}{d}}{2 c \sqrt {c+d x^2}}-\frac {a^2}{2 c x^2 \sqrt {c+d x^2}}-\frac {a (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 c^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 103, 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 = {446, 89, 78, 63, 208} \[ \frac {-\frac {3 a^2 d}{c}+4 a b-\frac {2 b^2 c}{d}}{2 c \sqrt {c+d x^2}}-\frac {a^2}{2 c x^2 \sqrt {c+d x^2}}-\frac {a (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 78

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

Rule 89

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 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^3 \left (c+d x^2\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^2}{x^2 (c+d x)^{3/2}} \, dx,x,x^2\right )\\ &=-\frac {a^2}{2 c x^2 \sqrt {c+d x^2}}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} a (4 b c-3 a d)+b^2 c x}{x (c+d x)^{3/2}} \, dx,x,x^2\right )}{2 c}\\ &=\frac {4 a b-\frac {2 b^2 c}{d}-\frac {3 a^2 d}{c}}{2 c \sqrt {c+d x^2}}-\frac {a^2}{2 c x^2 \sqrt {c+d x^2}}+\frac {(a (4 b c-3 a d)) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{4 c^2}\\ &=\frac {4 a b-\frac {2 b^2 c}{d}-\frac {3 a^2 d}{c}}{2 c \sqrt {c+d x^2}}-\frac {a^2}{2 c x^2 \sqrt {c+d x^2}}+\frac {(a (4 b c-3 a d)) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 c^2 d}\\ &=\frac {4 a b-\frac {2 b^2 c}{d}-\frac {3 a^2 d}{c}}{2 c \sqrt {c+d x^2}}-\frac {a^2}{2 c x^2 \sqrt {c+d x^2}}-\frac {a (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 96, normalized size = 0.93 \[ \frac {a (3 a d-4 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 c^{5/2}}-\frac {a^2 d \left (c+3 d x^2\right )-4 a b c d x^2+2 b^2 c^2 x^2}{2 c^2 d x^2 \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.69, size = 292, normalized size = 2.83 \[ \left [-\frac {{\left ({\left (4 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{4} + {\left (4 \, a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {c} \log \left (-\frac {d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (a^{2} c^{2} d + {\left (2 \, b^{2} c^{3} - 4 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{4 \, {\left (c^{3} d^{2} x^{4} + c^{4} d x^{2}\right )}}, \frac {{\left ({\left (4 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{4} + {\left (4 \, a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - {\left (a^{2} c^{2} d + {\left (2 \, b^{2} c^{3} - 4 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{2 \, {\left (c^{3} d^{2} x^{4} + c^{4} d x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.39, size = 140, normalized size = 1.36 \[ \frac {{\left (4 \, a b c - 3 \, a^{2} d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{2 \, \sqrt {-c} c^{2}} - \frac {2 \, {\left (d x^{2} + c\right )} b^{2} c^{2} - 2 \, b^{2} c^{3} - 4 \, {\left (d x^{2} + c\right )} a b c d + 4 \, a b c^{2} d + 3 \, {\left (d x^{2} + c\right )} a^{2} d^{2} - 2 \, a^{2} c d^{2}}{2 \, {\left ({\left (d x^{2} + c\right )}^{\frac {3}{2}} - \sqrt {d x^{2} + c} c\right )} c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 135, normalized size = 1.31 \[ \frac {3 a^{2} d \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{2 c^{\frac {5}{2}}}-\frac {2 a b \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{c^{\frac {3}{2}}}-\frac {3 a^{2} d}{2 \sqrt {d \,x^{2}+c}\, c^{2}}+\frac {2 a b}{\sqrt {d \,x^{2}+c}\, c}-\frac {b^{2}}{\sqrt {d \,x^{2}+c}\, d}-\frac {a^{2}}{2 \sqrt {d \,x^{2}+c}\, c \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.86, size = 112, normalized size = 1.09 \[ -\frac {2 \, a b \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{c^{\frac {3}{2}}} + \frac {3 \, a^{2} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{2 \, c^{\frac {5}{2}}} + \frac {2 \, a b}{\sqrt {d x^{2} + c} c} - \frac {b^{2}}{\sqrt {d x^{2} + c} d} - \frac {3 \, a^{2} d}{2 \, \sqrt {d x^{2} + c} c^{2}} - \frac {a^{2}}{2 \, \sqrt {d x^{2} + c} c x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.08, size = 119, normalized size = 1.16 \[ \frac {\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{c}-\frac {\left (d\,x^2+c\right )\,\left (3\,a^2\,d^2-4\,a\,b\,c\,d+2\,b^2\,c^2\right )}{2\,c^2}}{d\,{\left (d\,x^2+c\right )}^{3/2}-c\,d\,\sqrt {d\,x^2+c}}+\frac {a\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )\,\left (3\,a\,d-4\,b\,c\right )}{2\,c^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{2}\right )^{2}}{x^{3} \left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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