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

Optimal. Leaf size=91 \[ -\frac {a^2}{c x \sqrt {c+d x^2}}-\frac {x \left (b^2 c^2-2 a d (b c-a d)\right )}{c^2 d \sqrt {c+d x^2}}+\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{d^{3/2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 87, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {462, 385, 217, 206} \begin {gather*} -\frac {a^2}{c x \sqrt {c+d x^2}}-\frac {x \left (\frac {b^2}{d}-\frac {2 a (b c-a d)}{c^2}\right )}{\sqrt {c+d x^2}}+\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 385

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

Rule 462

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

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 81, normalized size = 0.89 \begin {gather*} \frac {b^2 \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{d^{3/2}}-\frac {\sqrt {c+d x^2} \left (a^2+\frac {x^2 (b c-a d)^2}{d \left (c+d x^2\right )}\right )}{c^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.15, size = 92, normalized size = 1.01 \begin {gather*} \frac {-a^2 c d-2 a^2 d^2 x^2+2 a b c d x^2-b^2 c^2 x^2}{c^2 d x \sqrt {c+d x^2}}-\frac {b^2 \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.55, size = 239, normalized size = 2.63 \begin {gather*} \left [\frac {{\left (b^{2} c^{2} d x^{3} + b^{2} c^{3} x\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 2 \, {\left (a^{2} c d^{2} + {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + 2 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{2 \, {\left (c^{2} d^{3} x^{3} + c^{3} d^{2} x\right )}}, -\frac {{\left (b^{2} c^{2} d x^{3} + b^{2} c^{3} x\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (a^{2} c d^{2} + {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + 2 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{c^{2} d^{3} x^{3} + c^{3} d^{2} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.47, size = 104, normalized size = 1.14 \begin {gather*} -\frac {b^{2} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{2 \, d^{\frac {3}{2}}} + \frac {2 \, a^{2} \sqrt {d}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )} c} - \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x}{\sqrt {d x^{2} + c} c^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 99, normalized size = 1.09 \begin {gather*} -\frac {2 a^{2} d x}{\sqrt {d \,x^{2}+c}\, c^{2}}+\frac {2 a b x}{\sqrt {d \,x^{2}+c}\, c}-\frac {b^{2} x}{\sqrt {d \,x^{2}+c}\, d}+\frac {b^{2} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}-\frac {a^{2}}{\sqrt {d \,x^{2}+c}\, c x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.90, size = 91, normalized size = 1.00 \begin {gather*} \frac {2 \, a b x}{\sqrt {d x^{2} + c} c} - \frac {b^{2} x}{\sqrt {d x^{2} + c} d} - \frac {2 \, a^{2} d x}{\sqrt {d x^{2} + c} c^{2}} + \frac {b^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{d^{\frac {3}{2}}} - \frac {a^{2}}{\sqrt {d x^{2} + c} c x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^2}{x^2\,{\left (d\,x^2+c\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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