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

Optimal. Leaf size=89 \[ -\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} c^{3/2}}-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 c x^2} \]

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Rubi [A]  time = 0.07, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {446, 94, 93, 208} \begin {gather*} -\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} c^{3/2}}-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 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 {\sqrt {a+b x^2}}{x^3 \sqrt {c+d x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^2 \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 c x^2}+\frac {(b c-a d) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{4 c}\\ &=-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 c x^2}+\frac {(b c-a d) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x^2}}{\sqrt {c+d x^2}}\right )}{2 c}\\ &=-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 c x^2}-\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 89, normalized size = 1.00 \begin {gather*} \frac {1}{2} \left (-\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} c^{3/2}}-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{c x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.91, size = 117, normalized size = 1.31 \begin {gather*} \frac {(a d-b c) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x^2}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{2 \sqrt {a} c^{3/2}}-\frac {\sqrt {c+d x^2} (b c-a d)}{2 c \sqrt {a+b x^2} \left (c-\frac {a \left (c+d x^2\right )}{a+b x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((b*c - a*d)*Sqrt[c + d*x^2])/(c*Sqrt[a + b*x^2]*(c - (a*(c + d*x^2))/(a + b*x^2))) + ((-(b*c) + a*d)*Arc
Tanh[(Sqrt[a]*Sqrt[c + d*x^2])/(Sqrt[c]*Sqrt[a + b*x^2])])/(2*Sqrt[a]*c^(3/2))

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fricas [A]  time = 1.64, size = 280, normalized size = 3.15 \begin {gather*} \left [-\frac {\sqrt {a c} {\left (b c - a d\right )} x^{2} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2} + 4 \, {\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {a c}}{x^{4}}\right ) + 4 \, \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} a c}{8 \, a c^{2} x^{2}}, \frac {\sqrt {-a c} {\left (b c - a d\right )} x^{2} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-a c}}{2 \, {\left (a b c d x^{4} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x^{2}\right )}}\right ) - 2 \, \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} a c}{4 \, a c^{2} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(sqrt(a*c)*(b*c - a*d)*x^2*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2 + 8*(a*b*c^2 + a^2*c*d)*
x^2 + 4*((b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(a*c))/x^4) + 4*sqrt(b*x^2 + a)*sqrt(d*x
^2 + c)*a*c)/(a*c^2*x^2), 1/4*(sqrt(-a*c)*(b*c - a*d)*x^2*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)
*sqrt(d*x^2 + c)*sqrt(-a*c)/(a*b*c*d*x^4 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x^2)) - 2*sqrt(b*x^2 + a)*sqrt(d*x^2
+ c)*a*c)/(a*c^2*x^2)]

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giac [B]  time = 2.10, size = 434, normalized size = 4.88 \begin {gather*} -\frac {b {\left (\frac {{\left (\sqrt {b d} b^{2} c - \sqrt {b d} a b d\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} b c} + \frac {2 \, {\left (\sqrt {b d} b^{4} c^{2} - 2 \, \sqrt {b d} a b^{3} c d + \sqrt {b d} a^{2} b^{2} d^{2} - \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} b^{2} c - \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a b d\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4}\right )} c}\right )}}{2 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.03, size = 207, normalized size = 2.33 \begin {gather*} \frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (a d \,x^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )-b c \,x^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )-2 \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\right )}{4 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, c \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [B]  time = 6.41, size = 477, normalized size = 5.36 \begin {gather*} \frac {\frac {\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )\,\left (\frac {c\,b^2}{8}+\frac {a\,d\,b}{8}\right )}{\sqrt {a}\,c^{3/2}\,d\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}-\frac {b^2}{8\,c\,d}+\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^2\,\left (\frac {a^2\,d^2}{8}-\frac {3\,a\,b\,c\,d}{8}+\frac {b^2\,c^2}{8}\right )}{a\,c^2\,d\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^2}}{\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^3}{{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^3}+\frac {b\,\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}{d\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}-\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^2\,\left (a\,d+b\,c\right )}{\sqrt {a}\,\sqrt {c}\,d\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^2}}-\frac {d\,\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}{8\,c\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}-\frac {\ln \left (\frac {\sqrt {b\,x^2+a}-\sqrt {a}}{\sqrt {d\,x^2+c}-\sqrt {c}}\right )\,\left (\sqrt {a}\,b\,c^{3/2}-a^{3/2}\,\sqrt {c}\,d\right )}{4\,a\,c^2}+\frac {\ln \left (\frac {\left (\sqrt {c}\,\sqrt {b\,x^2+a}-\sqrt {a}\,\sqrt {d\,x^2+c}\right )\,\left (b\,\sqrt {c}-\frac {\sqrt {a}\,d\,\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}{\sqrt {d\,x^2+c}-\sqrt {c}}\right )}{\sqrt {d\,x^2+c}-\sqrt {c}}\right )\,\left (\sqrt {a}\,b\,c^{3/2}-a^{3/2}\,\sqrt {c}\,d\right )}{4\,a\,c^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((((a + b*x^2)^(1/2) - a^(1/2))*((b^2*c)/8 + (a*b*d)/8))/(a^(1/2)*c^(3/2)*d*((c + d*x^2)^(1/2) - c^(1/2))) - b
^2/(8*c*d) + (((a + b*x^2)^(1/2) - a^(1/2))^2*((a^2*d^2)/8 + (b^2*c^2)/8 - (3*a*b*c*d)/8))/(a*c^2*d*((c + d*x^
2)^(1/2) - c^(1/2))^2))/(((a + b*x^2)^(1/2) - a^(1/2))^3/((c + d*x^2)^(1/2) - c^(1/2))^3 + (b*((a + b*x^2)^(1/
2) - a^(1/2)))/(d*((c + d*x^2)^(1/2) - c^(1/2))) - (((a + b*x^2)^(1/2) - a^(1/2))^2*(a*d + b*c))/(a^(1/2)*c^(1
/2)*d*((c + d*x^2)^(1/2) - c^(1/2))^2)) - (d*((a + b*x^2)^(1/2) - a^(1/2)))/(8*c*((c + d*x^2)^(1/2) - c^(1/2))
) - (log(((a + b*x^2)^(1/2) - a^(1/2))/((c + d*x^2)^(1/2) - c^(1/2)))*(a^(1/2)*b*c^(3/2) - a^(3/2)*c^(1/2)*d))
/(4*a*c^2) + (log(((c^(1/2)*(a + b*x^2)^(1/2) - a^(1/2)*(c + d*x^2)^(1/2))*(b*c^(1/2) - (a^(1/2)*d*((a + b*x^2
)^(1/2) - a^(1/2)))/((c + d*x^2)^(1/2) - c^(1/2))))/((c + d*x^2)^(1/2) - c^(1/2)))*(a^(1/2)*b*c^(3/2) - a^(3/2
)*c^(1/2)*d))/(4*a*c^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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