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

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

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Rubi [A]  time = 0.06, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {444, 47, 63, 208} \begin {gather*} -\frac {d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{3/2} \sqrt {b c-a d}}-\frac {\sqrt {c+d x^2}}{2 b \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 80, normalized size = 1.00 \begin {gather*} \frac {d \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {a d-b c}}\right )}{2 b^{3/2} \sqrt {a d-b c}}-\frac {\sqrt {c+d x^2}}{2 b \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*Sqrt[c + d*x^2]/(b*(a + b*x^2)) + (d*ArcTan[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[-(b*c) + a*d]])/(2*b^(3/2)*Sqr
t[-(b*c) + a*d])

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IntegrateAlgebraic [A]  time = 0.17, size = 90, normalized size = 1.12 \begin {gather*} -\frac {d \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2} \sqrt {a d-b c}}{b c-a d}\right )}{2 b^{3/2} \sqrt {a d-b c}}-\frac {\sqrt {c+d x^2}}{2 b \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.11, size = 356, normalized size = 4.45 \begin {gather*} \left [\frac {{\left (b d x^{2} + a d\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {b^{2} c - a b d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left (b^{2} c - a b d\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a b^{3} c - a^{2} b^{2} d + {\left (b^{4} c - a b^{3} d\right )} x^{2}\right )}}, -\frac {{\left (b d x^{2} + a d\right )} \sqrt {-b^{2} c + a b d} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {-b^{2} c + a b d} \sqrt {d x^{2} + c}}{2 \, {\left (b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (b^{2} c - a b d\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a b^{3} c - a^{2} b^{2} d + {\left (b^{4} c - a b^{3} d\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*((b*d*x^2 + a*d)*sqrt(b^2*c - a*b*d)*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d -
3*a*b*d^2)*x^2 - 4*(b*d*x^2 + 2*b*c - a*d)*sqrt(b^2*c - a*b*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) -
4*(b^2*c - a*b*d)*sqrt(d*x^2 + c))/(a*b^3*c - a^2*b^2*d + (b^4*c - a*b^3*d)*x^2), -1/4*((b*d*x^2 + a*d)*sqrt(
-b^2*c + a*b*d)*arctan(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(-b^2*c + a*b*d)*sqrt(d*x^2 + c)/(b^2*c^2 - a*b*c*d +
(b^2*c*d - a*b*d^2)*x^2)) + 2*(b^2*c - a*b*d)*sqrt(d*x^2 + c))/(a*b^3*c - a^2*b^2*d + (b^4*c - a*b^3*d)*x^2)]

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giac [A]  time = 0.40, size = 79, normalized size = 0.99 \begin {gather*} \frac {d \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, \sqrt {-b^{2} c + a b d} b} - \frac {\sqrt {d x^{2} + c} d}{2 \, {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 1617, normalized size = 20.21

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(x*(d*x^2+c)^(1/2)/(b*x^2+a)^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 positive or negative?

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mupad [B]  time = 0.74, size = 70, normalized size = 0.88 \begin {gather*} \frac {d\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}}{\sqrt {a\,d-b\,c}}\right )}{2\,b^{3/2}\,\sqrt {a\,d-b\,c}}-\frac {d\,\sqrt {d\,x^2+c}}{2\,\left (d\,b^2\,x^2+a\,d\,b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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