3.732 \(\int \frac{x^3 \sqrt{c+d x^2}}{(a+b x^2)^2} \, dx\)
Optimal. Leaf size=136 \[ \frac{\sqrt{c+d x^2} (2 b c-3 a d)}{2 b^2 (b c-a d)}-\frac{(2 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 b^{5/2} \sqrt{b c-a d}}+\frac{a \left (c+d x^2\right )^{3/2}}{2 b \left (a+b x^2\right ) (b c-a d)} \]
[Out]
((2*b*c - 3*a*d)*Sqrt[c + d*x^2])/(2*b^2*(b*c - a*d)) + (a*(c + d*x^2)^(3/2))/(2*b*(b*c - a*d)*(a + b*x^2)) -
((2*b*c - 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(2*b^(5/2)*Sqrt[b*c - a*d])
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Rubi [A] time = 0.110655, antiderivative size = 136, normalized size of antiderivative = 1.,
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, 78, 50, 63, 208} \[ \frac{\sqrt{c+d x^2} (2 b c-3 a d)}{2 b^2 (b c-a d)}-\frac{(2 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 b^{5/2} \sqrt{b c-a d}}+\frac{a \left (c+d x^2\right )^{3/2}}{2 b \left (a+b x^2\right ) (b c-a d)} \]
Antiderivative was successfully verified.
[In]
Int[(x^3*Sqrt[c + d*x^2])/(a + b*x^2)^2,x]
[Out]
((2*b*c - 3*a*d)*Sqrt[c + d*x^2])/(2*b^2*(b*c - a*d)) + (a*(c + d*x^2)^(3/2))/(2*b*(b*c - a*d)*(a + b*x^2)) -
((2*b*c - 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(2*b^(5/2)*Sqrt[b*c - a*d])
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]]
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 50
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 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]
Rubi steps
\begin{align*} \int \frac{x^3 \sqrt{c+d x^2}}{\left (a+b x^2\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x \sqrt{c+d x}}{(a+b x)^2} \, dx,x,x^2\right )\\ &=\frac{a \left (c+d x^2\right )^{3/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac{(2 b c-3 a d) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{a+b x} \, dx,x,x^2\right )}{4 b (b c-a d)}\\ &=\frac{(2 b c-3 a d) \sqrt{c+d x^2}}{2 b^2 (b c-a d)}+\frac{a \left (c+d x^2\right )^{3/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac{(2 b c-3 a d) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^2\right )}{4 b^2}\\ &=\frac{(2 b c-3 a d) \sqrt{c+d x^2}}{2 b^2 (b c-a d)}+\frac{a \left (c+d x^2\right )^{3/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac{(2 b c-3 a d) \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^2 d}\\ &=\frac{(2 b c-3 a d) \sqrt{c+d x^2}}{2 b^2 (b c-a d)}+\frac{a \left (c+d x^2\right )^{3/2}}{2 b (b c-a d) \left (a+b x^2\right )}-\frac{(2 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 b^{5/2} \sqrt{b c-a d}}\\ \end{align*}
Mathematica [A] time = 0.151748, size = 117, normalized size = 0.86 \[ \frac{\frac{(2 b c-3 a d) \left (\sqrt{b} \sqrt{c+d x^2}-\sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )\right )}{b^{3/2}}+\frac{a \left (c+d x^2\right )^{3/2}}{a+b x^2}}{2 b (b c-a d)} \]
Antiderivative was successfully verified.
[In]
Integrate[(x^3*Sqrt[c + d*x^2])/(a + b*x^2)^2,x]
[Out]
((a*(c + d*x^2)^(3/2))/(a + b*x^2) + ((2*b*c - 3*a*d)*(Sqrt[b]*Sqrt[c + d*x^2] - Sqrt[b*c - a*d]*ArcTanh[(Sqrt
[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]]))/b^(3/2))/(2*b*(b*c - a*d))
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Maple [B] time = 0.011, size = 2543, normalized size = 18.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^3*(d*x^2+c)^(1/2)/(b*x^2+a)^2,x)
[Out]
-1/4/b^2*(-a*b)^(1/2)/(a*d-b*c)/(x+1/b*(-a*b)^(1/2))*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b
)^(1/2))-(a*d-b*c)/b)^(3/2)+1/4/b^2*a*d/(a*d-b*c)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(
1/2))-(a*d-b*c)/b)^(1/2)-1/4/b^3*(-a*b)^(1/2)*d^(3/2)*a/(a*d-b*c)*ln((-d*(-a*b)^(1/2)/b+(x+1/b*(-a*b)^(1/2))*d
)/d^(1/2)+((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))+1/4/b^3*a^2*d^
2/(a*d-b*c)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(
1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^(1/2)
))-1/4/b^2*a*d/(a*d-b*c)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))+2*(-(
a*d-b*c)/b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x+1/b
*(-a*b)^(1/2)))*c+1/4/b^2*(-a*b)^(1/2)*d/(a*d-b*c)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^
(1/2))-(a*d-b*c)/b)^(1/2)*x+1/4/b^2*(-a*b)^(1/2)*d^(1/2)/(a*d-b*c)*ln((-d*(-a*b)^(1/2)/b+(x+1/b*(-a*b)^(1/2))*
d)/d^(1/2)+((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))*c+1/2/b^2*((x
+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)-1/2/b^3*d^(1/2)*(-a*b)^(1/2)
*ln((-d*(-a*b)^(1/2)/b+(x+1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a
*b)^(1/2))-(a*d-b*c)/b)^(1/2))+1/2/b^3/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b
)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)
^(1/2))/(x+1/b*(-a*b)^(1/2)))*a*d-1/2/b^2/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-
a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)
/b)^(1/2))/(x+1/b*(-a*b)^(1/2)))*c+1/2/b^2*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(
a*d-b*c)/b)^(1/2)+1/2/b^3*d^(1/2)*(-a*b)^(1/2)*ln((d*(-a*b)^(1/2)/b+(x-1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x-1/b*(-
a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))+1/2/b^3/(-(a*d-b*c)/b)^(1/2)*ln((-
2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a
*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2)))*a*d-1/2/b^2/(-(a*d-b*c)/b)^(1/2)*ln
((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*
(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2)))*c+1/4/b^2*(-a*b)^(1/2)/(a*d-b*c)
/(x-1/b*(-a*b)^(1/2))*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)+1/4
/b^2*a*d/(a*d-b*c)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+1/4/b^
3*(-a*b)^(1/2)*d^(3/2)*a/(a*d-b*c)*ln((d*(-a*b)^(1/2)/b+(x-1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x-1/b*(-a*b)^(1/2))^
2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))+1/4/b^3*a^2*d^2/(a*d-b*c)/(-(a*d-b*c)/b)^(1/2)
*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2
*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2)))-1/4/b^2*a*d/(a*d-b*c)/(-(a*d-
b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)
^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2)))*c-1/4/b^2*(-a*b)
^(1/2)*d/(a*d-b*c)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*x-1/4/
b^2*(-a*b)^(1/2)*d^(1/2)/(a*d-b*c)*ln((d*(-a*b)^(1/2)/b+(x-1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x-1/b*(-a*b)^(1/2))^
2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))*c
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(d*x^2+c)^(1/2)/(b*x^2+a)^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 1.83694, size = 921, normalized size = 6.77 \begin{align*} \left [-\frac{{\left (2 \, a b c - 3 \, a^{2} d +{\left (2 \, b^{2} c - 3 \, a b d\right )} x^{2}\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 (3 \, a b^{2} c - 3 \, a^{2} b d + 2 \,{\left (b^{3} c - a b^{2} d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{8 \,{\left (a b^{4} c - a^{2} b^{3} d +{\left (b^{5} c - a b^{4} d\right )} x^{2}\right )}}, -\frac{{\left (2 \, a b c - 3 \, a^{2} d +{\left (2 \, b^{2} c - 3 \, a b d\right )} x^{2}\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 (3 \, a b^{2} c - 3 \, a^{2} b d + 2 \,{\left (b^{3} c - a b^{2} d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{4 \,{\left (a b^{4} c - a^{2} b^{3} d +{\left (b^{5} c - a b^{4} d\right )} x^{2}\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(d*x^2+c)^(1/2)/(b*x^2+a)^2,x, algorithm="fricas")
[Out]
[-1/8*((2*a*b*c - 3*a^2*d + (2*b^2*c - 3*a*b*d)*x^2)*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*(3*a*b^2*c - 3*a^2*b*d + 2*(b^3*c - a*b^2*d)*x^2)*sqrt(d*x^2 + c))/(a*b^4*c
- a^2*b^3*d + (b^5*c - a*b^4*d)*x^2), -1/4*((2*a*b*c - 3*a^2*d + (2*b^2*c - 3*a*b*d)*x^2)*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*(3*a*b^2*c - 3*a^2*b*d + 2*(b^3*c - a*b^2*d)*x^2)*sqrt(d*x^2 + c))/(a*b^4*c - a^2*b^3*d + (b^5*
c - a*b^4*d)*x^2)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \sqrt{c + d x^{2}}}{\left (a + b x^{2}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**3*(d*x**2+c)**(1/2)/(b*x**2+a)**2,x)
[Out]
Integral(x**3*sqrt(c + d*x**2)/(a + b*x**2)**2, x)
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Giac [A] time = 1.15483, size = 150, normalized size = 1.1 \begin{align*} \frac{\frac{\sqrt{d x^{2} + c} a d^{2}}{{\left ({\left (d x^{2} + c\right )} b - b c + a d\right )} b^{2}} + \frac{2 \, \sqrt{d x^{2} + c} d}{b^{2}} + \frac{{\left (2 \, b c d - 3 \, a d^{2}\right )} \arctan \left (\frac{\sqrt{d x^{2} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} b^{2}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(d*x^2+c)^(1/2)/(b*x^2+a)^2,x, algorithm="giac")
[Out]
1/2*(sqrt(d*x^2 + c)*a*d^2/(((d*x^2 + c)*b - b*c + a*d)*b^2) + 2*sqrt(d*x^2 + c)*d/b^2 + (2*b*c*d - 3*a*d^2)*a
rctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^2))/d