3.8.32 \(\int \frac {x^3 \sqrt {c+d x^2}}{(a+b x^2)^2} \, dx\) [732]
Optimal. Leaf size=136 \[ \frac {(2 b c-3 a d) \sqrt {c+d x^2}}{2 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{3/2}}{2 b (b c-a d) \left (a+b x^2\right )}-\frac {(2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{5/2} \sqrt {b c-a d}} \]
[Out]
1/2*a*(d*x^2+c)^(3/2)/b/(-a*d+b*c)/(b*x^2+a)-1/2*(-3*a*d+2*b*c)*arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*d+b*c)^(1/
2))/b^(5/2)/(-a*d+b*c)^(1/2)+1/2*(-3*a*d+2*b*c)*(d*x^2+c)^(1/2)/b^2/(-a*d+b*c)
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Rubi [A]
time = 0.08, antiderivative size = 136, 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 = {457, 79, 52, 65,
214} \begin {gather*} -\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 {\sqrt {c+d x^2} (2 b c-3 a d)}{2 b^2 (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)} \end {gather*}
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 52
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 65
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 79
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] || L
tQ[p, n]))))
Rule 214
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]
Rule 457
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 {x^3 \sqrt {c+d x^2}}{\left (a+b x^2\right )^2} \, dx &=\frac {1}{2} \text {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) \text {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) \text {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) \text {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*}
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Mathematica [A]
time = 0.25, size = 98, normalized size = 0.72 \begin {gather*} \frac {\frac {\sqrt {b} \left (3 a+2 b x^2\right ) \sqrt {c+d x^2}}{a+b x^2}+\frac {(2 b c-3 a d) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{\sqrt {-b c+a d}}}{2 b^{5/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(x^3*Sqrt[c + d*x^2])/(a + b*x^2)^2,x]
[Out]
((Sqrt[b]*(3*a + 2*b*x^2)*Sqrt[c + d*x^2])/(a + b*x^2) + ((2*b*c - 3*a*d)*ArcTan[(Sqrt[b]*Sqrt[c + d*x^2])/Sqr
t[-(b*c) + a*d]])/Sqrt[-(b*c) + a*d])/(2*b^(5/2))
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1958\) vs.
\(2(116)=232\).
time = 0.12, size = 1959, normalized size = 14.40
| | |
| method |
result |
size |
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| risch |
\(\text {Expression too large to display}\) |
\(1645\) |
| default |
\(\text {Expression too large to display}\) |
\(1959\) |
| | |
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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,method=_RETURNVERBOSE)
[Out]
-1/4/b^3*(-a*b)^(1/2)*(1/(a*d-b*c)*b/(x+1/b*(-a*b)^(1/2))*(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*
(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)+d*(-a*b)^(1/2)/(a*d-b*c)*((d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b
*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)-d^(1/2)*(-a*b)^(1/2)/b*ln((-d*(-a*b)^(1/2)/b+d*(x+1/b*(-a*b)^(1/2)))/d^(1/2)
+(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))+(a*d-b*c)/b/(-(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)*(d*(x+1/b*(-a*b)^(
1/2))^2-2*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))))-2*d/(a*d-b*c)*b*(1/
4*(2*d*(x+1/b*(-a*b)^(1/2))-2*d*(-a*b)^(1/2)/b)/d*(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(
1/2))-(a*d-b*c)/b)^(1/2)+1/8*(-4*d*(a*d-b*c)/b+4*d^2*a/b)/d^(3/2)*ln((-d*(-a*b)^(1/2)/b+d*(x+1/b*(-a*b)^(1/2))
)/d^(1/2)+(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))))+1/2/b^2*((d*
(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+d^(1/2)*(-a*b)^(1/2)/b*ln((d
*(-a*b)^(1/2)/b+d*(x-1/b*(-a*b)^(1/2)))/d^(1/2)+(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/
2))-(a*d-b*c)/b)^(1/2))+(a*d-b*c)/b/(-(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)*(d*(x-1/b*(-a*b)^(1/2))^2+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))))+1/2/b^2*((d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*
c)/b)^(1/2)-d^(1/2)*(-a*b)^(1/2)/b*ln((-d*(-a*b)^(1/2)/b+d*(x+1/b*(-a*b)^(1/2)))/d^(1/2)+(d*(x+1/b*(-a*b)^(1/2
))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))+(a*d-b*c)/b/(-(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)*(d*(x+1/b*(-a*b)^(1/2))^2-2*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^3*(-a*b)^(1/2)*(1/(a*d-b*c)*b/(x-1/
b*(-a*b)^(1/2))*(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)-d*(-a*b)^
(1/2)/(a*d-b*c)*((d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+d^(1/2)*
(-a*b)^(1/2)/b*ln((d*(-a*b)^(1/2)/b+d*(x-1/b*(-a*b)^(1/2)))/d^(1/2)+(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)
/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))+(a*d-b*c)/b/(-(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)*(d*(x-1/b*(-a*b)^(1/2))^2+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))))-2*d/(a*d-b*c)*b*(1/4*(2*d*(x-1/b*(-a*b)^(1/2))+2*d*(-a*b)^(1/2
)/b)/d*(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+1/8*(-4*d*(a*d-b*c
)/b+4*d^2*a/b)/d^(3/2)*ln((d*(-a*b)^(1/2)/b+d*(x-1/b*(-a*b)^(1/2)))/d^(1/2)+(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*
b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(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(x^3*(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 detail
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Fricas [A]
time = 1.28, size = 436, normalized size = 3.21 \begin {gather*} \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 {gather*}
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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \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**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 = 0.61, size = 101, normalized size = 0.74 \begin {gather*} \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 {{\left (2 \, b c - 3 \, a d\right )} \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^{2}} + \frac {\sqrt {d x^{2} + c}}{b^{2}} \end {gather*}
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/(((d*x^2 + c)*b - b*c + a*d)*b^2) + 1/2*(2*b*c - 3*a*d)*arctan(sqrt(d*x^2 + c)*b/sqrt(
-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^2) + sqrt(d*x^2 + c)/b^2
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Mupad [B]
time = 0.51, size = 102, normalized size = 0.75 \begin {gather*} \frac {\sqrt {d\,x^2+c}}{b^2}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}}{\sqrt {a\,d-b\,c}}\right )\,\left (3\,a\,d-2\,b\,c\right )}{2\,b^{5/2}\,\sqrt {a\,d-b\,c}}+\frac {a\,d\,\sqrt {d\,x^2+c}}{2\,\left (b^3\,\left (d\,x^2+c\right )-b^3\,c+a\,b^2\,d\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((x^3*(c + d*x^2)^(1/2))/(a + b*x^2)^2,x)
[Out]
(c + d*x^2)^(1/2)/b^2 - (atan((b^(1/2)*(c + d*x^2)^(1/2))/(a*d - b*c)^(1/2))*(3*a*d - 2*b*c))/(2*b^(5/2)*(a*d
- b*c)^(1/2)) + (a*d*(c + d*x^2)^(1/2))/(2*(b^3*(c + d*x^2) - b^3*c + a*b^2*d))
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