3.8.43 \(\int \frac {x (c+d x^2)^{3/2}}{(a+b x^2)^2} \, dx\) [743]
Optimal. Leaf size=99 \[ \frac {3 d \sqrt {c+d x^2}}{2 b^2}-\frac {\left (c+d x^2\right )^{3/2}}{2 b \left (a+b x^2\right )}-\frac {3 d \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{5/2}} \]
[Out]
-1/2*(d*x^2+c)^(3/2)/b/(b*x^2+a)-3/2*d*arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*d+b*c)^(1/2))*(-a*d+b*c)^(1/2)/b^(5
/2)+3/2*d*(d*x^2+c)^(1/2)/b^2
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Rubi [A]
time = 0.06, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {455, 43, 52, 65,
214} \begin {gather*} -\frac {3 d \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{5/2}}-\frac {\left (c+d x^2\right )^{3/2}}{2 b \left (a+b x^2\right )}+\frac {3 d \sqrt {c+d x^2}}{2 b^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(x*(c + d*x^2)^(3/2))/(a + b*x^2)^2,x]
[Out]
(3*d*Sqrt[c + d*x^2])/(2*b^2) - (c + d*x^2)^(3/2)/(2*b*(a + b*x^2)) - (3*d*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sq
rt[c + d*x^2])/Sqrt[b*c - a*d]])/(2*b^(5/2))
Rule 43
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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] && !IntegerQ[n] && GtQ[n, 0]
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 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 455
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 \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx,x,x^2\right )\\ &=-\frac {\left (c+d x^2\right )^{3/2}}{2 b \left (a+b x^2\right )}+\frac {(3 d) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^2\right )}{4 b}\\ &=\frac {3 d \sqrt {c+d x^2}}{2 b^2}-\frac {\left (c+d x^2\right )^{3/2}}{2 b \left (a+b x^2\right )}+\frac {(3 d (b c-a d)) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{4 b^2}\\ &=\frac {3 d \sqrt {c+d x^2}}{2 b^2}-\frac {\left (c+d x^2\right )^{3/2}}{2 b \left (a+b x^2\right )}+\frac {(3 (b c-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}\\ &=\frac {3 d \sqrt {c+d x^2}}{2 b^2}-\frac {\left (c+d x^2\right )^{3/2}}{2 b \left (a+b x^2\right )}-\frac {3 d \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{5/2}}\\ \end {align*}
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Mathematica [A]
time = 0.26, size = 96, normalized size = 0.97 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-b c+3 a d+2 b d x^2\right )}{2 b^2 \left (a+b x^2\right )}-\frac {3 d \sqrt {-b c+a d} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{2 b^{5/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(x*(c + d*x^2)^(3/2))/(a + b*x^2)^2,x]
[Out]
(Sqrt[c + d*x^2]*(-(b*c) + 3*a*d + 2*b*d*x^2))/(2*b^2*(a + b*x^2)) - (3*d*Sqrt[-(b*c) + a*d]*ArcTan[(Sqrt[b]*S
qrt[c + d*x^2])/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. \(2151\) vs.
\(2(79)=158\).
time = 0.12, size = 2152, normalized size = 21.74
| | |
| method |
result |
size |
| | |
| default |
\(\text {Expression too large to display}\) |
\(2152\) |
| risch |
\(\text {Expression too large to display}\) |
\(2185\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*(d*x^2+c)^(3/2)/(b*x^2+a)^2,x,method=_RETURNVERBOSE)
[Out]
1/4*(-a*b)^(1/2)/a/b^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)^(5/2)+3*d*(-a*b)^(1/2)/(a*d-b*c)*(1/3*(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)/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)))-(a*d-b*c)/b*((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)))))-4*d/(a*d-b*c)*b*
(1/8*(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)^(3/2)+3/16*(-4*d*(a*d-b*c)/b+4*d^2*a/b)/d*(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)))))-1/4*(-a*b)^(1/2)/a/b^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)^(5/2)-3*d*(-a*b)^(1/2)/(a*
d-b*c)*(1/3*(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
)/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)))-(a*d-b*
c)/b*((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)))))-4*d/(a*d-b*c)*b*(1/8*(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)^(3/2)+3/16*(-4*d*(a*d-b*c)/b+4*d^2
*a/b)/d*(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*(d*x^2+c)^(3/2)/(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 = 1.40, size = 333, normalized size = 3.36 \begin {gather*} \left [\frac {3 \, {\left (b d x^{2} + a d\right )} \sqrt {\frac {b c - a d}{b}} \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^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (2 \, b d x^{2} - b c + 3 \, a d\right )} \sqrt {d x^{2} + c}}{8 \, {\left (b^{3} x^{2} + a b^{2}\right )}}, -\frac {3 \, {\left (b d x^{2} + a d\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (2 \, b d x^{2} - b c + 3 \, a d\right )} \sqrt {d x^{2} + c}}{4 \, {\left (b^{3} x^{2} + a b^{2}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(d*x^2+c)^(3/2)/(b*x^2+a)^2,x, algorithm="fricas")
[Out]
[1/8*(3*(b*d*x^2 + a*d)*sqrt((b*c - a*d)/b)*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^2*d*x^2 + 2*b^2*c - a*b*d)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/b))/(b^2*x^4 + 2*a*b*x^2 +
a^2)) + 4*(2*b*d*x^2 - b*c + 3*a*d)*sqrt(d*x^2 + c))/(b^3*x^2 + a*b^2), -1/4*(3*(b*d*x^2 + a*d)*sqrt(-(b*c -
a*d)/b)*arctan(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/b)/(b*c^2 - a*c*d + (b*c*d - a*d
^2)*x^2)) - 2*(2*b*d*x^2 - b*c + 3*a*d)*sqrt(d*x^2 + c))/(b^3*x^2 + a*b^2)]
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (c + d x^{2}\right )^{\frac {3}{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)**(3/2)/(b*x**2+a)**2,x)
[Out]
Integral(x*(c + d*x**2)**(3/2)/(a + b*x**2)**2, x)
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Giac [A]
time = 0.57, size = 122, normalized size = 1.23 \begin {gather*} \frac {\sqrt {d x^{2} + c} d}{b^{2}} + \frac {3 \, {\left (b c d - a d^{2}\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 c d - \sqrt {d x^{2} + c} a d^{2}}{2 \, {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )} b^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(d*x^2+c)^(3/2)/(b*x^2+a)^2,x, algorithm="giac")
[Out]
sqrt(d*x^2 + c)*d/b^2 + 3/2*(b*c*d - a*d^2)*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*
d)*b^2) - 1/2*(sqrt(d*x^2 + c)*b*c*d - sqrt(d*x^2 + c)*a*d^2)/(((d*x^2 + c)*b - b*c + a*d)*b^2)
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Mupad [B]
time = 0.51, size = 117, normalized size = 1.18 \begin {gather*} \frac {\sqrt {d\,x^2+c}\,\left (\frac {a\,d^2}{2}-\frac {b\,c\,d}{2}\right )}{b^3\,\left (d\,x^2+c\right )-b^3\,c+a\,b^2\,d}+\frac {d\,\sqrt {d\,x^2+c}}{b^2}-\frac {3\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,d\,\sqrt {d\,x^2+c}\,\sqrt {a\,d-b\,c}}{a\,d^2-b\,c\,d}\right )\,\sqrt {a\,d-b\,c}}{2\,b^{5/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((x*(c + d*x^2)^(3/2))/(a + b*x^2)^2,x)
[Out]
((c + d*x^2)^(1/2)*((a*d^2)/2 - (b*c*d)/2))/(b^3*(c + d*x^2) - b^3*c + a*b^2*d) + (d*(c + d*x^2)^(1/2))/b^2 -
(3*d*atan((b^(1/2)*d*(c + d*x^2)^(1/2)*(a*d - b*c)^(1/2))/(a*d^2 - b*c*d))*(a*d - b*c)^(1/2))/(2*b^(5/2))
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