3.8.52 \(\int \frac {x (c+d x^2)^{5/2}}{(a+b x^2)^2} \, dx\) [752]
Optimal. Leaf size=126 \[ \frac {5 d (b c-a d) \sqrt {c+d x^2}}{2 b^3}+\frac {5 d \left (c+d x^2\right )^{3/2}}{6 b^2}-\frac {\left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}-\frac {5 d (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{7/2}} \]
[Out]
5/6*d*(d*x^2+c)^(3/2)/b^2-1/2*(d*x^2+c)^(5/2)/b/(b*x^2+a)-5/2*d*(-a*d+b*c)^(3/2)*arctanh(b^(1/2)*(d*x^2+c)^(1/
2)/(-a*d+b*c)^(1/2))/b^(7/2)+5/2*d*(-a*d+b*c)*(d*x^2+c)^(1/2)/b^3
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Rubi [A]
time = 0.07, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps
used = 6, 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 {5 d (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{7/2}}+\frac {5 d \sqrt {c+d x^2} (b c-a d)}{2 b^3}-\frac {\left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {5 d \left (c+d x^2\right )^{3/2}}{6 b^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(x*(c + d*x^2)^(5/2))/(a + b*x^2)^2,x]
[Out]
(5*d*(b*c - a*d)*Sqrt[c + d*x^2])/(2*b^3) + (5*d*(c + d*x^2)^(3/2))/(6*b^2) - (c + d*x^2)^(5/2)/(2*b*(a + b*x^
2)) - (5*d*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(2*b^(7/2))
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 )^{5/2}}{\left (a+b x^2\right )^2} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(c+d x)^{5/2}}{(a+b x)^2} \, dx,x,x^2\right )\\ &=-\frac {\left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {(5 d) \text {Subst}\left (\int \frac {(c+d x)^{3/2}}{a+b x} \, dx,x,x^2\right )}{4 b}\\ &=\frac {5 d \left (c+d x^2\right )^{3/2}}{6 b^2}-\frac {\left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {(5 d (b c-a d)) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^2\right )}{4 b^2}\\ &=\frac {5 d (b c-a d) \sqrt {c+d x^2}}{2 b^3}+\frac {5 d \left (c+d x^2\right )^{3/2}}{6 b^2}-\frac {\left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {\left (5 d (b c-a d)^2\right ) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{4 b^3}\\ &=\frac {5 d (b c-a d) \sqrt {c+d x^2}}{2 b^3}+\frac {5 d \left (c+d x^2\right )^{3/2}}{6 b^2}-\frac {\left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {\left (5 (b c-a d)^2\right ) \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^3}\\ &=\frac {5 d (b c-a d) \sqrt {c+d x^2}}{2 b^3}+\frac {5 d \left (c+d x^2\right )^{3/2}}{6 b^2}-\frac {\left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}-\frac {5 d (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{7/2}}\\ \end {align*}
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Mathematica [A]
time = 0.30, size = 128, normalized size = 1.02 \begin {gather*} -\frac {\sqrt {c+d x^2} \left (15 a^2 d^2+10 a b d \left (-2 c+d x^2\right )+b^2 \left (3 c^2-14 c d x^2-2 d^2 x^4\right )\right )}{6 b^3 \left (a+b x^2\right )}+\frac {5 d (-b c+a d)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{2 b^{7/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(x*(c + d*x^2)^(5/2))/(a + b*x^2)^2,x]
[Out]
-1/6*(Sqrt[c + d*x^2]*(15*a^2*d^2 + 10*a*b*d*(-2*c + d*x^2) + b^2*(3*c^2 - 14*c*d*x^2 - 2*d^2*x^4)))/(b^3*(a +
b*x^2)) + (5*d*(-(b*c) + a*d)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[-(b*c) + a*d]])/(2*b^(7/2))
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(3223\) vs.
\(2(102)=204\).
time = 0.13, size = 3224, normalized size = 25.59
| | |
| method |
result |
size |
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| risch |
\(\text {Expression too large to display}\) |
\(3059\) |
| default |
\(\text {Expression too large to display}\) |
\(3224\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*(d*x^2+c)^(5/2)/(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)^(7/2)+5*d*(-a*b)^(1/2)/(a*d-b*c)*(1/5*(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)-d*(-a*b)^(1/2)/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))
))-(a*d-b*c)/b*(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))))))-6*d/(a*d-b*c)*b*(1/12*(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)^(5/2)+5/24*(-4*d*(a*d-
b*c)/b+4*d^2*a/b)/d*(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)^(7/2)
-5*d*(-a*b)^(1/2)/(a*d-b*c)*(1/5*(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)+d*(-a*b)^(1/2)/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))))-(a*d-b*c)/b*(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))))))-6*d/(a*d-b*c)*b*(1/12*(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)^(5/2)+5/24*(-4*d*(a*d-b*c)/b+4*d^2*a/b)/d*(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)^(5/2)/(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 = 1.96, size = 453, normalized size = 3.60 \begin {gather*} \left [-\frac {15 \, {\left (a b c d - a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\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^{2} d^{2} x^{4} - 3 \, b^{2} c^{2} + 20 \, a b c d - 15 \, a^{2} d^{2} + 2 \, {\left (7 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{24 \, {\left (b^{4} x^{2} + a b^{3}\right )}}, -\frac {15 \, {\left (a b c d - a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\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^{2} d^{2} x^{4} - 3 \, b^{2} c^{2} + 20 \, a b c d - 15 \, a^{2} d^{2} + 2 \, {\left (7 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, {\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(d*x^2+c)^(5/2)/(b*x^2+a)^2,x, algorithm="fricas")
[Out]
[-1/24*(15*(a*b*c*d - a^2*d^2 + (b^2*c*d - a*b*d^2)*x^2)*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^2*d^2*x^4 - 3*b^2*c^2 + 20*a*b*c*d - 15*a^2*d^2 + 2*(7*b^2*c*
d - 5*a*b*d^2)*x^2)*sqrt(d*x^2 + c))/(b^4*x^2 + a*b^3), -1/12*(15*(a*b*c*d - a^2*d^2 + (b^2*c*d - a*b*d^2)*x^2
)*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^2*d^2*x^4 - 3*b^2*c^2 + 20*a*b*c*d - 15*a^2*d^2 + 2*(7*b^2*c*d - 5*a*b*d^2)*
x^2)*sqrt(d*x^2 + c))/(b^4*x^2 + a*b^3)]
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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 {5}{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)**(5/2)/(b*x**2+a)**2,x)
[Out]
Integral(x*(c + d*x**2)**(5/2)/(a + b*x**2)**2, x)
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Giac [A]
time = 0.52, size = 197, normalized size = 1.56 \begin {gather*} \frac {5 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\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^{3}} - \frac {\sqrt {d x^{2} + c} b^{2} c^{2} d - 2 \, \sqrt {d x^{2} + c} a b c d^{2} + \sqrt {d x^{2} + c} a^{2} d^{3}}{2 \, {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )} b^{3}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{4} d + 6 \, \sqrt {d x^{2} + c} b^{4} c d - 6 \, \sqrt {d x^{2} + c} a b^{3} d^{2}}{3 \, b^{6}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(d*x^2+c)^(5/2)/(b*x^2+a)^2,x, algorithm="giac")
[Out]
5/2*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b
^3) - 1/2*(sqrt(d*x^2 + c)*b^2*c^2*d - 2*sqrt(d*x^2 + c)*a*b*c*d^2 + sqrt(d*x^2 + c)*a^2*d^3)/(((d*x^2 + c)*b
- b*c + a*d)*b^3) + 1/3*((d*x^2 + c)^(3/2)*b^4*d + 6*sqrt(d*x^2 + c)*b^4*c*d - 6*sqrt(d*x^2 + c)*a*b^3*d^2)/b^
6
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Mupad [B]
time = 0.57, size = 172, normalized size = 1.37 \begin {gather*} \frac {d\,{\left (d\,x^2+c\right )}^{3/2}}{3\,b^2}-\frac {\sqrt {d\,x^2+c}\,\left (\frac {a^2\,d^3}{2}-a\,b\,c\,d^2+\frac {b^2\,c^2\,d}{2}\right )}{b^4\,\left (d\,x^2+c\right )-b^4\,c+a\,b^3\,d}+\frac {5\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,d\,\sqrt {d\,x^2+c}\,{\left (a\,d-b\,c\right )}^{3/2}}{a^2\,d^3-2\,a\,b\,c\,d^2+b^2\,c^2\,d}\right )\,{\left (a\,d-b\,c\right )}^{3/2}}{2\,b^{7/2}}+\frac {d\,\sqrt {d\,x^2+c}\,\left (2\,b^2\,c-2\,a\,b\,d\right )}{b^4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((x*(c + d*x^2)^(5/2))/(a + b*x^2)^2,x)
[Out]
(d*(c + d*x^2)^(3/2))/(3*b^2) - ((c + d*x^2)^(1/2)*((a^2*d^3)/2 + (b^2*c^2*d)/2 - a*b*c*d^2))/(b^4*(c + d*x^2)
- b^4*c + a*b^3*d) + (5*d*atan((b^(1/2)*d*(c + d*x^2)^(1/2)*(a*d - b*c)^(3/2))/(a^2*d^3 + b^2*c^2*d - 2*a*b*c
*d^2))*(a*d - b*c)^(3/2))/(2*b^(7/2)) + (d*(c + d*x^2)^(1/2)*(2*b^2*c - 2*a*b*d))/b^4
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