3.752 \(\int \frac{x (c+d x^2)^{5/2}}{(a+b x^2)^2} \, dx\)
Optimal. Leaf size=126 \[ \frac{5 d \sqrt{c+d x^2} (b c-a d)}{2 b^3}-\frac{5 d (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 b^{7/2}}-\frac{\left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac{5 d \left (c+d x^2\right )^{3/2}}{6 b^2} \]
[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))
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Rubi [A] time = 0.102801, antiderivative size = 126, normalized size of antiderivative = 1.,
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 =
{444, 47, 50, 63, 208} \[ \frac{5 d \sqrt{c+d x^2} (b c-a d)}{2 b^3}-\frac{5 d (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 b^{7/2}}-\frac{\left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac{5 d \left (c+d x^2\right )^{3/2}}{6 b^2} \]
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 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]
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 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 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx &=\frac{1}{2} \operatorname{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) \operatorname{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)) \operatorname{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 ) \operatorname{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 ) \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^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*}
Mathematica [C] time = 0.0200995, size = 54, normalized size = 0.43 \[ \frac{d \left (c+d x^2\right )^{7/2} \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};-\frac{b \left (d x^2+c\right )}{a d-b c}\right )}{7 (a d-b c)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(x*(c + d*x^2)^(5/2))/(a + b*x^2)^2,x]
[Out]
(d*(c + d*x^2)^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, -((b*(c + d*x^2))/(-(b*c) + a*d))])/(7*(-(b*c) + a*d)^2)
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Maple [B] time = 0.011, size = 4363, normalized size = 34.6 \begin{align*} \text{output too large to display} \end{align*}
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)
[Out]
-1/4/b*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)^(5/2)-1/4/b*
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)^(5/2)+5/12*a/b^2*d^
2/(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)^(3/2)-5/12/b*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)^(3/2)*c+5/2*a/b^2*d^2/(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)*c+5/4*(-a*b)^(1/2)
*a^2/b^4*d^(7/2)/(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))-5/4*a^3/b^4*d^4/(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)))-5/16*(-a*b)^(1/2)/b^2*d^2/(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)^(3/2)*x-75/32*(-a*b)^(1/2)/b^2*
d^(3/2)/(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^2+5/2*a/b^2*d^2/(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)*c-5/4*(-a*b)^(1/2)*a^2/b^4*d^(7/2)/(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))-5/4*a^3/b^4*d^4/(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)))+5/4/b*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^3+1/4*(-a*b)^(1/2)/a/b/(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)^(7/2)+5/16*(-a*b)^(1/2)/b^2*d^2/(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)^(3/2)*x+75/32*(-a*b)^(1/2)/b^2*d^(
3/2)/(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^2-1/4*(-a*b)^(1/2)/a/b/(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)^(7/2)+5/4/b*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^3-5/4*a^2/b^3*d^3/(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)-5/4/b*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)*c^2+5/12*a/b^2*d^2/(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)^(3/2)-5/12/b*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)^(3/2)*c-5/4*a^2/b^3*d^3/(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)-5/4/b*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)*c^2-5/8*(-a*b)^(1/2)*a/b^3*d^3/(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-25/8*(-a*b)^(1/
2)*a/b^3*d^(5/2)/(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+15/4*a^2/b^3*d^3/(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+15/4*a^2/b^3*d^3/(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-15/4*a/b^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)))*c^2-1/4*(-a*b)^(1/2)/a/b*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)^(5/2)*x-15/32*(-a*b)^(1/2)/a/b*d^(1/2)/(a*d-b*c)*c^3*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))-15/4*a/b^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)))*
c^2+1/4*(-a*b)^(1/2)/a/b*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)^(5/2)*x+15/32*(-a*b)^(1/2)/a/b*d^(1/2)/(a*d-b*c)*c^3*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))-35/32*(-a*b)^(1/2)/b^2
*d^2/(a*d-b*c)*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+5/8*(-
a*b)^(1/2)*a/b^3*d^3/(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+25/8*(-a*b)^(1/2)*a/b^3*d^(5/2)/(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+35/32*(-a*b)^(1/2)/b^2*d^2/(a
*d-b*c)*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+5/16*(-a*b)^(
1/2)/a/b*d/(a*d-b*c)*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)^(3/2)*x+
15/32*(-a*b)^(1/2)/a/b*d/(a*d-b*c)*c^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-5/16*(-a*b)^(1/2)/a/b*d/(a*d-b*c)*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)^(3/2)*x-15/32*(-a*b)^(1/2)/a/b*d/(a*d-b*c)*c^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
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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*(d*x^2+c)^(5/2)/(b*x^2+a)^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 2.06908, size = 950, normalized size = 7.54 \begin{align*} \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{align*}
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(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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]
Timed out
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Giac [A] time = 1.1492, size = 257, normalized size = 2.04 \begin{align*} \frac{1}{6} \, d{\left (\frac{15 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} 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^{3}} - \frac{3 \,{\left (\sqrt{d x^{2} + c} b^{2} c^{2} - 2 \, \sqrt{d x^{2} + c} a b c d + \sqrt{d x^{2} + c} a^{2} d^{2}\right )}}{{\left ({\left (d x^{2} + c\right )} b - b c + a d\right )} b^{3}} + \frac{2 \,{\left ({\left (d x^{2} + c\right )}^{\frac{3}{2}} b^{4} + 6 \, \sqrt{d x^{2} + c} b^{4} c - 6 \, \sqrt{d x^{2} + c} a b^{3} d\right )}}{b^{6}}\right )} \end{align*}
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]
1/6*d*(15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)
*b^3) - 3*(sqrt(d*x^2 + c)*b^2*c^2 - 2*sqrt(d*x^2 + c)*a*b*c*d + sqrt(d*x^2 + c)*a^2*d^2)/(((d*x^2 + c)*b - b*
c + a*d)*b^3) + 2*((d*x^2 + c)^(3/2)*b^4 + 6*sqrt(d*x^2 + c)*b^4*c - 6*sqrt(d*x^2 + c)*a*b^3*d)/b^6)