3.8.68 \(\int \frac {x^3}{(a+b x^2)^2 (c+d x^2)^{3/2}} \, dx\) [768]

Optimal. Leaf size=134 \[ \frac {2 b c+a d}{2 b (b c-a d)^2 \sqrt {c+d x^2}}+\frac {a}{2 b (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {(2 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 \sqrt {b} (b c-a d)^{5/2}} \]

[Out]

-1/2*(a*d+2*b*c)*arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*d+b*c)^(1/2))/(-a*d+b*c)^(5/2)/b^(1/2)+1/2*(a*d+2*b*c)/b/
(-a*d+b*c)^2/(d*x^2+c)^(1/2)+1/2*a/b/(-a*d+b*c)/(b*x^2+a)/(d*x^2+c)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 134, 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, 53, 65, 214} \begin {gather*} \frac {a}{2 b \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}+\frac {a d+2 b c}{2 b \sqrt {c+d x^2} (b c-a d)^2}-\frac {(a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 \sqrt {b} (b c-a d)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/((a + b*x^2)^2*(c + d*x^2)^(3/2)),x]

[Out]

(2*b*c + a*d)/(2*b*(b*c - a*d)^2*Sqrt[c + d*x^2]) + a/(2*b*(b*c - a*d)*(a + b*x^2)*Sqrt[c + d*x^2]) - ((2*b*c
+ a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(2*Sqrt[b]*(b*c - a*d)^(5/2))

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x}{(a+b x)^2 (c+d x)^{3/2}} \, dx,x,x^2\right )\\ &=\frac {a}{2 b (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}+\frac {(2 b c+a d) \text {Subst}\left (\int \frac {1}{(a+b x) (c+d x)^{3/2}} \, dx,x,x^2\right )}{4 b (b c-a d)}\\ &=\frac {2 b c+a d}{2 b (b c-a d)^2 \sqrt {c+d x^2}}+\frac {a}{2 b (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}+\frac {(2 b c+a d) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{4 (b c-a d)^2}\\ &=\frac {2 b c+a d}{2 b (b c-a d)^2 \sqrt {c+d x^2}}+\frac {a}{2 b (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}+\frac {(2 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 d (b c-a d)^2}\\ &=\frac {2 b c+a d}{2 b (b c-a d)^2 \sqrt {c+d x^2}}+\frac {a}{2 b (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {(2 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 \sqrt {b} (b c-a d)^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 110, normalized size = 0.82 \begin {gather*} \frac {1}{2} \left (\frac {3 a c+2 b c x^2+a d x^2}{(b c-a d)^2 \left (a+b x^2\right ) \sqrt {c+d x^2}}+\frac {(2 b c+a d) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{\sqrt {b} (-b c+a d)^{5/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/((a + b*x^2)^2*(c + d*x^2)^(3/2)),x]

[Out]

((3*a*c + 2*b*c*x^2 + a*d*x^2)/((b*c - a*d)^2*(a + b*x^2)*Sqrt[c + d*x^2]) + ((2*b*c + a*d)*ArcTan[(Sqrt[b]*Sq
rt[c + d*x^2])/Sqrt[-(b*c) + a*d]])/(Sqrt[b]*(-(b*c) + a*d)^(5/2)))/2

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1921\) vs. \(2(114)=228\).
time = 0.09, size = 1922, normalized size = 14.34

method	result	size
default	\(\text {Expression too large to display}\)	\(1922\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b*x^2+a)^2/(d*x^2+c)^(3/2),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)^(1/2)-3*d*(-a*b)^(1/2)/(a*d-b*c)*(-1/(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)-2*d*(-a*b)^(1/2)/(a*d-b*c)*(2*d*(x+1/b*(-a*b)^(1/2))-2*d*(-
a*b)^(1/2)/b)/(-4*d*(a*d-b*c)/b+4*d^2*a/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)+1/(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*(2*d*(x+1/b*(-a*b)^(1/2))-2*d*(-a*b)^(1/2)/b)/(-4*d*(a*d-b*c)/b+4*d^2*
a/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))+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)^(1/2)+3*d*(-a*b)^(1/2)/(a*d-b*c)*(-1/(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)+2*d*(-a*b)^(1/2)/(a*d-b*c)*(2*d*(x-1/b*(-a*b)^(1/2))+2*d*(-a*b)^(1/2)/b)/(-4*d*
(a*d-b*c)/b+4*d^2*a/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)+1/
(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*(2*d*(x-1/b*(-a*b)^(1/2))+2*d*(-a*b)^(1/2)/b)/(-4*d*(a*d-b*c)/b+4*d^2*a/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))+1/2/b^2*(-1/(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)+2*d*(-a*b)^(1/2)/(a*d-b*c)*(2*d*(x-1/b*(
-a*b)^(1/2))+2*d*(-a*b)^(1/2)/b)/(-4*d*(a*d-b*c)/b+4*d^2*a/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)+1/(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*(-1/(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)-2*d*(-a*b)^(1/2)/(a*d-b*c)*(2*d*(x+1/b*(-a*b)^(1/2))-2*d*(-a*b)^(1/2
)/b)/(-4*d*(a*d-b*c)/b+4*d^2*a/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)+1/(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))))

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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/(b*x^2+a)^2/(d*x^2+c)^(3/2),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 [B] Leaf count of result is larger than twice the leaf count of optimal. 346 vs. \(2 (114) = 228\).
time = 1.61, size = 732, normalized size = 5.46 \begin {gather*} \left [\frac {{\left ({\left (2 \, b^{2} c d + a b d^{2}\right )} x^{4} + 2 \, a b c^{2} + a^{2} c d + {\left (2 \, b^{2} c^{2} + 3 \, a b c d + a^{2} d^{2}\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^{2} - 3 \, a^{2} b c d + {\left (2 \, b^{3} c^{2} - a b^{2} c d - a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a b^{4} c^{4} - 3 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} - a^{4} b c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{4} + {\left (b^{5} c^{4} - 2 \, a b^{4} c^{3} d + 2 \, a^{3} b^{2} c d^{3} - a^{4} b d^{4}\right )} x^{2}\right )}}, -\frac {{\left ({\left (2 \, b^{2} c d + a b d^{2}\right )} x^{4} + 2 \, a b c^{2} + a^{2} c d + {\left (2 \, b^{2} c^{2} + 3 \, a b c d + a^{2} d^{2}\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^{2} - 3 \, a^{2} b c d + {\left (2 \, b^{3} c^{2} - a b^{2} c d - a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a b^{4} c^{4} - 3 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} - a^{4} b c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{4} + {\left (b^{5} c^{4} - 2 \, a b^{4} c^{3} d + 2 \, a^{3} b^{2} c d^{3} - a^{4} b d^{4}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(((2*b^2*c*d + a*b*d^2)*x^4 + 2*a*b*c^2 + a^2*c*d + (2*b^2*c^2 + 3*a*b*c*d + a^2*d^2)*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^2 - 3*a^2*b*c*d + (2*b
^3*c^2 - a*b^2*c*d - a^2*b*d^2)*x^2)*sqrt(d*x^2 + c))/(a*b^4*c^4 - 3*a^2*b^3*c^3*d + 3*a^3*b^2*c^2*d^2 - a^4*b
*c*d^3 + (b^5*c^3*d - 3*a*b^4*c^2*d^2 + 3*a^2*b^3*c*d^3 - a^3*b^2*d^4)*x^4 + (b^5*c^4 - 2*a*b^4*c^3*d + 2*a^3*
b^2*c*d^3 - a^4*b*d^4)*x^2), -1/4*(((2*b^2*c*d + a*b*d^2)*x^4 + 2*a*b*c^2 + a^2*c*d + (2*b^2*c^2 + 3*a*b*c*d +
a^2*d^2)*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^2 - 3*a^2*b*c*d + (2*b^3*c^2 - a*b^2*c*d - a^2*b*
d^2)*x^2)*sqrt(d*x^2 + c))/(a*b^4*c^4 - 3*a^2*b^3*c^3*d + 3*a^3*b^2*c^2*d^2 - a^4*b*c*d^3 + (b^5*c^3*d - 3*a*b
^4*c^2*d^2 + 3*a^2*b^3*c*d^3 - a^3*b^2*d^4)*x^4 + (b^5*c^4 - 2*a*b^4*c^3*d + 2*a^3*b^2*c*d^3 - a^4*b*d^4)*x^2)
]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\left (a + b x^{2}\right )^{2} \left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(b*x**2+a)**2/(d*x**2+c)**(3/2),x)

[Out]

Integral(x**3/((a + b*x**2)**2*(c + d*x**2)**(3/2)), x)

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Giac [A]
time = 0.53, size = 181, normalized size = 1.35 \begin {gather*} \frac {\frac {{\left (2 \, b c d + a d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, {\left (d x^{2} + c\right )} b c d - 2 \, b c^{2} d + {\left (d x^{2} + c\right )} a d^{2} + 2 \, a c d^{2}}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left ({\left (d x^{2} + c\right )}^{\frac {3}{2}} b - \sqrt {d x^{2} + c} b c + \sqrt {d x^{2} + c} a d\right )}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

1/2*((2*b*c*d + a*d^2)*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-b
^2*c + a*b*d)) + (2*(d*x^2 + c)*b*c*d - 2*b*c^2*d + (d*x^2 + c)*a*d^2 + 2*a*c*d^2)/((b^2*c^2 - 2*a*b*c*d + a^2
*d^2)*((d*x^2 + c)^(3/2)*b - sqrt(d*x^2 + c)*b*c + sqrt(d*x^2 + c)*a*d)))/d

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Mupad [B]
time = 0.62, size = 142, normalized size = 1.06 \begin {gather*} \frac {\frac {c}{a\,d-b\,c}+\frac {\left (d\,x^2+c\right )\,\left (a\,d+2\,b\,c\right )}{2\,{\left (a\,d-b\,c\right )}^2}}{b\,{\left (d\,x^2+c\right )}^{3/2}+\sqrt {d\,x^2+c}\,\left (a\,d-b\,c\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{{\left (a\,d-b\,c\right )}^{5/2}}\right )\,\left (a\,d+2\,b\,c\right )}{2\,\sqrt {b}\,{\left (a\,d-b\,c\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/((a + b*x^2)^2*(c + d*x^2)^(3/2)),x)

[Out]

(c/(a*d - b*c) + ((c + d*x^2)*(a*d + 2*b*c))/(2*(a*d - b*c)^2))/(b*(c + d*x^2)^(3/2) + (c + d*x^2)^(1/2)*(a*d
- b*c)) + (atan((b^(1/2)*(c + d*x^2)^(1/2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/(a*d - b*c)^(5/2))*(a*d + 2*b*c))/
(2*b^(1/2)*(a*d - b*c)^(5/2))

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