3.332 \(\int x^3 (a+\frac{b}{c+d x^2})^{3/2} \, dx\)

Optimal. Leaf size=172 \[ \frac{a \left (c+d x^2\right )^2 \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{4 d^2}+\frac{(5 b-4 a c) \left (c+d x^2\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{8 d^2}+\frac{b c \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{d^2}+\frac{3 b (b-4 a c) \tanh ^{-1}\left (\frac{\sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{\sqrt{a}}\right )}{8 \sqrt{a} d^2} \]

[Out]

(b*c*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/d^2 + ((5*b - 4*a*c)*(c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x
^2)])/(8*d^2) + (a*(c + d*x^2)^2*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(4*d^2) + (3*b*(b - 4*a*c)*ArcTanh[Sqr
t[(b + a*c + a*d*x^2)/(c + d*x^2)]/Sqrt[a]])/(8*Sqrt[a]*d^2)

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Rubi [A]  time = 0.535715, antiderivative size = 222, normalized size of antiderivative = 1.29, number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {6722, 1975, 446, 78, 50, 63, 217, 206} \[ \frac{c \sqrt{a+\frac{b}{c+d x^2}} \left (a \left (c+d x^2\right )+b\right )^2}{b d^2}+\frac{(b-4 a c) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (a \left (c+d x^2\right )+b\right )}{4 b d^2}+\frac{3 (b-4 a c) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}}}{8 d^2}+\frac{3 b (b-4 a c) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c+d x^2}}{\sqrt{a \left (c+d x^2\right )+b}}\right )}{8 \sqrt{a} d^2 \sqrt{a \left (c+d x^2\right )+b}} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b/(c + d*x^2))^(3/2),x]

[Out]

(3*(b - 4*a*c)*(c + d*x^2)*Sqrt[a + b/(c + d*x^2)])/(8*d^2) + ((b - 4*a*c)*(c + d*x^2)*Sqrt[a + b/(c + d*x^2)]
*(b + a*(c + d*x^2)))/(4*b*d^2) + (c*Sqrt[a + b/(c + d*x^2)]*(b + a*(c + d*x^2))^2)/(b*d^2) + (3*b*(b - 4*a*c)
*Sqrt[c + d*x^2]*Sqrt[a + b/(c + d*x^2)]*ArcTanh[(Sqrt[a]*Sqrt[c + d*x^2])/Sqrt[b + a*(c + d*x^2)]])/(8*Sqrt[a
]*d^2*Sqrt[b + a*(c + d*x^2)])

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/
v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[p] && ILtQ[n, 0] &
& BinomialQ[v, x] &&  !LinearQ[v, x]

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 446

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]]

Rule 78

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] || LtQ
[p, n]))))

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int x^3 \left (a+\frac{b}{c+d x^2}\right )^{3/2} \, dx &=\frac{\left (\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{x^3 \left (b+a \left (c+d x^2\right )\right )^{3/2}}{\left (c+d x^2\right )^{3/2}} \, dx}{\sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{\left (\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{x^3 \left (b+a c+a d x^2\right )^{3/2}}{\left (c+d x^2\right )^{3/2}} \, dx}{\sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{\left (\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \operatorname{Subst}\left (\int \frac{x (b+a c+a d x)^{3/2}}{(c+d x)^{3/2}} \, dx,x,x^2\right )}{2 \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{c \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )^2}{b d^2}+\frac{\left ((b-4 a c) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \operatorname{Subst}\left (\int \frac{(b+a c+a d x)^{3/2}}{\sqrt{c+d x}} \, dx,x,x^2\right )}{2 b d \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{(b-4 a c) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{4 b d^2}+\frac{c \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )^2}{b d^2}+\frac{\left (3 (b-4 a c) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{b+a c+a d x}}{\sqrt{c+d x}} \, dx,x,x^2\right )}{8 d \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{3 (b-4 a c) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}}}{8 d^2}+\frac{(b-4 a c) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{4 b d^2}+\frac{c \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )^2}{b d^2}+\frac{\left (3 b (b-4 a c) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x} \sqrt{b+a c+a d x}} \, dx,x,x^2\right )}{16 d \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{3 (b-4 a c) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}}}{8 d^2}+\frac{(b-4 a c) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{4 b d^2}+\frac{c \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )^2}{b d^2}+\frac{\left (3 b (b-4 a c) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+a x^2}} \, dx,x,\sqrt{c+d x^2}\right )}{8 d^2 \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{3 (b-4 a c) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}}}{8 d^2}+\frac{(b-4 a c) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{4 b d^2}+\frac{c \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )^2}{b d^2}+\frac{\left (3 b (b-4 a c) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{\sqrt{c+d x^2}}{\sqrt{b+a \left (c+d x^2\right )}}\right )}{8 d^2 \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{3 (b-4 a c) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}}}{8 d^2}+\frac{(b-4 a c) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{4 b d^2}+\frac{c \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )^2}{b d^2}+\frac{3 b (b-4 a c) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c+d x^2}}{\sqrt{b+a \left (c+d x^2\right )}}\right )}{8 \sqrt{a} d^2 \sqrt{b+a \left (c+d x^2\right )}}\\ \end{align*}

Mathematica [A]  time = 0.207054, size = 104, normalized size = 0.6 \[ \frac{\sqrt{a} \sqrt{\frac{a c+a d x^2+b}{c+d x^2}} \left (-2 a c^2+2 a d^2 x^4+13 b c+5 b d x^2\right )+3 b (b-4 a c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{c+d x^2}}}{\sqrt{a}}\right )}{8 \sqrt{a} d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(13*b*c - 2*a*c^2 + 5*b*d*x^2 + 2*a*d^2*x^4) + 3*b*(b - 4*a*c)*
ArcTanh[Sqrt[a + b/(c + d*x^2)]/Sqrt[a]])/(8*Sqrt[a]*d^2)

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Maple [B]  time = 0.016, size = 593, normalized size = 3.5 \begin{align*}{\frac{1}{16\,{d}^{2}} \left ( 4\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}{x}^{4}a{d}^{2}-12\,\ln \left ( 1/2\,{\frac{2\,a{d}^{2}{x}^{2}+2\,acd+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}+bd}{\sqrt{a{d}^{2}}}} \right ){x}^{2}abc{d}^{2}+3\,\ln \left ( 1/2\,{\frac{2\,a{d}^{2}{x}^{2}+2\,acd+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}+bd}{\sqrt{a{d}^{2}}}} \right ){x}^{2}{b}^{2}{d}^{2}+10\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}{x}^{2}bd-12\,\ln \left ( 1/2\,{\frac{2\,a{d}^{2}{x}^{2}+2\,acd+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}+bd}{\sqrt{a{d}^{2}}}} \right ) ab{c}^{2}d-4\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}a{c}^{2}+3\,\ln \left ( 1/2\,{\frac{2\,a{d}^{2}{x}^{2}+2\,acd+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}+bd}{\sqrt{a{d}^{2}}}} \right ){b}^{2}cd+10\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}bc+16\,\sqrt{a{d}^{2}}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( ad{x}^{2}+ac+b \right ) }bc \right ) \sqrt{{\frac{ad{x}^{2}+ac+b}{d{x}^{2}+c}}}{\frac{1}{\sqrt{a{d}^{2}}}}{\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( ad{x}^{2}+ac+b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)*x^4*a*d^2-12*ln(1/2*(2*a*d^2*x^2+2*a*c*d
+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*x^2*a*b*c*d^2+3*ln(1/2*(2
*a*d^2*x^2+2*a*c*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*x^2*b^2
*d^2+10*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)*x^2*b*d-12*ln(1/2*(2*a*d^2*x^2+2*a*c*d+2
*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*a*b*c^2*d-4*(a*d^2*x^4+2*a*
c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)*a*c^2+3*ln(1/2*(2*a*d^2*x^2+2*a*c*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*
d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*b^2*c*d+10*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^
(1/2)*(a*d^2)^(1/2)*b*c+16*(a*d^2)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*b*c)/d^2*((a*d*x^2+a*c+b)/(d*x^2+c)
)^(1/2)/(a*d^2)^(1/2)/((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.49546, size = 757, normalized size = 4.4 \begin{align*} \left [\frac{3 \,{\left (4 \, a b c - b^{2}\right )} \sqrt{a} \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \,{\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} - 4 \,{\left (2 \, a d^{2} x^{4} +{\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt{a} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 4 \,{\left (2 \, a^{2} d^{2} x^{4} + 5 \, a b d x^{2} - 2 \, a^{2} c^{2} + 13 \, a b c\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{32 \, a d^{2}}, \frac{3 \,{\left (4 \, a b c - b^{2}\right )} \sqrt{-a} \arctan \left (\frac{{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt{-a} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{2 \,{\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) + 2 \,{\left (2 \, a^{2} d^{2} x^{4} + 5 \, a b d x^{2} - 2 \, a^{2} c^{2} + 13 \, a b c\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{16 \, a d^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(3*(4*a*b*c - b^2)*sqrt(a)*log(8*a^2*d^2*x^4 + 8*a^2*c^2 + 8*(2*a^2*c + a*b)*d*x^2 + 8*a*b*c + b^2 - 4*(
2*a*d^2*x^4 + (4*a*c + b)*d*x^2 + 2*a*c^2 + b*c)*sqrt(a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))) + 4*(2*a^2*d^2
*x^4 + 5*a*b*d*x^2 - 2*a^2*c^2 + 13*a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a*d^2), 1/16*(3*(4*a*b*c -
b^2)*sqrt(-a)*arctan(1/2*(2*a*d*x^2 + 2*a*c + b)*sqrt(-a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(a^2*d*x^2 + a
^2*c + a*b)) + 2*(2*a^2*d^2*x^4 + 5*a*b*d*x^2 - 2*a^2*c^2 + 13*a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(
a*d^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 3.18521, size = 598, normalized size = 3.48 \begin{align*} \frac{1}{16} \,{\left (2 \, \sqrt{a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}{\left (\frac{2 \, a x^{2}}{d} - \frac{2 \, a^{2} c d^{2} - 5 \, a b d^{2}}{a d^{4}}\right )} + \frac{{\left (4 \, a^{\frac{3}{2}} b c - \sqrt{a} b^{2}\right )} \log \left ({\left | -2 \, a^{\frac{5}{2}} c^{3} d - 6 \,{\left (\sqrt{a d^{2}} x^{2} - \sqrt{a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a^{2} c^{2}{\left | d \right |} - 6 \,{\left (\sqrt{a d^{2}} x^{2} - \sqrt{a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{2} a^{\frac{3}{2}} c d - a^{\frac{3}{2}} b c^{2} d - 2 \,{\left (\sqrt{a d^{2}} x^{2} - \sqrt{a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{3} a{\left | d \right |} - 2 \,{\left (\sqrt{a d^{2}} x^{2} - \sqrt{a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a b c{\left | d \right |} -{\left (\sqrt{a d^{2}} x^{2} - \sqrt{a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{2} \sqrt{a} b d \right |}\right )}{a d{\left | d \right |}} + \frac{2 \,{\left (4 \, a^{\frac{3}{2}} b c{\left | d \right |} - \sqrt{a} b^{2}{\left | d \right |}\right )} \log \left (48 \, d^{2}{\left | a \right |}\right )}{a d^{3}}\right )} \mathrm{sgn}\left (d x^{2} + c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c)*(2*a*x^2/d - (2*a^2*c*d^2 - 5*a*b*d^2)/(a*d^4))
+ (4*a^(3/2)*b*c - sqrt(a)*b^2)*log(abs(-2*a^(5/2)*c^3*d - 6*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 +
b*d*x^2 + a*c^2 + b*c))*a^2*c^2*abs(d) - 6*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2
+ b*c))^2*a^(3/2)*c*d - a^(3/2)*b*c^2*d - 2*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2
+ b*c))^3*a*abs(d) - 2*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))*a*b*c*abs(d)
- (sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))^2*sqrt(a)*b*d))/(a*d*abs(d)) + 2*(
4*a^(3/2)*b*c*abs(d) - sqrt(a)*b^2*abs(d))*log(48*d^2*abs(a))/(a*d^3))*sgn(d*x^2 + c)