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

Optimal. Leaf size=187 \[ \frac{\left (c+d x^2\right )^2 \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{4 a^2 d^2}-\frac{(4 a c+7 b) \left (c+d x^2\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{8 a^3 d^2}-\frac{b (a c+b)}{a^3 d^2 \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}+\frac{3 b (4 a c+5 b) \tanh ^{-1}\left (\frac{\sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{\sqrt{a}}\right )}{8 a^{7/2} d^2} \]

[Out]

-((b*(b + a*c))/(a^3*d^2*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])) - ((7*b + 4*a*c)*(c + d*x^2)*Sqrt[(b + a*c +
a*d*x^2)/(c + d*x^2)])/(8*a^3*d^2) + ((c + d*x^2)^2*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(4*a^2*d^2) + (3*b*
(5*b + 4*a*c)*ArcTanh[Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]/Sqrt[a]])/(8*a^(7/2)*d^2)

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Rubi [A]  time = 0.573866, antiderivative size = 242, 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} \[ \frac{(4 a c+5 b) \left (c+d x^2\right ) \left (a \left (c+d x^2\right )+b\right )}{4 a^2 b d^2 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{3 (4 a c+5 b) \left (a \left (c+d x^2\right )+b\right )}{8 a^3 d^2 \sqrt{a+\frac{b}{c+d x^2}}}+\frac{3 b (4 a c+5 b) \sqrt{a \left (c+d x^2\right )+b} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c+d x^2}}{\sqrt{a \left (c+d x^2\right )+b}}\right )}{8 a^{7/2} d^2 \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}-\frac{(a c+b) \left (c+d x^2\right )^2}{a b d^2 \sqrt{a+\frac{b}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b/(c + d*x^2))^(3/2),x]

[Out]

-(((b + a*c)*(c + d*x^2)^2)/(a*b*d^2*Sqrt[a + b/(c + d*x^2)])) - (3*(5*b + 4*a*c)*(b + a*(c + d*x^2)))/(8*a^3*
d^2*Sqrt[a + b/(c + d*x^2)]) + ((5*b + 4*a*c)*(c + d*x^2)*(b + a*(c + d*x^2)))/(4*a^2*b*d^2*Sqrt[a + b/(c + d*
x^2)]) + (3*b*(5*b + 4*a*c)*Sqrt[b + a*(c + d*x^2)]*ArcTanh[(Sqrt[a]*Sqrt[c + d*x^2])/Sqrt[b + a*(c + d*x^2)]]
)/(8*a^(7/2)*d^2*Sqrt[c + d*x^2]*Sqrt[a + b/(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 \frac{x^3}{\left (a+\frac{b}{c+d x^2}\right )^{3/2}} \, dx &=\frac{\sqrt{b+a \left (c+d x^2\right )} \int \frac{x^3 \left (c+d x^2\right )^{3/2}}{\left (b+a \left (c+d x^2\right )\right )^{3/2}} \, dx}{\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}\\ &=\frac{\sqrt{b+a \left (c+d x^2\right )} \int \frac{x^3 \left (c+d x^2\right )^{3/2}}{\left (b+a c+a d x^2\right )^{3/2}} \, dx}{\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}\\ &=\frac{\sqrt{b+a \left (c+d x^2\right )} \operatorname{Subst}\left (\int \frac{x (c+d x)^{3/2}}{(b+a c+a d x)^{3/2}} \, dx,x,x^2\right )}{2 \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}\\ &=-\frac{(b+a c) \left (c+d x^2\right )^2}{a b d^2 \sqrt{a+\frac{b}{c+d x^2}}}+\frac{\left ((5 b+4 a c) \sqrt{b+a \left (c+d x^2\right )}\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{\sqrt{b+a c+a d x}} \, dx,x,x^2\right )}{2 a b d \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}\\ &=-\frac{(b+a c) \left (c+d x^2\right )^2}{a b d^2 \sqrt{a+\frac{b}{c+d x^2}}}+\frac{(5 b+4 a c) \left (c+d x^2\right ) \left (b+a \left (c+d x^2\right )\right )}{4 a^2 b d^2 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{\left (3 (5 b+4 a c) \sqrt{b+a \left (c+d x^2\right )}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{\sqrt{b+a c+a d x}} \, dx,x,x^2\right )}{8 a^2 d \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}\\ &=-\frac{(b+a c) \left (c+d x^2\right )^2}{a b d^2 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{3 (5 b+4 a c) \left (b+a \left (c+d x^2\right )\right )}{8 a^3 d^2 \sqrt{a+\frac{b}{c+d x^2}}}+\frac{(5 b+4 a c) \left (c+d x^2\right ) \left (b+a \left (c+d x^2\right )\right )}{4 a^2 b d^2 \sqrt{a+\frac{b}{c+d x^2}}}+\frac{\left (3 b (5 b+4 a c) \sqrt{b+a \left (c+d x^2\right )}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x} \sqrt{b+a c+a d x}} \, dx,x,x^2\right )}{16 a^3 d \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}\\ &=-\frac{(b+a c) \left (c+d x^2\right )^2}{a b d^2 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{3 (5 b+4 a c) \left (b+a \left (c+d x^2\right )\right )}{8 a^3 d^2 \sqrt{a+\frac{b}{c+d x^2}}}+\frac{(5 b+4 a c) \left (c+d x^2\right ) \left (b+a \left (c+d x^2\right )\right )}{4 a^2 b d^2 \sqrt{a+\frac{b}{c+d x^2}}}+\frac{\left (3 b (5 b+4 a c) \sqrt{b+a \left (c+d x^2\right )}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+a x^2}} \, dx,x,\sqrt{c+d x^2}\right )}{8 a^3 d^2 \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}\\ &=-\frac{(b+a c) \left (c+d x^2\right )^2}{a b d^2 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{3 (5 b+4 a c) \left (b+a \left (c+d x^2\right )\right )}{8 a^3 d^2 \sqrt{a+\frac{b}{c+d x^2}}}+\frac{(5 b+4 a c) \left (c+d x^2\right ) \left (b+a \left (c+d x^2\right )\right )}{4 a^2 b d^2 \sqrt{a+\frac{b}{c+d x^2}}}+\frac{\left (3 b (5 b+4 a c) \sqrt{b+a \left (c+d x^2\right )}\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 a^3 d^2 \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}\\ &=-\frac{(b+a c) \left (c+d x^2\right )^2}{a b d^2 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{3 (5 b+4 a c) \left (b+a \left (c+d x^2\right )\right )}{8 a^3 d^2 \sqrt{a+\frac{b}{c+d x^2}}}+\frac{(5 b+4 a c) \left (c+d x^2\right ) \left (b+a \left (c+d x^2\right )\right )}{4 a^2 b d^2 \sqrt{a+\frac{b}{c+d x^2}}}+\frac{3 b (5 b+4 a c) \sqrt{b+a \left (c+d x^2\right )} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c+d x^2}}{\sqrt{b+a \left (c+d x^2\right )}}\right )}{8 a^{7/2} d^2 \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}\\ \end{align*}

Mathematica [A]  time = 0.275619, size = 133, normalized size = 0.71 \[ \frac{3 b (4 a c+5 b) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}} \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{c+d x^2}}}{\sqrt{a}}\right )-\sqrt{a} \left (2 a^2 \left (c^2-d^2 x^4\right )+a b \left (17 c+5 d x^2\right )+15 b^2\right )}{8 a^{7/2} d^2 \sqrt{a+\frac{b}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[a]*(15*b^2 + a*b*(17*c + 5*d*x^2) + 2*a^2*(c^2 - d^2*x^4))) + 3*b*(5*b + 4*a*c)*Sqrt[(b + a*c + a*d*x^
2)/(c + d*x^2)]*ArcTanh[Sqrt[a + b/(c + d*x^2)]/Sqrt[a]])/(8*a^(7/2)*d^2*Sqrt[a + b/(c + d*x^2)])

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Maple [B]  time = 0.014, size = 783, normalized size = 4.2 \begin{align*}{\frac{d{x}^{2}+c}{16\,{a}^{3}{d}^{2} \left ( ad{x}^{2}+ac+b \right ) }\sqrt{{\frac{ad{x}^{2}+ac+b}{d{x}^{2}+c}}} \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}^{2}{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}{a}^{2}bc{d}^{2}+15\,\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}a{b}^{2}{d}^{2}-10\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}{x}^{2}bad\sqrt{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 ){a}^{2}b{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}^{2}{c}^{2}+27\,\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}cad-18\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}cba\sqrt{a{d}^{2}}+15\,\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}^{3}d-16\,\sqrt{a{d}^{2}}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( ad{x}^{2}+ac+b \right ) }abc-14\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}{b}^{2}\sqrt{a{d}^{2}}-16\,\sqrt{a{d}^{2}}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( ad{x}^{2}+ac+b \right ) }{b}^{2} \right ){\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( ad{x}^{2}+ac+b \right ) }}}{\frac{1}{\sqrt{a{d}^{2}}}}} \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*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)*(d*x^2+c)/a^3/d^2*(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^2*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^2*b*c*d^2+15*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^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)*x^2*b*a*d*(a*d^2)^(1/2)+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))*a^2*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^2*c^2+27*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*a*d-18*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*c*b*a*(a*d^2)^(1/2)+15*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^3*d-16*(a
*d^2)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*a*b*c-14*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*b^2*(a*
d^2)^(1/2)-16*(a*d^2)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*b^2)/((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)/(a*d^2)^(
1/2)/(a*d*x^2+a*c+b)

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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 = 2.6916, size = 1172, normalized size = 6.27 \begin{align*} \left [\frac{3 \,{\left (4 \, a^{2} b c^{2} + 9 \, a b^{2} c +{\left (4 \, a^{2} b c + 5 \, a b^{2}\right )} d x^{2} + 5 \, b^{3}\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^{3} d^{3} x^{6} +{\left (2 \, a^{3} c - 5 \, a^{2} b\right )} d^{2} x^{4} - 2 \, a^{3} c^{3} - 17 \, a^{2} b c^{2} - 15 \, a b^{2} c -{\left (2 \, a^{3} c^{2} + 22 \, a^{2} b c + 15 \, a b^{2}\right )} d x^{2}\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{32 \,{\left (a^{5} d^{3} x^{2} +{\left (a^{5} c + a^{4} b\right )} d^{2}\right )}}, -\frac{3 \,{\left (4 \, a^{2} b c^{2} + 9 \, a b^{2} c +{\left (4 \, a^{2} b c + 5 \, a b^{2}\right )} d x^{2} + 5 \, b^{3}\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^{3} d^{3} x^{6} +{\left (2 \, a^{3} c - 5 \, a^{2} b\right )} d^{2} x^{4} - 2 \, a^{3} c^{3} - 17 \, a^{2} b c^{2} - 15 \, a b^{2} c -{\left (2 \, a^{3} c^{2} + 22 \, a^{2} b c + 15 \, a b^{2}\right )} d x^{2}\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{16 \,{\left (a^{5} d^{3} x^{2} +{\left (a^{5} c + a^{4} b\right )} d^{2}\right )}}\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^2*b*c^2 + 9*a*b^2*c + (4*a^2*b*c + 5*a*b^2)*d*x^2 + 5*b^3)*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^3*d^3*x^6 + (2*a^3*c - 5*a^2*b)*d^2*x^4 - 2*a^3*c^3 - 17*a^2*b*c^2
- 15*a*b^2*c - (2*a^3*c^2 + 22*a^2*b*c + 15*a*b^2)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^5*d^3*x^2
+ (a^5*c + a^4*b)*d^2), -1/16*(3*(4*a^2*b*c^2 + 9*a*b^2*c + (4*a^2*b*c + 5*a*b^2)*d*x^2 + 5*b^3)*sqrt(-a)*arct
an(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^3*d^3*x^6 + (2*a^3*c - 5*a^2*b)*d^2*x^4 - 2*a^3*c^3 - 17*a^2*b*c^2 - 15*a*b^2*c - (2*a^3*c^2 + 22*a^2*b*c
+ 15*a*b^2)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^5*d^3*x^2 + (a^5*c + a^4*b)*d^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{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 = 8.96026, size = 767, normalized size = 4.1 \begin{align*} \frac{1}{8} \, \sqrt{a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}{\left (\frac{2 \, x^{2}}{a^{2} d \mathrm{sgn}\left (d x^{2} + c\right )} - \frac{2 \, a^{6} c d^{2} + 7 \, a^{5} b d^{2}}{a^{8} d^{4} \mathrm{sgn}\left (d x^{2} + c\right )}\right )} - \frac{{\left (4 \, a^{\frac{3}{2}} b c + 5 \, \sqrt{a} b^{2}\right )} \log \left ({\left | -2 \, a^{\frac{7}{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^{3} 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{5}{2}} c d - 5 \, a^{\frac{5}{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^{2}{\left | d \right |} - 10 \,{\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} b c{\left | d \right |} - 5 \,{\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}} b d - 4 \, a^{\frac{3}{2}} b^{2} c d - 4 \,{\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^{2}{\left | d \right |} - \sqrt{a} b^{3} d \right |}\right )}{16 \, a^{4} d{\left | d \right |} \mathrm{sgn}\left (d x^{2} + c\right )} - \frac{{\left (4 \, a^{\frac{3}{2}} b c{\left | d \right |} + 5 \, \sqrt{a} b^{2}{\left | d \right |}\right )} \log \left (48 \, a^{4} d^{2}{\left | a \right |}{\left | \mathrm{sgn}\left (d x^{2} + c\right ) \right |}\right )}{8 \, a^{4} d^{3} \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]

1/8*sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c)*(2*x^2/(a^2*d*sgn(d*x^2 + c)) - (2*a^6*c*d^2 + 7*a^5
*b*d^2)/(a^8*d^4*sgn(d*x^2 + c))) - 1/16*(4*a^(3/2)*b*c + 5*sqrt(a)*b^2)*log(abs(-2*a^(7/2)*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^3*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^(5/2)*c*d - 5*a^(5/2)*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^2*abs(d) - 10*(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*b*c*abs(d) - 5*(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)*b*d - 4*a^(3/2)*b^2*c*d - 4*(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^2*abs(d) - sqrt(a)*b^3*d))/(a^4*d*abs(d)*sgn(d*x^2 + c)) - 1/8*(4*a^(3/2)*b*c*abs(d) + 5
*sqrt(a)*b^2*abs(d))*log(48*a^4*d^2*abs(a)*abs(sgn(d*x^2 + c)))/(a^4*d^3*sgn(d*x^2 + c))