3.4.32 \(\int x^3 (a+\frac {b}{c+d x^2})^{3/2} \, dx\) [332]

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

[Out]

3/8*b*(-4*a*c+b)*arctanh(((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/a^(1/2))/d^2/a^(1/2)+b*c*((a*d*x^2+a*c+b)/(d*x^2+c)
)^(1/2)/d^2+1/8*(-4*a*c+5*b)*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/d^2+1/4*a*(d*x^2+c)^2*((a*d*x^2+a*c+b
)/(d*x^2+c))^(1/2)/d^2

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Rubi [A]
time = 0.22, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1985, 1981, 1980, 466, 1171, 396, 214} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b/(c + d*x^2))^(3/2),x]

[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)

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 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 466

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1171

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1
)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 1980

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_Symbol] :> With[{q = Denominator[p]}
, Dist[q*e*(b*c - a*d), Subst[Int[x^(q*(p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a
+ b*x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] && IntegerQ[m]

Rule 1981

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Dist[1/n, Sub
st[Int[x^(Simplify[(m + 1)/n] - 1)*(e*((a + b*x)/(c + d*x)))^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n,
p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1985

Int[(u_.)*((a_) + (b_.)/((c_) + (d_.)*(x_)^(n_)))^(p_), x_Symbol] :> Int[u*((b + a*c + a*d*x^n)/(c + d*x^n))^p
, x] /; FreeQ[{a, b, c, d, n, p}, x]

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.18, size = 113, normalized size = 0.66 \begin {gather*} \frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}} \left (13 b c-2 a c^2+5 b d x^2+2 a d^2 x^4\right )}{8 d^2}-\frac {3 b (-b+4 a c) \tanh ^{-1}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{8 \sqrt {a} d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(592\) vs. \(2(154)=308\).
time = 0.10, size = 593, normalized size = 3.45

method	result	size
risch	\(-\frac {\left (-2 a d \,x^{2}+2 a c -5 b \right ) \left (d \,x^{2}+c \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{8 d^{2}}+\frac {\left (-\frac {3 b \ln \left (\frac {a c d +\frac {1}{2} b d +a \,d^{2} x^{2}}{\sqrt {a \,d^{2}}}+\sqrt {c^{2} a +b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}\right ) a c}{4 d \sqrt {a \,d^{2}}}+\frac {3 b^{2} \ln \left (\frac {a c d +\frac {1}{2} b d +a \,d^{2} x^{2}}{\sqrt {a \,d^{2}}}+\sqrt {c^{2} a +b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}\right )}{16 d \sqrt {a \,d^{2}}}+\frac {b c \,x^{2} a}{d \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}}+\frac {b \,c^{2} a}{d^{2} \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}}+\frac {b^{2} c}{d^{2} \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}}\right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}}{a d \,x^{2}+a c +b}\)	\(385\)
default	\(-\frac {\left (-4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}\, a \,d^{2} x^{4}+12 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a b c \,d^{2} x^{2}-3 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) b^{2} d^{2} x^{2}-10 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}\, b d \,x^{2}+12 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a b \,c^{2} d +4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}\, a \,c^{2}-3 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) b^{2} c d -10 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}\, b c -16 \sqrt {a \,d^{2}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, b c \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{16 d^{2} \sqrt {a \,d^{2}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}}\)	\(593\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)*a*d^2*x^4+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*d^2*x^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*d
^2*x^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)*b*d*x^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*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 [A]
time = 0.52, size = 247, normalized size = 1.44 \begin {gather*} \frac {b c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{d^{2}} + \frac {3 \, {\left (4 \, a c - b\right )} b \log \left (-\frac {\sqrt {a} - \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{\sqrt {a} + \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}\right )}{16 \, \sqrt {a} d^{2}} - \frac {{\left (4 \, a b c - 5 \, b^{2}\right )} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}} - {\left (4 \, a^{2} b c - 3 \, a b^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, {\left (a^{2} d^{2} - \frac {2 \, {\left (a d x^{2} + a c + b\right )} a d^{2}}{d x^{2} + c} + \frac {{\left (a d x^{2} + a c + b\right )}^{2} d^{2}}{{\left (d x^{2} + c\right )}^{2}}\right )}} \end {gather*}

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]

b*c*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/d^2 + 3/16*(4*a*c - b)*b*log(-(sqrt(a) - sqrt((a*d*x^2 + a*c + b)/(d
*x^2 + c)))/(sqrt(a) + sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))))/(sqrt(a)*d^2) - 1/8*((4*a*b*c - 5*b^2)*((a*d*x^
2 + a*c + b)/(d*x^2 + c))^(3/2) - (4*a^2*b*c - 3*a*b^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^2*d^2 - 2*(a
*d*x^2 + a*c + b)*a*d^2/(d*x^2 + c) + (a*d*x^2 + a*c + b)^2*d^2/(d*x^2 + c)^2)

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Fricas [A]
time = 0.42, size = 335, normalized size = 1.95 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \left (\frac {a c + a d x^{2} + b}{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*(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] Leaf count of result is larger than twice the leaf count of optimal. 438 vs. \(2 (156) = 312\).
time = 4.03, size = 438, normalized size = 2.55 \begin {gather*} \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 \, a x^{2} \mathrm {sgn}\left (d x^{2} + c\right )}{d} - \frac {2 \, a^{2} c d^{2} \mathrm {sgn}\left (d x^{2} + c\right ) - 5 \, a b d^{2} \mathrm {sgn}\left (d x^{2} + c\right )}{a d^{4}}\right )} + \frac {{\left (4 \, a^{\frac {3}{2}} b c \mathrm {sgn}\left (d x^{2} + c\right ) - \sqrt {a} b^{2} \mathrm {sgn}\left (d x^{2} + c\right )\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 )}{16 \, a d {\left | d \right |}} \end {gather*}

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*a*x^2*sgn(d*x^2 + c)/d - (2*a^2*c*d^2*sgn(d*x^2 +
c) - 5*a*b*d^2*sgn(d*x^2 + c))/(a*d^4)) + 1/16*(4*a^(3/2)*b*c*sgn(d*x^2 + c) - sqrt(a)*b^2*sgn(d*x^2 + c))*lo
g(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*a
bs(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))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,{\left (a+\frac {b}{d\,x^2+c}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b/(c + d*x^2))^(3/2),x)

[Out]

int(x^3*(a + b/(c + d*x^2))^(3/2), x)

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