3.4.59 \(\int \frac {1}{x^3 (a+\frac {b}{c+d x^2})^{3/2}} \, dx\) [359]

Optimal. Leaf size=146 \[ \frac {3 b d}{2 (b+a c)^2 \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}-\frac {c+d x^2}{2 (b+a c) x^2 \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}-\frac {3 b \sqrt {c} d \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {b+a c}}\right )}{2 (b+a c)^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1985, 1981, 1980, 296, 331, 214} \begin {gather*} \frac {3 b d}{2 (a c+b)^2 \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}-\frac {c+d x^2}{2 x^2 (a c+b) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}-\frac {3 b \sqrt {c} d \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a c+b}}\right )}{2 (a c+b)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b/(c + d*x^2))^(3/2)),x]

[Out]

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

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 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 \frac {1}{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 {\left (c+d x^2\right )^{3/2}}{x^3 \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 {\left (c+d x^2\right )^{3/2}}{x^3 \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 )} \text {Subst}\left (\int \frac {(c+d x)^{3/2}}{x^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 {c+d x^2}{2 (b+a c) x^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (3 b d \sqrt {b+a \left (c+d x^2\right )}\right ) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{x (b+a c+a d x)^{3/2}} \, dx,x,x^2\right )}{4 (b+a c) \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=\frac {3 b d}{2 (b+a c)^2 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {c+d x^2}{2 (b+a c) x^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (3 b c d \sqrt {b+a \left (c+d x^2\right )}\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x} \sqrt {b+a c+a d x}} \, dx,x,x^2\right )}{4 (b+a c)^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=\frac {3 b d}{2 (b+a c)^2 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {c+d x^2}{2 (b+a c) x^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (3 b c d \sqrt {b+a \left (c+d x^2\right )}\right ) \text {Subst}\left (\int \frac {1}{-c-(-b-a c) x^2} \, dx,x,\frac {\sqrt {c+d x^2}}{\sqrt {b+a \left (c+d x^2\right )}}\right )}{2 (b+a c)^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=\frac {3 b d}{2 (b+a c)^2 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {c+d x^2}{2 (b+a c) x^2 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {3 b \sqrt {c} d \sqrt {b+a \left (c+d x^2\right )} \tanh ^{-1}\left (\frac {\sqrt {b+a c} \sqrt {c+d x^2}}{\sqrt {c} \sqrt {b+a \left (c+d x^2\right )}}\right )}{2 (b+a c)^{5/2} \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 148, normalized size = 1.01 \begin {gather*} -\frac {\left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}} \left (b \left (c-2 d x^2\right )+a c \left (c+d x^2\right )\right )}{2 (b+a c)^2 x^2 \left (b+a \left (c+d x^2\right )\right )}+\frac {3 b \sqrt {c} d \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {-b-a c}}\right )}{2 (-b-a c)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b/(c + d*x^2))^(3/2)),x]

[Out]

-1/2*((c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(b*(c - 2*d*x^2) + a*c*(c + d*x^2)))/((b + a*c)^2*x^2*
(b + a*(c + d*x^2))) + (3*b*Sqrt[c]*d*ArcTan[(Sqrt[c]*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/Sqrt[-b - a*c]])/
(2*(-b - a*c)^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1087\) vs. \(2(129)=258\).
time = 0.11, size = 1088, normalized size = 7.45

method	result	size
risch	\(-\frac {c \left (a d \,x^{2}+a c +b \right )}{2 \left (a c +b \right )^{2} x^{2} \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}+\frac {\left (-\frac {3 b d c \ln \left (\frac {2 c^{2} a +2 b c +\left (2 a c d +b d \right ) x^{2}+2 \sqrt {c^{2} a +b c}\, \sqrt {c^{2} a +b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}}{x^{2}}\right )}{4 \left (a c +b \right )^{2} \sqrt {c^{2} a +b c}}+\frac {b \,d^{2} x^{2}}{\left (a c +b \right )^{2} \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}}+\frac {b d c}{\left (a c +b \right )^{2} \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}}\right ) \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}}{\sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\)	\(293\)
default	\(-\frac {\sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {c^{2} a +b c}\, a^{2} d^{3} x^{6}+3 \ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 c^{2} a +2 \sqrt {c^{2} a +b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}+2 b c}{x^{2}}\right ) a^{2} b \,c^{2} d^{2} x^{4}-6 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {c^{2} a +b c}\, a^{2} c \,d^{2} x^{4}+3 \ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 c^{2} a +2 \sqrt {c^{2} a +b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}+2 b c}{x^{2}}\right ) a \,b^{2} c \,d^{2} x^{4}-4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {c^{2} a +b c}\, a b \,d^{2} x^{4}+3 \ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 c^{2} a +2 \sqrt {c^{2} a +b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}+2 b c}{x^{2}}\right ) a^{2} b \,c^{3} d \,x^{2}-4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {c^{2} a +b c}\, a^{2} c^{2} d \,x^{2}+6 \ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 c^{2} a +2 \sqrt {c^{2} a +b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}+2 b c}{x^{2}}\right ) a \,b^{2} c^{2} d \,x^{2}+2 \left (a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c \right )^{\frac {3}{2}} \sqrt {c^{2} a +b c}\, a d \,x^{2}-6 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {c^{2} a +b c}\, a b c d \,x^{2}-4 \sqrt {c^{2} a +b c}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, a b c d \,x^{2}+3 \ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 c^{2} a +2 \sqrt {c^{2} a +b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}+2 b c}{x^{2}}\right ) b^{3} c d \,x^{2}-2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {c^{2} a +b c}\, b^{2} d \,x^{2}-4 \sqrt {c^{2} a +b c}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, b^{2} d \,x^{2}+2 \left (a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c \right )^{\frac {3}{2}} \sqrt {c^{2} a +b c}\, a c +2 \left (a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c \right )^{\frac {3}{2}} \sqrt {c^{2} a +b c}\, b \right )}{4 \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \left (a c +b \right )^{3} x^{2} \sqrt {c^{2} a +b c}\, \left (a d \,x^{2}+a c +b \right )}\)	\(1088\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(a+b/(d*x^2+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/4*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)*(d*x^2+c)*(-2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*
c)^(1/2)*a^2*d^3*x^6+3*ln((2*a*c*d*x^2+b*d*x^2+2*c^2*a+2*(a*c^2+b*c)^(1/2)*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^
2+b*c)^(1/2)+2*b*c)/x^2)*a^2*b*c^2*d^2*x^4-6*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(1/2)
*a^2*c*d^2*x^4+3*ln((2*a*c*d*x^2+b*d*x^2+2*c^2*a+2*(a*c^2+b*c)^(1/2)*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)
^(1/2)+2*b*c)/x^2)*a*b^2*c*d^2*x^4-4*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(1/2)*a*b*d^2
*x^4+3*ln((2*a*c*d*x^2+b*d*x^2+2*c^2*a+2*(a*c^2+b*c)^(1/2)*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)+2*b
*c)/x^2)*a^2*b*c^3*d*x^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*c^2+b*c)^(1/2)*a^2*c^2*d*x^2+6*l
n((2*a*c*d*x^2+b*d*x^2+2*c^2*a+2*(a*c^2+b*c)^(1/2)*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)+2*b*c)/x^2)
*a*b^2*c^2*d*x^2+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(3/2)*(a*c^2+b*c)^(1/2)*a*d*x^2-6*(a*d^2*x^4+2*a*
c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(1/2)*a*b*c*d*x^2-4*(a*c^2+b*c)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))
^(1/2)*a*b*c*d*x^2+3*ln((2*a*c*d*x^2+b*d*x^2+2*c^2*a+2*(a*c^2+b*c)^(1/2)*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+
b*c)^(1/2)+2*b*c)/x^2)*b^3*c*d*x^2-2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(1/2)*b^2*d*x
^2-4*(a*c^2+b*c)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*b^2*d*x^2+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)
^(3/2)*(a*c^2+b*c)^(1/2)*a*c+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(3/2)*(a*c^2+b*c)^(1/2)*b)/((d*x^2+c)
*(a*d*x^2+a*c+b))^(1/2)/(a*c+b)^3/x^2/(a*c^2+b*c)^(1/2)/(a*d*x^2+a*c+b)

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Maxima [A]
time = 0.52, size = 247, normalized size = 1.69 \begin {gather*} \frac {3 \, b c d \log \left (\frac {c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} - \sqrt {{\left (a c + b\right )} c}}{c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} + \sqrt {{\left (a c + b\right )} c}}\right )}{4 \, {\left (a^{2} c^{2} + 2 \, a b c + b^{2}\right )} \sqrt {{\left (a c + b\right )} c}} + \frac {\frac {3 \, {\left (a d x^{2} + a c + b\right )} b c d}{d x^{2} + c} - 2 \, {\left (a b c + b^{2}\right )} d}{2 \, {\left ({\left (a^{2} c^{3} + 2 \, a b c^{2} + b^{2} c\right )} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}} - {\left (a^{3} c^{3} + 3 \, a^{2} b c^{2} + 3 \, a b^{2} c + b^{3}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b/(d*x^2+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.48, size = 599, normalized size = 4.10 \begin {gather*} \left [\frac {3 \, {\left (a b d^{2} x^{4} + {\left (a b c + b^{2}\right )} d x^{2}\right )} \sqrt {\frac {c}{a c + b}} \log \left (\frac {{\left (8 \, a^{2} c^{2} + 8 \, a b c + b^{2}\right )} d^{2} x^{4} + 8 \, a^{2} c^{4} + 16 \, a b c^{3} + 8 \, b^{2} c^{2} + 8 \, {\left (2 \, a^{2} c^{3} + 3 \, a b c^{2} + b^{2} c\right )} d x^{2} - 4 \, {\left ({\left (2 \, a^{2} c^{2} + 3 \, a b c + b^{2}\right )} d^{2} x^{4} + 2 \, a^{2} c^{4} + 4 \, a b c^{3} + 2 \, b^{2} c^{2} + {\left (4 \, a^{2} c^{3} + 7 \, a b c^{2} + 3 \, b^{2} c\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} \sqrt {\frac {c}{a c + b}}}{x^{4}}\right ) - 4 \, {\left ({\left (a c - 2 \, b\right )} d^{2} x^{4} + a c^{3} + {\left (2 \, a c^{2} - b c\right )} d x^{2} + b c^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, {\left ({\left (a^{3} c^{2} + 2 \, a^{2} b c + a b^{2}\right )} d x^{4} + {\left (a^{3} c^{3} + 3 \, a^{2} b c^{2} + 3 \, a b^{2} c + b^{3}\right )} x^{2}\right )}}, \frac {3 \, {\left (a b d^{2} x^{4} + {\left (a b c + b^{2}\right )} d x^{2}\right )} \sqrt {-\frac {c}{a c + b}} \arctan \left (\frac {{\left ({\left (2 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + 2 \, b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} \sqrt {-\frac {c}{a c + b}}}{2 \, {\left (a c d x^{2} + a c^{2} + b c\right )}}\right ) - 2 \, {\left ({\left (a c - 2 \, b\right )} d^{2} x^{4} + a c^{3} + {\left (2 \, a c^{2} - b c\right )} d x^{2} + b c^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{4 \, {\left ({\left (a^{3} c^{2} + 2 \, a^{2} b c + a b^{2}\right )} d x^{4} + {\left (a^{3} c^{3} + 3 \, a^{2} b c^{2} + 3 \, a b^{2} c + b^{3}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b/(d*x^2+c))^(3/2),x, algorithm="fricas")

[Out]

[1/8*(3*(a*b*d^2*x^4 + (a*b*c + b^2)*d*x^2)*sqrt(c/(a*c + b))*log(((8*a^2*c^2 + 8*a*b*c + b^2)*d^2*x^4 + 8*a^2
*c^4 + 16*a*b*c^3 + 8*b^2*c^2 + 8*(2*a^2*c^3 + 3*a*b*c^2 + b^2*c)*d*x^2 - 4*((2*a^2*c^2 + 3*a*b*c + b^2)*d^2*x
^4 + 2*a^2*c^4 + 4*a*b*c^3 + 2*b^2*c^2 + (4*a^2*c^3 + 7*a*b*c^2 + 3*b^2*c)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*
x^2 + c))*sqrt(c/(a*c + b)))/x^4) - 4*((a*c - 2*b)*d^2*x^4 + a*c^3 + (2*a*c^2 - b*c)*d*x^2 + b*c^2)*sqrt((a*d*
x^2 + a*c + b)/(d*x^2 + c)))/((a^3*c^2 + 2*a^2*b*c + a*b^2)*d*x^4 + (a^3*c^3 + 3*a^2*b*c^2 + 3*a*b^2*c + b^3)*
x^2), 1/4*(3*(a*b*d^2*x^4 + (a*b*c + b^2)*d*x^2)*sqrt(-c/(a*c + b))*arctan(1/2*((2*a*c + b)*d*x^2 + 2*a*c^2 +
2*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))*sqrt(-c/(a*c + b))/(a*c*d*x^2 + a*c^2 + b*c)) - 2*((a*c - 2*b)*d^
2*x^4 + a*c^3 + (2*a*c^2 - b*c)*d*x^2 + b*c^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/((a^3*c^2 + 2*a^2*b*c +
a*b^2)*d*x^4 + (a^3*c^3 + 3*a^2*b*c^2 + 3*a*b^2*c + b^3)*x^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{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(1/x**3/(a+b/(d*x**2+c))**(3/2),x)

[Out]

Integral(1/(x**3*((a*c + a*d*x**2 + b)/(c + d*x**2))**(3/2)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b/(d*x^2+c))^(3/2),x, algorithm="giac")

[Out]

undef

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

[Out]

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

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