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

Optimal. Leaf size=138 \[ \frac {3 b d \sqrt {a c+b} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a c+b}}\right )}{2 c^{5/2}}-\frac {3 b d \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{2 c^2}-\frac {\left (c+d x^2\right ) \left (\frac {a c+a d x^2+b}{c+d x^2}\right )^{3/2}}{2 c x^2} \]

[Out]

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

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Rubi [A]  time = 0.53, antiderivative size = 170, normalized size of antiderivative = 1.23, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6722, 1975, 446, 94, 93, 208} \[ -\frac {3 b d \sqrt {a+\frac {b}{c+d x^2}}}{2 c^2}+\frac {3 b d \sqrt {a c+b} \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}} \tanh ^{-1}\left (\frac {\sqrt {a c+b} \sqrt {c+d x^2}}{\sqrt {c} \sqrt {a \left (c+d x^2\right )+b}}\right )}{2 c^{5/2} \sqrt {a \left (c+d x^2\right )+b}}-\frac {\sqrt {a+\frac {b}{c+d x^2}} \left (a \left (c+d x^2\right )+b\right )}{2 c x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/(c + d*x^2))^(3/2)/x^3,x]

[Out]

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 208

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

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.58, size = 256, normalized size = 1.86 \[ \frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \left (-2 \sqrt {c (a c+b)} \left (a^2 c \left (c+d x^2\right )^2+a b \left (2 c^2+5 c d x^2+3 d^2 x^4\right )+b^2 \left (c+3 d x^2\right )\right )-6 b d x^2 \log (x) (a c+b) \sqrt {\left (c+d x^2\right ) \left (a \left (c+d x^2\right )+b\right )}+3 b d x^2 (a c+b) \sqrt {\left (c+d x^2\right ) \left (a \left (c+d x^2\right )+b\right )} \log \left (2 \sqrt {c (a c+b)} \sqrt {\left (c+d x^2\right ) \left (a c+a d x^2+b\right )}+2 a c \left (c+d x^2\right )+b \left (2 c+d x^2\right )\right )\right )}{4 c^2 x^2 \sqrt {c (a c+b)} \left (a \left (c+d x^2\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(-2*Sqrt[c*(b + a*c)]*(a^2*c*(c + d*x^2)^2 + b^2*(c + 3*d*x^2) + a*b*(2
*c^2 + 5*c*d*x^2 + 3*d^2*x^4)) - 6*b*(b + a*c)*d*x^2*Sqrt[(c + d*x^2)*(b + a*(c + d*x^2))]*Log[x] + 3*b*(b + a
*c)*d*x^2*Sqrt[(c + d*x^2)*(b + a*(c + d*x^2))]*Log[2*a*c*(c + d*x^2) + b*(2*c + d*x^2) + 2*Sqrt[c*(b + a*c)]*
Sqrt[(c + d*x^2)*(b + a*c + a*d*x^2)]]))/(4*c^2*Sqrt[c*(b + a*c)]*x^2*(b + a*(c + d*x^2)))

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fricas [A]  time = 1.07, size = 404, normalized size = 2.93 \[ \left [\frac {3 \, b d x^{2} \sqrt {\frac {a c + b}{c}} \log \left (\frac {{\left (8 \, a^{2} c^{2} + 8 \, a b c + b^{2}\right )} d^{2} x^{4} + 8 \, a^{2} c^{4} + 16 \, a b c^{3} + 8 \, b^{2} c^{2} + 8 \, {\left (2 \, a^{2} c^{3} + 3 \, a b c^{2} + b^{2} c\right )} d x^{2} + 4 \, {\left ({\left (2 \, a c^{2} + b c\right )} d^{2} x^{4} + 2 \, a c^{4} + 2 \, b c^{3} + {\left (4 \, a c^{3} + 3 \, b c^{2}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} \sqrt {\frac {a c + b}{c}}}{x^{4}}\right ) - 4 \, {\left ({\left (a c + 3 \, b\right )} d x^{2} + a c^{2} + b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, c^{2} x^{2}}, -\frac {3 \, b d x^{2} \sqrt {-\frac {a c + b}{c}} \arctan \left (\frac {{\left ({\left (2 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + 2 \, b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} \sqrt {-\frac {a c + b}{c}}}{2 \, {\left (a^{2} c^{2} + {\left (a^{2} c + a b\right )} d x^{2} + 2 \, a b c + b^{2}\right )}}\right ) + 2 \, {\left ({\left (a c + 3 \, b\right )} d x^{2} + a c^{2} + b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{4 \, c^{2} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [B]  time = 0.07, size = 820, normalized size = 5.94 \[ -\frac {\sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (-3 a b \,c^{2} d^{2} x^{4} \ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 a \,c^{2}+2 b c +2 \sqrt {a \,c^{2}+b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}}{x^{2}}\right )-2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,c^{2}+b c}\, a \,d^{3} x^{6}-3 b^{2} c \,d^{2} x^{4} \ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 a \,c^{2}+2 b c +2 \sqrt {a \,c^{2}+b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}}{x^{2}}\right )-3 a b \,c^{3} d \,x^{2} \ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 a \,c^{2}+2 b c +2 \sqrt {a \,c^{2}+b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}}{x^{2}}\right )-6 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,c^{2}+b c}\, a c \,d^{2} x^{4}-3 b^{2} c^{2} d \,x^{2} \ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 a \,c^{2}+2 b c +2 \sqrt {a \,c^{2}+b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}}{x^{2}}\right )-2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,c^{2}+b c}\, b \,d^{2} x^{4}-4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,c^{2}+b c}\, a \,c^{2} d \,x^{2}+4 \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {a \,c^{2}+b c}\, b c d \,x^{2}-2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,c^{2}+b c}\, b c d \,x^{2}+2 \left (a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c \right )^{\frac {3}{2}} \sqrt {a \,c^{2}+b c}\, d \,x^{2}+2 \left (a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c \right )^{\frac {3}{2}} \sqrt {a \,c^{2}+b c}\, c \right )}{4 \sqrt {a \,c^{2}+b c}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, c^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.41, size = 202, normalized size = 1.46 \[ -\frac {{\left (a b c + b^{2}\right )} d \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a c^{3} + b c^{2} - \frac {{\left (a d x^{2} + a c + b\right )} c^{3}}{d x^{2} + c}\right )}} - \frac {b d \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{c^{2}} - \frac {3 \, {\left (a b c + b^{2}\right )} d \log \left (\frac {c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} - \sqrt {{\left (a c + b\right )} c}}{c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} + \sqrt {{\left (a c + b\right )} c}}\right )}{4 \, \sqrt {{\left (a c + b\right )} c} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/(c + d*x^2))^(3/2)/x^3,x)

[Out]

int((a + b/(c + d*x^2))^(3/2)/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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