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

Optimal. Leaf size=130 \[ -\frac {2 \sqrt {a+b x} \left (a^2 d^2+b^2 c^2\right )}{b^2 d \sqrt {c+d x} (b c-a d)^2}-\frac {2 a^2}{b^2 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} d^{3/2}} \]

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Rubi [A]  time = 0.09, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {89, 78, 63, 217, 206} \begin {gather*} -\frac {2 \sqrt {a+b x} \left (a^2 d^2+b^2 c^2\right )}{b^2 d \sqrt {c+d x} (b c-a d)^2}-\frac {2 a^2}{b^2 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.33, size = 143, normalized size = 1.10 \begin {gather*} \frac {2 \sqrt {a+b x} (b c-a d)^{5/2} \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )-2 b \sqrt {d} \left (a^2 d (c+d x)+a b c^2+b^2 c^2 x\right )}{b^2 d^{3/2} \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.19, size = 102, normalized size = 0.78 \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{3/2} d^{3/2}}-\frac {2 \sqrt {a+b x} \left (\frac {a^2 d (c+d x)}{a+b x}+b c^2\right )}{b d \sqrt {c+d x} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.91, size = 710, normalized size = 5.46 \begin {gather*} \left [\frac {{\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{2} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (a b^{2} c^{2} d + a^{2} b c d^{2} + {\left (b^{3} c^{2} d + a^{2} b d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b^{4} c^{3} d^{2} - 2 \, a^{2} b^{3} c^{2} d^{3} + a^{3} b^{2} c d^{4} + {\left (b^{5} c^{2} d^{3} - 2 \, a b^{4} c d^{4} + a^{2} b^{3} d^{5}\right )} x^{2} + {\left (b^{5} c^{3} d^{2} - a b^{4} c^{2} d^{3} - a^{2} b^{3} c d^{4} + a^{3} b^{2} d^{5}\right )} x\right )}}, -\frac {{\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{2} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (a b^{2} c^{2} d + a^{2} b c d^{2} + {\left (b^{3} c^{2} d + a^{2} b d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{a b^{4} c^{3} d^{2} - 2 \, a^{2} b^{3} c^{2} d^{3} + a^{3} b^{2} c d^{4} + {\left (b^{5} c^{2} d^{3} - 2 \, a b^{4} c d^{4} + a^{2} b^{3} d^{5}\right )} x^{2} + {\left (b^{5} c^{3} d^{2} - a b^{4} c^{2} d^{3} - a^{2} b^{3} c d^{4} + a^{3} b^{2} d^{5}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^2 + (b^3*c^3 - a*b^2*
c^2*d - a^2*b*c*d^2 + a^3*d^3)*x)*sqrt(b*d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b
*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(a*b^2*c^2*d + a^2*b*c*d^2 + (b
^3*c^2*d + a^2*b*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*b^4*c^3*d^2 - 2*a^2*b^3*c^2*d^3 + a^3*b^2*c*d^4 + (b^
5*c^2*d^3 - 2*a*b^4*c*d^4 + a^2*b^3*d^5)*x^2 + (b^5*c^3*d^2 - a*b^4*c^2*d^3 - a^2*b^3*c*d^4 + a^3*b^2*d^5)*x),
-((a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^2 + (b^3*c^3 - a*b^2*c^2
*d - a^2*b*c*d^2 + a^3*d^3)*x)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x +
c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) + 2*(a*b^2*c^2*d + a^2*b*c*d^2 + (b^3*c^2*d + a^2*b*d^3)*
x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*b^4*c^3*d^2 - 2*a^2*b^3*c^2*d^3 + a^3*b^2*c*d^4 + (b^5*c^2*d^3 - 2*a*b^4*c*
d^4 + a^2*b^3*d^5)*x^2 + (b^5*c^3*d^2 - a*b^4*c^2*d^3 - a^2*b^3*c*d^4 + a^3*b^2*d^5)*x)]

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giac [A]  time = 2.46, size = 202, normalized size = 1.55 \begin {gather*} -\frac {2 \, \sqrt {b x + a} b^{2} c^{2} {\left | b \right |}}{{\left (b^{4} c^{2} d - 2 \, a b^{3} c d^{2} + a^{2} b^{2} d^{3}\right )} \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} - \frac {4 \, \sqrt {b d} a^{2}}{{\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )} {\left (b c {\left | b \right |} - a d {\left | b \right |}\right )}} - \frac {\log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{\sqrt {b d} d {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.03, size = 654, normalized size = 5.03 \begin {gather*} \frac {a^{2} b \,d^{3} x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-2 a \,b^{2} c \,d^{2} x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+b^{3} c^{2} d \,x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+a^{3} d^{3} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-a^{2} b c \,d^{2} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-a \,b^{2} c^{2} d x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+b^{3} c^{3} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+a^{3} c \,d^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-2 a^{2} b \,c^{2} d \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+a \,b^{2} c^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-2 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2} x -2 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2} x -2 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c d -2 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,c^{2}}{\sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \left (a d -b c \right )^{2} \sqrt {b d}\, \sqrt {b x +a}\, \sqrt {d x +c}\, b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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