3.8.9 \(\int \frac {x^2}{\sqrt {a+b x} \sqrt {c+d x}} \, dx\)

Optimal. Leaf size=127 \[ -\frac {\left (4 a b c d-3 (a d+b c)^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} d^{5/2}}-\frac {3 \sqrt {a+b x} \sqrt {c+d x} (a d+b c)}{4 b^2 d^2}+\frac {x \sqrt {a+b x} \sqrt {c+d x}}{2 b d} \]

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Rubi [A]  time = 0.09, antiderivative size = 127, 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 = {90, 80, 63, 217, 206} \begin {gather*} -\frac {3 \sqrt {a+b x} \sqrt {c+d x} (a d+b c)}{4 b^2 d^2}-\frac {\left (4 a b c d-3 (a d+b c)^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} d^{5/2}}+\frac {x \sqrt {a+b x} \sqrt {c+d x}}{2 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 80

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

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 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])

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

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Mathematica [A]  time = 0.25, size = 142, normalized size = 1.12 \begin {gather*} \frac {\sqrt {b c-a d} \left (3 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \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 )+b \sqrt {d} \sqrt {a+b x} (c+d x) (-3 a d-3 b c+2 b d x)}{4 b^3 d^{5/2} \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.26, size = 203, normalized size = 1.60 \begin {gather*} \frac {\left (3 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{4 b^{5/2} d^{5/2}}-\frac {\sqrt {c+d x} \left (-\frac {5 a^2 b d^2 (c+d x)}{a+b x}+3 a^2 d^3+\frac {3 b^3 c^2 (c+d x)}{a+b x}+\frac {2 a b^2 c d (c+d x)}{a+b x}+2 a b c d^2-5 b^2 c^2 d\right )}{4 b^2 d^2 \sqrt {a+b x} \left (\frac {b (c+d x)}{a+b x}-d\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(Sqrt[c + d*x]*(-5*b^2*c^2*d + 2*a*b*c*d^2 + 3*a^2*d^3 + (3*b^3*c^2*(c + d*x))/(a + b*x) + (2*a*b^2*c*d*(
c + d*x))/(a + b*x) - (5*a^2*b*d^2*(c + d*x))/(a + b*x)))/(b^2*d^2*Sqrt[a + b*x]*(-d + (b*(c + d*x))/(a + b*x)
)^2) + ((3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(4*b^(5/
2)*d^(5/2))

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fricas [A]  time = 1.31, size = 308, normalized size = 2.43 \begin {gather*} \left [\frac {{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\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 (2 \, b^{2} d^{2} x - 3 \, b^{2} c d - 3 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, b^{3} d^{3}}, -\frac {{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\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 (2 \, b^{2} d^{2} x - 3 \, b^{2} c d - 3 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, b^{3} d^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.13, size = 150, normalized size = 1.18 \begin {gather*} \frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )}}{b^{3} d} - \frac {3 \, b^{6} c d + 5 \, a b^{5} d^{2}}{b^{8} d^{3}}\right )} - \frac {{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \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 |}\right )}{\sqrt {b d} b^{2} d^{2}}\right )} b}{4 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 251, normalized size = 1.98 \begin {gather*} \frac {\left (3 a^{2} 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 b c 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 )+3 b^{2} c^{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 )+4 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b d x -6 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a d -6 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b c \right ) \sqrt {b x +a}\, \sqrt {d x +c}}{8 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b*d*x+3*a^2*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))+2*a*b*c*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))
+3*b^2*c^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))-6*(b*d)^(1/2)*((b*x+a)*
(d*x+c))^(1/2)*a*d-6*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/(b*d)^(1/2)/d^2/b^2/
((b*x+a)*(d*x+c))^(1/2)

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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)^(1/2)/(d*x+c)^(1/2),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 [B]  time = 16.38, size = 508, normalized size = 4.00 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {\sqrt {d}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}\right )\,\left (3\,a^2\,d^2+2\,a\,b\,c\,d+3\,b^2\,c^2\right )}{2\,b^{5/2}\,d^{5/2}}-\frac {\frac {\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (\frac {3\,a^2\,b\,d^2}{2}+a\,b^2\,c\,d+\frac {3\,b^3\,c^2}{2}\right )}{d^6\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (\frac {11\,a^2\,d^2}{2}+25\,a\,b\,c\,d+\frac {11\,b^2\,c^2}{2}\right )}{d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7\,\left (\frac {3\,a^2\,d^2}{2}+a\,b\,c\,d+\frac {3\,b^2\,c^2}{2}\right )}{b^2\,d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^7}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5\,\left (\frac {11\,a^2\,d^2}{2}+25\,a\,b\,c\,d+\frac {11\,b^2\,c^2}{2}\right )}{b\,d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}+\frac {\sqrt {a}\,\sqrt {c}\,\left (32\,a\,d+32\,b\,c\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}}{\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}+\frac {b^4}{d^4}-\frac {4\,b^3\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}+\frac {6\,b^2\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}-\frac {4\,b\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}{d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((d^(1/2)*((a + b*x)^(1/2) - a^(1/2)))/(b^(1/2)*((c + d*x)^(1/2) - c^(1/2))))*(3*a^2*d^2 + 3*b^2*c^2 + 2
*a*b*c*d))/(2*b^(5/2)*d^(5/2)) - ((((a + b*x)^(1/2) - a^(1/2))*((3*b^3*c^2)/2 + (3*a^2*b*d^2)/2 + a*b^2*c*d))/
(d^6*((c + d*x)^(1/2) - c^(1/2))) - (((a + b*x)^(1/2) - a^(1/2))^3*((11*a^2*d^2)/2 + (11*b^2*c^2)/2 + 25*a*b*c
*d))/(d^5*((c + d*x)^(1/2) - c^(1/2))^3) + (((a + b*x)^(1/2) - a^(1/2))^7*((3*a^2*d^2)/2 + (3*b^2*c^2)/2 + a*b
*c*d))/(b^2*d^3*((c + d*x)^(1/2) - c^(1/2))^7) - (((a + b*x)^(1/2) - a^(1/2))^5*((11*a^2*d^2)/2 + (11*b^2*c^2)
/2 + 25*a*b*c*d))/(b*d^4*((c + d*x)^(1/2) - c^(1/2))^5) + (a^(1/2)*c^(1/2)*(32*a*d + 32*b*c)*((a + b*x)^(1/2)
- a^(1/2))^4)/(d^4*((c + d*x)^(1/2) - c^(1/2))^4))/(((a + b*x)^(1/2) - a^(1/2))^8/((c + d*x)^(1/2) - c^(1/2))^
8 + b^4/d^4 - (4*b^3*((a + b*x)^(1/2) - a^(1/2))^2)/(d^3*((c + d*x)^(1/2) - c^(1/2))^2) + (6*b^2*((a + b*x)^(1
/2) - a^(1/2))^4)/(d^2*((c + d*x)^(1/2) - c^(1/2))^4) - (4*b*((a + b*x)^(1/2) - a^(1/2))^6)/(d*((c + d*x)^(1/2
) - c^(1/2))^6))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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