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

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

[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))

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Rubi [A]  time = 0.0914174, antiderivative size = 127, normalized size of antiderivative = 1., 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} \[ -\frac{3 \sqrt{a+b x} \sqrt{c+d x} (a d+b c)}{4 b^2 d^2}-\frac{\left (4 a b c d-3 (a d+b c)^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{5/2} d^{5/2}}+\frac{x \sqrt{a+b x} \sqrt{c+d x}}{2 b d} \]

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 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 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 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 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]

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])

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*}

Mathematica [A]  time = 0.232894, size = 142, normalized size = 1.12 \[ \frac{\sqrt{b c-a d} \left (3 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )+b \sqrt{d} \sqrt{a+b x} (c+d x) (-3 a d-3 b c+2 b d x)}{4 b^3 d^{5/2} \sqrt{c+d x}} \]

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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Maple [B]  time = 0.019, size = 251, normalized size = 2. \begin{align*}{\frac{1}{8\,{b}^{2}{d}^{2}} \left ( 3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}{d}^{2}+2\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) abcd+3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){b}^{2}{c}^{2}+4\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}xbd-6\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}ad-6\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }bc \right ) \sqrt{bx+a}\sqrt{dx+c}{\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}}} \end{align*}

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*(3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*d^2+2*ln(1/2*(2*b*d*x+2
*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b*c*d+3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)
*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^2*c^2+4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x*b*d-6*((b*x+a)*(d*x+c))^(1/
2)*(b*d)^(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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [A]  time = 2.15189, size = 722, normalized size = 5.69 \begin{align*} \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{align*}

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

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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Giac [A]  time = 1.31089, size = 203, normalized size = 1.6 \begin{align*} \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{align*}

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)