3.553 \(\int \frac{\sqrt{a+b x} \sqrt{c+d x}}{x^3} \, dx\)

Optimal. Leaf size=122 \[ \frac{(b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{3/2} c^{3/2}}-\frac{\sqrt{a+b x} (c+d x)^{3/2}}{2 c x^2}-\frac{\sqrt{a+b x} \sqrt{c+d x} (b c-a d)}{4 a c x} \]

[Out]

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

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Rubi [A]  time = 0.0478093, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {94, 93, 208} \[ \frac{(b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{3/2} c^{3/2}}-\frac{\sqrt{a+b x} (c+d x)^{3/2}}{2 c x^2}-\frac{\sqrt{a+b x} \sqrt{c+d x} (b c-a d)}{4 a c x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.066321, size = 99, normalized size = 0.81 \[ \frac{(b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{3/2} c^{3/2}}-\frac{\sqrt{a+b x} \sqrt{c+d x} (2 a c+a d x+b c x)}{4 a c x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.015, size = 305, normalized size = 2.5 \begin{align*}{\frac{1}{8\,ac{x}^{2}}\sqrt{bx+a}\sqrt{dx+c} \left ( \ln \left ({\frac{1}{x} \left ( adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac \right ) } \right ){x}^{2}{a}^{2}{d}^{2}-2\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{2}abcd+\ln \left ({\frac{1}{x} \left ( adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac \right ) } \right ){x}^{2}{b}^{2}{c}^{2}-2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{ac}xad-2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{ac}xbc-4\,\sqrt{d{x}^{2}b+adx+bcx+ac}ac\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{d{x}^{2}b+adx+bcx+ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a/c*(ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x
^2*a^2*d^2-2*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x^2*a*b*c*d+ln((a*d*x+b*c
*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x^2*b^2*c^2-2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(a*c)
^(1/2)*x*a*d-2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(a*c)^(1/2)*x*b*c-4*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a*c*(a*c)^(
1/2))/(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)/x^2/(a*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((b*x+a)^(1/2)*(d*x+c)^(1/2)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.28637, size = 744, normalized size = 6.1 \begin{align*} \left [\frac{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{a c} x^{2} \log \left (\frac{8 \, a^{2} c^{2} +{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} + 4 \,{\left (2 \, a c +{\left (b c + a d\right )} x\right )} \sqrt{a c} \sqrt{b x + a} \sqrt{d x + c} + 8 \,{\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \,{\left (2 \, a^{2} c^{2} +{\left (a b c^{2} + a^{2} c d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{16 \, a^{2} c^{2} x^{2}}, -\frac{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{-a c} x^{2} \arctan \left (\frac{{\left (2 \, a c +{\left (b c + a d\right )} x\right )} \sqrt{-a c} \sqrt{b x + a} \sqrt{d x + c}}{2 \,{\left (a b c d x^{2} + a^{2} c^{2} +{\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) + 2 \,{\left (2 \, a^{2} c^{2} +{\left (a b c^{2} + a^{2} c d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{8 \, a^{2} c^{2} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(a*c)*x^2*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 + 4*
(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(2*a^2*c^2 +
(a*b*c^2 + a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^2*x^2), -1/8*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt
(-a*c)*x^2*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 +
(a*b*c^2 + a^2*c*d)*x)) + 2*(2*a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^2*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + b x} \sqrt{c + d x}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError