3.549 \(\int x \sqrt{a+b x} \sqrt{c+d x} \, dx\)

Optimal. Leaf size=163 \[ \frac{1}{8} \sqrt{a+b x} \sqrt{c+d x} \left (\frac{a^2}{b^2}-\frac{c^2}{d^2}\right )+\frac{(a d+b c) (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{5/2} d^{5/2}}-\frac{(a+b x)^{3/2} \sqrt{c+d x} (a d+b c)}{4 b^2 d}+\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{3 b d} \]

[Out]

((a^2/b^2 - c^2/d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/8 - ((b*c + a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(4*b^2*d) +
((a + b*x)^(3/2)*(c + d*x)^(3/2))/(3*b*d) + ((b*c - a*d)^2*(b*c + a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b
]*Sqrt[c + d*x])])/(8*b^(5/2)*d^(5/2))

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Rubi [A]  time = 0.0905002, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {80, 50, 63, 217, 206} \[ \frac{1}{8} \sqrt{a+b x} \sqrt{c+d x} \left (\frac{a^2}{b^2}-\frac{c^2}{d^2}\right )+\frac{(a d+b c) (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{5/2} d^{5/2}}-\frac{(a+b x)^{3/2} \sqrt{c+d x} (a d+b c)}{4 b^2 d}+\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a + b*x]*Sqrt[c + d*x],x]

[Out]

((a^2/b^2 - c^2/d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/8 - ((b*c + a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(4*b^2*d) +
((a + b*x)^(3/2)*(c + d*x)^(3/2))/(3*b*d) + ((b*c - a*d)^2*(b*c + a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b
]*Sqrt[c + d*x])])/(8*b^(5/2)*d^(5/2))

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 x \sqrt{a+b x} \sqrt{c+d x} \, dx &=\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{3 b d}-\frac{(b c+a d) \int \sqrt{a+b x} \sqrt{c+d x} \, dx}{2 b d}\\ &=-\frac{(b c+a d) (a+b x)^{3/2} \sqrt{c+d x}}{4 b^2 d}+\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{3 b d}-\frac{\left (c^2-\frac{a^2 d^2}{b^2}\right ) \int \frac{\sqrt{a+b x}}{\sqrt{c+d x}} \, dx}{8 d}\\ &=-\frac{\left (c^2-\frac{a^2 d^2}{b^2}\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 d^2}-\frac{(b c+a d) (a+b x)^{3/2} \sqrt{c+d x}}{4 b^2 d}+\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{3 b d}+\frac{\left ((b c-a d)^2 (b c+a d)\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{16 b^2 d^2}\\ &=-\frac{\left (c^2-\frac{a^2 d^2}{b^2}\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 d^2}-\frac{(b c+a d) (a+b x)^{3/2} \sqrt{c+d x}}{4 b^2 d}+\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{3 b d}+\frac{\left ((b c-a d)^2 (b c+a d)\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 )}{8 b^3 d^2}\\ &=-\frac{\left (c^2-\frac{a^2 d^2}{b^2}\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 d^2}-\frac{(b c+a d) (a+b x)^{3/2} \sqrt{c+d x}}{4 b^2 d}+\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{3 b d}+\frac{\left ((b c-a d)^2 (b c+a d)\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 )}{8 b^3 d^2}\\ &=-\frac{\left (c^2-\frac{a^2 d^2}{b^2}\right ) \sqrt{a+b x} \sqrt{c+d x}}{8 d^2}-\frac{(b c+a d) (a+b x)^{3/2} \sqrt{c+d x}}{4 b^2 d}+\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{3 b d}+\frac{(b c-a d)^2 (b c+a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{5/2} d^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.471125, size = 156, normalized size = 0.96 \[ \frac{3 (b c-a d)^{5/2} (a d+b c) \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )-b \sqrt{d} \sqrt{a+b x} (c+d x) \left (3 a^2 d^2-2 a b d (c+d x)+b^2 \left (3 c^2-2 c d x-8 d^2 x^2\right )\right )}{24 b^3 d^{5/2} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[a + b*x]*Sqrt[c + d*x],x]

[Out]

(-(b*Sqrt[d]*Sqrt[a + b*x]*(c + d*x)*(3*a^2*d^2 - 2*a*b*d*(c + d*x) + b^2*(3*c^2 - 2*c*d*x - 8*d^2*x^2))) + 3*
(b*c - a*d)^(5/2)*(b*c + a*d)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]]
)/(24*b^3*d^(5/2)*Sqrt[c + d*x])

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Maple [B]  time = 0.011, size = 472, normalized size = 2.9 \begin{align*}{\frac{1}{48\,{b}^{2}{d}^{2}}\sqrt{bx+a}\sqrt{dx+c} \left ( 16\,{x}^{2}{b}^{2}{d}^{2}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{3}{d}^{3}-3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}bc{d}^{2}-3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) a{b}^{2}{c}^{2}d+3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){b}^{3}{c}^{3}+4\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}xab{d}^{2}+4\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}x{b}^{2}cd-6\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}{a}^{2}{d}^{2}+4\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}abcd-6\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}{b}^{2}{c}^{2} \right ){\frac{1}{\sqrt{d{x}^{2}b+adx+bcx+ac}}}{\frac{1}{\sqrt{bd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(16*x^2*b^2*d^2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+3*ln(1/2*(2*b*d*x
+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^3*d^3-3*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*
x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*b*c*d^2-3*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)
^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b^2*c^2*d+3*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)
^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^3*c^3+4*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*x*a*b*d^2+4*(b*d*x^2+a*d*x+
b*c*x+a*c)^(1/2)*(b*d)^(1/2)*x*b^2*c*d-6*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*a^2*d^2+4*(b*d*x^2+a*d*x+
b*c*x+a*c)^(1/2)*(b*d)^(1/2)*a*b*c*d-6*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*b^2*c^2)/(b*d*x^2+a*d*x+b*c
*x+a*c)^(1/2)/b^2/d^2/(b*d)^(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*(b*x+a)^(1/2)*(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.92442, size = 902, normalized size = 5.53 \begin{align*} \left [\frac{3 \,{\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\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 (8 \, b^{3} d^{3} x^{2} - 3 \, b^{3} c^{2} d + 2 \, a b^{2} c d^{2} - 3 \, a^{2} b d^{3} + 2 \,{\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{96 \, b^{3} d^{3}}, -\frac{3 \,{\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\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 (8 \, b^{3} d^{3} x^{2} - 3 \, b^{3} c^{2} d + 2 \, a b^{2} c d^{2} - 3 \, a^{2} b d^{3} + 2 \,{\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{48 \, b^{3} d^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(3*(b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*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*(8*b^3*d
^3*x^2 - 3*b^3*c^2*d + 2*a*b^2*c*d^2 - 3*a^2*b*d^3 + 2*(b^3*c*d^2 + a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))
/(b^3*d^3), -1/48*(3*(b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*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*(8*b^3*d^3*x^2
- 3*b^3*c^2*d + 2*a*b^2*c*d^2 - 3*a^2*b*d^3 + 2*(b^3*c*d^2 + a*b^2*d^3)*x)*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 x \sqrt{a + b x} \sqrt{c + d x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.23676, size = 259, normalized size = 1.59 \begin{align*} \frac{{\left (\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \sqrt{b x + a}{\left (2 \,{\left (b x + a\right )}{\left (\frac{4 \,{\left (b x + a\right )}}{b^{6} d^{2}} + \frac{b c d^{3} - 7 \, a d^{4}}{b^{6} d^{6}}\right )} - \frac{3 \,{\left (b^{2} c^{2} d^{2} - a^{2} d^{4}\right )}}{b^{6} d^{6}}\right )} - \frac{3 \,{\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\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^{5} d^{4}}\right )}{\left | b \right |}}{1920 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/(b^6*d^2) + (b*c*d^3 - 7*a
*d^4)/(b^6*d^6)) - 3*(b^2*c^2*d^2 - a^2*d^4)/(b^6*d^6)) - 3*(b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*lo
g(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^5*d^4))*abs(b)/b^4