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

Optimal. Leaf size=237 \[ -\frac{\sqrt{a+b x} \sqrt{c+d x} \left (4 a b c d-5 (a d+b c)^2\right ) (b c-a d)}{64 b^3 d^3}-\frac{5 (a+b x)^{3/2} (c+d x)^{3/2} (a d+b c)}{24 b^2 d^2}-\frac{(a+b x)^{3/2} \sqrt{c+d x} \left (4 a b c d-5 (a d+b c)^2\right )}{32 b^3 d^2}+\frac{\left (4 a b c d-5 (a d+b c)^2\right ) (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{7/2} d^{7/2}}+\frac{x (a+b x)^{3/2} (c+d x)^{3/2}}{4 b d} \]

[Out]

-((b*c - a*d)*(4*a*b*c*d - 5*(b*c + a*d)^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b^3*d^3) - ((4*a*b*c*d - 5*(b*c +
a*d)^2)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(32*b^3*d^2) - (5*(b*c + a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2))/(24*b^2
*d^2) + (x*(a + b*x)^(3/2)*(c + d*x)^(3/2))/(4*b*d) + ((b*c - a*d)^2*(4*a*b*c*d - 5*(b*c + a*d)^2)*ArcTanh[(Sq
rt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(7/2)*d^(7/2))

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Rubi [A]  time = 0.195934, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {90, 80, 50, 63, 217, 206} \[ -\frac{\sqrt{a+b x} \sqrt{c+d x} \left (4 a b c d-5 (a d+b c)^2\right ) (b c-a d)}{64 b^3 d^3}-\frac{5 (a+b x)^{3/2} (c+d x)^{3/2} (a d+b c)}{24 b^2 d^2}-\frac{(a+b x)^{3/2} \sqrt{c+d x} \left (4 a b c d-5 (a d+b c)^2\right )}{32 b^3 d^2}+\frac{\left (4 a b c d-5 (a d+b c)^2\right ) (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{7/2} d^{7/2}}+\frac{x (a+b x)^{3/2} (c+d x)^{3/2}}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c - a*d)*(4*a*b*c*d - 5*(b*c + a*d)^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b^3*d^3) - ((4*a*b*c*d - 5*(b*c +
a*d)^2)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(32*b^3*d^2) - (5*(b*c + a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2))/(24*b^2
*d^2) + (x*(a + b*x)^(3/2)*(c + d*x)^(3/2))/(4*b*d) + ((b*c - a*d)^2*(4*a*b*c*d - 5*(b*c + a*d)^2)*ArcTanh[(Sq
rt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(7/2)*d^(7/2))

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 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^2 \sqrt{a+b x} \sqrt{c+d x} \, dx &=\frac{x (a+b x)^{3/2} (c+d x)^{3/2}}{4 b d}+\frac{\int \sqrt{a+b x} \sqrt{c+d x} \left (-a c-\frac{5}{2} (b c+a d) x\right ) \, dx}{4 b d}\\ &=-\frac{5 (b c+a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 b^2 d^2}+\frac{x (a+b x)^{3/2} (c+d x)^{3/2}}{4 b d}-\frac{\left (4 a b c d-5 (b c+a d)^2\right ) \int \sqrt{a+b x} \sqrt{c+d x} \, dx}{16 b^2 d^2}\\ &=-\frac{\left (4 a b c d-5 (b c+a d)^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{32 b^3 d^2}-\frac{5 (b c+a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 b^2 d^2}+\frac{x (a+b x)^{3/2} (c+d x)^{3/2}}{4 b d}-\frac{\left ((b c-a d) \left (4 a b c d-5 (b c+a d)^2\right )\right ) \int \frac{\sqrt{a+b x}}{\sqrt{c+d x}} \, dx}{64 b^3 d^2}\\ &=-\frac{(b c-a d) \left (4 a b c d-5 (b c+a d)^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{64 b^3 d^3}-\frac{\left (4 a b c d-5 (b c+a d)^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{32 b^3 d^2}-\frac{5 (b c+a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 b^2 d^2}+\frac{x (a+b x)^{3/2} (c+d x)^{3/2}}{4 b d}+\frac{\left ((b c-a d)^2 \left (4 a b c d-5 (b c+a d)^2\right )\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{128 b^3 d^3}\\ &=-\frac{(b c-a d) \left (4 a b c d-5 (b c+a d)^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{64 b^3 d^3}-\frac{\left (4 a b c d-5 (b c+a d)^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{32 b^3 d^2}-\frac{5 (b c+a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 b^2 d^2}+\frac{x (a+b x)^{3/2} (c+d x)^{3/2}}{4 b d}+\frac{\left ((b c-a d)^2 \left (4 a b c d-5 (b c+a d)^2\right )\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 )}{64 b^4 d^3}\\ &=-\frac{(b c-a d) \left (4 a b c d-5 (b c+a d)^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{64 b^3 d^3}-\frac{\left (4 a b c d-5 (b c+a d)^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{32 b^3 d^2}-\frac{5 (b c+a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 b^2 d^2}+\frac{x (a+b x)^{3/2} (c+d x)^{3/2}}{4 b d}+\frac{\left ((b c-a d)^2 \left (4 a b c d-5 (b c+a d)^2\right )\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 )}{64 b^4 d^3}\\ &=-\frac{(b c-a d) \left (4 a b c d-5 (b c+a d)^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{64 b^3 d^3}-\frac{\left (4 a b c d-5 (b c+a d)^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{32 b^3 d^2}-\frac{5 (b c+a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 b^2 d^2}+\frac{x (a+b x)^{3/2} (c+d x)^{3/2}}{4 b d}+\frac{(b c-a d)^2 \left (4 a b c d-5 (b c+a d)^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{7/2} d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.494562, size = 214, normalized size = 0.9 \[ \frac{b \sqrt{d} \sqrt{a+b x} (c+d x) \left (-a^2 b d^2 (7 c+10 d x)+15 a^3 d^3+a b^2 d \left (-7 c^2+4 c d x+8 d^2 x^2\right )+b^3 \left (-10 c^2 d x+15 c^3+8 c d^2 x^2+48 d^3 x^3\right )\right )-3 (b c-a d)^{5/2} \left (5 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{192 b^4 d^{7/2} \sqrt{c+d x}} \]

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)*(15*a^3*d^3 - a^2*b*d^2*(7*c + 10*d*x) + a*b^2*d*(-7*c^2 + 4*c*d*x + 8*d^2*
x^2) + b^3*(15*c^3 - 10*c^2*d*x + 8*c*d^2*x^2 + 48*d^3*x^3)) - 3*(b*c - a*d)^(5/2)*(5*b^2*c^2 + 6*a*b*c*d + 5*
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]])/(192*b^4*d^(7/2)*Sq
rt[c + d*x])

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Maple [B]  time = 0.013, size = 686, normalized size = 2.9 \begin{align*} -{\frac{1}{384\,{b}^{3}{d}^{3}}\sqrt{bx+a}\sqrt{dx+c} \left ( -96\,{x}^{3}{b}^{3}{d}^{3}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}-16\,{x}^{2}a{b}^{2}{d}^{3}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}-16\,{x}^{2}{b}^{3}c{d}^{2}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+15\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{4}{d}^{4}-12\,\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}bc{d}^{3}-6\,\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}{b}^{2}{c}^{2}{d}^{2}-12\,\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}^{3}{c}^{3}d+15\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){b}^{4}{c}^{4}+20\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}x{a}^{2}b{d}^{3}-8\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}xa{b}^{2}c{d}^{2}+20\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}x{b}^{3}{c}^{2}d-30\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}{a}^{3}{d}^{3}+14\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}{a}^{2}bc{d}^{2}+14\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}a{b}^{2}{c}^{2}d-30\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}{b}^{3}{c}^{3} \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^2*(b*x+a)^(1/2)*(d*x+c)^(1/2),x)

[Out]

-1/384*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(-96*x^3*b^3*d^3*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)-16*x^2*a*b^2*d
^3*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)-16*x^2*b^3*c*d^2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+15
*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^4*d^4-12*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*b*c*d^3-6*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^2*c^2*d^2-12*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^3*c^3*d+15*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^4*c^4+20*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a^2*b*d^3-8*(
b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a*b^2*c*d^2+20*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*b^3*
c^2*d-30*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^3*d^3+14*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^
2*b*c*d^2+14*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a*b^2*c^2*d-30*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^
(1/2)*b^3*c^3)/(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)/b^3/d^3/(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^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 = 3.04908, size = 1192, normalized size = 5.03 \begin{align*} \left [\frac{3 \,{\left (5 \, b^{4} c^{4} - 4 \, a b^{3} c^{3} d - 2 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + 5 \, a^{4} d^{4}\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 (48 \, b^{4} d^{4} x^{3} + 15 \, b^{4} c^{3} d - 7 \, a b^{3} c^{2} d^{2} - 7 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} + 8 \,{\left (b^{4} c d^{3} + a b^{3} d^{4}\right )} x^{2} - 2 \,{\left (5 \, b^{4} c^{2} d^{2} - 2 \, a b^{3} c d^{3} + 5 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{768 \, b^{4} d^{4}}, \frac{3 \,{\left (5 \, b^{4} c^{4} - 4 \, a b^{3} c^{3} d - 2 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + 5 \, a^{4} d^{4}\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 (48 \, b^{4} d^{4} x^{3} + 15 \, b^{4} c^{3} d - 7 \, a b^{3} c^{2} d^{2} - 7 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} + 8 \,{\left (b^{4} c d^{3} + a b^{3} d^{4}\right )} x^{2} - 2 \,{\left (5 \, b^{4} c^{2} d^{2} - 2 \, a b^{3} c d^{3} + 5 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{384 \, b^{4} d^{4}}\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]

[1/768*(3*(5*b^4*c^4 - 4*a*b^3*c^3*d - 2*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + 5*a^4*d^4)*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*(48*b^4*d^4*x^3 + 15*b^4*c^3*d - 7*a*b^3*c^2*d^2 - 7*a^2*b^2*c*d^3 + 15*a^3*b*d^4 + 8*(b^
4*c*d^3 + a*b^3*d^4)*x^2 - 2*(5*b^4*c^2*d^2 - 2*a*b^3*c*d^3 + 5*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(
b^4*d^4), 1/384*(3*(5*b^4*c^4 - 4*a*b^3*c^3*d - 2*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + 5*a^4*d^4)*sqrt(-b*d)*arct
an(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*(48*b^4*d^4*x^3 + 15*b^4*c^3*d - 7*a*b^3*c^2*d^2 - 7*a^2*b^2*c*d^3 + 15*a^3*b*d^4 + 8*(b^4*c*d^3 +
a*b^3*d^4)*x^2 - 2*(5*b^4*c^2*d^2 - 2*a*b^3*c*d^3 + 5*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4*d^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int 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 = 1.35884, size = 386, normalized size = 1.63 \begin{align*} \frac{{\left (\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}{\left (2 \,{\left (b x + a\right )}{\left (4 \,{\left (b x + a\right )}{\left (\frac{6 \,{\left (b x + a\right )}}{b^{2}} + \frac{b^{7} c d^{5} - 17 \, a b^{6} d^{6}}{b^{8} d^{6}}\right )} - \frac{5 \, b^{8} c^{2} d^{4} + 6 \, a b^{7} c d^{5} - 59 \, a^{2} b^{6} d^{6}}{b^{8} d^{6}}\right )} + \frac{3 \,{\left (5 \, b^{9} c^{3} d^{3} + a b^{8} c^{2} d^{4} - a^{2} b^{7} c d^{5} - 5 \, a^{3} b^{6} d^{6}\right )}}{b^{8} d^{6}}\right )} \sqrt{b x + a} + \frac{3 \,{\left (5 \, b^{4} c^{4} - 4 \, a b^{3} c^{3} d - 2 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + 5 \, a^{4} d^{4}\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 d^{3}}\right )}{\left | b \right |}}{192 \, b^{3}} \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]

1/192*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/b^2 + (b^7*c*d^5 - 17*a*b^6*
d^6)/(b^8*d^6)) - (5*b^8*c^2*d^4 + 6*a*b^7*c*d^5 - 59*a^2*b^6*d^6)/(b^8*d^6)) + 3*(5*b^9*c^3*d^3 + a*b^8*c^2*d
^4 - a^2*b^7*c*d^5 - 5*a^3*b^6*d^6)/(b^8*d^6))*sqrt(b*x + a) + 3*(5*b^4*c^4 - 4*a*b^3*c^3*d - 2*a^2*b^2*c^2*d^
2 - 4*a^3*b*c*d^3 + 5*a^4*d^4)*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*d^3))*abs(b)/b^3