3.6.48 \(\int x^2 \sqrt {a+b x} \sqrt {c+d x} \, dx\) [548]

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

[Out]

-5/24*(a*d+b*c)*(b*x+a)^(3/2)*(d*x+c)^(3/2)/b^2/d^2+1/4*x*(b*x+a)^(3/2)*(d*x+c)^(3/2)/b/d+1/64*(-a*d+b*c)^2*(4
*a*b*c*d-5*(a*d+b*c)^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(7/2)/d^(7/2)-1/32*(4*a*b*c*d-5
*(a*d+b*c)^2)*(b*x+a)^(3/2)*(d*x+c)^(1/2)/b^3/d^2-1/64*(-a*d+b*c)*(4*a*b*c*d-5*(a*d+b*c)^2)*(b*x+a)^(1/2)*(d*x
+c)^(1/2)/b^3/d^3

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Rubi [A]
time = 0.13, antiderivative size = 237, normalized size of antiderivative = 1.00, 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 = {92, 81, 52, 65, 223, 212} \begin {gather*} \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 {\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 {(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 {5 (a+b x)^{3/2} (c+d x)^{3/2} (a d+b c)}{24 b^2 d^2}+\frac {x (a+b x)^{3/2} (c+d x)^{3/2}}{4 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/64*((b*c - a*d)*(4*a*b*c*d - 5*(b*c + a*d)^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(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[(
Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(7/2)*d^(7/2))

Rule 52

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 65

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 81

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 92

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.47, size = 191, normalized size = 0.81 \begin {gather*} \frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^3 d^3-a^2 b d^2 (7 c+10 d x)+a b^2 d \left (-7 c^2+4 c d x+8 d^2 x^2\right )+b^3 \left (15 c^3-10 c^2 d x+8 c d^2 x^2+48 d^3 x^3\right )\right )}{192 b^3 d^3}-\frac {(b c-a d)^2 \left (5 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{64 b^{7/2} d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*Sqrt[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)))/(192*b^3*d^3) - ((b*c - a*d)^2*(5*b^2*c^2 + 6*a*b*c*d
+ 5*a^2*d^2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(64*b^(7/2)*d^(7/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(573\) vs. \(2(199)=398\).
time = 0.08, size = 574, normalized size = 2.42

method	result	size
default	\(-\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (-96 b^{3} d^{3} x^{3} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}-16 a \,b^{2} d^{3} x^{2} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}-16 b^{3} c \,d^{2} x^{2} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+15 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{4} d^{4}-12 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} b c \,d^{3}-6 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b^{2} c^{2} d^{2}-12 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{3} c^{3} d +15 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{4} c^{4}+20 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{2} b \,d^{3} x -8 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a \,b^{2} c \,d^{2} x +20 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b^{3} c^{2} d x -30 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{3} d^{3}+14 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{2} b c \,d^{2}+14 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a \,b^{2} c^{2} d -30 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b^{3} c^{3}\right )}{384 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b^{3} d^{3} \sqrt {b d}}\)	\(574\)

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,method=_RETURNVERBOSE)

[Out]

-1/384*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(-96*b^3*d^3*x^3*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)-16*a*b^2*d^3*x^2*((d*x
+c)*(b*x+a))^(1/2)*(b*d)^(1/2)-16*b^3*c*d^2*x^2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+15*ln(1/2*(2*b*d*x+2*((d*x
+c)*(b*x+a))^(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*((d*x+c)*(b*x+a))^(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*((d*x+c)*(b*x+a))^(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*((d*x+c)*(b*x+a))^(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*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^4*c^4+20*(b*d)^(1/2)*(
(d*x+c)*(b*x+a))^(1/2)*a^2*b*d^3*x-8*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a*b^2*c*d^2*x+20*(b*d)^(1/2)*((d*x+c)
*(b*x+a))^(1/2)*b^3*c^2*d*x-30*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a^3*d^3+14*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1
/2)*a^2*b*c*d^2+14*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a*b^2*c^2*d-30*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*b^3*
c^3)/((d*x+c)*(b*x+a))^(1/2)/b^3/d^3/(b*d)^(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 detail

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Fricas [A]
time = 1.21, size = 542, normalized size = 2.29 \begin {gather*} \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 {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]

[1/768*(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int 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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 500 vs. \(2 (199) = 398\).
time = 1.65, size = 500, normalized size = 2.11 \begin {gather*} \frac {\frac {8 \, {\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^{2}} + \frac {b^{6} c d^{3} - 13 \, a b^{5} d^{4}}{b^{7} d^{4}}\right )} - \frac {3 \, {\left (b^{7} c^{2} d^{2} + 2 \, a b^{6} c d^{3} - 11 \, a^{2} b^{5} d^{4}\right )}}{b^{7} d^{4}}\right )} - \frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, 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 d^{2}}\right )} a {\left | b \right |}}{b^{2}} + \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^{3}} + \frac {b^{12} c d^{5} - 25 \, a b^{11} d^{6}}{b^{14} d^{6}}\right )} - \frac {5 \, b^{13} c^{2} d^{4} + 14 \, a b^{12} c d^{5} - 163 \, a^{2} b^{11} d^{6}}{b^{14} d^{6}}\right )} + \frac {3 \, {\left (5 \, b^{14} c^{3} d^{3} + 9 \, a b^{13} c^{2} d^{4} + 15 \, a^{2} b^{12} c d^{5} - 93 \, a^{3} b^{11} d^{6}\right )}}{b^{14} d^{6}}\right )} \sqrt {b x + a} + \frac {3 \, {\left (5 \, b^{4} c^{4} + 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} + 20 \, a^{3} b c d^{3} - 35 \, 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^{2} d^{3}}\right )} {\left | b \right |}}{b}}{192 \, b} \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]

1/192*(8*(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^2 + (b^6*c*d^3 - 13*a*
b^5*d^4)/(b^7*d^4)) - 3*(b^7*c^2*d^2 + 2*a*b^6*c*d^3 - 11*a^2*b^5*d^4)/(b^7*d^4)) - 3*(b^3*c^3 + a*b^2*c^2*d +
3*a^2*b*c*d^2 - 5*a^3*d^3)*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^2))*a*abs(b)/b^2 + (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^3 + (b^
12*c*d^5 - 25*a*b^11*d^6)/(b^14*d^6)) - (5*b^13*c^2*d^4 + 14*a*b^12*c*d^5 - 163*a^2*b^11*d^6)/(b^14*d^6)) + 3*
(5*b^14*c^3*d^3 + 9*a*b^13*c^2*d^4 + 15*a^2*b^12*c*d^5 - 93*a^3*b^11*d^6)/(b^14*d^6))*sqrt(b*x + a) + 3*(5*b^4
*c^4 + 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 + 20*a^3*b*c*d^3 - 35*a^4*d^4)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqr
t(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^2*d^3))*abs(b)/b)/b

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \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]

\text{Hanged}

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