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

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

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Rubi [A]  time = 0.09, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {80, 50, 63, 217, 206} \begin {gather*} \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} \end {gather*}

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 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 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 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])

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]

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*}

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Mathematica [A]  time = 0.57, size = 156, normalized size = 0.96 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.35, size = 213, normalized size = 1.31 \begin {gather*} \frac {(b c-a d)^2 (a d+b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 b^{5/2} d^{5/2}}-\frac {\sqrt {c+d x} (b c-a d)^2 \left (\frac {3 b^3 c (c+d x)^2}{(a+b x)^2}+\frac {3 a b^2 d (c+d x)^2}{(a+b x)^2}-\frac {8 b^2 c d (c+d x)}{a+b x}+\frac {8 a b d^2 (c+d x)}{a+b x}-3 a d^3-3 b c d^2\right )}{24 b^2 d^2 \sqrt {a+b x} \left (\frac {b (c+d x)}{a+b x}-d\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*((b*c - a*d)^2*Sqrt[c + d*x]*(-3*b*c*d^2 - 3*a*d^3 - (8*b^2*c*d*(c + d*x))/(a + b*x) + (8*a*b*d^2*(c + d
*x))/(a + b*x) + (3*b^3*c*(c + d*x)^2)/(a + b*x)^2 + (3*a*b^2*d*(c + d*x)^2)/(a + b*x)^2))/(b^2*d^2*Sqrt[a + b
*x]*(-d + (b*(c + d*x))/(a + b*x))^3) + ((b*c - a*d)^2*(b*c + a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sq
rt[a + b*x])])/(8*b^(5/2)*d^(5/2))

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fricas [A]  time = 1.05, size = 406, normalized size = 2.49 \begin {gather*} \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 {gather*}

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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giac [B]  time = 1.61, size = 345, normalized size = 2.12 \begin {gather*} \frac {\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^{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 )} {\left | b \right |}}{b} + \frac {6 \, {\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, b x + 2 \, a + \frac {b c d - 5 \, a d^{2}}{d^{2}}\right )} \sqrt {b x + a} + \frac {{\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b 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} d}\right )} a {\left | b \right |}}{b^{3}}}{24 \, b} \end {gather*}

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]

1/24*((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))*abs(b)/b + 6*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*b*x + 2*a + (b*c*d - 5*a*d^2)/d^2)*sqrt(b*x + a) +
(b^3*c^2 + 2*a*b^2*c*d - 3*a^2*b*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)*d))*a*abs(b)/b^3)/b

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maple [B]  time = 0.02, size = 472, normalized size = 2.90 \begin {gather*} \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (3 a^{3} d^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-3 a^{2} b c \,d^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-3 a \,b^{2} c^{2} d \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 b^{3} c^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+16 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, b^{2} d^{2} x^{2}+4 \sqrt {b d}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a b \,d^{2} x +4 \sqrt {b d}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b^{2} c d x -6 \sqrt {b d}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{2} d^{2}+4 \sqrt {b d}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a b c d -6 \sqrt {b d}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b^{2} c^{2}\right )}{48 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, b^{2} d^{2}} \end {gather*}

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
+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^3*d^3-3*ln(1/2*(2*b*d*x+a*d+b*c+2*(b*d*
x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^2*b*c*d^2-3*ln(1/2*(2*b*d*x+a*d+b*c+2*(b*d*x^2+a*d*x+b*
c*x+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a*b^2*c^2*d+3*ln(1/2*(2*b*d*x+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/
2)*(b*d)^(1/2))/(b*d)^(1/2))*b^3*c^3+4*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a*b*d^2+4*(b*d)^(1/2)*(b*
d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*b^2*c*d-6*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^2*d^2+4*(b*d)^(1/2)*(b*
d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a*b*c*d-6*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(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 = 0.00, 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*(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 details)Is a*d-b*c zero or nonzero?

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mupad [B]  time = 78.40, size = 1077, normalized size = 6.61 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {\sqrt {d}\,\left (a\,d+b\,c\right )\,{\left (a\,d-b\,c\right )}^2\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )\,\left (a^3\,d^3-a^2\,b\,c\,d^2-a\,b^2\,c^2\,d+b^3\,c^3\right )}\right )\,\left (a\,d+b\,c\right )\,{\left (a\,d-b\,c\right )}^2}{4\,b^{5/2}\,d^{5/2}}-\frac {\frac {\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (\frac {a^3\,b^3\,d^3}{4}-\frac {a^2\,b^4\,c\,d^2}{4}-\frac {a\,b^5\,c^2\,d}{4}+\frac {b^6\,c^3}{4}\right )}{d^8\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (\frac {17\,a^3\,b^2\,d^3}{12}+\frac {101\,a^2\,b^3\,c\,d^2}{4}+\frac {101\,a\,b^4\,c^2\,d}{4}+\frac {17\,b^5\,c^3}{12}\right )}{d^7\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7\,\left (\frac {19\,a^3\,d^3}{2}+\frac {269\,a^2\,b\,c\,d^2}{2}+\frac {269\,a\,b^2\,c^2\,d}{2}+\frac {19\,b^3\,c^3}{2}\right )}{d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^7}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5\,\left (\frac {19\,a^3\,b\,d^3}{2}+\frac {269\,a^2\,b^2\,c\,d^2}{2}+\frac {269\,a\,b^3\,c^2\,d}{2}+\frac {19\,b^4\,c^3}{2}\right )}{d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}+\frac {8\,a^{3/2}\,c^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{10}}{d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{10}}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{11}\,\left (\frac {a^3\,d^3}{4}-\frac {a^2\,b\,c\,d^2}{4}-\frac {a\,b^2\,c^2\,d}{4}+\frac {b^3\,c^3}{4}\right )}{b^2\,d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{11}}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^9\,\left (\frac {17\,a^3\,d^3}{12}+\frac {101\,a^2\,b\,c\,d^2}{4}+\frac {101\,a\,b^2\,c^2\,d}{4}+\frac {17\,b^3\,c^3}{12}\right )}{b\,d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^9}+\frac {\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8\,\left (32\,a^2\,d^2+96\,a\,b\,c\,d+32\,b^2\,c^2\right )}{d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}+\frac {\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4\,\left (32\,a^2\,b^2\,d^2+96\,a\,b^3\,c\,d+32\,b^4\,c^2\right )}{d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}+\frac {8\,a^{3/2}\,b^4\,c^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}+\frac {\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6\,\left (64\,a^2\,b\,d^2+\frac {656\,a\,b^2\,c\,d}{3}+64\,b^3\,c^2\right )}{d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}}{\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{12}}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{12}}+\frac {b^6}{d^6}-\frac {6\,b^5\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}+\frac {15\,b^4\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}-\frac {20\,b^3\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}{d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}+\frac {15\,b^2\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}{d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}-\frac {6\,b\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{10}}{d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{10}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((d^(1/2)*(a*d + b*c)*(a*d - b*c)^2*((a + b*x)^(1/2) - a^(1/2)))/(b^(1/2)*((c + d*x)^(1/2) - c^(1/2))*(a
^3*d^3 + b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2)))*(a*d + b*c)*(a*d - b*c)^2)/(4*b^(5/2)*d^(5/2)) - ((((a + b*x)^
(1/2) - a^(1/2))*((b^6*c^3)/4 + (a^3*b^3*d^3)/4 - (a^2*b^4*c*d^2)/4 - (a*b^5*c^2*d)/4))/(d^8*((c + d*x)^(1/2)
- c^(1/2))) - (((a + b*x)^(1/2) - a^(1/2))^3*((17*b^5*c^3)/12 + (17*a^3*b^2*d^3)/12 + (101*a^2*b^3*c*d^2)/4 +
(101*a*b^4*c^2*d)/4))/(d^7*((c + d*x)^(1/2) - c^(1/2))^3) - (((a + b*x)^(1/2) - a^(1/2))^7*((19*a^3*d^3)/2 + (
19*b^3*c^3)/2 + (269*a*b^2*c^2*d)/2 + (269*a^2*b*c*d^2)/2))/(d^5*((c + d*x)^(1/2) - c^(1/2))^7) - (((a + b*x)^
(1/2) - a^(1/2))^5*((19*b^4*c^3)/2 + (19*a^3*b*d^3)/2 + (269*a^2*b^2*c*d^2)/2 + (269*a*b^3*c^2*d)/2))/(d^6*((c
+ d*x)^(1/2) - c^(1/2))^5) + (8*a^(3/2)*c^(3/2)*((a + b*x)^(1/2) - a^(1/2))^10)/(d^2*((c + d*x)^(1/2) - c^(1/
2))^10) + (((a + b*x)^(1/2) - a^(1/2))^11*((a^3*d^3)/4 + (b^3*c^3)/4 - (a*b^2*c^2*d)/4 - (a^2*b*c*d^2)/4))/(b^
2*d^3*((c + d*x)^(1/2) - c^(1/2))^11) - (((a + b*x)^(1/2) - a^(1/2))^9*((17*a^3*d^3)/12 + (17*b^3*c^3)/12 + (1
01*a*b^2*c^2*d)/4 + (101*a^2*b*c*d^2)/4))/(b*d^4*((c + d*x)^(1/2) - c^(1/2))^9) + (a^(1/2)*c^(1/2)*((a + b*x)^
(1/2) - a^(1/2))^8*(32*a^2*d^2 + 32*b^2*c^2 + 96*a*b*c*d))/(d^4*((c + d*x)^(1/2) - c^(1/2))^8) + (a^(1/2)*c^(1
/2)*((a + b*x)^(1/2) - a^(1/2))^4*(32*b^4*c^2 + 32*a^2*b^2*d^2 + 96*a*b^3*c*d))/(d^6*((c + d*x)^(1/2) - c^(1/2
))^4) + (8*a^(3/2)*b^4*c^(3/2)*((a + b*x)^(1/2) - a^(1/2))^2)/(d^6*((c + d*x)^(1/2) - c^(1/2))^2) + (a^(1/2)*c
^(1/2)*((a + b*x)^(1/2) - a^(1/2))^6*(64*b^3*c^2 + 64*a^2*b*d^2 + (656*a*b^2*c*d)/3))/(d^5*((c + d*x)^(1/2) -
c^(1/2))^6))/(((a + b*x)^(1/2) - a^(1/2))^12/((c + d*x)^(1/2) - c^(1/2))^12 + b^6/d^6 - (6*b^5*((a + b*x)^(1/2
) - a^(1/2))^2)/(d^5*((c + d*x)^(1/2) - c^(1/2))^2) + (15*b^4*((a + b*x)^(1/2) - a^(1/2))^4)/(d^4*((c + d*x)^(
1/2) - c^(1/2))^4) - (20*b^3*((a + b*x)^(1/2) - a^(1/2))^6)/(d^3*((c + d*x)^(1/2) - c^(1/2))^6) + (15*b^2*((a
+ b*x)^(1/2) - a^(1/2))^8)/(d^2*((c + d*x)^(1/2) - c^(1/2))^8) - (6*b*((a + b*x)^(1/2) - a^(1/2))^10)/(d*((c +
d*x)^(1/2) - c^(1/2))^10))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \sqrt {a + b x} \sqrt {c + d x}\, dx \end {gather*}

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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