\(\int \sqrt {x} \sqrt {a+b x} (A+B x) \, dx\) [481]

Optimal result

Integrand size = 20, antiderivative size = 126 \[ \int \sqrt {x} \sqrt {a+b x} (A+B x) \, dx=\frac {a (2 A b-a B) \sqrt {x} \sqrt {a+b x}}{8 b^2}+\frac {(2 A b-a B) x^{3/2} \sqrt {a+b x}}{4 b}+\frac {B x^{3/2} (a+b x)^{3/2}}{3 b}-\frac {a^2 (2 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{5/2}} \]

[Out]

1/3*B*x^(3/2)*(b*x+a)^(3/2)/b-1/8*a^2*(2*A*b-B*a)*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(5/2)+1/4*(2*A*b-B*
a)*x^(3/2)*(b*x+a)^(1/2)/b+1/8*a*(2*A*b-B*a)*x^(1/2)*(b*x+a)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {81, 52, 65, 223, 212} \[ \int \sqrt {x} \sqrt {a+b x} (A+B x) \, dx=-\frac {a^2 (2 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{5/2}}+\frac {a \sqrt {x} \sqrt {a+b x} (2 A b-a B)}{8 b^2}+\frac {x^{3/2} \sqrt {a+b x} (2 A b-a B)}{4 b}+\frac {B x^{3/2} (a+b x)^{3/2}}{3 b} \]

[In]

Int[Sqrt[x]*Sqrt[a + b*x]*(A + B*x),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.83 \[ \int \sqrt {x} \sqrt {a+b x} (A+B x) \, dx=\frac {\sqrt {x} \sqrt {a+b x} \left (-3 a^2 B+2 a b (3 A+B x)+4 b^2 x (3 A+2 B x)\right )}{24 b^2}+\frac {a^2 (-2 A b+a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{4 b^{5/2}} \]

[In]

Integrate[Sqrt[x]*Sqrt[a + b*x]*(A + B*x),x]

[Out]

(Sqrt[x]*Sqrt[a + b*x]*(-3*a^2*B + 2*a*b*(3*A + B*x) + 4*b^2*x*(3*A + 2*B*x)))/(24*b^2) + (a^2*(-2*A*b + a*B)*
ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])])/(4*b^(5/2))

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.88

method	result	size
risch	\(\frac {\left (8 b^{2} B \,x^{2}+12 A \,b^{2} x +2 B a b x +6 a b A -3 a^{2} B \right ) \sqrt {x}\, \sqrt {b x +a}}{24 b^{2}}-\frac {a^{2} \left (2 A b -B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{16 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {b x +a}}\)	\(111\)
default	\(-\frac {\sqrt {x}\, \sqrt {b x +a}\, \left (-16 B \,b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}-24 A \,b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, x -4 B \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a x +6 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b -12 A \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a -3 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3}+6 B \sqrt {b}\, \sqrt {x \left (b x +a \right )}\, a^{2}\right )}{48 b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}}\)	\(176\)

[In]

int((B*x+A)*x^(1/2)*(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/24*(8*B*b^2*x^2+12*A*b^2*x+2*B*a*b*x+6*A*a*b-3*B*a^2)*x^(1/2)*(b*x+a)^(1/2)/b^2-1/16*a^2*(2*A*b-B*a)/b^(5/2)
*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))*(x*(b*x+a))^(1/2)/x^(1/2)/(b*x+a)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.56 \[ \int \sqrt {x} \sqrt {a+b x} (A+B x) \, dx=\left [-\frac {3 \, {\left (B a^{3} - 2 \, A a^{2} b\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (8 \, B b^{3} x^{2} - 3 \, B a^{2} b + 6 \, A a b^{2} + 2 \, {\left (B a b^{2} + 6 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{48 \, b^{3}}, -\frac {3 \, {\left (B a^{3} - 2 \, A a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (8 \, B b^{3} x^{2} - 3 \, B a^{2} b + 6 \, A a b^{2} + 2 \, {\left (B a b^{2} + 6 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{24 \, b^{3}}\right ] \]

[In]

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

[Out]

[-1/48*(3*(B*a^3 - 2*A*a^2*b)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(8*B*b^3*x^2 - 3*B*
a^2*b + 6*A*a*b^2 + 2*(B*a*b^2 + 6*A*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^3, -1/24*(3*(B*a^3 - 2*A*a^2*b)*sqrt(-b)
*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (8*B*b^3*x^2 - 3*B*a^2*b + 6*A*a*b^2 + 2*(B*a*b^2 + 6*A*b^3)*x)*
sqrt(b*x + a)*sqrt(x))/b^3]

Sympy [A] (verification not implemented)

Time = 0.82 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.64 \[ \int \sqrt {x} \sqrt {a+b x} (A+B x) \, dx=2 A \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{8 b} + \sqrt {a + b x} \left (\frac {a \sqrt {x}}{8 b} + \frac {x^{\frac {3}{2}}}{4}\right ) & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{\frac {3}{2}}}{3} & \text {otherwise} \end {cases}\right ) + 2 B \left (\begin {cases} \frac {a^{3} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{16 b^{2}} + \sqrt {a + b x} \left (- \frac {a^{2} \sqrt {x}}{16 b^{2}} + \frac {a x^{\frac {3}{2}}}{24 b} + \frac {x^{\frac {5}{2}}}{6}\right ) & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{\frac {5}{2}}}{5} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((B*x+A)*x**(1/2)*(b*x+a)**(1/2),x)

[Out]

2*A*Piecewise((-a**2*Piecewise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)), (sqrt(x)*log(sq
rt(x))/sqrt(b*x), True))/(8*b) + sqrt(a + b*x)*(a*sqrt(x)/(8*b) + x**(3/2)/4), Ne(b, 0)), (sqrt(a)*x**(3/2)/3,
True)) + 2*B*Piecewise((a**3*Piecewise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)), (sqrt(
x)*log(sqrt(x))/sqrt(b*x), True))/(16*b**2) + sqrt(a + b*x)*(-a**2*sqrt(x)/(16*b**2) + a*x**(3/2)/(24*b) + x**
(5/2)/6), Ne(b, 0)), (sqrt(a)*x**(5/2)/5, True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.22 \[ \int \sqrt {x} \sqrt {a+b x} (A+B x) \, dx=\frac {1}{2} \, \sqrt {b x^{2} + a x} A x - \frac {\sqrt {b x^{2} + a x} B a x}{4 \, b} + \frac {B a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {5}{2}}} - \frac {A a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {3}{2}}} - \frac {\sqrt {b x^{2} + a x} B a^{2}}{8 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B}{3 \, b} + \frac {\sqrt {b x^{2} + a x} A a}{4 \, b} \]

[In]

integrate((B*x+A)*x^(1/2)*(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(b*x^2 + a*x)*A*x - 1/4*sqrt(b*x^2 + a*x)*B*a*x/b + 1/16*B*a^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqr
t(b))/b^(5/2) - 1/8*A*a^2*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(3/2) - 1/8*sqrt(b*x^2 + a*x)*B*a^2/b
^2 + 1/3*(b*x^2 + a*x)^(3/2)*B/b + 1/4*sqrt(b*x^2 + a*x)*A*a/b

Giac [F(-1)]

Timed out. \[ \int \sqrt {x} \sqrt {a+b x} (A+B x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 10.63 (sec) , antiderivative size = 399, normalized size of antiderivative = 3.17 \[ \int \sqrt {x} \sqrt {a+b x} (A+B x) \, dx=\frac {\frac {x^{11/2}\,\left (\frac {A\,a^2\,b^4}{2}-\frac {B\,a^3\,b^3}{4}\right )}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{11}}+\frac {x^{9/2}\,\left (\frac {17\,B\,a^3\,b^2}{12}+\frac {5\,A\,a^2\,b^3}{2}\right )}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^9}-\frac {x^{7/2}\,\left (3\,A\,a^2\,b^2-\frac {19\,B\,a^3\,b}{2}\right )}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7}+\frac {x^{5/2}\,\left (\frac {19\,B\,a^3}{2}-3\,A\,a^2\,b\right )}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5}-\frac {\sqrt {x}\,\left (B\,a^3-2\,A\,a^2\,b\right )}{4\,b^2\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}+\frac {x^{3/2}\,\left (17\,B\,a^3+30\,A\,b\,a^2\right )}{12\,b\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}}{\frac {15\,b^2\,x^2}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}-\frac {20\,b^3\,x^3}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}+\frac {15\,b^4\,x^4}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}-\frac {6\,b^5\,x^5}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{10}}+\frac {b^6\,x^6}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{12}}-\frac {6\,b\,x}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}+1}-\frac {a^2\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a+b\,x}-\sqrt {a}}\right )\,\left (2\,A\,b-B\,a\right )}{4\,b^{5/2}} \]

[In]

int(x^(1/2)*(A + B*x)*(a + b*x)^(1/2),x)

[Out]

((x^(11/2)*((A*a^2*b^4)/2 - (B*a^3*b^3)/4))/((a + b*x)^(1/2) - a^(1/2))^11 + (x^(9/2)*((5*A*a^2*b^3)/2 + (17*B
*a^3*b^2)/12))/((a + b*x)^(1/2) - a^(1/2))^9 - (x^(7/2)*(3*A*a^2*b^2 - (19*B*a^3*b)/2))/((a + b*x)^(1/2) - a^(
1/2))^7 + (x^(5/2)*((19*B*a^3)/2 - 3*A*a^2*b))/((a + b*x)^(1/2) - a^(1/2))^5 - (x^(1/2)*(B*a^3 - 2*A*a^2*b))/(
4*b^2*((a + b*x)^(1/2) - a^(1/2))) + (x^(3/2)*(17*B*a^3 + 30*A*a^2*b))/(12*b*((a + b*x)^(1/2) - a^(1/2))^3))/(
(15*b^2*x^2)/((a + b*x)^(1/2) - a^(1/2))^4 - (20*b^3*x^3)/((a + b*x)^(1/2) - a^(1/2))^6 + (15*b^4*x^4)/((a + b
*x)^(1/2) - a^(1/2))^8 - (6*b^5*x^5)/((a + b*x)^(1/2) - a^(1/2))^10 + (b^6*x^6)/((a + b*x)^(1/2) - a^(1/2))^12
- (6*b*x)/((a + b*x)^(1/2) - a^(1/2))^2 + 1) - (a^2*atanh((b^(1/2)*x^(1/2))/((a + b*x)^(1/2) - a^(1/2)))*(2*A
*b - B*a))/(4*b^(5/2))