\(\int \frac {\sqrt {x} (A+B x)}{(a+b x)^{3/2}} \, dx\) [529]

Optimal result

Integrand size = 20, antiderivative size = 98 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {2 (A b-a B) x^{3/2}}{a b \sqrt {a+b x}}-\frac {(2 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{a b^2}+\frac {(2 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {79, 52, 65, 223, 212} \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {(2 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{5/2}}-\frac {\sqrt {x} \sqrt {a+b x} (2 A b-3 a B)}{a b^2}+\frac {2 x^{3/2} (A b-a B)}{a b \sqrt {a+b x}} \]

[In]

Int[(Sqrt[x]*(A + B*x))/(a + b*x)^(3/2),x]

[Out]

(2*(A*b - a*B)*x^(3/2))/(a*b*Sqrt[a + b*x]) - ((2*A*b - 3*a*B)*Sqrt[x]*Sqrt[a + b*x])/(a*b^2) + ((2*A*b - 3*a*
B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {\sqrt {x} (-2 A b+3 a B+b B x)}{b^2 \sqrt {a+b x}}+\frac {2 (2 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{b^{5/2}} \]

[In]

Integrate[(Sqrt[x]*(A + B*x))/(a + b*x)^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.52

method	result	size
risch	\(\frac {B \sqrt {x}\, \sqrt {b x +a}}{b^{2}}+\frac {\left (2 A \sqrt {b}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )-\frac {3 B a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{\sqrt {b}}-\frac {4 \left (A b -B a \right ) \sqrt {b \left (x +\frac {a}{b}\right )^{2}-\left (x +\frac {a}{b}\right ) a}}{b \left (x +\frac {a}{b}\right )}\right ) \sqrt {x \left (b x +a \right )}}{2 b^{2} \sqrt {x}\, \sqrt {b x +a}}\)	\(149\)
default	\(\frac {\left (2 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) b^{2} x -3 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a b x +2 B \,b^{\frac {3}{2}} x \sqrt {x \left (b x +a \right )}+2 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a b -4 A \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}-3 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2}+6 B a \sqrt {b}\, \sqrt {x \left (b x +a \right )}\right ) \sqrt {x}}{2 \sqrt {x \left (b x +a \right )}\, b^{\frac {5}{2}} \sqrt {b x +a}}\)	\(201\)

[In]

int((B*x+A)*x^(1/2)/(b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.99 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^{3/2}} \, dx=\left [-\frac {{\left (3 \, B a^{2} - 2 \, A a b + {\left (3 \, B a b - 2 \, A b^{2}\right )} x\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (B b^{2} x + 3 \, B a b - 2 \, A b^{2}\right )} \sqrt {b x + a} \sqrt {x}}{2 \, {\left (b^{4} x + a b^{3}\right )}}, \frac {{\left (3 \, B a^{2} - 2 \, A a b + {\left (3 \, B a b - 2 \, A b^{2}\right )} x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (B b^{2} x + 3 \, B a b - 2 \, A b^{2}\right )} \sqrt {b x + a} \sqrt {x}}{b^{4} x + a b^{3}}\right ] \]

[In]

integrate((B*x+A)*x^(1/2)/(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/2*((3*B*a^2 - 2*A*a*b + (3*B*a*b - 2*A*b^2)*x)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) -
2*(B*b^2*x + 3*B*a*b - 2*A*b^2)*sqrt(b*x + a)*sqrt(x))/(b^4*x + a*b^3), ((3*B*a^2 - 2*A*a*b + (3*B*a*b - 2*A*b
^2)*x)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (B*b^2*x + 3*B*a*b - 2*A*b^2)*sqrt(b*x + a)*sqrt(
x))/(b^4*x + a*b^3)]

Sympy [A] (verification not implemented)

Time = 5.67 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.24 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^{3/2}} \, dx=A \left (\frac {2 \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{b^{\frac {3}{2}}} - \frac {2 \sqrt {x}}{\sqrt {a} b \sqrt {1 + \frac {b x}{a}}}\right ) + B \left (\frac {3 \sqrt {a} \sqrt {x}}{b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{b^{\frac {5}{2}}} + \frac {x^{\frac {3}{2}}}{\sqrt {a} b \sqrt {1 + \frac {b x}{a}}}\right ) \]

[In]

integrate((B*x+A)*x**(1/2)/(b*x+a)**(3/2),x)

[Out]

A*(2*asinh(sqrt(b)*sqrt(x)/sqrt(a))/b**(3/2) - 2*sqrt(x)/(sqrt(a)*b*sqrt(1 + b*x/a))) + B*(3*sqrt(a)*sqrt(x)/(
b**2*sqrt(1 + b*x/a)) - 3*a*asinh(sqrt(b)*sqrt(x)/sqrt(a))/b**(5/2) + x**(3/2)/(sqrt(a)*b*sqrt(1 + b*x/a)))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.37 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {2 \, \sqrt {b x^{2} + a x} B a}{b^{3} x + a b^{2}} - \frac {2 \, \sqrt {b x^{2} + a x} A}{b^{2} x + a b} - \frac {3 \, B a \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{2 \, b^{\frac {5}{2}}} + \frac {A \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{b^{\frac {3}{2}}} + \frac {\sqrt {b x^{2} + a x} B}{b^{2}} \]

[In]

integrate((B*x+A)*x^(1/2)/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

2*sqrt(b*x^2 + a*x)*B*a/(b^3*x + a*b^2) - 2*sqrt(b*x^2 + a*x)*A/(b^2*x + a*b) - 3/2*B*a*log(2*x + a/b + 2*sqrt
(b*x^2 + a*x)/sqrt(b))/b^(5/2) + A*log(2*x + a/b + 2*sqrt(b*x^2 + a*x)/sqrt(b))/b^(3/2) + sqrt(b*x^2 + a*x)*B/
b^2

Giac [A] (verification not implemented)

none

Time = 15.71 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.37 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {\sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} B {\left | b \right |}}{b^{4}} + \frac {{\left (3 \, B a {\left | b \right |} - 2 \, A b {\left | b \right |}\right )} \log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{2 \, b^{\frac {7}{2}}} + \frac {4 \, {\left (B a^{2} {\left | b \right |} - A a b {\left | b \right |}\right )}}{{\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} b^{\frac {5}{2}}} \]

[In]

integrate((B*x+A)*x^(1/2)/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*B*abs(b)/b^4 + 1/2*(3*B*a*abs(b) - 2*A*b*abs(b))*log((sqrt(b*x + a)*sqrt
(b) - sqrt((b*x + a)*b - a*b))^2)/b^(7/2) + 4*(B*a^2*abs(b) - A*a*b*abs(b))/(((sqrt(b*x + a)*sqrt(b) - sqrt((b
*x + a)*b - a*b))^2 + a*b)*b^(5/2))

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^{3/2}} \, dx=\int \frac {\sqrt {x}\,\left (A+B\,x\right )}{{\left (a+b\,x\right )}^{3/2}} \,d x \]

[In]

int((x^(1/2)*(A + B*x))/(a + b*x)^(3/2),x)

[Out]

int((x^(1/2)*(A + B*x))/(a + b*x)^(3/2), x)