\(\int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx\) [516]

Optimal result

Integrand size = 20, antiderivative size = 126 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx=-\frac {a (6 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{8 b^3}+\frac {(6 A b-5 a B) x^{3/2} \sqrt {a+b x}}{12 b^2}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}+\frac {a^2 (6 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{7/2}} \]

[Out]

1/8*a^2*(6*A*b-5*B*a)*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(7/2)+1/12*(6*A*b-5*B*a)*x^(3/2)*(b*x+a)^(1/2)/
b^2+1/3*B*x^(5/2)*(b*x+a)^(1/2)/b-1/8*a*(6*A*b-5*B*a)*x^(1/2)*(b*x+a)^(1/2)/b^3

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} \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx=\frac {a^2 (6 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{7/2}}-\frac {a \sqrt {x} \sqrt {a+b x} (6 A b-5 a B)}{8 b^3}+\frac {x^{3/2} \sqrt {a+b x} (6 A b-5 a B)}{12 b^2}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b} \]

[In]

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

[Out]

-1/8*(a*(6*A*b - 5*a*B)*Sqrt[x]*Sqrt[a + b*x])/b^3 + ((6*A*b - 5*a*B)*x^(3/2)*Sqrt[a + b*x])/(12*b^2) + (B*x^(
5/2)*Sqrt[a + b*x])/(3*b) + (a^2*(6*A*b - 5*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(8*b^(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 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^{5/2} \sqrt {a+b x}}{3 b}+\frac {\left (3 A b-\frac {5 a B}{2}\right ) \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx}{3 b} \\ & = \frac {(6 A b-5 a B) x^{3/2} \sqrt {a+b x}}{12 b^2}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}-\frac {(a (6 A b-5 a B)) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{8 b^2} \\ & = -\frac {a (6 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{8 b^3}+\frac {(6 A b-5 a B) x^{3/2} \sqrt {a+b x}}{12 b^2}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}+\frac {\left (a^2 (6 A b-5 a B)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{16 b^3} \\ & = -\frac {a (6 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{8 b^3}+\frac {(6 A b-5 a B) x^{3/2} \sqrt {a+b x}}{12 b^2}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}+\frac {\left (a^2 (6 A b-5 a B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{8 b^3} \\ & = -\frac {a (6 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{8 b^3}+\frac {(6 A b-5 a B) x^{3/2} \sqrt {a+b x}}{12 b^2}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}+\frac {\left (a^2 (6 A b-5 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^3} \\ & = -\frac {a (6 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{8 b^3}+\frac {(6 A b-5 a B) x^{3/2} \sqrt {a+b x}}{12 b^2}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}+\frac {a^2 (6 A b-5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.86 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx=\frac {\sqrt {x} \sqrt {a+b x} \left (-18 a A b+15 a^2 B+12 A b^2 x-10 a b B x+8 b^2 B x^2\right )}{24 b^3}-\frac {a^2 (-6 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{4 b^{7/2}} \]

[In]

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

[Out]

(Sqrt[x]*Sqrt[a + b*x]*(-18*a*A*b + 15*a^2*B + 12*A*b^2*x - 10*a*b*B*x + 8*b^2*B*x^2))/(24*b^3) - (a^2*(-6*A*b
+ 5*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])])/(4*b^(7/2))

Maple [A] (verified)

Time = 1.45 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.88

method	result	size
risch	\(-\frac {\left (-8 b^{2} B \,x^{2}-12 A \,b^{2} x +10 B a b x +18 a b A -15 a^{2} B \right ) \sqrt {x}\, \sqrt {b x +a}}{24 b^{3}}+\frac {a^{2} \left (6 A b -5 B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{16 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {b x +a}}\)	\(111\)
default	\(\frac {\sqrt {x}\, \sqrt {b x +a}\, \left (16 B \,b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}+24 A \,b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, x -20 B \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a x +18 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b -36 A \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a -15 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3}+30 B \sqrt {b}\, \sqrt {x \left (b x +a \right )}\, a^{2}\right )}{48 b^{\frac {7}{2}} \sqrt {x \left (b x +a \right )}}\)	\(176\)

[In]

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

[Out]

-1/24*(-8*B*b^2*x^2-12*A*b^2*x+10*B*a*b*x+18*A*a*b-15*B*a^2)*x^(1/2)*(b*x+a)^(1/2)/b^3+1/16*a^2*(6*A*b-5*B*a)/
b^(7/2)*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 = 0.24 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.59 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx=\left [-\frac {3 \, {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (8 \, B b^{3} x^{2} + 15 \, B a^{2} b - 18 \, A a b^{2} - 2 \, {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{48 \, b^{4}}, \frac {3 \, {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (8 \, B b^{3} x^{2} + 15 \, B a^{2} b - 18 \, A a b^{2} - 2 \, {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{24 \, b^{4}}\right ] \]

[In]

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

[Out]

[-1/48*(3*(5*B*a^3 - 6*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 + 15
*B*a^2*b - 18*A*a*b^2 - 2*(5*B*a*b^2 - 6*A*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^4, 1/24*(3*(5*B*a^3 - 6*A*a^2*b)*s
qrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (8*B*b^3*x^2 + 15*B*a^2*b - 18*A*a*b^2 - 2*(5*B*a*b^2 - 6
*A*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^4]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (117) = 234\).

Time = 11.05 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.94 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx=- \frac {3 A a^{\frac {3}{2}} \sqrt {x}}{4 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {A \sqrt {a} x^{\frac {3}{2}}}{4 b \sqrt {1 + \frac {b x}{a}}} + \frac {3 A a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4 b^{\frac {5}{2}}} + \frac {A x^{\frac {5}{2}}}{2 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} + \frac {5 B a^{\frac {5}{2}} \sqrt {x}}{8 b^{3} \sqrt {1 + \frac {b x}{a}}} + \frac {5 B a^{\frac {3}{2}} x^{\frac {3}{2}}}{24 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {B \sqrt {a} x^{\frac {5}{2}}}{12 b \sqrt {1 + \frac {b x}{a}}} - \frac {5 B a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 b^{\frac {7}{2}}} + \frac {B x^{\frac {7}{2}}}{3 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \]

[In]

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

[Out]

-3*A*a**(3/2)*sqrt(x)/(4*b**2*sqrt(1 + b*x/a)) - A*sqrt(a)*x**(3/2)/(4*b*sqrt(1 + b*x/a)) + 3*A*a**2*asinh(sqr
t(b)*sqrt(x)/sqrt(a))/(4*b**(5/2)) + A*x**(5/2)/(2*sqrt(a)*sqrt(1 + b*x/a)) + 5*B*a**(5/2)*sqrt(x)/(8*b**3*sqr
t(1 + b*x/a)) + 5*B*a**(3/2)*x**(3/2)/(24*b**2*sqrt(1 + b*x/a)) - B*sqrt(a)*x**(5/2)/(12*b*sqrt(1 + b*x/a)) -
5*B*a**3*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(7/2)) + B*x**(7/2)/(3*sqrt(a)*sqrt(1 + b*x/a))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.27 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx=\frac {\sqrt {b x^{2} + a x} B x^{2}}{3 \, b} - \frac {5 \, \sqrt {b x^{2} + a x} B a x}{12 \, b^{2}} + \frac {\sqrt {b x^{2} + a x} A x}{2 \, b} - \frac {5 \, B a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {7}{2}}} + \frac {3 \, A a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {5}{2}}} + \frac {5 \, \sqrt {b x^{2} + a x} B a^{2}}{8 \, b^{3}} - \frac {3 \, \sqrt {b x^{2} + a x} A a}{4 \, b^{2}} \]

[In]

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

[Out]

1/3*sqrt(b*x^2 + a*x)*B*x^2/b - 5/12*sqrt(b*x^2 + a*x)*B*a*x/b^2 + 1/2*sqrt(b*x^2 + a*x)*A*x/b - 5/16*B*a^3*lo
g(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(7/2) + 3/8*A*a^2*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^
(5/2) + 5/8*sqrt(b*x^2 + a*x)*B*a^2/b^3 - 3/4*sqrt(b*x^2 + a*x)*A*a/b^2

Giac [A] (verification not implemented)

none

Time = 152.40 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.45 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx=\frac {\frac {{\left (\frac {15 \, a^{3} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {3}{2}}} + \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{2}} - \frac {13 \, a}{b^{2}}\right )} + \frac {33 \, a^{2}}{b^{2}}\right )}\right )} B {\left | b \right |}}{b^{2}} - \frac {6 \, {\left (3 \, a^{2} \sqrt {b} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right ) - \sqrt {{\left (b x + a\right )} b - a b} {\left (2 \, b x - 3 \, a\right )} \sqrt {b x + a}\right )} A {\left | b \right |}}{b^{3}}}{24 \, b} \]

[In]

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

[Out]

1/24*((15*a^3*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b)))/b^(3/2) + sqrt((b*x + a)*b - a*b)*sqr
t(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/b^2 - 13*a/b^2) + 33*a^2/b^2))*B*abs(b)/b^2 - 6*(3*a^2*sqrt(b)*log(abs(-s
qrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b))) - sqrt((b*x + a)*b - a*b)*(2*b*x - 3*a)*sqrt(b*x + a))*A*abs(
b)/b^3)/b

Mupad [F(-1)]

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

[In]

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

[Out]

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