Integrand size = 20, antiderivative size = 126 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx=-\frac {a (6 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{8 b^3}+\frac {(6 A b-5 a B) x^{3/2} \sqrt {a+b x}}{12 b^2}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}+\frac {a^2 (6 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{7/2}} \] Output:
-1/8*a*(6*A*b-5*B*a)*x^(1/2)*(b*x+a)^(1/2)/b^3+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^2*(6*A*b-5*B*a)*arct anh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(7/2)
Time = 0.39 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.86 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx=\frac {\sqrt {x} \sqrt {a+b x} \left (-18 a A b+15 a^2 B+12 A b^2 x-10 a b B x+8 b^2 B x^2\right )}{24 b^3}-\frac {a^2 (-6 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{4 b^{7/2}} \] Input:
Integrate[(x^(3/2)*(A + B*x))/Sqrt[a + b*x],x]
Output:
(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))
Time = 0.20 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {90, 60, 60, 65, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {(6 A b-5 a B) \int \frac {x^{3/2}}{\sqrt {a+b x}}dx}{6 b}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(6 A b-5 a B) \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \int \frac {\sqrt {x}}{\sqrt {a+b x}}dx}{4 b}\right )}{6 b}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(6 A b-5 a B) \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \int \frac {1}{\sqrt {x} \sqrt {a+b x}}dx}{2 b}\right )}{4 b}\right )}{6 b}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}\) |
\(\Big \downarrow \) 65 |
\(\displaystyle \frac {(6 A b-5 a B) \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \int \frac {1}{1-\frac {b x}{a+b x}}d\frac {\sqrt {x}}{\sqrt {a+b x}}}{b}\right )}{4 b}\right )}{6 b}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(6 A b-5 a B) \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{3/2}}\right )}{4 b}\right )}{6 b}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}\) |
Input:
Int[(x^(3/2)*(A + B*x))/Sqrt[a + b*x],x]
Output:
(B*x^(5/2)*Sqrt[a + b*x])/(3*b) + ((6*A*b - 5*a*B)*((x^(3/2)*Sqrt[a + b*x] )/(2*b) - (3*a*((Sqrt[x]*Sqrt[a + b*x])/b - (a*ArcTanh[(Sqrt[b]*Sqrt[x])/S qrt[a + b*x]])/b^(3/2)))/(4*b)))/(6*b)
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2 Sub st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d }, x] && !GtQ[c, 0]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(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]
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] && (Gt Q[a, 0] || LtQ[b, 0])
Time = 0.15 (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 | \(\frac {\sqrt {x}\, \sqrt {b x +a}\, \left (16 B \,b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}+24 A \sqrt {x \left (b x +a \right )}\, b^{\frac {5}{2}} x -20 B \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a x +18 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b -36 A \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a -15 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3}+30 B \sqrt {x \left (b x +a \right )}\, \sqrt {b}\, a^{2}\right )}{48 b^{\frac {7}{2}} \sqrt {x \left (b x +a \right )}}\) | \(176\) |
Input:
int(x^(3/2)*(B*x+A)/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-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)
Time = 0.11 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.56 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx=\left [-\frac {3 \, {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (8 \, B b^{3} x^{2} + 15 \, B a^{2} b - 18 \, A a b^{2} - 2 \, {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{48 \, b^{4}}, \frac {3 \, {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) + {\left (8 \, B b^{3} x^{2} + 15 \, B a^{2} b - 18 \, A a b^{2} - 2 \, {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{24 \, b^{4}}\right ] \] Input:
integrate(x^(3/2)*(B*x+A)/(b*x+a)^(1/2),x, algorithm="fricas")
Output:
[-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)*sqr t(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(b*x + a)) + (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]
Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (117) = 234\).
Time = 10.98 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.94 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx=- \frac {3 A a^{\frac {3}{2}} \sqrt {x}}{4 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {A \sqrt {a} x^{\frac {3}{2}}}{4 b \sqrt {1 + \frac {b x}{a}}} + \frac {3 A a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4 b^{\frac {5}{2}}} + \frac {A x^{\frac {5}{2}}}{2 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} + \frac {5 B a^{\frac {5}{2}} \sqrt {x}}{8 b^{3} \sqrt {1 + \frac {b x}{a}}} + \frac {5 B a^{\frac {3}{2}} x^{\frac {3}{2}}}{24 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {B \sqrt {a} x^{\frac {5}{2}}}{12 b \sqrt {1 + \frac {b x}{a}}} - \frac {5 B a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 b^{\frac {7}{2}}} + \frac {B x^{\frac {7}{2}}}{3 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \] Input:
integrate(x**(3/2)*(B*x+A)/(b*x+a)**(1/2),x)
Output:
-3*A*a**(3/2)*sqrt(x)/(4*b**2*sqrt(1 + b*x/a)) - A*sqrt(a)*x**(3/2)/(4*b*s qrt(1 + b*x/a)) + 3*A*a**2*asinh(sqrt(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*sqrt( 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))
Time = 0.03 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.27 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx=\frac {\sqrt {b x^{2} + a x} B x^{2}}{3 \, b} - \frac {5 \, \sqrt {b x^{2} + a x} B a x}{12 \, b^{2}} + \frac {\sqrt {b x^{2} + a x} A x}{2 \, b} - \frac {5 \, B a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {7}{2}}} + \frac {3 \, A a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {5}{2}}} + \frac {5 \, \sqrt {b x^{2} + a x} B a^{2}}{8 \, b^{3}} - \frac {3 \, \sqrt {b x^{2} + a x} A a}{4 \, b^{2}} \] Input:
integrate(x^(3/2)*(B*x+A)/(b*x+a)^(1/2),x, algorithm="maxima")
Output:
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*sqr t(b*x^2 + a*x)*A*x/b - 5/16*B*a^3*log(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
Time = 150.32 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.37 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx=-\frac {\frac {6 \, {\left (3 \, a^{2} \sqrt {b} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right ) - \sqrt {{\left (b x + a\right )} b - a b} {\left (2 \, b x - 3 \, a\right )} \sqrt {b x + a}\right )} A {\left | b \right |}}{b^{3}} - \frac {{\left (15 \, a^{3} \sqrt {b} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right ) + {\left (2 \, {\left (4 \, b x - 9 \, a\right )} {\left (b x + a\right )} + 33 \, a^{2}\right )} \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a}\right )} B {\left | b \right |}}{b^{4}}}{24 \, b} \] Input:
integrate(x^(3/2)*(B*x+A)/(b*x+a)^(1/2),x, algorithm="giac")
Output:
-1/24*(6*(3*a^2*sqrt(b)*log(abs(-sqrt(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 - (15*a^3*sqrt(b)*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a *b))) + (2*(4*b*x - 9*a)*(b*x + a) + 33*a^2)*sqrt((b*x + a)*b - a*b)*sqrt( b*x + a))*B*abs(b)/b^4)/b
Timed out. \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx=\int \frac {x^{3/2}\,\left (A+B\,x\right )}{\sqrt {a+b\,x}} \,d x \] Input:
int((x^(3/2)*(A + B*x))/(a + b*x)^(1/2),x)
Output:
int((x^(3/2)*(A + B*x))/(a + b*x)^(1/2), x)
Time = 0.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.60 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a+b x}} \, dx=\frac {-3 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b +2 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} x +8 \sqrt {x}\, \sqrt {b x +a}\, b^{3} x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{3}}{24 b^{3}} \] Input:
int(x^(3/2)*(B*x+A)/(b*x+a)^(1/2),x)
Output:
( - 3*sqrt(x)*sqrt(a + b*x)*a**2*b + 2*sqrt(x)*sqrt(a + b*x)*a*b**2*x + 8* sqrt(x)*sqrt(a + b*x)*b**3*x**2 + 3*sqrt(b)*log((sqrt(a + b*x) + sqrt(x)*s qrt(b))/sqrt(a))*a**3)/(24*b**3)