Integrand size = 15, antiderivative size = 99 \[ \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx=-\frac {2 a^2 \sqrt {x}}{b^3 \sqrt {a+b x}}-\frac {7 a \sqrt {x} \sqrt {a+b x}}{4 b^3}+\frac {x^{3/2} \sqrt {a+b x}}{2 b^2}+\frac {15 a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{7/2}} \] Output:
-2*a^2*x^(1/2)/b^3/(b*x+a)^(1/2)-7/4*a*x^(1/2)*(b*x+a)^(1/2)/b^3+1/2*x^(3/ 2)*(b*x+a)^(1/2)/b^2+15/4*a^2*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(7/ 2)
Time = 0.22 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.85 \[ \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx=\frac {\sqrt {x} \left (-15 a^2-5 a b x+2 b^2 x^2\right )}{4 b^3 \sqrt {a+b x}}+\frac {15 a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{2 b^{7/2}} \] Input:
Integrate[x^(5/2)/(a + b*x)^(3/2),x]
Output:
(Sqrt[x]*(-15*a^2 - 5*a*b*x + 2*b^2*x^2))/(4*b^3*Sqrt[a + b*x]) + (15*a^2* ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])])/(2*b^(7/2))
Time = 0.17 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {57, 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.
\(\displaystyle \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {5 \int \frac {x^{3/2}}{\sqrt {a+b x}}dx}{b}-\frac {2 x^{5/2}}{b \sqrt {a+b x}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5 \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \int \frac {\sqrt {x}}{\sqrt {a+b x}}dx}{4 b}\right )}{b}-\frac {2 x^{5/2}}{b \sqrt {a+b x}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5 \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \int \frac {1}{\sqrt {x} \sqrt {a+b x}}dx}{2 b}\right )}{4 b}\right )}{b}-\frac {2 x^{5/2}}{b \sqrt {a+b x}}\) |
\(\Big \downarrow \) 65 |
\(\displaystyle \frac {5 \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \int \frac {1}{1-\frac {b x}{a+b x}}d\frac {\sqrt {x}}{\sqrt {a+b x}}}{b}\right )}{4 b}\right )}{b}-\frac {2 x^{5/2}}{b \sqrt {a+b x}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5 \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{3/2}}\right )}{4 b}\right )}{b}-\frac {2 x^{5/2}}{b \sqrt {a+b x}}\) |
Input:
Int[x^(5/2)/(a + b*x)^(3/2),x]
Output:
(-2*x^(5/2))/(b*Sqrt[a + b*x]) + (5*((x^(3/2)*Sqrt[a + b*x])/(2*b) - (3*a* ((Sqrt[x]*Sqrt[a + b*x])/b - (a*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/ b^(3/2)))/(4*b)))/b
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
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_)^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 = 0.08 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.20
method | result | size |
risch | \(-\frac {\left (-2 b x +7 a \right ) \sqrt {x}\, \sqrt {b x +a}}{4 b^{3}}+\frac {\left (\frac {15 a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {7}{2}}}-\frac {2 a^{2} \sqrt {b \left (x +\frac {a}{b}\right )^{2}-\left (x +\frac {a}{b}\right ) a}}{b^{4} \left (x +\frac {a}{b}\right )}\right ) \sqrt {x \left (b x +a \right )}}{\sqrt {x}\, \sqrt {b x +a}}\) | \(119\) |
Input:
int(x^(5/2)/(b*x+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/4*(-2*b*x+7*a)*x^(1/2)*(b*x+a)^(1/2)/b^3+(15/8*a^2/b^(7/2)*ln((1/2*a+b* x)/b^(1/2)+(b*x^2+a*x)^(1/2))-2*a^2/b^4/(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)
Time = 0.09 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.74 \[ \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx=\left [\frac {15 \, {\left (a^{2} b x + a^{3}\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (2 \, b^{3} x^{2} - 5 \, a b^{2} x - 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{8 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {15 \, {\left (a^{2} b x + a^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) - {\left (2 \, b^{3} x^{2} - 5 \, a b^{2} x - 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{4 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \] Input:
integrate(x^(5/2)/(b*x+a)^(3/2),x, algorithm="fricas")
Output:
[1/8*(15*(a^2*b*x + a^3)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt( x) + a) + 2*(2*b^3*x^2 - 5*a*b^2*x - 15*a^2*b)*sqrt(b*x + a)*sqrt(x))/(b^5 *x + a*b^4), -1/4*(15*(a^2*b*x + a^3)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqr t(b*x + a)) - (2*b^3*x^2 - 5*a*b^2*x - 15*a^2*b)*sqrt(b*x + a)*sqrt(x))/(b ^5*x + a*b^4)]
Time = 4.25 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.06 \[ \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx=- \frac {15 a^{\frac {3}{2}} \sqrt {x}}{4 b^{3} \sqrt {1 + \frac {b x}{a}}} - \frac {5 \sqrt {a} x^{\frac {3}{2}}}{4 b^{2} \sqrt {1 + \frac {b x}{a}}} + \frac {15 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4 b^{\frac {7}{2}}} + \frac {x^{\frac {5}{2}}}{2 \sqrt {a} b \sqrt {1 + \frac {b x}{a}}} \] Input:
integrate(x**(5/2)/(b*x+a)**(3/2),x)
Output:
-15*a**(3/2)*sqrt(x)/(4*b**3*sqrt(1 + b*x/a)) - 5*sqrt(a)*x**(3/2)/(4*b**2 *sqrt(1 + b*x/a)) + 15*a**2*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(4*b**(7/2)) + x**(5/2)/(2*sqrt(a)*b*sqrt(1 + b*x/a))
Time = 0.13 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.32 \[ \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx=-\frac {8 \, a^{2} b^{2} - \frac {25 \, {\left (b x + a\right )} a^{2} b}{x} + \frac {15 \, {\left (b x + a\right )}^{2} a^{2}}{x^{2}}}{4 \, {\left (\frac {\sqrt {b x + a} b^{5}}{\sqrt {x}} - \frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {{\left (b x + a\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}}\right )}} - \frac {15 \, a^{2} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{8 \, b^{\frac {7}{2}}} \] Input:
integrate(x^(5/2)/(b*x+a)^(3/2),x, algorithm="maxima")
Output:
-1/4*(8*a^2*b^2 - 25*(b*x + a)*a^2*b/x + 15*(b*x + a)^2*a^2/x^2)/(sqrt(b*x + a)*b^5/sqrt(x) - 2*(b*x + a)^(3/2)*b^4/x^(3/2) + (b*x + a)^(5/2)*b^3/x^ (5/2)) - 15/8*a^2*log(-(sqrt(b) - sqrt(b*x + a)/sqrt(x))/(sqrt(b) + sqrt(b *x + a)/sqrt(x)))/b^(7/2)
Time = 15.23 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.32 \[ \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx=-\frac {{\left (\frac {32 \, a^{3} \sqrt {b}}{{\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b} + \frac {15 \, a^{2} \log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{\sqrt {b}} - 2 \, \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )}}{b} - \frac {9 \, a}{b}\right )}\right )} {\left | b \right |}}{8 \, b^{4}} \] Input:
integrate(x^(5/2)/(b*x+a)^(3/2),x, algorithm="giac")
Output:
-1/8*(32*a^3*sqrt(b)/((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b) + 15*a^2*log((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2)/s qrt(b) - 2*sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2*(b*x + a)/b - 9*a/b))* abs(b)/b^4
Timed out. \[ \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx=\int \frac {x^{5/2}}{{\left (a+b\,x\right )}^{3/2}} \,d x \] Input:
int(x^(5/2)/(a + b*x)^(3/2),x)
Output:
int(x^(5/2)/(a + b*x)^(3/2), x)
Time = 0.16 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.86 \[ \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx=\frac {15 \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{2}-10 \sqrt {b}\, \sqrt {b x +a}\, a^{2}-15 \sqrt {x}\, a^{2} b -5 \sqrt {x}\, a \,b^{2} x +2 \sqrt {x}\, b^{3} x^{2}}{4 \sqrt {b x +a}\, b^{4}} \] Input:
int(x^(5/2)/(b*x+a)^(3/2),x)
Output:
(15*sqrt(b)*sqrt(a + b*x)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a **2 - 10*sqrt(b)*sqrt(a + b*x)*a**2 - 15*sqrt(x)*a**2*b - 5*sqrt(x)*a*b**2 *x + 2*sqrt(x)*b**3*x**2)/(4*sqrt(a + b*x)*b**4)