Integrand size = 17, antiderivative size = 94 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^3} \, dx=-\frac {5}{8} a (a+2 b x) \sqrt {a x+b x^2}-\frac {5}{3} b \left (a x+b x^2\right )^{3/2}+\frac {2 \left (a x+b x^2\right )^{5/2}}{x^2}+\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{8 \sqrt {b}} \]
-5/3*b*(b*x^2+a*x)^(3/2)+2*(b*x^2+a*x)^(5/2)/x^2+5/8*a^3*arctanh(x*b^(1/2) /(b*x^2+a*x)^(1/2))/b^(1/2)-5/8*a*(2*b*x+a)*(b*x^2+a*x)^(1/2)
Time = 0.06 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^3} \, dx=\frac {1}{24} \sqrt {x (a+b x)} \left (33 a^2+26 a b x+8 b^2 x^2-\frac {15 a^3 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{\sqrt {b} \sqrt {x} \sqrt {a+b x}}\right ) \]
(Sqrt[x*(a + b*x)]*(33*a^2 + 26*a*b*x + 8*b^2*x^2 - (15*a^3*Log[-(Sqrt[b]* Sqrt[x]) + Sqrt[a + b*x]])/(Sqrt[b]*Sqrt[x]*Sqrt[a + b*x])))/24
Time = 0.24 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {1130, 1131, 1087, 1091, 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.
\(\displaystyle \int \frac {\left (a x+b x^2\right )^{5/2}}{x^3} \, dx\) |
\(\Big \downarrow \) 1130 |
\(\displaystyle \frac {2 \left (a x+b x^2\right )^{5/2}}{x^2}-5 b \int \frac {\left (b x^2+a x\right )^{3/2}}{x}dx\) |
\(\Big \downarrow \) 1131 |
\(\displaystyle \frac {2 \left (a x+b x^2\right )^{5/2}}{x^2}-5 b \left (\frac {1}{2} a \int \sqrt {b x^2+a x}dx+\frac {1}{3} \left (a x+b x^2\right )^{3/2}\right )\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {2 \left (a x+b x^2\right )^{5/2}}{x^2}-5 b \left (\frac {1}{2} a \left (\frac {(a+2 b x) \sqrt {a x+b x^2}}{4 b}-\frac {a^2 \int \frac {1}{\sqrt {b x^2+a x}}dx}{8 b}\right )+\frac {1}{3} \left (a x+b x^2\right )^{3/2}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {2 \left (a x+b x^2\right )^{5/2}}{x^2}-5 b \left (\frac {1}{2} a \left (\frac {(a+2 b x) \sqrt {a x+b x^2}}{4 b}-\frac {a^2 \int \frac {1}{1-\frac {b x^2}{b x^2+a x}}d\frac {x}{\sqrt {b x^2+a x}}}{4 b}\right )+\frac {1}{3} \left (a x+b x^2\right )^{3/2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 \left (a x+b x^2\right )^{5/2}}{x^2}-5 b \left (\frac {1}{2} a \left (\frac {(a+2 b x) \sqrt {a x+b x^2}}{4 b}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{4 b^{3/2}}\right )+\frac {1}{3} \left (a x+b x^2\right )^{3/2}\right )\) |
(2*(a*x + b*x^2)^(5/2))/x^2 - 5*b*((a*x + b*x^2)^(3/2)/3 + (a*(((a + 2*b*x )*Sqrt[a*x + b*x^2])/(4*b) - (a^2*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/ (4*b^(3/2))))/2)
3.1.29.3.1 Defintions of rubi rules used
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])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + p + 1))), x] - Simp[c*(p/(e^2*(m + p + 1))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] & & IntegerQ[2*p]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x ] - Simp[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1))) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b *d*e + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && Ne Q[m + 2*p + 1, 0] && IntegerQ[2*p]
Time = 2.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.68
method | result | size |
pseudoelliptic | \(\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right ) a^{3}}{8 \sqrt {b}}+\frac {11 \left (\sqrt {b}\, a^{2}+\frac {26 b^{\frac {3}{2}} a x}{33}+\frac {8 b^{\frac {5}{2}} x^{2}}{33}\right ) \sqrt {x \left (b x +a \right )}}{8 \sqrt {b}}\) | \(64\) |
risch | \(\frac {\left (8 b^{2} x^{2}+26 a b x +33 a^{2}\right ) x \left (b x +a \right )}{24 \sqrt {x \left (b x +a \right )}}+\frac {5 a^{3} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{16 \sqrt {b}}\) | \(70\) |
default | \(\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{a \,x^{3}}-\frac {8 b \left (\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{3 a \,x^{2}}-\frac {10 b \left (\frac {\left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{5}+\frac {a \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{8 b}-\frac {3 a^{2} \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{16 b}\right )}{2}\right )}{3 a}\right )}{a}\) | \(156\) |
5/8/b^(1/2)*arctanh((x*(b*x+a))^(1/2)/x/b^(1/2))*a^3+11/8*(b^(1/2)*a^2+26/ 33*b^(3/2)*a*x+8/33*b^(5/2)*x^2)/b^(1/2)*(x*(b*x+a))^(1/2)
Time = 0.26 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.57 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^3} \, dx=\left [\frac {15 \, a^{3} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (8 \, b^{3} x^{2} + 26 \, a b^{2} x + 33 \, a^{2} b\right )} \sqrt {b x^{2} + a x}}{48 \, b}, -\frac {15 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) - {\left (8 \, b^{3} x^{2} + 26 \, a b^{2} x + 33 \, a^{2} b\right )} \sqrt {b x^{2} + a x}}{24 \, b}\right ] \]
[1/48*(15*a^3*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(8* b^3*x^2 + 26*a*b^2*x + 33*a^2*b)*sqrt(b*x^2 + a*x))/b, -1/24*(15*a^3*sqrt( -b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) - (8*b^3*x^2 + 26*a*b^2*x + 3 3*a^2*b)*sqrt(b*x^2 + a*x))/b]
\[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^3} \, dx=\int \frac {\left (x \left (a + b x\right )\right )^{\frac {5}{2}}}{x^{3}}\, dx \]
Time = 0.18 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^3} \, dx=\frac {5 \, a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, \sqrt {b}} + \frac {5}{8} \, \sqrt {b x^{2} + a x} a^{2} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a}{12 \, x} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}}}{3 \, x^{2}} \]
5/16*a^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/sqrt(b) + 5/8*sqrt(b *x^2 + a*x)*a^2 + 5/12*(b*x^2 + a*x)^(3/2)*a/x + 1/3*(b*x^2 + a*x)^(5/2)/x ^2
Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^3} \, dx=-\frac {5 \, a^{3} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{16 \, \sqrt {b}} + \frac {1}{24} \, \sqrt {b x^{2} + a x} {\left (33 \, a^{2} + 2 \, {\left (4 \, b^{2} x + 13 \, a b\right )} x\right )} \]
-5/16*a^3*log(abs(2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a))/sqrt(b) + 1/24*sqrt(b*x^2 + a*x)*(33*a^2 + 2*(4*b^2*x + 13*a*b)*x)
Timed out. \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^3} \, dx=\int \frac {{\left (b\,x^2+a\,x\right )}^{5/2}}{x^3} \,d x \]