Integrand size = 22, antiderivative size = 95 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^5} \, dx=-\frac {2 B c \sqrt {b x+c x^2}}{x}-\frac {2 B \left (b x+c x^2\right )^{3/2}}{3 x^3}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}+2 B c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \]
-2/3*B*(c*x^2+b*x)^(3/2)/x^3-2/5*A*(c*x^2+b*x)^(5/2)/b/x^5+2*B*c^(3/2)*arc tanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))-2*B*c*(c*x^2+b*x)^(1/2)/x
Time = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^5} \, dx=-\frac {2 \sqrt {x (b+c x)} \left (\sqrt {b+c x} \left (3 A (b+c x)^2+5 b B x (b+4 c x)\right )+15 b B c^{3/2} x^{5/2} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )\right )}{15 b x^3 \sqrt {b+c x}} \]
(-2*Sqrt[x*(b + c*x)]*(Sqrt[b + c*x]*(3*A*(b + c*x)^2 + 5*b*B*x*(b + 4*c*x )) + 15*b*B*c^(3/2)*x^(5/2)*Log[-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]]))/(15* b*x^3*Sqrt[b + c*x])
Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1220, 1130, 1125, 25, 27, 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.
\(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^5} \, dx\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle B \int \frac {\left (c x^2+b x\right )^{3/2}}{x^4}dx-\frac {2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}\) |
\(\Big \downarrow \) 1130 |
\(\displaystyle B \left (c \int \frac {\sqrt {c x^2+b x}}{x^2}dx-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}\right )-\frac {2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}\) |
\(\Big \downarrow \) 1125 |
\(\displaystyle B \left (c \left (-\int -\frac {c}{\sqrt {c x^2+b x}}dx-\frac {2 \sqrt {b x+c x^2}}{x}\right )-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}\right )-\frac {2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle B \left (c \left (\int \frac {c}{\sqrt {c x^2+b x}}dx-\frac {2 \sqrt {b x+c x^2}}{x}\right )-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}\right )-\frac {2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle B \left (c \left (c \int \frac {1}{\sqrt {c x^2+b x}}dx-\frac {2 \sqrt {b x+c x^2}}{x}\right )-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}\right )-\frac {2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle B \left (c \left (2 c \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}-\frac {2 \sqrt {b x+c x^2}}{x}\right )-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}\right )-\frac {2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle B \left (c \left (2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )-\frac {2 \sqrt {b x+c x^2}}{x}\right )-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}\right )-\frac {2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}\) |
(-2*A*(b*x + c*x^2)^(5/2))/(5*b*x^5) + B*((-2*(b*x + c*x^2)^(3/2))/(3*x^3) + c*((-2*Sqrt[b*x + c*x^2])/x + 2*Sqrt[c]*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]]))
3.1.87.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[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_ Symbol] :> Simp[-2*e^(2*m + 3)*(Sqrt[a + b*x + c*x^2]/((-2*c*d + b*e)^(m + 2)*(d + e*x))), x] - Simp[e^(2*m + 2) Int[(1/Sqrt[a + b*x + c*x^2])*Expan dToSum[((-2*c*d + b*e)^(-m - 1) - ((-c)*d + b*e + c*e*x)^(-m - 1))/(d + e*x ), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ILtQ[m, 0] && EqQ[m + p, -3/2]
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_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x ^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e *f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)) Int[(d + e*x )^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0 ]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 ]
Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.82
method | result | size |
pseudoelliptic | \(\frac {2 B b \,c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) x^{3}-\frac {2 \left (\left (\frac {5 B x}{3}+A \right ) b^{2}+2 c x \left (\frac {10 B x}{3}+A \right ) b +A \,c^{2} x^{2}\right ) \sqrt {x \left (c x +b \right )}}{5}}{b \,x^{3}}\) | \(78\) |
risch | \(-\frac {2 \left (c x +b \right ) \left (3 A \,c^{2} x^{2}+20 B b c \,x^{2}+6 A b c x +5 b^{2} B x +3 A \,b^{2}\right )}{15 x^{2} \sqrt {x \left (c x +b \right )}\, b}+B \,c^{\frac {3}{2}} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )\) | \(90\) |
default | \(-\frac {2 A \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5 b \,x^{5}}+B \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{3 b \,x^{4}}+\frac {2 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{3}}+\frac {4 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{2}}-\frac {6 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2}\right )}{b}\right )}{b}\right )}{3 b}\right )\) | \(174\) |
2/5*(5*B*b*c^(3/2)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))*x^3-((5/3*B*x+A)*b ^2+2*c*x*(10/3*B*x+A)*b+A*c^2*x^2)*(x*(c*x+b))^(1/2))/b/x^3
Time = 0.26 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.98 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^5} \, dx=\left [\frac {15 \, B b c^{\frac {3}{2}} x^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (3 \, A b^{2} + {\left (20 \, B b c + 3 \, A c^{2}\right )} x^{2} + {\left (5 \, B b^{2} + 6 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x}}{15 \, b x^{3}}, -\frac {2 \, {\left (15 \, B b \sqrt {-c} c x^{3} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (3 \, A b^{2} + {\left (20 \, B b c + 3 \, A c^{2}\right )} x^{2} + {\left (5 \, B b^{2} + 6 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x}\right )}}{15 \, b x^{3}}\right ] \]
[1/15*(15*B*b*c^(3/2)*x^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2 *(3*A*b^2 + (20*B*b*c + 3*A*c^2)*x^2 + (5*B*b^2 + 6*A*b*c)*x)*sqrt(c*x^2 + b*x))/(b*x^3), -2/15*(15*B*b*sqrt(-c)*c*x^3*arctan(sqrt(c*x^2 + b*x)*sqrt (-c)/(c*x)) + (3*A*b^2 + (20*B*b*c + 3*A*c^2)*x^2 + (5*B*b^2 + 6*A*b*c)*x) *sqrt(c*x^2 + b*x))/(b*x^3)]
\[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^5} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (A + B x\right )}{x^{5}}\, dx \]
Time = 0.18 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.66 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^5} \, dx=B c^{\frac {3}{2}} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - \frac {7 \, \sqrt {c x^{2} + b x} B c}{3 \, x} - \frac {2 \, \sqrt {c x^{2} + b x} A c^{2}}{5 \, b x} - \frac {\sqrt {c x^{2} + b x} B b}{3 \, x^{2}} + \frac {\sqrt {c x^{2} + b x} A c}{5 \, x^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B}{3 \, x^{3}} + \frac {3 \, \sqrt {c x^{2} + b x} A b}{5 \, x^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A}{x^{4}} \]
B*c^(3/2)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 7/3*sqrt(c*x^2 + b*x)*B*c/x - 2/5*sqrt(c*x^2 + b*x)*A*c^2/(b*x) - 1/3*sqrt(c*x^2 + b*x)*B*b /x^2 + 1/5*sqrt(c*x^2 + b*x)*A*c/x^2 - 1/3*(c*x^2 + b*x)^(3/2)*B/x^3 + 3/5 *sqrt(c*x^2 + b*x)*A*b/x^3 - (c*x^2 + b*x)^(3/2)*A/x^4
Leaf count of result is larger than twice the leaf count of optimal. 259 vs. \(2 (79) = 158\).
Time = 0.28 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.73 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^5} \, dx=-B c^{\frac {3}{2}} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right ) + \frac {2 \, {\left (30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b c + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A c^{2} + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{2} \sqrt {c} + 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b c^{\frac {3}{2}} + 5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{3} + 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{2} c + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{3} \sqrt {c} + 3 \, A b^{4}\right )}}{15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5}} \]
-B*c^(3/2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b)) + 2/15* (30*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b*c + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*A*c^2 + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^2*sqrt(c) + 30*( sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b*c^(3/2) + 5*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^3 + 30*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^2*c + 15*(sqrt(c )*x - sqrt(c*x^2 + b*x))*A*b^3*sqrt(c) + 3*A*b^4)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^5
Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^5} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (A+B\,x\right )}{x^5} \,d x \]