Integrand size = 22, antiderivative size = 120 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^4} \, dx=\frac {c (3 b B+2 A c) \sqrt {b x+c x^2}}{b}-\frac {2 (3 b B+2 A c) \left (b x+c x^2\right )^{3/2}}{3 b x^2}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{3 b x^4}+\sqrt {c} (3 b B+2 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \]
-2/3*(2*A*c+3*B*b)*(c*x^2+b*x)^(3/2)/b/x^2-2/3*A*(c*x^2+b*x)^(5/2)/b/x^4+( 2*A*c+3*B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))*c^(1/2)+c*(2*A*c+3*B*b)* (c*x^2+b*x)^(1/2)/b
Time = 0.23 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.92 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^4} \, dx=\frac {\sqrt {x (b+c x)} \left (\sqrt {b+c x} (3 B x (-2 b+c x)-2 A (b+4 c x))+6 \sqrt {c} (3 b B+2 A c) x^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )\right )}{3 x^2 \sqrt {b+c x}} \]
(Sqrt[x*(b + c*x)]*(Sqrt[b + c*x]*(3*B*x*(-2*b + c*x) - 2*A*(b + 4*c*x)) + 6*Sqrt[c]*(3*b*B + 2*A*c)*x^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + S qrt[b + c*x])]))/(3*x^2*Sqrt[b + c*x])
Time = 0.26 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.88, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1220, 1125, 25, 27, 1160, 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.
| \(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^4} \, dx\) |
| \(\Big \downarrow \) 1220 |
| \(\displaystyle \frac {(2 A c+3 b B) \int \frac {\left (c x^2+b x\right )^{3/2}}{x^3}dx}{3 b}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{3 b x^4}\) |
| \(\Big \downarrow \) 1125 |
| \(\displaystyle \frac {(2 A c+3 b B) \left (-\int -\frac {c (2 b+c x)}{\sqrt {c x^2+b x}}dx-\frac {2 b \sqrt {b x+c x^2}}{x}\right )}{3 b}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{3 b x^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(2 A c+3 b B) \left (\int \frac {c (2 b+c x)}{\sqrt {c x^2+b x}}dx-\frac {2 b \sqrt {b x+c x^2}}{x}\right )}{3 b}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{3 b x^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(2 A c+3 b B) \left (c \int \frac {2 b+c x}{\sqrt {c x^2+b x}}dx-\frac {2 b \sqrt {b x+c x^2}}{x}\right )}{3 b}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{3 b x^4}\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {(2 A c+3 b B) \left (c \left (\frac {3}{2} b \int \frac {1}{\sqrt {c x^2+b x}}dx+\sqrt {b x+c x^2}\right )-\frac {2 b \sqrt {b x+c x^2}}{x}\right )}{3 b}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{3 b x^4}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {(2 A c+3 b B) \left (c \left (3 b \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}+\sqrt {b x+c x^2}\right )-\frac {2 b \sqrt {b x+c x^2}}{x}\right )}{3 b}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{3 b x^4}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(2 A c+3 b B) \left (c \left (\frac {3 b \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c}}+\sqrt {b x+c x^2}\right )-\frac {2 b \sqrt {b x+c x^2}}{x}\right )}{3 b}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{3 b x^4}\) |
(-2*A*(b*x + c*x^2)^(5/2))/(3*b*x^4) + ((3*b*B + 2*A*c)*((-2*b*Sqrt[b*x + c*x^2])/x + c*(Sqrt[b*x + c*x^2] + (3*b*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x ^2]])/Sqrt[c])))/(3*b)
3.1.86.3.1 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_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
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 = 0.17 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.65
| method | result | size |
| pseudoelliptic | \(\frac {2 c \,x^{2} \left (A c +\frac {3 B b}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )-\frac {2 \sqrt {x \left (c x +b \right )}\, \left (\left (-\frac {3}{2} B \,x^{2}+4 A x \right ) c^{\frac {3}{2}}+b \sqrt {c}\, \left (3 B x +A \right )\right )}{3}}{x^{2} \sqrt {c}}\) | \(78\) |
| risch | \(-\frac {\left (c x +b \right ) \left (-3 B c \,x^{2}+8 A c x +6 B b x +2 A b \right )}{3 x \sqrt {x \left (c x +b \right )}}+\frac {\left (2 A c +3 B b \right ) \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2}\) | \(81\) |
| default | \(B \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{3}}+\frac {4 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{2}}-\frac {6 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2}\right )}{b}\right )}{b}\right )+A \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{3 b \,x^{4}}+\frac {2 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{3}}+\frac {4 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{2}}-\frac {6 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2}\right )}{b}\right )}{b}\right )}{3 b}\right )\) | \(280\) |
2*(c*x^2*(A*c+3/2*B*b)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))-1/3*(x*(c*x+b) )^(1/2)*((-3/2*B*x^2+4*A*x)*c^(3/2)+b*c^(1/2)*(3*B*x+A)))/c^(1/2)/x^2
Time = 0.25 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.42 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^4} \, dx=\left [\frac {3 \, {\left (3 \, B b + 2 \, A c\right )} \sqrt {c} x^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (3 \, B c x^{2} - 2 \, A b - 2 \, {\left (3 \, B b + 4 \, A c\right )} x\right )} \sqrt {c x^{2} + b x}}{6 \, x^{2}}, -\frac {3 \, {\left (3 \, B b + 2 \, A c\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (3 \, B c x^{2} - 2 \, A b - 2 \, {\left (3 \, B b + 4 \, A c\right )} x\right )} \sqrt {c x^{2} + b x}}{3 \, x^{2}}\right ] \]
[1/6*(3*(3*B*b + 2*A*c)*sqrt(c)*x^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sq rt(c)) + 2*(3*B*c*x^2 - 2*A*b - 2*(3*B*b + 4*A*c)*x)*sqrt(c*x^2 + b*x))/x^ 2, -1/3*(3*(3*B*b + 2*A*c)*sqrt(-c)*x^2*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/ (c*x)) - (3*B*c*x^2 - 2*A*b - 2*(3*B*b + 4*A*c)*x)*sqrt(c*x^2 + b*x))/x^2]
\[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^4} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (A + B x\right )}{x^{4}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.22 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^4} \, dx=\frac {3}{2} \, B b \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + A c^{\frac {3}{2}} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - \frac {3 \, \sqrt {c x^{2} + b x} B b}{x} - \frac {7 \, \sqrt {c x^{2} + b x} A c}{3 \, x} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B}{x^{2}} - \frac {\sqrt {c x^{2} + b x} A b}{3 \, x^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A}{3 \, x^{3}} \]
3/2*B*b*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + A*c^(3/2)*l og(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 3*sqrt(c*x^2 + b*x)*B*b/x - 7/3*sqrt(c*x^2 + b*x)*A*c/x + (c*x^2 + b*x)^(3/2)*B/x^2 - 1/3*sqrt(c*x^2 + b*x)*A*b/x^2 - 1/3*(c*x^2 + b*x)^(3/2)*A/x^3
Time = 0.28 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.42 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^4} \, dx=\sqrt {c x^{2} + b x} B c - \frac {{\left (3 \, B b c + 2 \, A c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{2 \, \sqrt {c}} + \frac {2 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{2} + 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b c + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{2} \sqrt {c} + A b^{3}\right )}}{3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3}} \]
sqrt(c*x^2 + b*x)*B*c - 1/2*(3*B*b*c + 2*A*c^2)*log(abs(2*(sqrt(c)*x - sqr t(c*x^2 + b*x))*sqrt(c) + b))/sqrt(c) + 2/3*(3*(sqrt(c)*x - sqrt(c*x^2 + b *x))^2*B*b^2 + 6*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b*c + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^2*sqrt(c) + A*b^3)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^ 3
Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (A+B\,x\right )}{x^4} \,d x \]