Integrand size = 22, antiderivative size = 155 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^3} \, dx=\frac {5 b (b B-8 A c) (b+2 c x) \sqrt {b x+c x^2}}{64 c}+\frac {5}{24} (b B-8 A c) \left (b x+c x^2\right )^{3/2}+\frac {(b B-8 A c) \left (b x+c x^2\right )^{5/2}}{4 b x}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^3}-\frac {5 b^3 (b B-8 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{3/2}} \]
5/24*(-8*A*c+B*b)*(c*x^2+b*x)^(3/2)+1/4*(-8*A*c+B*b)*(c*x^2+b*x)^(5/2)/b/x +2*A*(c*x^2+b*x)^(7/2)/b/x^3-5/64*b^3*(-8*A*c+B*b)*arctanh(x*c^(1/2)/(c*x^ 2+b*x)^(1/2))/c^(3/2)+5/64*b*(-8*A*c+B*b)*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c
Time = 0.23 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.83 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^3} \, dx=\frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (15 b^3 B+16 c^3 x^2 (4 A+3 B x)+8 b c^2 x (26 A+17 B x)+2 b^2 c (132 A+59 B x)\right )+\frac {15 b^3 (b B-8 A c) \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{\sqrt {x} \sqrt {b+c x}}\right )}{192 c^{3/2}} \]
(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^3*B + 16*c^3*x^2*(4*A + 3*B*x) + 8*b*c^2 *x*(26*A + 17*B*x) + 2*b^2*c*(132*A + 59*B*x)) + (15*b^3*(b*B - 8*A*c)*Log [-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]])/(Sqrt[x]*Sqrt[b + c*x])))/(192*c^(3/ 2))
Time = 0.30 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1220, 1131, 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.
| \(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^3} \, dx\) |
| \(\Big \downarrow \) 1220 |
| \(\displaystyle \frac {(b B-8 A c) \int \frac {\left (c x^2+b x\right )^{5/2}}{x^2}dx}{b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^3}\) |
| \(\Big \downarrow \) 1131 |
| \(\displaystyle \frac {(b B-8 A c) \left (\frac {5}{8} b \int \frac {\left (c x^2+b x\right )^{3/2}}{x}dx+\frac {\left (b x+c x^2\right )^{5/2}}{4 x}\right )}{b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^3}\) |
| \(\Big \downarrow \) 1131 |
| \(\displaystyle \frac {(b B-8 A c) \left (\frac {5}{8} b \left (\frac {1}{2} b \int \sqrt {c x^2+b x}dx+\frac {1}{3} \left (b x+c x^2\right )^{3/2}\right )+\frac {\left (b x+c x^2\right )^{5/2}}{4 x}\right )}{b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^3}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {(b B-8 A c) \left (\frac {5}{8} b \left (\frac {1}{2} b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )+\frac {1}{3} \left (b x+c x^2\right )^{3/2}\right )+\frac {\left (b x+c x^2\right )^{5/2}}{4 x}\right )}{b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^3}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {(b B-8 A c) \left (\frac {5}{8} b \left (\frac {1}{2} b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )+\frac {1}{3} \left (b x+c x^2\right )^{3/2}\right )+\frac {\left (b x+c x^2\right )^{5/2}}{4 x}\right )}{b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^3}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(b B-8 A c) \left (\frac {5}{8} b \left (\frac {1}{2} b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right )+\frac {1}{3} \left (b x+c x^2\right )^{3/2}\right )+\frac {\left (b x+c x^2\right )^{5/2}}{4 x}\right )}{b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^3}\) |
(2*A*(b*x + c*x^2)^(7/2))/(b*x^3) + ((b*B - 8*A*c)*((b*x + c*x^2)^(5/2)/(4 *x) + (5*b*((b*x + c*x^2)^(3/2)/3 + (b*(((b + 2*c*x)*Sqrt[b*x + c*x^2])/(4 *c) - (b^2*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(3/2))))/2))/8))/b
3.1.99.3.1 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 + 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]
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 = 0.18 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.63
| method | result | size |
| pseudoelliptic | \(\frac {\frac {5 \left (A c -\frac {B b}{8}\right ) b^{3} \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )}{8}+\frac {13 \left (\frac {33 \left (\frac {59 B x}{132}+A \right ) b^{2} c^{\frac {3}{2}}}{26}+b x \left (\frac {17 B x}{26}+A \right ) c^{\frac {5}{2}}+\frac {4 x^{2} \left (\frac {3 B x}{4}+A \right ) c^{\frac {7}{2}}}{13}+\frac {15 B \sqrt {c}\, b^{3}}{208}\right ) \sqrt {x \left (c x +b \right )}}{12}}{c^{\frac {3}{2}}}\) | \(97\) |
| risch | \(\frac {\left (48 B \,c^{3} x^{3}+64 A \,c^{3} x^{2}+136 B b \,c^{2} x^{2}+208 A b \,c^{2} x +118 B \,b^{2} c x +264 A \,b^{2} c +15 B \,b^{3}\right ) x \left (c x +b \right )}{192 c \sqrt {x \left (c x +b \right )}}+\frac {5 b^{3} \left (8 A c -B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {3}{2}}}\) | \(121\) |
| default | \(A \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{b \,x^{3}}-\frac {8 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{3 b \,x^{2}}-\frac {10 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5}+\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2}\right )}{3 b}\right )}{b}\right )+B \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{3 b \,x^{2}}-\frac {10 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5}+\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2}\right )}{3 b}\right )\) | \(290\) |
13/12*(15/26*(A*c-1/8*B*b)*b^3*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))+(33/26 *(59/132*B*x+A)*b^2*c^(3/2)+b*x*(17/26*B*x+A)*c^(5/2)+4/13*x^2*(3/4*B*x+A) *c^(7/2)+15/208*B*c^(1/2)*b^3)*(x*(c*x+b))^(1/2))/c^(3/2)
Time = 0.27 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.63 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^3} \, dx=\left [-\frac {15 \, {\left (B b^{4} - 8 \, A b^{3} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (48 \, B c^{4} x^{3} + 15 \, B b^{3} c + 264 \, A b^{2} c^{2} + 8 \, {\left (17 \, B b c^{3} + 8 \, A c^{4}\right )} x^{2} + 2 \, {\left (59 \, B b^{2} c^{2} + 104 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{384 \, c^{2}}, \frac {15 \, {\left (B b^{4} - 8 \, A b^{3} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (48 \, B c^{4} x^{3} + 15 \, B b^{3} c + 264 \, A b^{2} c^{2} + 8 \, {\left (17 \, B b c^{3} + 8 \, A c^{4}\right )} x^{2} + 2 \, {\left (59 \, B b^{2} c^{2} + 104 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{192 \, c^{2}}\right ] \]
[-1/384*(15*(B*b^4 - 8*A*b^3*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x )*sqrt(c)) - 2*(48*B*c^4*x^3 + 15*B*b^3*c + 264*A*b^2*c^2 + 8*(17*B*b*c^3 + 8*A*c^4)*x^2 + 2*(59*B*b^2*c^2 + 104*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^2, 1/192*(15*(B*b^4 - 8*A*b^3*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/ (c*x)) + (48*B*c^4*x^3 + 15*B*b^3*c + 264*A*b^2*c^2 + 8*(17*B*b*c^3 + 8*A* c^4)*x^2 + 2*(59*B*b^2*c^2 + 104*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^2]
\[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^3} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {5}{2}} \left (A + B x\right )}{x^{3}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.21 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^3} \, dx=\frac {5}{32} \, \sqrt {c x^{2} + b x} B b^{2} x - \frac {5 \, B b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {3}{2}}} + \frac {5 \, A b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, \sqrt {c}} + \frac {5}{24} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b + \frac {5}{8} \, \sqrt {c x^{2} + b x} A b^{2} + \frac {5 \, \sqrt {c x^{2} + b x} B b^{3}}{64 \, c} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B}{4 \, x} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b}{12 \, x} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A}{3 \, x^{2}} \]
5/32*sqrt(c*x^2 + b*x)*B*b^2*x - 5/128*B*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) + 5/16*A*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*s qrt(c))/sqrt(c) + 5/24*(c*x^2 + b*x)^(3/2)*B*b + 5/8*sqrt(c*x^2 + b*x)*A*b ^2 + 5/64*sqrt(c*x^2 + b*x)*B*b^3/c + 1/4*(c*x^2 + b*x)^(5/2)*B/x + 5/12*( c*x^2 + b*x)^(3/2)*A*b/x + 1/3*(c*x^2 + b*x)^(5/2)*A/x^2
Time = 0.28 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^3} \, dx=\frac {1}{192} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, B c^{2} x + \frac {17 \, B b c^{4} + 8 \, A c^{5}}{c^{3}}\right )} x + \frac {59 \, B b^{2} c^{3} + 104 \, A b c^{4}}{c^{3}}\right )} x + \frac {3 \, {\left (5 \, B b^{3} c^{2} + 88 \, A b^{2} c^{3}\right )}}{c^{3}}\right )} + \frac {5 \, {\left (B b^{4} - 8 \, A b^{3} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {3}{2}}} \]
1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*B*c^2*x + (17*B*b*c^4 + 8*A*c^5)/c^3)*x + (59*B*b^2*c^3 + 104*A*b*c^4)/c^3)*x + 3*(5*B*b^3*c^2 + 88*A*b^2*c^3)/c^3) + 5/128*(B*b^4 - 8*A*b^3*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqr t(c) + b))/c^(3/2)
Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^3} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right )}{x^3} \,d x \]