Integrand size = 22, antiderivative size = 153 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^4} \, dx=-\frac {5}{8} (b B+6 A c) (b+2 c x) \sqrt {b x+c x^2}-\frac {5 c (b B+6 A c) \left (b x+c x^2\right )^{3/2}}{3 b}+\frac {2 (b B+6 A c) \left (b x+c x^2\right )^{5/2}}{b x^2}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^4}+\frac {5 b^2 (b B+6 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 \sqrt {c}} \]
-5/3*c*(6*A*c+B*b)*(c*x^2+b*x)^(3/2)/b+2*(6*A*c+B*b)*(c*x^2+b*x)^(5/2)/b/x ^2-2*A*(c*x^2+b*x)^(7/2)/b/x^4+5/8*b^2*(6*A*c+B*b)*arctanh(x*c^(1/2)/(c*x^ 2+b*x)^(1/2))/c^(1/2)-5/8*(6*A*c+B*b)*(2*c*x+b)*(c*x^2+b*x)^(1/2)
Time = 0.45 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.82 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^4} \, dx=\frac {\sqrt {x (b+c x)} \left (-6 A \left (8 b^2-9 b c x-2 c^2 x^2\right )+B x \left (33 b^2+26 b c x+8 c^2 x^2\right )+\frac {30 b^2 (b B+6 A c) \sqrt {x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )}{\sqrt {c} \sqrt {b+c x}}\right )}{24 x} \]
(Sqrt[x*(b + c*x)]*(-6*A*(8*b^2 - 9*b*c*x - 2*c^2*x^2) + B*x*(33*b^2 + 26* b*c*x + 8*c^2*x^2) + (30*b^2*(b*B + 6*A*c)*Sqrt[x]*ArcTanh[(Sqrt[c]*Sqrt[x ])/(-Sqrt[b] + Sqrt[b + c*x])])/(Sqrt[c]*Sqrt[b + c*x])))/(24*x)
Time = 0.30 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1220, 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.
| \(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^4} \, dx\) |
| \(\Big \downarrow \) 1220 |
| \(\displaystyle \frac {(6 A c+b B) \int \frac {\left (c x^2+b x\right )^{5/2}}{x^3}dx}{b}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^4}\) |
| \(\Big \downarrow \) 1130 |
| \(\displaystyle \frac {(6 A c+b B) \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{x^2}-5 c \int \frac {\left (c x^2+b x\right )^{3/2}}{x}dx\right )}{b}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^4}\) |
| \(\Big \downarrow \) 1131 |
| \(\displaystyle \frac {(6 A c+b B) \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{x^2}-5 c \left (\frac {1}{2} b \int \sqrt {c x^2+b x}dx+\frac {1}{3} \left (b x+c x^2\right )^{3/2}\right )\right )}{b}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^4}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {(6 A c+b B) \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{x^2}-5 c \left (\frac {1}{2} b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )+\frac {1}{3} \left (b x+c x^2\right )^{3/2}\right )\right )}{b}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^4}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {(6 A c+b B) \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{x^2}-5 c \left (\frac {1}{2} b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )+\frac {1}{3} \left (b x+c x^2\right )^{3/2}\right )\right )}{b}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^4}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(6 A c+b B) \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{x^2}-5 c \left (\frac {1}{2} b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right )+\frac {1}{3} \left (b x+c x^2\right )^{3/2}\right )\right )}{b}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^4}\) |
(-2*A*(b*x + c*x^2)^(7/2))/(b*x^4) + ((b*B + 6*A*c)*((2*(b*x + c*x^2)^(5/2 ))/x^2 - 5*c*((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)))/b
3.1.100.3.1 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]
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 = 0.20 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.60
| method | result | size |
| pseudoelliptic | \(-\frac {2 \left (-\frac {15 x \left (A c +\frac {B b}{6}\right ) b^{2} \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )}{8}+\sqrt {x \left (c x +b \right )}\, \left (-\frac {9 x \left (\frac {13 B x}{27}+A \right ) b \,c^{\frac {3}{2}}}{8}-\frac {x^{2} \left (\frac {2 B x}{3}+A \right ) c^{\frac {5}{2}}}{4}+b^{2} \sqrt {c}\, \left (-\frac {11 B x}{16}+A \right )\right )\right )}{\sqrt {c}\, x}\) | \(92\) |
| risch | \(-\frac {\left (c x +b \right ) \left (-8 B \,c^{2} x^{3}-12 A \,c^{2} x^{2}-26 B b c \,x^{2}-54 A b c x -33 b^{2} B x +48 A \,b^{2}\right )}{24 \sqrt {x \left (c x +b \right )}}+\frac {5 \left (6 A c +B b \right ) b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 \sqrt {c}}\) | \(104\) |
| default | \(B \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{b \,x^{3}}-\frac {8 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{3 b \,x^{2}}-\frac {10 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5}+\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2}\right )}{3 b}\right )}{b}\right )+A \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{b \,x^{4}}+\frac {6 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{b \,x^{3}}-\frac {8 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{3 b \,x^{2}}-\frac {10 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5}+\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2}\right )}{3 b}\right )}{b}\right )}{b}\right )\) | \(342\) |
-2*(-15/8*x*(A*c+1/6*B*b)*b^2*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))+(x*(c*x +b))^(1/2)*(-9/8*x*(13/27*B*x+A)*b*c^(3/2)-1/4*x^2*(2/3*B*x+A)*c^(5/2)+b^2 *c^(1/2)*(-11/16*B*x+A)))/c^(1/2)/x
Time = 0.27 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.57 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^4} \, dx=\left [\frac {15 \, {\left (B b^{3} + 6 \, A b^{2} c\right )} \sqrt {c} x \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (8 \, B c^{3} x^{3} - 48 \, A b^{2} c + 2 \, {\left (13 \, B b c^{2} + 6 \, A c^{3}\right )} x^{2} + 3 \, {\left (11 \, B b^{2} c + 18 \, A b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{48 \, c x}, -\frac {15 \, {\left (B b^{3} + 6 \, A b^{2} c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (8 \, B c^{3} x^{3} - 48 \, A b^{2} c + 2 \, {\left (13 \, B b c^{2} + 6 \, A c^{3}\right )} x^{2} + 3 \, {\left (11 \, B b^{2} c + 18 \, A b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, c x}\right ] \]
[1/48*(15*(B*b^3 + 6*A*b^2*c)*sqrt(c)*x*log(2*c*x + b + 2*sqrt(c*x^2 + b*x )*sqrt(c)) + 2*(8*B*c^3*x^3 - 48*A*b^2*c + 2*(13*B*b*c^2 + 6*A*c^3)*x^2 + 3*(11*B*b^2*c + 18*A*b*c^2)*x)*sqrt(c*x^2 + b*x))/(c*x), -1/24*(15*(B*b^3 + 6*A*b^2*c)*sqrt(-c)*x*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (8*B*c^ 3*x^3 - 48*A*b^2*c + 2*(13*B*b*c^2 + 6*A*c^3)*x^2 + 3*(11*B*b^2*c + 18*A*b *c^2)*x)*sqrt(c*x^2 + b*x))/(c*x)]
\[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^4} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {5}{2}} \left (A + B x\right )}{x^{4}}\, dx \]
Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^4} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.91 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^4} \, dx=\frac {2 \, A b^{3}}{\sqrt {c} x - \sqrt {c x^{2} + b x}} + \frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, B c^{2} x + \frac {13 \, B b c^{3} + 6 \, A c^{4}}{c^{2}}\right )} x + \frac {3 \, {\left (11 \, B b^{2} c^{2} + 18 \, A b c^{3}\right )}}{c^{2}}\right )} - \frac {5 \, {\left (B b^{3} + 6 \, A b^{2} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{16 \, \sqrt {c}} \]
2*A*b^3/(sqrt(c)*x - sqrt(c*x^2 + b*x)) + 1/24*sqrt(c*x^2 + b*x)*(2*(4*B*c ^2*x + (13*B*b*c^3 + 6*A*c^4)/c^2)*x + 3*(11*B*b^2*c^2 + 18*A*b*c^3)/c^2) - 5/16*(B*b^3 + 6*A*b^2*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt( c) + b))/sqrt(c)
Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^4} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right )}{x^4} \,d x \]