Integrand size = 22, antiderivative size = 169 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^2} \, dx=-\frac {b^2 (3 b B-10 A c) (b+2 c x) \sqrt {b x+c x^2}}{128 c^2}+\frac {(3 b B-10 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{48 c}+\frac {(3 b B-10 A c) \left (b x+c x^2\right )^{5/2}}{15 b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}+\frac {b^4 (3 b B-10 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{5/2}} \]
1/48*(-10*A*c+3*B*b)*(2*c*x+b)*(c*x^2+b*x)^(3/2)/c+1/15*(-10*A*c+3*B*b)*(c *x^2+b*x)^(5/2)/b+2/3*A*(c*x^2+b*x)^(7/2)/b/x^2+1/128*b^4*(-10*A*c+3*B*b)* arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(5/2)-1/128*b^2*(-10*A*c+3*B*b)*(2* c*x+b)*(c*x^2+b*x)^(1/2)/c^2
Time = 0.74 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^2} \, dx=\frac {\sqrt {x} \sqrt {b+c x} \left (\sqrt {c} \sqrt {x} \sqrt {b+c x} \left (-45 b^4 B+30 b^3 c (5 A+B x)+96 c^4 x^3 (5 A+4 B x)+16 b c^3 x^2 (85 A+63 B x)+4 b^2 c^2 x (295 A+186 B x)\right )+300 A b^4 c \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )+90 b^5 B \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )\right )}{1920 c^{5/2} \sqrt {x (b+c x)}} \]
(Sqrt[x]*Sqrt[b + c*x]*(Sqrt[c]*Sqrt[x]*Sqrt[b + c*x]*(-45*b^4*B + 30*b^3* c*(5*A + B*x) + 96*c^4*x^3*(5*A + 4*B*x) + 16*b*c^3*x^2*(85*A + 63*B*x) + 4*b^2*c^2*x*(295*A + 186*B*x)) + 300*A*b^4*c*ArcTanh[(Sqrt[c]*Sqrt[x])/(Sq rt[b] - Sqrt[b + c*x])] + 90*b^5*B*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + S qrt[b + c*x])]))/(1920*c^(5/2)*Sqrt[x*(b + c*x)])
Time = 0.31 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1220, 1131, 1087, 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.
| \(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^2} \, dx\) |
| \(\Big \downarrow \) 1220 |
| \(\displaystyle \frac {(3 b B-10 A c) \int \frac {\left (c x^2+b x\right )^{5/2}}{x}dx}{3 b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}\) |
| \(\Big \downarrow \) 1131 |
| \(\displaystyle \frac {(3 b B-10 A c) \left (\frac {1}{2} b \int \left (c x^2+b x\right )^{3/2}dx+\frac {1}{5} \left (b x+c x^2\right )^{5/2}\right )}{3 b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {(3 b B-10 A c) \left (\frac {1}{2} b \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \int \sqrt {c x^2+b x}dx}{16 c}\right )+\frac {1}{5} \left (b x+c x^2\right )^{5/2}\right )}{3 b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {(3 b B-10 A c) \left (\frac {1}{2} b \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )}{16 c}\right )+\frac {1}{5} \left (b x+c x^2\right )^{5/2}\right )}{3 b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {(3 b B-10 A c) \left (\frac {1}{2} b \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )}{16 c}\right )+\frac {1}{5} \left (b x+c x^2\right )^{5/2}\right )}{3 b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(3 b B-10 A c) \left (\frac {1}{2} b \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right )}{16 c}\right )+\frac {1}{5} \left (b x+c x^2\right )^{5/2}\right )}{3 b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}\) |
(2*A*(b*x + c*x^2)^(7/2))/(3*b*x^2) + ((3*b*B - 10*A*c)*((b*x + c*x^2)^(5/ 2)/5 + (b*(((b + 2*c*x)*(b*x + c*x^2)^(3/2))/(8*c) - (3*b^2*(((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))))/(16*c)))/2))/(3*b)
3.1.98.3.1 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 = 116, normalized size of antiderivative = 0.69
| method | result | size |
| pseudoelliptic | \(\frac {\frac {59 \left (-\frac {15}{118} A \,b^{4} c +\frac {9}{236} B \,b^{5}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )}{96}+\frac {59 \sqrt {x \left (c x +b \right )}\, \left (\frac {15 \left (\frac {B x}{5}+A \right ) b^{3} c^{\frac {3}{2}}}{118}+b^{2} x \left (\frac {186 B x}{295}+A \right ) c^{\frac {5}{2}}+\frac {68 \left (\frac {63 B x}{85}+A \right ) x^{2} b \,c^{\frac {7}{2}}}{59}+\frac {24 \left (\frac {4 B x}{5}+A \right ) x^{3} c^{\frac {9}{2}}}{59}-\frac {9 B \,b^{4} \sqrt {c}}{236}\right )}{96}}{c^{\frac {5}{2}}}\) | \(116\) |
| risch | \(\frac {\left (384 B \,c^{4} x^{4}+480 A \,c^{4} x^{3}+1008 B b \,c^{3} x^{3}+1360 A b \,c^{3} x^{2}+744 B \,b^{2} c^{2} x^{2}+1180 A \,b^{2} c^{2} x +30 B \,b^{3} c x +150 A \,b^{3} c -45 b^{4} B \right ) x \left (c x +b \right )}{1920 c^{2} \sqrt {x \left (c x +b \right )}}-\frac {b^{4} \left (10 A c -3 B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {5}{2}}}\) | \(145\) |
| default | \(B \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5}+\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2}\right )+A \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{3 b \,x^{2}}-\frac {10 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5}+\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2}\right )}{3 b}\right )\) | \(238\) |
59/96*((-15/118*A*b^4*c+9/236*B*b^5)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))+ (x*(c*x+b))^(1/2)*(15/118*(1/5*B*x+A)*b^3*c^(3/2)+b^2*x*(186/295*B*x+A)*c^ (5/2)+68/59*(63/85*B*x+A)*x^2*b*c^(7/2)+24/59*(4/5*B*x+A)*x^3*c^(9/2)-9/23 6*B*b^4*c^(1/2)))/c^(5/2)
Time = 0.28 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.80 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^2} \, dx=\left [-\frac {15 \, {\left (3 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (384 \, B c^{5} x^{4} - 45 \, B b^{4} c + 150 \, A b^{3} c^{2} + 48 \, {\left (21 \, B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 8 \, {\left (93 \, B b^{2} c^{3} + 170 \, A b c^{4}\right )} x^{2} + 10 \, {\left (3 \, B b^{3} c^{2} + 118 \, A b^{2} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{3840 \, c^{3}}, -\frac {15 \, {\left (3 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (384 \, B c^{5} x^{4} - 45 \, B b^{4} c + 150 \, A b^{3} c^{2} + 48 \, {\left (21 \, B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 8 \, {\left (93 \, B b^{2} c^{3} + 170 \, A b c^{4}\right )} x^{2} + 10 \, {\left (3 \, B b^{3} c^{2} + 118 \, A b^{2} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{1920 \, c^{3}}\right ] \]
[-1/3840*(15*(3*B*b^5 - 10*A*b^4*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(384*B*c^5*x^4 - 45*B*b^4*c + 150*A*b^3*c^2 + 48*(21*B* b*c^4 + 10*A*c^5)*x^3 + 8*(93*B*b^2*c^3 + 170*A*b*c^4)*x^2 + 10*(3*B*b^3*c ^2 + 118*A*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^3, -1/1920*(15*(3*B*b^5 - 10*A *b^4*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (384*B*c^5*x^4 - 45*B*b^4*c + 150*A*b^3*c^2 + 48*(21*B*b*c^4 + 10*A*c^5)*x^3 + 8*(93*B*b ^2*c^3 + 170*A*b*c^4)*x^2 + 10*(3*B*b^3*c^2 + 118*A*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^3]
Time = 0.81 (sec) , antiderivative size = 796, normalized size of antiderivative = 4.71 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^2} \, dx=\text {Too large to display} \]
A*b**2*Piecewise((-b**2*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/ (2*c) + x)**2), True))/(8*c) + (b/(4*c) + x/2)*sqrt(b*x + c*x**2), Ne(c, 0 )), (2*(b*x)**(3/2)/(3*b), Ne(b, 0)), (0, True)) + 2*A*b*c*Piecewise((b**3 *Piecewise((log(b + 2*sqrt(c)*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2 /c, 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True))/ (16*c**2) + sqrt(b*x + c*x**2)*(-b**2/(8*c**2) + b*x/(12*c) + x**2/3), Ne( c, 0)), (2*(b*x)**(5/2)/(5*b**2), Ne(b, 0)), (0, True)) + A*c**2*Piecewise ((-5*b**4*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c) , Ne(b**2/c, 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2) , True))/(128*c**3) + sqrt(b*x + c*x**2)*(5*b**3/(64*c**3) - 5*b**2*x/(96* c**2) + b*x**2/(24*c) + x**3/4), Ne(c, 0)), (2*(b*x)**(7/2)/(7*b**3), Ne(b , 0)), (0, True)) + B*b**2*Piecewise((b**3*Piecewise((log(b + 2*sqrt(c)*sq rt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x)*log(b/(2 *c) + x)/sqrt(c*(b/(2*c) + x)**2), True))/(16*c**2) + sqrt(b*x + c*x**2)*( -b**2/(8*c**2) + b*x/(12*c) + x**2/3), Ne(c, 0)), (2*(b*x)**(5/2)/(5*b**2) , Ne(b, 0)), (0, True)) + 2*B*b*c*Piecewise((-5*b**4*Piecewise((log(b + 2* sqrt(c)*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x )*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True))/(128*c**3) + sqrt(b*x + c*x**2)*(5*b**3/(64*c**3) - 5*b**2*x/(96*c**2) + b*x**2/(24*c) + x**3...
Time = 0.19 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.34 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^2} \, dx=\frac {1}{8} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b x + \frac {5}{32} \, \sqrt {c x^{2} + b x} A b^{2} x - \frac {3 \, \sqrt {c x^{2} + b x} B b^{3} x}{64 \, c} + \frac {3 \, B b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {5}{2}}} - \frac {5 \, A b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {3}{2}}} + \frac {1}{5} \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} B + \frac {5}{24} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b - \frac {3 \, \sqrt {c x^{2} + b x} B b^{4}}{128 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{2}}{16 \, c} + \frac {5 \, \sqrt {c x^{2} + b x} A b^{3}}{64 \, c} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A}{4 \, x} \]
1/8*(c*x^2 + b*x)^(3/2)*B*b*x + 5/32*sqrt(c*x^2 + b*x)*A*b^2*x - 3/64*sqrt (c*x^2 + b*x)*B*b^3*x/c + 3/256*B*b^5*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)* sqrt(c))/c^(5/2) - 5/128*A*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c) )/c^(3/2) + 1/5*(c*x^2 + b*x)^(5/2)*B + 5/24*(c*x^2 + b*x)^(3/2)*A*b - 3/1 28*sqrt(c*x^2 + b*x)*B*b^4/c^2 + 1/16*(c*x^2 + b*x)^(3/2)*B*b^2/c + 5/64*s qrt(c*x^2 + b*x)*A*b^3/c + 1/4*(c*x^2 + b*x)^(5/2)*A/x
Time = 0.28 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.99 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^2} \, dx=\frac {1}{1920} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, B c^{2} x + \frac {21 \, B b c^{5} + 10 \, A c^{6}}{c^{4}}\right )} x + \frac {93 \, B b^{2} c^{4} + 170 \, A b c^{5}}{c^{4}}\right )} x + \frac {5 \, {\left (3 \, B b^{3} c^{3} + 118 \, A b^{2} c^{4}\right )}}{c^{4}}\right )} x - \frac {15 \, {\left (3 \, B b^{4} c^{2} - 10 \, A b^{3} c^{3}\right )}}{c^{4}}\right )} - \frac {{\left (3 \, B b^{5} - 10 \, A b^{4} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{256 \, c^{\frac {5}{2}}} \]
1/1920*sqrt(c*x^2 + b*x)*(2*(4*(6*(8*B*c^2*x + (21*B*b*c^5 + 10*A*c^6)/c^4 )*x + (93*B*b^2*c^4 + 170*A*b*c^5)/c^4)*x + 5*(3*B*b^3*c^3 + 118*A*b^2*c^4 )/c^4)*x - 15*(3*B*b^4*c^2 - 10*A*b^3*c^3)/c^4) - 1/256*(3*B*b^5 - 10*A*b^ 4*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(5/2)
Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^2} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right )}{x^2} \,d x \]