Integrand size = 22, antiderivative size = 170 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x} \, dx=\frac {b^3 (5 b B-12 A c) (b+2 c x) \sqrt {b x+c x^2}}{512 c^3}-\frac {b (5 b B-12 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}-\frac {(5 b B-12 A c) \left (b x+c x^2\right )^{5/2}}{60 c}+\frac {B \left (b x+c x^2\right )^{7/2}}{6 c x}-\frac {b^5 (5 b B-12 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{512 c^{7/2}} \]
-1/192*b*(-12*A*c+5*B*b)*(2*c*x+b)*(c*x^2+b*x)^(3/2)/c^2-1/60*(-12*A*c+5*B *b)*(c*x^2+b*x)^(5/2)/c+1/6*B*(c*x^2+b*x)^(7/2)/c/x-1/512*b^5*(-12*A*c+5*B *b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(7/2)+1/512*b^3*(-12*A*c+5*B*b) *(2*c*x+b)*(c*x^2+b*x)^(1/2)/c^3
Time = 0.94 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.25 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x} \, dx=\frac {(x (b+c x))^{5/2} \left (75 b^5 B-180 A b^4 c-50 b^4 B c x+120 A b^3 c^2 x+40 b^3 B c^2 x^2+2976 A b^2 c^3 x^2+2160 b^2 B c^3 x^3+4032 A b c^4 x^3+3200 b B c^4 x^4+1536 A c^5 x^4+1280 B c^5 x^5\right )}{7680 c^3 x^2 (b+c x)^2}-\frac {b^5 (5 b B-12 A c) (x (b+c x))^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )}{256 c^{7/2} x^{5/2} (b+c x)^{5/2}} \]
((x*(b + c*x))^(5/2)*(75*b^5*B - 180*A*b^4*c - 50*b^4*B*c*x + 120*A*b^3*c^ 2*x + 40*b^3*B*c^2*x^2 + 2976*A*b^2*c^3*x^2 + 2160*b^2*B*c^3*x^3 + 4032*A* b*c^4*x^3 + 3200*b*B*c^4*x^4 + 1536*A*c^5*x^4 + 1280*B*c^5*x^5))/(7680*c^3 *x^2*(b + c*x)^2) - (b^5*(5*b*B - 12*A*c)*(x*(b + c*x))^(5/2)*ArcTanh[(Sqr t[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])])/(256*c^(7/2)*x^(5/2)*(b + c*x)^ (5/2))
Time = 0.29 (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 = {1221, 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.
| \(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x} \, dx\) |
| \(\Big \downarrow \) 1221 |
| \(\displaystyle \frac {B \left (b x+c x^2\right )^{7/2}}{6 c x}-\frac {(5 b B-12 A c) \int \frac {\left (c x^2+b x\right )^{5/2}}{x}dx}{12 c}\) |
| \(\Big \downarrow \) 1131 |
| \(\displaystyle \frac {B \left (b x+c x^2\right )^{7/2}}{6 c x}-\frac {(5 b B-12 A c) \left (\frac {1}{2} b \int \left (c x^2+b x\right )^{3/2}dx+\frac {1}{5} \left (b x+c x^2\right )^{5/2}\right )}{12 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {B \left (b x+c x^2\right )^{7/2}}{6 c x}-\frac {(5 b B-12 A c) \left (\frac {1}{2} b \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \int \sqrt {c x^2+b x}dx}{16 c}\right )+\frac {1}{5} \left (b x+c x^2\right )^{5/2}\right )}{12 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {B \left (b x+c x^2\right )^{7/2}}{6 c x}-\frac {(5 b B-12 A c) \left (\frac {1}{2} b \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )}{16 c}\right )+\frac {1}{5} \left (b x+c x^2\right )^{5/2}\right )}{12 c}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {B \left (b x+c x^2\right )^{7/2}}{6 c x}-\frac {(5 b B-12 A c) \left (\frac {1}{2} b \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )}{16 c}\right )+\frac {1}{5} \left (b x+c x^2\right )^{5/2}\right )}{12 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {B \left (b x+c x^2\right )^{7/2}}{6 c x}-\frac {(5 b B-12 A c) \left (\frac {1}{2} b \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right )}{16 c}\right )+\frac {1}{5} \left (b x+c x^2\right )^{5/2}\right )}{12 c}\) |
(B*(b*x + c*x^2)^(7/2))/(6*c*x) - ((5*b*B - 12*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))/(12*c)
3.1.97.3.1 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[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 )/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c *f - b*g))/(c*e*(m + 2*p + 2)) Int[(d + e*x)^m*(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] && NeQ[m + 2*p + 2, 0]
Time = 0.11 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.99
| method | result | size |
| risch | \(-\frac {\left (-1280 B \,c^{5} x^{5}-1536 A \,c^{5} x^{4}-3200 B b \,c^{4} x^{4}-4032 A b \,c^{4} x^{3}-2160 B \,b^{2} c^{3} x^{3}-2976 A \,b^{2} c^{3} x^{2}-40 B \,b^{3} c^{2} x^{2}-120 A \,b^{3} c^{2} x +50 B \,b^{4} c x +180 A \,b^{4} c -75 B \,b^{5}\right ) x \left (c x +b \right )}{7680 c^{3} \sqrt {x \left (c x +b \right )}}+\frac {b^{5} \left (12 A c -5 B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{1024 c^{\frac {7}{2}}}\) | \(169\) |
| default | \(B \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{12 c}-\frac {5 b^{2} \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{24 c}\right )+A \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5}+\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2}\right )\) | \(226\) |
-1/7680/c^3*(-1280*B*c^5*x^5-1536*A*c^5*x^4-3200*B*b*c^4*x^4-4032*A*b*c^4* x^3-2160*B*b^2*c^3*x^3-2976*A*b^2*c^3*x^2-40*B*b^3*c^2*x^2-120*A*b^3*c^2*x +50*B*b^4*c*x+180*A*b^4*c-75*B*b^5)*x*(c*x+b)/(x*(c*x+b))^(1/2)+1/1024*b^5 *(12*A*c-5*B*b)/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))
Time = 0.27 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.06 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x} \, dx=\left [-\frac {15 \, {\left (5 \, B b^{6} - 12 \, A b^{5} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (1280 \, B c^{6} x^{5} + 75 \, B b^{5} c - 180 \, A b^{4} c^{2} + 128 \, {\left (25 \, B b c^{5} + 12 \, A c^{6}\right )} x^{4} + 144 \, {\left (15 \, B b^{2} c^{4} + 28 \, A b c^{5}\right )} x^{3} + 8 \, {\left (5 \, B b^{3} c^{3} + 372 \, A b^{2} c^{4}\right )} x^{2} - 10 \, {\left (5 \, B b^{4} c^{2} - 12 \, A b^{3} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{15360 \, c^{4}}, \frac {15 \, {\left (5 \, B b^{6} - 12 \, A b^{5} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (1280 \, B c^{6} x^{5} + 75 \, B b^{5} c - 180 \, A b^{4} c^{2} + 128 \, {\left (25 \, B b c^{5} + 12 \, A c^{6}\right )} x^{4} + 144 \, {\left (15 \, B b^{2} c^{4} + 28 \, A b c^{5}\right )} x^{3} + 8 \, {\left (5 \, B b^{3} c^{3} + 372 \, A b^{2} c^{4}\right )} x^{2} - 10 \, {\left (5 \, B b^{4} c^{2} - 12 \, A b^{3} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{7680 \, c^{4}}\right ] \]
[-1/15360*(15*(5*B*b^6 - 12*A*b^5*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(1280*B*c^6*x^5 + 75*B*b^5*c - 180*A*b^4*c^2 + 128*(25 *B*b*c^5 + 12*A*c^6)*x^4 + 144*(15*B*b^2*c^4 + 28*A*b*c^5)*x^3 + 8*(5*B*b^ 3*c^3 + 372*A*b^2*c^4)*x^2 - 10*(5*B*b^4*c^2 - 12*A*b^3*c^3)*x)*sqrt(c*x^2 + b*x))/c^4, 1/7680*(15*(5*B*b^6 - 12*A*b^5*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (1280*B*c^6*x^5 + 75*B*b^5*c - 180*A*b^4*c^2 + 1 28*(25*B*b*c^5 + 12*A*c^6)*x^4 + 144*(15*B*b^2*c^4 + 28*A*b*c^5)*x^3 + 8*( 5*B*b^3*c^3 + 372*A*b^2*c^4)*x^2 - 10*(5*B*b^4*c^2 - 12*A*b^3*c^3)*x)*sqrt (c*x^2 + b*x))/c^4]
Time = 3.13 (sec) , antiderivative size = 886, normalized size of antiderivative = 5.21 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x} \, dx=\text {Too large to display} \]
A*b**2*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, Tru e)) + 2*A*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)/s qrt(c*(b/(2*c) + x)**2), True))/(128*c**3) + sqrt(b*x + c*x**2)*(5*b**3/(6 4*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)) + A*c**2*Piecewise((7*b**5*Piecew ise((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))/(256*c* *4) + sqrt(b*x + c*x**2)*(-7*b**4/(128*c**4) + 7*b**3*x/(192*c**3) - 7*b** 2*x**2/(240*c**2) + b*x**3/(40*c) + x**4/5), Ne(c, 0)), (2*(b*x)**(9/2)/(9 *b**4), Ne(b, 0)), (0, True)) + B*b**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)) + 2*B*b* c*Piecewise((7*b**5*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/(...
Time = 0.19 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.59 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x} \, dx=\frac {1}{6} \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} B x + \frac {1}{8} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b x + \frac {5 \, \sqrt {c x^{2} + b x} B b^{4} x}{256 \, c^{2}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{2} x}{96 \, c} - \frac {3 \, \sqrt {c x^{2} + b x} A b^{3} x}{64 \, c} - \frac {5 \, B b^{6} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{1024 \, c^{\frac {7}{2}}} + \frac {3 \, A b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {5}{2}}} + \frac {1}{5} \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} A + \frac {5 \, \sqrt {c x^{2} + b x} B b^{5}}{512 \, c^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{3}}{192 \, c^{2}} - \frac {3 \, \sqrt {c x^{2} + b x} A b^{4}}{128 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B b}{12 \, c} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b^{2}}{16 \, c} \]
1/6*(c*x^2 + b*x)^(5/2)*B*x + 1/8*(c*x^2 + b*x)^(3/2)*A*b*x + 5/256*sqrt(c *x^2 + b*x)*B*b^4*x/c^2 - 5/96*(c*x^2 + b*x)^(3/2)*B*b^2*x/c - 3/64*sqrt(c *x^2 + b*x)*A*b^3*x/c - 5/1024*B*b^6*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*s qrt(c))/c^(7/2) + 3/256*A*b^5*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) /c^(5/2) + 1/5*(c*x^2 + b*x)^(5/2)*A + 5/512*sqrt(c*x^2 + b*x)*B*b^5/c^3 - 5/192*(c*x^2 + b*x)^(3/2)*B*b^3/c^2 - 3/128*sqrt(c*x^2 + b*x)*A*b^4/c^2 + 1/12*(c*x^2 + b*x)^(5/2)*B*b/c + 1/16*(c*x^2 + b*x)^(3/2)*A*b^2/c
Time = 0.30 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.15 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x} \, dx=\frac {1}{7680} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, B c^{2} x + \frac {25 \, B b c^{6} + 12 \, A c^{7}}{c^{5}}\right )} x + \frac {9 \, {\left (15 \, B b^{2} c^{5} + 28 \, A b c^{6}\right )}}{c^{5}}\right )} x + \frac {5 \, B b^{3} c^{4} + 372 \, A b^{2} c^{5}}{c^{5}}\right )} x - \frac {5 \, {\left (5 \, B b^{4} c^{3} - 12 \, A b^{3} c^{4}\right )}}{c^{5}}\right )} x + \frac {15 \, {\left (5 \, B b^{5} c^{2} - 12 \, A b^{4} c^{3}\right )}}{c^{5}}\right )} + \frac {{\left (5 \, B b^{6} - 12 \, A b^{5} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{1024 \, c^{\frac {7}{2}}} \]
1/7680*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*(10*B*c^2*x + (25*B*b*c^6 + 12*A*c^7) /c^5)*x + 9*(15*B*b^2*c^5 + 28*A*b*c^6)/c^5)*x + (5*B*b^3*c^4 + 372*A*b^2* c^5)/c^5)*x - 5*(5*B*b^4*c^3 - 12*A*b^3*c^4)/c^5)*x + 15*(5*B*b^5*c^2 - 12 *A*b^4*c^3)/c^5) + 1/1024*(5*B*b^6 - 12*A*b^5*c)*log(abs(2*(sqrt(c)*x - sq rt(c*x^2 + b*x))*sqrt(c) + b))/c^(7/2)
Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right )}{x} \,d x \]