Integrand size = 22, antiderivative size = 197 \[ \int \frac {x^4 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {7 b^3 (9 b B-10 A c) \sqrt {b x+c x^2}}{128 c^5}-\frac {7 b^2 (9 b B-10 A c) x \sqrt {b x+c x^2}}{192 c^4}+\frac {7 b (9 b B-10 A c) x^2 \sqrt {b x+c x^2}}{240 c^3}-\frac {(9 b B-10 A c) x^3 \sqrt {b x+c x^2}}{40 c^2}+\frac {B x^4 \sqrt {b x+c x^2}}{5 c}-\frac {7 b^4 (9 b B-10 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{11/2}} \]
-7/128*b^4*(-10*A*c+9*B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(11/2)+7 /128*b^3*(-10*A*c+9*B*b)*(c*x^2+b*x)^(1/2)/c^5-7/192*b^2*(-10*A*c+9*B*b)*x *(c*x^2+b*x)^(1/2)/c^4+7/240*b*(-10*A*c+9*B*b)*x^2*(c*x^2+b*x)^(1/2)/c^3-1 /40*(-10*A*c+9*B*b)*x^3*(c*x^2+b*x)^(1/2)/c^2+1/5*B*x^4*(c*x^2+b*x)^(1/2)/ c
Time = 0.63 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.83 \[ \int \frac {x^4 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c} x (b+c x) \left (945 b^4 B-210 b^3 c (5 A+3 B x)+96 c^4 x^3 (5 A+4 B x)+28 b^2 c^2 x (25 A+18 B x)-16 b c^3 x^2 (35 A+27 B x)\right )+210 b^4 (9 b B-10 A c) \sqrt {x} \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )}{1920 c^{11/2} \sqrt {x (b+c x)}} \]
(Sqrt[c]*x*(b + c*x)*(945*b^4*B - 210*b^3*c*(5*A + 3*B*x) + 96*c^4*x^3*(5* A + 4*B*x) + 28*b^2*c^2*x*(25*A + 18*B*x) - 16*b*c^3*x^2*(35*A + 27*B*x)) + 210*b^4*(9*b*B - 10*A*c)*Sqrt[x]*Sqrt[b + c*x]*ArcTanh[(Sqrt[c]*Sqrt[x]) /(Sqrt[b] - Sqrt[b + c*x])])/(1920*c^(11/2)*Sqrt[x*(b + c*x)])
Time = 0.35 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1221, 1134, 1134, 1134, 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.
| \(\displaystyle \int \frac {x^4 (A+B x)}{\sqrt {b x+c x^2}} \, dx\) |
| \(\Big \downarrow \) 1221 |
| \(\displaystyle \frac {B x^4 \sqrt {b x+c x^2}}{5 c}-\frac {(9 b B-10 A c) \int \frac {x^4}{\sqrt {c x^2+b x}}dx}{10 c}\) |
| \(\Big \downarrow \) 1134 |
| \(\displaystyle \frac {B x^4 \sqrt {b x+c x^2}}{5 c}-\frac {(9 b B-10 A c) \left (\frac {x^3 \sqrt {b x+c x^2}}{4 c}-\frac {7 b \int \frac {x^3}{\sqrt {c x^2+b x}}dx}{8 c}\right )}{10 c}\) |
| \(\Big \downarrow \) 1134 |
| \(\displaystyle \frac {B x^4 \sqrt {b x+c x^2}}{5 c}-\frac {(9 b B-10 A c) \left (\frac {x^3 \sqrt {b x+c x^2}}{4 c}-\frac {7 b \left (\frac {x^2 \sqrt {b x+c x^2}}{3 c}-\frac {5 b \int \frac {x^2}{\sqrt {c x^2+b x}}dx}{6 c}\right )}{8 c}\right )}{10 c}\) |
| \(\Big \downarrow \) 1134 |
| \(\displaystyle \frac {B x^4 \sqrt {b x+c x^2}}{5 c}-\frac {(9 b B-10 A c) \left (\frac {x^3 \sqrt {b x+c x^2}}{4 c}-\frac {7 b \left (\frac {x^2 \sqrt {b x+c x^2}}{3 c}-\frac {5 b \left (\frac {x \sqrt {b x+c x^2}}{2 c}-\frac {3 b \int \frac {x}{\sqrt {c x^2+b x}}dx}{4 c}\right )}{6 c}\right )}{8 c}\right )}{10 c}\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {B x^4 \sqrt {b x+c x^2}}{5 c}-\frac {(9 b B-10 A c) \left (\frac {x^3 \sqrt {b x+c x^2}}{4 c}-\frac {7 b \left (\frac {x^2 \sqrt {b x+c x^2}}{3 c}-\frac {5 b \left (\frac {x \sqrt {b x+c x^2}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x+c x^2}}{c}-\frac {b \int \frac {1}{\sqrt {c x^2+b x}}dx}{2 c}\right )}{4 c}\right )}{6 c}\right )}{8 c}\right )}{10 c}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {B x^4 \sqrt {b x+c x^2}}{5 c}-\frac {(9 b B-10 A c) \left (\frac {x^3 \sqrt {b x+c x^2}}{4 c}-\frac {7 b \left (\frac {x^2 \sqrt {b x+c x^2}}{3 c}-\frac {5 b \left (\frac {x \sqrt {b x+c x^2}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x+c x^2}}{c}-\frac {b \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{c}\right )}{4 c}\right )}{6 c}\right )}{8 c}\right )}{10 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {B x^4 \sqrt {b x+c x^2}}{5 c}-\frac {(9 b B-10 A c) \left (\frac {x^3 \sqrt {b x+c x^2}}{4 c}-\frac {7 b \left (\frac {x^2 \sqrt {b x+c x^2}}{3 c}-\frac {5 b \left (\frac {x \sqrt {b x+c x^2}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x+c x^2}}{c}-\frac {b \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}}\right )}{4 c}\right )}{6 c}\right )}{8 c}\right )}{10 c}\) |
(B*x^4*Sqrt[b*x + c*x^2])/(5*c) - ((9*b*B - 10*A*c)*((x^3*Sqrt[b*x + c*x^2 ])/(4*c) - (7*b*((x^2*Sqrt[b*x + c*x^2])/(3*c) - (5*b*((x*Sqrt[b*x + c*x^2 ])/(2*c) - (3*b*(Sqrt[b*x + c*x^2]/c - (b*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c *x^2]])/c^(3/2)))/(4*c)))/(6*c)))/(8*c)))/(10*c)
3.2.9.3.1 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[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[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1))) Int[(d + e*x)^ (m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[ c*d^2 - b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2 *p]
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[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 = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.59
| method | result | size |
| pseudoelliptic | \(-\frac {7 \left (\left (-\frac {15}{8} A \,b^{4} c +\frac {27}{16} B \,b^{5}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )+\sqrt {x \left (c x +b \right )}\, \left (\frac {15 \left (\frac {3 B x}{5}+A \right ) b^{3} c^{\frac {3}{2}}}{8}-\frac {5 \left (\frac {18 B x}{25}+A \right ) x \,b^{2} c^{\frac {5}{2}}}{4}+b \,x^{2} \left (\frac {27 B x}{35}+A \right ) c^{\frac {7}{2}}-\frac {6 \left (\frac {4 B x}{5}+A \right ) x^{3} c^{\frac {9}{2}}}{7}-\frac {27 B \,b^{4} \sqrt {c}}{16}\right )\right )}{24 c^{\frac {11}{2}}}\) | \(116\) |
| risch | \(-\frac {\left (-384 B \,c^{4} x^{4}-480 A \,c^{4} x^{3}+432 B b \,c^{3} x^{3}+560 A b \,c^{3} x^{2}-504 B \,b^{2} c^{2} x^{2}-700 A \,b^{2} c^{2} x +630 B \,b^{3} c x +1050 A \,b^{3} c -945 b^{4} B \right ) x \left (c x +b \right )}{1920 c^{5} \sqrt {x \left (c x +b \right )}}+\frac {7 b^{4} \left (10 A c -9 B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {11}{2}}}\) | \(145\) |
| default | \(B \left (\frac {x^{4} \sqrt {c \,x^{2}+b x}}{5 c}-\frac {9 b \left (\frac {x^{3} \sqrt {c \,x^{2}+b x}}{4 c}-\frac {7 b \left (\frac {x^{2} \sqrt {c \,x^{2}+b x}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )}{6 c}\right )}{8 c}\right )}{10 c}\right )+A \left (\frac {x^{3} \sqrt {c \,x^{2}+b x}}{4 c}-\frac {7 b \left (\frac {x^{2} \sqrt {c \,x^{2}+b x}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )}{6 c}\right )}{8 c}\right )\) | \(276\) |
-7/24/c^(11/2)*((-15/8*A*b^4*c+27/16*B*b^5)*arctanh((x*(c*x+b))^(1/2)/x/c^ (1/2))+(x*(c*x+b))^(1/2)*(15/8*(3/5*B*x+A)*b^3*c^(3/2)-5/4*(18/25*B*x+A)*x *b^2*c^(5/2)+b*x^2*(27/35*B*x+A)*c^(7/2)-6/7*(4/5*B*x+A)*x^3*c^(9/2)-27/16 *B*b^4*c^(1/2)))
Time = 0.28 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.54 \[ \int \frac {x^4 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\left [-\frac {105 \, {\left (9 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (384 \, B c^{5} x^{4} + 945 \, B b^{4} c - 1050 \, A b^{3} c^{2} - 48 \, {\left (9 \, B b c^{4} - 10 \, A c^{5}\right )} x^{3} + 56 \, {\left (9 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} x^{2} - 70 \, {\left (9 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{3840 \, c^{6}}, \frac {105 \, {\left (9 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (384 \, B c^{5} x^{4} + 945 \, B b^{4} c - 1050 \, A b^{3} c^{2} - 48 \, {\left (9 \, B b c^{4} - 10 \, A c^{5}\right )} x^{3} + 56 \, {\left (9 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} x^{2} - 70 \, {\left (9 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{1920 \, c^{6}}\right ] \]
[-1/3840*(105*(9*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 + 945*B*b^4*c - 1050*A*b^3*c^2 - 48*(9* B*b*c^4 - 10*A*c^5)*x^3 + 56*(9*B*b^2*c^3 - 10*A*b*c^4)*x^2 - 70*(9*B*b^3* c^2 - 10*A*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^6, 1/1920*(105*(9*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 + 945*B*b^4*c - 1050*A*b^3*c^2 - 48*(9*B*b*c^4 - 10*A*c^5)*x^3 + 56*(9*B* b^2*c^3 - 10*A*b*c^4)*x^2 - 70*(9*B*b^3*c^2 - 10*A*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^6]
Time = 0.50 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.16 \[ \int \frac {x^4 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\begin {cases} \frac {35 b^{4} \left (A - \frac {9 B b}{10 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{128 c^{4}} + \sqrt {b x + c x^{2}} \left (\frac {B x^{4}}{5 c} - \frac {35 b^{3} \left (A - \frac {9 B b}{10 c}\right )}{64 c^{4}} + \frac {35 b^{2} x \left (A - \frac {9 B b}{10 c}\right )}{96 c^{3}} - \frac {7 b x^{2} \left (A - \frac {9 B b}{10 c}\right )}{24 c^{2}} + \frac {x^{3} \left (A - \frac {9 B b}{10 c}\right )}{4 c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {A \left (b x\right )^{\frac {9}{2}}}{9} + \frac {B \left (b x\right )^{\frac {11}{2}}}{11 b}\right )}{b^{5}} & \text {for}\: b \neq 0 \\\tilde {\infty } \left (\frac {A x^{5}}{5} + \frac {B x^{6}}{6}\right ) & \text {otherwise} \end {cases} \]
Piecewise((35*b**4*(A - 9*B*b/(10*c))*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**4) + sqrt(b*x + c*x**2)*(B*x* *4/(5*c) - 35*b**3*(A - 9*B*b/(10*c))/(64*c**4) + 35*b**2*x*(A - 9*B*b/(10 *c))/(96*c**3) - 7*b*x**2*(A - 9*B*b/(10*c))/(24*c**2) + x**3*(A - 9*B*b/( 10*c))/(4*c)), Ne(c, 0)), (2*(A*(b*x)**(9/2)/9 + B*(b*x)**(11/2)/(11*b))/b **5, Ne(b, 0)), (zoo*(A*x**5/5 + B*x**6/6), True))
Time = 0.19 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.28 \[ \int \frac {x^4 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c x^{2} + b x} B x^{4}}{5 \, c} - \frac {9 \, \sqrt {c x^{2} + b x} B b x^{3}}{40 \, c^{2}} + \frac {\sqrt {c x^{2} + b x} A x^{3}}{4 \, c} + \frac {21 \, \sqrt {c x^{2} + b x} B b^{2} x^{2}}{80 \, c^{3}} - \frac {7 \, \sqrt {c x^{2} + b x} A b x^{2}}{24 \, c^{2}} - \frac {21 \, \sqrt {c x^{2} + b x} B b^{3} x}{64 \, c^{4}} + \frac {35 \, \sqrt {c x^{2} + b x} A b^{2} x}{96 \, c^{3}} - \frac {63 \, B b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {11}{2}}} + \frac {35 \, A b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {9}{2}}} + \frac {63 \, \sqrt {c x^{2} + b x} B b^{4}}{128 \, c^{5}} - \frac {35 \, \sqrt {c x^{2} + b x} A b^{3}}{64 \, c^{4}} \]
1/5*sqrt(c*x^2 + b*x)*B*x^4/c - 9/40*sqrt(c*x^2 + b*x)*B*b*x^3/c^2 + 1/4*s qrt(c*x^2 + b*x)*A*x^3/c + 21/80*sqrt(c*x^2 + b*x)*B*b^2*x^2/c^3 - 7/24*sq rt(c*x^2 + b*x)*A*b*x^2/c^2 - 21/64*sqrt(c*x^2 + b*x)*B*b^3*x/c^4 + 35/96* sqrt(c*x^2 + b*x)*A*b^2*x/c^3 - 63/256*B*b^5*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(11/2) + 35/128*A*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x )*sqrt(c))/c^(9/2) + 63/128*sqrt(c*x^2 + b*x)*B*b^4/c^5 - 35/64*sqrt(c*x^2 + b*x)*A*b^3/c^4
Time = 0.29 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.83 \[ \int \frac {x^4 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {1}{1920} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, {\left (\frac {8 \, B x}{c} - \frac {9 \, B b c^{3} - 10 \, A c^{4}}{c^{5}}\right )} x + \frac {7 \, {\left (9 \, B b^{2} c^{2} - 10 \, A b c^{3}\right )}}{c^{5}}\right )} x - \frac {35 \, {\left (9 \, B b^{3} c - 10 \, A b^{2} c^{2}\right )}}{c^{5}}\right )} x + \frac {105 \, {\left (9 \, B b^{4} - 10 \, A b^{3} c\right )}}{c^{5}}\right )} + \frac {7 \, {\left (9 \, B b^{5} - 10 \, A b^{4} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{256 \, c^{\frac {11}{2}}} \]
1/1920*sqrt(c*x^2 + b*x)*(2*(4*(6*(8*B*x/c - (9*B*b*c^3 - 10*A*c^4)/c^5)*x + 7*(9*B*b^2*c^2 - 10*A*b*c^3)/c^5)*x - 35*(9*B*b^3*c - 10*A*b^2*c^2)/c^5 )*x + 105*(9*B*b^4 - 10*A*b^3*c)/c^5) + 7/256*(9*B*b^5 - 10*A*b^4*c)*log(a bs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(11/2)
Timed out. \[ \int \frac {x^4 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\int \frac {x^4\,\left (A+B\,x\right )}{\sqrt {c\,x^2+b\,x}} \,d x \]