Integrand size = 22, antiderivative size = 162 \[ \int \frac {x^3 (A+B x)}{\sqrt {b x+c x^2}} \, dx=-\frac {5 b^2 (7 b B-8 A c) \sqrt {b x+c x^2}}{64 c^4}+\frac {5 b (7 b B-8 A c) x \sqrt {b x+c x^2}}{96 c^3}-\frac {(7 b B-8 A c) x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {b x+c x^2}}{4 c}+\frac {5 b^3 (7 b B-8 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{9/2}} \]
5/64*b^3*(-8*A*c+7*B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(9/2)-5/64* b^2*(-8*A*c+7*B*b)*(c*x^2+b*x)^(1/2)/c^4+5/96*b*(-8*A*c+7*B*b)*x*(c*x^2+b* x)^(1/2)/c^3-1/24*(-8*A*c+7*B*b)*x^2*(c*x^2+b*x)^(1/2)/c^2+1/4*B*x^3*(c*x^ 2+b*x)^(1/2)/c
Time = 0.47 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.16 \[ \int \frac {x^3 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {x} (b+c x) \left (-105 b^3 B \sqrt {x}+120 A b^2 c \sqrt {x}+70 b^2 B c x^{3/2}-80 A b c^2 x^{3/2}-56 b B c^2 x^{5/2}+64 A c^3 x^{5/2}+48 B c^3 x^{7/2}\right )}{192 c^4 \sqrt {x (b+c x)}}+\frac {5 b^3 (7 b B-8 A c) \sqrt {x} \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )}{32 c^{9/2} \sqrt {x (b+c x)}} \]
(Sqrt[x]*(b + c*x)*(-105*b^3*B*Sqrt[x] + 120*A*b^2*c*Sqrt[x] + 70*b^2*B*c* x^(3/2) - 80*A*b*c^2*x^(3/2) - 56*b*B*c^2*x^(5/2) + 64*A*c^3*x^(5/2) + 48* B*c^3*x^(7/2)))/(192*c^4*Sqrt[x*(b + c*x)]) + (5*b^3*(7*b*B - 8*A*c)*Sqrt[ x]*Sqrt[b + c*x]*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])])/(3 2*c^(9/2)*Sqrt[x*(b + c*x)])
Time = 0.32 (sec) , antiderivative size = 150, 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 = {1221, 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.
| \(\displaystyle \int \frac {x^3 (A+B x)}{\sqrt {b x+c x^2}} \, dx\) |
| \(\Big \downarrow \) 1221 |
| \(\displaystyle \frac {B x^3 \sqrt {b x+c x^2}}{4 c}-\frac {(7 b B-8 A c) \int \frac {x^3}{\sqrt {c x^2+b x}}dx}{8 c}\) |
| \(\Big \downarrow \) 1134 |
| \(\displaystyle \frac {B x^3 \sqrt {b x+c x^2}}{4 c}-\frac {(7 b B-8 A c) \left (\frac {x^2 \sqrt {b x+c x^2}}{3 c}-\frac {5 b \int \frac {x^2}{\sqrt {c x^2+b x}}dx}{6 c}\right )}{8 c}\) |
| \(\Big \downarrow \) 1134 |
| \(\displaystyle \frac {B x^3 \sqrt {b x+c x^2}}{4 c}-\frac {(7 b B-8 A c) \left (\frac {x^2 \sqrt {b x+c x^2}}{3 c}-\frac {5 b \left (\frac {x \sqrt {b x+c x^2}}{2 c}-\frac {3 b \int \frac {x}{\sqrt {c x^2+b x}}dx}{4 c}\right )}{6 c}\right )}{8 c}\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {B x^3 \sqrt {b x+c x^2}}{4 c}-\frac {(7 b B-8 A c) \left (\frac {x^2 \sqrt {b x+c x^2}}{3 c}-\frac {5 b \left (\frac {x \sqrt {b x+c x^2}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x+c x^2}}{c}-\frac {b \int \frac {1}{\sqrt {c x^2+b x}}dx}{2 c}\right )}{4 c}\right )}{6 c}\right )}{8 c}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {B x^3 \sqrt {b x+c x^2}}{4 c}-\frac {(7 b B-8 A c) \left (\frac {x^2 \sqrt {b x+c x^2}}{3 c}-\frac {5 b \left (\frac {x \sqrt {b x+c x^2}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x+c x^2}}{c}-\frac {b \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{c}\right )}{4 c}\right )}{6 c}\right )}{8 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {B x^3 \sqrt {b x+c x^2}}{4 c}-\frac {(7 b B-8 A c) \left (\frac {x^2 \sqrt {b x+c x^2}}{3 c}-\frac {5 b \left (\frac {x \sqrt {b x+c x^2}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x+c x^2}}{c}-\frac {b \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}}\right )}{4 c}\right )}{6 c}\right )}{8 c}\) |
(B*x^3*Sqrt[b*x + c*x^2])/(4*c) - ((7*b*B - 8*A*c)*((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)
3.2.10.3.1 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 = 0.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.61
| method | result | size |
| pseudoelliptic | \(-\frac {5 \left (\frac {3 \left (A \,b^{3} c -\frac {7}{8} b^{4} B \right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )}{2}+\sqrt {x \left (c x +b \right )}\, \left (-\frac {3 \left (\frac {7 B x}{12}+A \right ) b^{2} c^{\frac {3}{2}}}{2}+b x \left (\frac {7 B x}{10}+A \right ) c^{\frac {5}{2}}-\frac {4 x^{2} \left (\frac {3 B x}{4}+A \right ) c^{\frac {7}{2}}}{5}+\frac {21 B \sqrt {c}\, b^{3}}{16}\right )\right )}{12 c^{\frac {9}{2}}}\) | \(99\) |
| risch | \(\frac {\left (48 B \,c^{3} x^{3}+64 A \,c^{3} x^{2}-56 B b \,c^{2} x^{2}-80 A b \,c^{2} x +70 B \,b^{2} c x +120 A \,b^{2} c -105 B \,b^{3}\right ) x \left (c x +b \right )}{192 c^{4} \sqrt {x \left (c x +b \right )}}-\frac {5 b^{3} \left (8 A c -7 B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {9}{2}}}\) | \(121\) |
| default | \(B \left (\frac {x^{3} \sqrt {c \,x^{2}+b x}}{4 c}-\frac {7 b \left (\frac {x^{2} \sqrt {c \,x^{2}+b x}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )}{6 c}\right )}{8 c}\right )+A \left (\frac {x^{2} \sqrt {c \,x^{2}+b x}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )}{6 c}\right )\) | \(224\) |
-5/12*(3/2*(A*b^3*c-7/8*b^4*B)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))+(x*(c* x+b))^(1/2)*(-3/2*(7/12*B*x+A)*b^2*c^(3/2)+b*x*(7/10*B*x+A)*c^(5/2)-4/5*x^ 2*(3/4*B*x+A)*c^(7/2)+21/16*B*c^(1/2)*b^3))/c^(9/2)
Time = 0.29 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.58 \[ \int \frac {x^3 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\left [-\frac {15 \, {\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (48 \, B c^{4} x^{3} - 105 \, B b^{3} c + 120 \, A b^{2} c^{2} - 8 \, {\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{2} + 10 \, {\left (7 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{384 \, c^{5}}, -\frac {15 \, {\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (48 \, B c^{4} x^{3} - 105 \, B b^{3} c + 120 \, A b^{2} c^{2} - 8 \, {\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{2} + 10 \, {\left (7 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{192 \, c^{5}}\right ] \]
[-1/384*(15*(7*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 - 105*B*b^3*c + 120*A*b^2*c^2 - 8*(7*B*b*c^ 3 - 8*A*c^4)*x^2 + 10*(7*B*b^2*c^2 - 8*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^5, -1/192*(15*(7*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 - 105*B*b^3*c + 120*A*b^2*c^2 - 8*(7*B*b*c^3 - 8 *A*c^4)*x^2 + 10*(7*B*b^2*c^2 - 8*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^5]
Time = 0.50 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.26 \[ \int \frac {x^3 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\begin {cases} - \frac {5 b^{3} \left (A - \frac {7 B b}{8 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{16 c^{3}} + \sqrt {b x + c x^{2}} \left (\frac {B x^{3}}{4 c} + \frac {5 b^{2} \left (A - \frac {7 B b}{8 c}\right )}{8 c^{3}} - \frac {5 b x \left (A - \frac {7 B b}{8 c}\right )}{12 c^{2}} + \frac {x^{2} \left (A - \frac {7 B b}{8 c}\right )}{3 c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {A \left (b x\right )^{\frac {7}{2}}}{7} + \frac {B \left (b x\right )^{\frac {9}{2}}}{9 b}\right )}{b^{4}} & \text {for}\: b \neq 0 \\\tilde {\infty } \left (\frac {A x^{4}}{4} + \frac {B x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]
Piecewise((-5*b**3*(A - 7*B*b/(8*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))/(16*c**3) + sqrt(b*x + c*x**2)*(B*x**3 /(4*c) + 5*b**2*(A - 7*B*b/(8*c))/(8*c**3) - 5*b*x*(A - 7*B*b/(8*c))/(12*c **2) + x**2*(A - 7*B*b/(8*c))/(3*c)), Ne(c, 0)), (2*(A*(b*x)**(7/2)/7 + B* (b*x)**(9/2)/(9*b))/b**4, Ne(b, 0)), (zoo*(A*x**4/4 + B*x**5/5), True))
Time = 0.18 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.27 \[ \int \frac {x^3 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c x^{2} + b x} B x^{3}}{4 \, c} - \frac {7 \, \sqrt {c x^{2} + b x} B b x^{2}}{24 \, c^{2}} + \frac {\sqrt {c x^{2} + b x} A x^{2}}{3 \, c} + \frac {35 \, \sqrt {c x^{2} + b x} B b^{2} x}{96 \, c^{3}} - \frac {5 \, \sqrt {c x^{2} + b x} A b x}{12 \, c^{2}} + \frac {35 \, B b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {9}{2}}} - \frac {5 \, A b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {7}{2}}} - \frac {35 \, \sqrt {c x^{2} + b x} B b^{3}}{64 \, c^{4}} + \frac {5 \, \sqrt {c x^{2} + b x} A b^{2}}{8 \, c^{3}} \]
1/4*sqrt(c*x^2 + b*x)*B*x^3/c - 7/24*sqrt(c*x^2 + b*x)*B*b*x^2/c^2 + 1/3*s qrt(c*x^2 + b*x)*A*x^2/c + 35/96*sqrt(c*x^2 + b*x)*B*b^2*x/c^3 - 5/12*sqrt (c*x^2 + b*x)*A*b*x/c^2 + 35/128*B*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x) *sqrt(c))/c^(9/2) - 5/16*A*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c) )/c^(7/2) - 35/64*sqrt(c*x^2 + b*x)*B*b^3/c^4 + 5/8*sqrt(c*x^2 + b*x)*A*b^ 2/c^3
Time = 0.29 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.83 \[ \int \frac {x^3 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {1}{192} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (\frac {6 \, B x}{c} - \frac {7 \, B b c^{2} - 8 \, A c^{3}}{c^{4}}\right )} x + \frac {5 \, {\left (7 \, B b^{2} c - 8 \, A b c^{2}\right )}}{c^{4}}\right )} x - \frac {15 \, {\left (7 \, B b^{3} - 8 \, A b^{2} c\right )}}{c^{4}}\right )} - \frac {5 \, {\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {9}{2}}} \]
1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*B*x/c - (7*B*b*c^2 - 8*A*c^3)/c^4)*x + 5* (7*B*b^2*c - 8*A*b*c^2)/c^4)*x - 15*(7*B*b^3 - 8*A*b^2*c)/c^4) - 5/128*(7* B*b^4 - 8*A*b^3*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b)) /c^(9/2)
Timed out. \[ \int \frac {x^3 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\int \frac {x^3\,\left (A+B\,x\right )}{\sqrt {c\,x^2+b\,x}} \,d x \]