Integrand size = 22, antiderivative size = 132 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x} \, dx=-\frac {b (3 b B-8 A c) (b+2 c x) \sqrt {b x+c x^2}}{64 c^2}-\frac {(3 b B-8 A c) \left (b x+c x^2\right )^{3/2}}{24 c}+\frac {B \left (b x+c x^2\right )^{5/2}}{4 c x}+\frac {b^3 (3 b B-8 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{5/2}} \]
-1/24*(-8*A*c+3*B*b)*(c*x^2+b*x)^(3/2)/c+1/4*B*(c*x^2+b*x)^(5/2)/c/x+1/64* b^3*(-8*A*c+3*B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(5/2)-1/64*b*(-8 *A*c+3*B*b)*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c^2
Time = 0.57 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x} \, dx=\frac {\sqrt {x} \sqrt {b+c x} \left (\sqrt {c} \sqrt {x} \sqrt {b+c x} \left (-9 b^3 B+6 b^2 c (4 A+B x)+16 c^3 x^2 (4 A+3 B x)+8 b c^2 x (14 A+9 B x)\right )+48 A b^3 c \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )+18 b^4 B \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )\right )}{192 c^{5/2} \sqrt {x (b+c x)}} \]
(Sqrt[x]*Sqrt[b + c*x]*(Sqrt[c]*Sqrt[x]*Sqrt[b + c*x]*(-9*b^3*B + 6*b^2*c* (4*A + B*x) + 16*c^3*x^2*(4*A + 3*B*x) + 8*b*c^2*x*(14*A + 9*B*x)) + 48*A* b^3*c*ArcTanh[(Sqrt[c]*Sqrt[x])/(Sqrt[b] - Sqrt[b + c*x])] + 18*b^4*B*ArcT anh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])]))/(192*c^(5/2)*Sqrt[x*(b + c*x)])
Time = 0.26 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1221, 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.
| \(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x} \, dx\) |
| \(\Big \downarrow \) 1221 |
| \(\displaystyle \frac {B \left (b x+c x^2\right )^{5/2}}{4 c x}-\frac {(3 b B-8 A c) \int \frac {\left (c x^2+b x\right )^{3/2}}{x}dx}{8 c}\) |
| \(\Big \downarrow \) 1131 |
| \(\displaystyle \frac {B \left (b x+c x^2\right )^{5/2}}{4 c x}-\frac {(3 b B-8 A c) \left (\frac {1}{2} b \int \sqrt {c x^2+b x}dx+\frac {1}{3} \left (b x+c x^2\right )^{3/2}\right )}{8 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {B \left (b x+c x^2\right )^{5/2}}{4 c x}-\frac {(3 b B-8 A c) \left (\frac {1}{2} b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )+\frac {1}{3} \left (b x+c x^2\right )^{3/2}\right )}{8 c}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {B \left (b x+c x^2\right )^{5/2}}{4 c x}-\frac {(3 b B-8 A c) \left (\frac {1}{2} b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )+\frac {1}{3} \left (b x+c x^2\right )^{3/2}\right )}{8 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {B \left (b x+c x^2\right )^{5/2}}{4 c x}-\frac {(3 b B-8 A c) \left (\frac {1}{2} b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right )+\frac {1}{3} \left (b x+c x^2\right )^{3/2}\right )}{8 c}\) |
(B*(b*x + c*x^2)^(5/2))/(4*c*x) - ((3*b*B - 8*A*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)/Sqr t[b*x + c*x^2]])/(4*c^(3/2))))/2))/(8*c)
3.1.83.3.1 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 = 0.15 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.75
| method | result | size |
| pseudoelliptic | \(\frac {\frac {7 \left (-\frac {3}{14} A \,b^{3} c +\frac {9}{112} b^{4} B \right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )}{12}+\frac {7 \left (\frac {3 \left (\frac {B x}{4}+A \right ) b^{2} c^{\frac {3}{2}}}{14}+b x \left (\frac {9 B x}{14}+A \right ) c^{\frac {5}{2}}+\frac {4 x^{2} \left (\frac {3 B x}{4}+A \right ) c^{\frac {7}{2}}}{7}-\frac {9 B \sqrt {c}\, b^{3}}{112}\right ) \sqrt {x \left (c x +b \right )}}{12}}{c^{\frac {5}{2}}}\) | \(99\) |
| risch | \(\frac {\left (48 B \,c^{3} x^{3}+64 A \,c^{3} x^{2}+72 B b \,c^{2} x^{2}+112 A b \,c^{2} x +6 B \,b^{2} c x +24 A \,b^{2} c -9 B \,b^{3}\right ) x \left (c x +b \right )}{192 c^{2} \sqrt {x \left (c x +b \right )}}-\frac {b^{3} \left (8 A c -3 B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {5}{2}}}\) | \(121\) |
| default | \(B \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )+A \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2}\right )\) | \(164\) |
7/12*((-3/14*A*b^3*c+9/112*b^4*B)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))+(3/ 14*(1/4*B*x+A)*b^2*c^(3/2)+b*x*(9/14*B*x+A)*c^(5/2)+4/7*x^2*(3/4*B*x+A)*c^ (7/2)-9/112*B*c^(1/2)*b^3)*(x*(c*x+b))^(1/2))/c^(5/2)
Time = 0.28 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.94 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x} \, dx=\left [-\frac {3 \, {\left (3 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (48 \, B c^{4} x^{3} - 9 \, B b^{3} c + 24 \, A b^{2} c^{2} + 8 \, {\left (9 \, B b c^{3} + 8 \, A c^{4}\right )} x^{2} + 2 \, {\left (3 \, B b^{2} c^{2} + 56 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{384 \, c^{3}}, -\frac {3 \, {\left (3 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (48 \, B c^{4} x^{3} - 9 \, B b^{3} c + 24 \, A b^{2} c^{2} + 8 \, {\left (9 \, B b c^{3} + 8 \, A c^{4}\right )} x^{2} + 2 \, {\left (3 \, B b^{2} c^{2} + 56 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{192 \, c^{3}}\right ] \]
[-1/384*(3*(3*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 - 9*B*b^3*c + 24*A*b^2*c^2 + 8*(9*B*b*c^3 + 8*A*c^4)*x^2 + 2*(3*B*b^2*c^2 + 56*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^3, -1/ 192*(3*(3*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 - 9*B*b^3*c + 24*A*b^2*c^2 + 8*(9*B*b*c^3 + 8*A*c^4)* x^2 + 2*(3*B*b^2*c^2 + 56*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^3]
Time = 2.12 (sec) , antiderivative size = 490, normalized size of antiderivative = 3.71 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x} \, dx=A b \left (\begin {cases} - \frac {b^{2} \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 )}{8 c} + \left (\frac {b}{4 c} + \frac {x}{2}\right ) \sqrt {b x + c x^{2}} & \text {for}\: c \neq 0 \\\frac {2 \left (b x\right )^{\frac {3}{2}}}{3 b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right ) + A c \left (\begin {cases} \frac {b^{3} \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^{2}} + \sqrt {b x + c x^{2}} \left (- \frac {b^{2}}{8 c^{2}} + \frac {b x}{12 c} + \frac {x^{2}}{3}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x\right )^{\frac {5}{2}}}{5 b^{2}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right ) + B b \left (\begin {cases} \frac {b^{3} \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^{2}} + \sqrt {b x + c x^{2}} \left (- \frac {b^{2}}{8 c^{2}} + \frac {b x}{12 c} + \frac {x^{2}}{3}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x\right )^{\frac {5}{2}}}{5 b^{2}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right ) + B c \left (\begin {cases} - \frac {5 b^{4} \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^{3}} + \sqrt {b x + c x^{2}} \cdot \left (\frac {5 b^{3}}{64 c^{3}} - \frac {5 b^{2} x}{96 c^{2}} + \frac {b x^{2}}{24 c} + \frac {x^{3}}{4}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x\right )^{\frac {7}{2}}}{7 b^{3}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right ) \]
A*b*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)) + A*c*Piecewise((b**3*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))/(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)) + B*b*Piecewise((b**3*Pie cewise((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)) + 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**3/4), Ne(c, 0)), (2*(b*x)**(7/2)/(7*b**3), Ne(b, 0)), (0, True))
Time = 0.19 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.43 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x} \, dx=\frac {1}{4} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B x + \frac {1}{4} \, \sqrt {c x^{2} + b x} A b x - \frac {3 \, \sqrt {c x^{2} + b x} B b^{2} x}{32 \, c} + \frac {3 \, B b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {5}{2}}} - \frac {A b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {3}{2}}} + \frac {1}{3} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A - \frac {3 \, \sqrt {c x^{2} + b x} B b^{3}}{64 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b}{8 \, c} + \frac {\sqrt {c x^{2} + b x} A b^{2}}{8 \, c} \]
1/4*(c*x^2 + b*x)^(3/2)*B*x + 1/4*sqrt(c*x^2 + b*x)*A*b*x - 3/32*sqrt(c*x^ 2 + b*x)*B*b^2*x/c + 3/128*B*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt( c))/c^(5/2) - 1/16*A*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3 /2) + 1/3*(c*x^2 + b*x)^(3/2)*A - 3/64*sqrt(c*x^2 + b*x)*B*b^3/c^2 + 1/8*( c*x^2 + b*x)^(3/2)*B*b/c + 1/8*sqrt(c*x^2 + b*x)*A*b^2/c
Time = 0.27 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.03 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x} \, dx=\frac {1}{192} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, B c x + \frac {9 \, B b c^{3} + 8 \, A c^{4}}{c^{3}}\right )} x + \frac {3 \, B b^{2} c^{2} + 56 \, A b c^{3}}{c^{3}}\right )} x - \frac {3 \, {\left (3 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )}}{c^{3}}\right )} - \frac {{\left (3 \, B b^{4} - 8 \, A b^{3} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {5}{2}}} \]
1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*B*c*x + (9*B*b*c^3 + 8*A*c^4)/c^3)*x + (3 *B*b^2*c^2 + 56*A*b*c^3)/c^3)*x - 3*(3*B*b^3*c - 8*A*b^2*c^2)/c^3) - 1/128 *(3*B*b^4 - 8*A*b^3*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(5/2)
Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (A+B\,x\right )}{x} \,d x \]