Integrand size = 19, antiderivative size = 137 \[ \int (d+e x) \left (b x+c x^2\right )^{3/2} \, dx=-\frac {3 b^2 (2 c d-b e) (b+2 c x) \sqrt {b x+c x^2}}{128 c^3}+\frac {(2 c d-b e) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {e \left (b x+c x^2\right )^{5/2}}{5 c}+\frac {3 b^4 (2 c d-b e) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{7/2}} \]
1/16*(-b*e+2*c*d)*(2*c*x+b)*(c*x^2+b*x)^(3/2)/c^2+1/5*e*(c*x^2+b*x)^(5/2)/ c+3/128*b^4*(-b*e+2*c*d)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(7/2)-3/12 8*b^2*(-b*e+2*c*d)*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c^3
Time = 1.14 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.22 \[ \int (d+e x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {(x (b+c x))^{3/2} \left (\frac {\sqrt {c} \sqrt {x} \left (15 b^4 e-10 b^3 c (3 d+e x)+4 b^2 c^2 x (5 d+2 e x)+32 c^4 x^3 (5 d+4 e x)+16 b c^3 x^2 (15 d+11 e x)\right )}{b+c x}+\frac {30 b^4 (-2 c d+b e) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )}{(b+c x)^{3/2}}\right )}{640 c^{7/2} x^{3/2}} \]
((x*(b + c*x))^(3/2)*((Sqrt[c]*Sqrt[x]*(15*b^4*e - 10*b^3*c*(3*d + e*x) + 4*b^2*c^2*x*(5*d + 2*e*x) + 32*c^4*x^3*(5*d + 4*e*x) + 16*b*c^3*x^2*(15*d + 11*e*x)))/(b + c*x) + (30*b^4*(-2*c*d + b*e)*ArcTanh[(Sqrt[c]*Sqrt[x])/( Sqrt[b] - Sqrt[b + c*x])])/(b + c*x)^(3/2)))/(640*c^(7/2)*x^(3/2))
Time = 0.25 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1160, 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.
| \(\displaystyle \int \left (b x+c x^2\right )^{3/2} (d+e x) \, dx\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {(2 c d-b e) \int \left (c x^2+b x\right )^{3/2}dx}{2 c}+\frac {e \left (b x+c x^2\right )^{5/2}}{5 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {(2 c d-b e) \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \int \sqrt {c x^2+b x}dx}{16 c}\right )}{2 c}+\frac {e \left (b x+c x^2\right )^{5/2}}{5 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {(2 c d-b e) \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )}{16 c}\right )}{2 c}+\frac {e \left (b x+c x^2\right )^{5/2}}{5 c}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {(2 c d-b e) \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )}{16 c}\right )}{2 c}+\frac {e \left (b x+c x^2\right )^{5/2}}{5 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right )}{16 c}\right ) (2 c d-b e)}{2 c}+\frac {e \left (b x+c x^2\right )^{5/2}}{5 c}\) |
(e*(b*x + c*x^2)^(5/2))/(5*c) + ((2*c*d - b*e)*(((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*ArcTa nh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(3/2))))/(16*c)))/(2*c)
3.3.97.3.1 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_))*((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]
Time = 1.87 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.85
| method | result | size |
| pseudoelliptic | \(-\frac {3 \left (\left (b^{5} e -2 b^{4} c d \right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )-\sqrt {x \left (c x +b \right )}\, \left (-2 \left (\frac {e x}{3}+d \right ) b^{3} c^{\frac {3}{2}}+\frac {4 \left (\frac {2 e x}{5}+d \right ) x \,b^{2} c^{\frac {5}{2}}}{3}+16 x^{2} \left (\frac {11 e x}{15}+d \right ) b \,c^{\frac {7}{2}}+\frac {32 x^{3} \left (\frac {4 e x}{5}+d \right ) c^{\frac {9}{2}}}{3}+\sqrt {c}\, b^{4} e \right )\right )}{128 c^{\frac {7}{2}}}\) | \(116\) |
| risch | \(\frac {\left (128 c^{4} e \,x^{4}+176 b \,c^{3} e \,x^{3}+160 c^{4} d \,x^{3}+8 b^{2} c^{2} e \,x^{2}+240 b \,c^{3} d \,x^{2}-10 b^{3} c e x +20 x \,b^{2} c^{2} d +15 b^{4} e -30 b^{3} d c \right ) x \left (c x +b \right )}{640 c^{3} \sqrt {x \left (c x +b \right )}}-\frac {3 b^{4} \left (b e -2 c d \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {7}{2}}}\) | \(144\) |
| default | \(d \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )+e \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )\) | \(201\) |
-3/128*((b^5*e-2*b^4*c*d)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))-(x*(c*x+b)) ^(1/2)*(-2*(1/3*e*x+d)*b^3*c^(3/2)+4/3*(2/5*e*x+d)*x*b^2*c^(5/2)+16*x^2*(1 1/15*e*x+d)*b*c^(7/2)+32/3*x^3*(4/5*e*x+d)*c^(9/2)+c^(1/2)*b^4*e))/c^(7/2)
Time = 0.30 (sec) , antiderivative size = 302, normalized size of antiderivative = 2.20 \[ \int (d+e x) \left (b x+c x^2\right )^{3/2} \, dx=\left [-\frac {15 \, {\left (2 \, b^{4} c d - b^{5} e\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (128 \, c^{5} e x^{4} - 30 \, b^{3} c^{2} d + 15 \, b^{4} c e + 16 \, {\left (10 \, c^{5} d + 11 \, b c^{4} e\right )} x^{3} + 8 \, {\left (30 \, b c^{4} d + b^{2} c^{3} e\right )} x^{2} + 10 \, {\left (2 \, b^{2} c^{3} d - b^{3} c^{2} e\right )} x\right )} \sqrt {c x^{2} + b x}}{1280 \, c^{4}}, -\frac {15 \, {\left (2 \, b^{4} c d - b^{5} e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (128 \, c^{5} e x^{4} - 30 \, b^{3} c^{2} d + 15 \, b^{4} c e + 16 \, {\left (10 \, c^{5} d + 11 \, b c^{4} e\right )} x^{3} + 8 \, {\left (30 \, b c^{4} d + b^{2} c^{3} e\right )} x^{2} + 10 \, {\left (2 \, b^{2} c^{3} d - b^{3} c^{2} e\right )} x\right )} \sqrt {c x^{2} + b x}}{640 \, c^{4}}\right ] \]
[-1/1280*(15*(2*b^4*c*d - b^5*e)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b* x)*sqrt(c)) - 2*(128*c^5*e*x^4 - 30*b^3*c^2*d + 15*b^4*c*e + 16*(10*c^5*d + 11*b*c^4*e)*x^3 + 8*(30*b*c^4*d + b^2*c^3*e)*x^2 + 10*(2*b^2*c^3*d - b^3 *c^2*e)*x)*sqrt(c*x^2 + b*x))/c^4, -1/640*(15*(2*b^4*c*d - b^5*e)*sqrt(-c) *arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (128*c^5*e*x^4 - 30*b^3*c^2*d + 15*b^4*c*e + 16*(10*c^5*d + 11*b*c^4*e)*x^3 + 8*(30*b*c^4*d + b^2*c^3*e) *x^2 + 10*(2*b^2*c^3*d - b^3*c^2*e)*x)*sqrt(c*x^2 + b*x))/c^4]
Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (126) = 252\).
Time = 0.48 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.38 \[ \int (d+e x) \left (b x+c x^2\right )^{3/2} \, dx=\begin {cases} \frac {3 b^{2} \left (b^{2} d - \frac {5 b \left (b^{2} e + 2 b c d - \frac {7 b \left (\frac {11 b c e}{10} + c^{2} d\right )}{8 c}\right )}{6 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8 c^{2}} + \sqrt {b x + c x^{2}} \left (- \frac {3 b \left (b^{2} d - \frac {5 b \left (b^{2} e + 2 b c d - \frac {7 b \left (\frac {11 b c e}{10} + c^{2} d\right )}{8 c}\right )}{6 c}\right )}{4 c^{2}} + \frac {c e x^{4}}{5} + \frac {x^{3} \cdot \left (\frac {11 b c e}{10} + c^{2} d\right )}{4 c} + \frac {x^{2} \left (b^{2} e + 2 b c d - \frac {7 b \left (\frac {11 b c e}{10} + c^{2} d\right )}{8 c}\right )}{3 c} + \frac {x \left (b^{2} d - \frac {5 b \left (b^{2} e + 2 b c d - \frac {7 b \left (\frac {11 b c e}{10} + c^{2} d\right )}{8 c}\right )}{6 c}\right )}{2 c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {d \left (b x\right )^{\frac {5}{2}}}{5} + \frac {e \left (b x\right )^{\frac {7}{2}}}{7 b}\right )}{b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((3*b**2*(b**2*d - 5*b*(b**2*e + 2*b*c*d - 7*b*(11*b*c*e/10 + c** 2*d)/(8*c))/(6*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))/(8*c**2) + sqrt(b*x + c*x**2)*(-3*b*(b**2*d - 5*b*(b**2* e + 2*b*c*d - 7*b*(11*b*c*e/10 + c**2*d)/(8*c))/(6*c))/(4*c**2) + c*e*x**4 /5 + x**3*(11*b*c*e/10 + c**2*d)/(4*c) + x**2*(b**2*e + 2*b*c*d - 7*b*(11* b*c*e/10 + c**2*d)/(8*c))/(3*c) + x*(b**2*d - 5*b*(b**2*e + 2*b*c*d - 7*b* (11*b*c*e/10 + c**2*d)/(8*c))/(6*c))/(2*c)), Ne(c, 0)), (2*(d*(b*x)**(5/2) /5 + e*(b*x)**(7/2)/(7*b))/b, Ne(b, 0)), (0, True))
Leaf count of result is larger than twice the leaf count of optimal. 236 vs. \(2 (117) = 234\).
Time = 0.19 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.72 \[ \int (d+e x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {1}{4} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} d x - \frac {3 \, \sqrt {c x^{2} + b x} b^{2} d x}{32 \, c} + \frac {3 \, \sqrt {c x^{2} + b x} b^{3} e x}{64 \, c^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} b e x}{8 \, c} + \frac {3 \, b^{4} d \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {5}{2}}} - \frac {3 \, b^{5} e \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {7}{2}}} - \frac {3 \, \sqrt {c x^{2} + b x} b^{3} d}{64 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} b d}{8 \, c} + \frac {3 \, \sqrt {c x^{2} + b x} b^{4} e}{128 \, c^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2} e}{16 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} e}{5 \, c} \]
1/4*(c*x^2 + b*x)^(3/2)*d*x - 3/32*sqrt(c*x^2 + b*x)*b^2*d*x/c + 3/64*sqrt (c*x^2 + b*x)*b^3*e*x/c^2 - 1/8*(c*x^2 + b*x)^(3/2)*b*e*x/c + 3/128*b^4*d* log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) - 3/256*b^5*e*log(2*c *x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) - 3/64*sqrt(c*x^2 + b*x)*b^3 *d/c^2 + 1/8*(c*x^2 + b*x)^(3/2)*b*d/c + 3/128*sqrt(c*x^2 + b*x)*b^4*e/c^3 - 1/16*(c*x^2 + b*x)^(3/2)*b^2*e/c^2 + 1/5*(c*x^2 + b*x)^(5/2)*e/c
Time = 0.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.19 \[ \int (d+e x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {1}{640} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, c e x + \frac {10 \, c^{5} d + 11 \, b c^{4} e}{c^{4}}\right )} x + \frac {30 \, b c^{4} d + b^{2} c^{3} e}{c^{4}}\right )} x + \frac {5 \, {\left (2 \, b^{2} c^{3} d - b^{3} c^{2} e\right )}}{c^{4}}\right )} x - \frac {15 \, {\left (2 \, b^{3} c^{2} d - b^{4} c e\right )}}{c^{4}}\right )} - \frac {3 \, {\left (2 \, b^{4} c d - b^{5} e\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{256 \, c^{\frac {7}{2}}} \]
1/640*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*c*e*x + (10*c^5*d + 11*b*c^4*e)/c^4)*x + (30*b*c^4*d + b^2*c^3*e)/c^4)*x + 5*(2*b^2*c^3*d - b^3*c^2*e)/c^4)*x - 15*(2*b^3*c^2*d - b^4*c*e)/c^4) - 3/256*(2*b^4*c*d - b^5*e)*log(abs(2*(sqr t(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(7/2)
Time = 9.61 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.52 \[ \int (d+e x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {e\,{\left (c\,x^2+b\,x\right )}^{5/2}}{5\,c}-\frac {3\,b^2\,d\,\left (\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )-\frac {b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}\right )}{16\,c}+\frac {d\,{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (\frac {b}{2}+c\,x\right )}{4\,c}-\frac {b\,e\,\left (\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4}+\frac {b\,{\left (c\,x^2+b\,x\right )}^{3/2}}{8\,c}-\frac {3\,b^2\,\left (\frac {\sqrt {c\,x^2+b\,x}\,\left (b+2\,c\,x\right )}{4\,c}-\frac {b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}\right )}{16\,c}\right )}{2\,c} \]
(e*(b*x + c*x^2)^(5/2))/(5*c) - (3*b^2*d*((b*x + c*x^2)^(1/2)*(x/2 + b/(4* c)) - (b^2*log((b/2 + c*x)/c^(1/2) + (b*x + c*x^2)^(1/2)))/(8*c^(3/2))))/( 16*c) + (d*(b*x + c*x^2)^(3/2)*(b/2 + c*x))/(4*c) - (b*e*((x*(b*x + c*x^2) ^(3/2))/4 + (b*(b*x + c*x^2)^(3/2))/(8*c) - (3*b^2*(((b*x + c*x^2)^(1/2)*( b + 2*c*x))/(4*c) - (b^2*log((b/2 + c*x)/c^(1/2) + (b*x + c*x^2)^(1/2)))/( 8*c^(3/2))))/(16*c)))/(2*c)