Integrand size = 21, antiderivative size = 210 \[ \int (d+e x)^3 \sqrt {b x+c x^2} \, dx=\frac {(2 c d-b e) \left (16 c^2 d^2-16 b c d e+7 b^2 e^2\right ) (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {e (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {e \left (192 c^2 d^2-150 b c d e+35 b^2 e^2+42 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}-\frac {b^2 (2 c d-b e) \left (16 c^2 d^2-16 b c d e+7 b^2 e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{9/2}} \]
1/5*e*(e*x+d)^2*(c*x^2+b*x)^(3/2)/c+1/240*e*(192*c^2*d^2-150*b*c*d*e+35*b^ 2*e^2+42*c*e*(-b*e+2*c*d)*x)*(c*x^2+b*x)^(3/2)/c^3-1/128*b^2*(-b*e+2*c*d)* (7*b^2*e^2-16*b*c*d*e+16*c^2*d^2)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^( 9/2)+1/128*(-b*e+2*c*d)*(7*b^2*e^2-16*b*c*d*e+16*c^2*d^2)*(2*c*x+b)*(c*x^2 +b*x)^(1/2)/c^4
Time = 1.78 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.34 \[ \int (d+e x)^3 \sqrt {b x+c x^2} \, dx=\frac {\sqrt {x (b+c x)} \left (480 b c^3 d^3-720 b^2 c^2 d^2 e+450 b^3 c d e^2-105 b^4 e^3+960 c^4 d^3 x+480 b c^3 d^2 e x-300 b^2 c^2 d e^2 x+70 b^3 c e^3 x+1920 c^4 d^2 e x^2+240 b c^3 d e^2 x^2-56 b^2 c^2 e^3 x^2+1440 c^4 d e^2 x^3+48 b c^3 e^3 x^3+384 c^4 e^3 x^4\right )}{1920 c^4}+\frac {b^2 \left (-32 c^3 d^3+48 b c^2 d^2 e-30 b^2 c d e^2+7 b^3 e^3\right ) \sqrt {x (b+c x)} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )}{64 c^{9/2} \sqrt {x} \sqrt {b+c x}} \]
(Sqrt[x*(b + c*x)]*(480*b*c^3*d^3 - 720*b^2*c^2*d^2*e + 450*b^3*c*d*e^2 - 105*b^4*e^3 + 960*c^4*d^3*x + 480*b*c^3*d^2*e*x - 300*b^2*c^2*d*e^2*x + 70 *b^3*c*e^3*x + 1920*c^4*d^2*e*x^2 + 240*b*c^3*d*e^2*x^2 - 56*b^2*c^2*e^3*x ^2 + 1440*c^4*d*e^2*x^3 + 48*b*c^3*e^3*x^3 + 384*c^4*e^3*x^4))/(1920*c^4) + (b^2*(-32*c^3*d^3 + 48*b*c^2*d^2*e - 30*b^2*c*d*e^2 + 7*b^3*e^3)*Sqrt[x* (b + c*x)]*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])])/(64*c^(9 /2)*Sqrt[x]*Sqrt[b + c*x])
Time = 0.39 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1166, 27, 1225, 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.
| \(\displaystyle \int \sqrt {b x+c x^2} (d+e x)^3 \, dx\) |
| \(\Big \downarrow \) 1166 |
| \(\displaystyle \frac {\int \frac {1}{2} (d+e x) (d (10 c d-3 b e)+7 e (2 c d-b e) x) \sqrt {c x^2+b x}dx}{5 c}+\frac {e \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (d+e x) (d (10 c d-3 b e)+7 e (2 c d-b e) x) \sqrt {c x^2+b x}dx}{10 c}+\frac {e \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c}\) |
| \(\Big \downarrow \) 1225 |
| \(\displaystyle \frac {\frac {5 (2 c d-b e) \left (7 b^2 e^2-16 b c d e+16 c^2 d^2\right ) \int \sqrt {c x^2+b x}dx}{16 c^2}+\frac {e \left (b x+c x^2\right )^{3/2} \left (35 b^2 e^2+42 c e x (2 c d-b e)-150 b c d e+192 c^2 d^2\right )}{24 c^2}}{10 c}+\frac {e \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {\frac {5 (2 c d-b e) \left (7 b^2 e^2-16 b c d e+16 c^2 d^2\right ) \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )}{16 c^2}+\frac {e \left (b x+c x^2\right )^{3/2} \left (35 b^2 e^2+42 c e x (2 c d-b e)-150 b c d e+192 c^2 d^2\right )}{24 c^2}}{10 c}+\frac {e \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {\frac {5 (2 c d-b e) \left (7 b^2 e^2-16 b c d e+16 c^2 d^2\right ) \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )}{16 c^2}+\frac {e \left (b x+c x^2\right )^{3/2} \left (35 b^2 e^2+42 c e x (2 c d-b e)-150 b c d e+192 c^2 d^2\right )}{24 c^2}}{10 c}+\frac {e \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {5 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right ) (2 c d-b e) \left (7 b^2 e^2-16 b c d e+16 c^2 d^2\right )}{16 c^2}+\frac {e \left (b x+c x^2\right )^{3/2} \left (35 b^2 e^2+42 c e x (2 c d-b e)-150 b c d e+192 c^2 d^2\right )}{24 c^2}}{10 c}+\frac {e \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c}\) |
(e*(d + e*x)^2*(b*x + c*x^2)^(3/2))/(5*c) + ((e*(192*c^2*d^2 - 150*b*c*d*e + 35*b^2*e^2 + 42*c*e*(2*c*d - b*e)*x)*(b*x + c*x^2)^(3/2))/(24*c^2) + (5 *(2*c*d - b*e)*(16*c^2*d^2 - 16*b*c*d*e + 7*b^2*e^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))/(10*c)
3.3.85.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[1/(c*(m + 2*p + 1)) Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* (a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat icQ[a, b, c, d, e, m, p, x]
Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c , d, e, f, g, p}, x] && !LeQ[p, -1]
Time = 1.99 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.90
| method | result | size |
| pseudoelliptic | \(\frac {\frac {7 \left (b^{2} e^{2}-\frac {16}{7} b c d e +\frac {16}{7} c^{2} d^{2}\right ) \left (b e -2 c d \right ) b^{2} \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )}{128}-\frac {7 \left (-\frac {32 \left (\frac {1}{10} e^{3} x^{3}+\frac {1}{2} d \,e^{2} x^{2}+d^{2} e x +d^{3}\right ) b \,c^{\frac {7}{2}}}{7}-\frac {64 x \left (\frac {2}{5} e^{3} x^{3}+\frac {3}{2} d \,e^{2} x^{2}+2 d^{2} e x +d^{3}\right ) c^{\frac {9}{2}}}{7}+\left (\left (\frac {8}{15} x^{2} e^{2}+\frac {20}{7} d e x +\frac {48}{7} d^{2}\right ) c^{\frac {5}{2}}+\left (\left (-\frac {2 e x}{3}-\frac {30 d}{7}\right ) c^{\frac {3}{2}}+\sqrt {c}\, b e \right ) e b \right ) e \,b^{2}\right ) \sqrt {x \left (c x +b \right )}}{128}}{c^{\frac {9}{2}}}\) | \(188\) |
| risch | \(-\frac {\left (-384 c^{4} e^{3} x^{4}-48 b \,c^{3} e^{3} x^{3}-1440 c^{4} d \,e^{2} x^{3}+56 b^{2} c^{2} e^{3} x^{2}-240 b \,c^{3} d \,e^{2} x^{2}-1920 c^{4} d^{2} e \,x^{2}-70 b^{3} c \,e^{3} x +300 b^{2} c^{2} d \,e^{2} x -480 b \,c^{3} d^{2} e x -960 c^{4} d^{3} x +105 b^{4} e^{3}-450 b^{3} c d \,e^{2}+720 b^{2} c^{2} d^{2} e -480 c^{3} b \,d^{3}\right ) x \left (c x +b \right )}{1920 c^{4} \sqrt {x \left (c x +b \right )}}+\frac {b^{2} \left (7 b^{3} e^{3}-30 b^{2} d \,e^{2} c +48 b \,c^{2} d^{2} e -32 c^{3} d^{3}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {9}{2}}}\) | \(248\) |
| default | \(d^{3} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )+e^{3} \left (\frac {x^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 c}-\frac {7 b \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}\right )}{10 c}\right )+3 d \,e^{2} \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}\right )+3 d^{2} e \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )\) | \(385\) |
7/128*((b^2*e^2-16/7*b*c*d*e+16/7*c^2*d^2)*(b*e-2*c*d)*b^2*arctanh((x*(c*x +b))^(1/2)/x/c^(1/2))-(-32/7*(1/10*e^3*x^3+1/2*d*e^2*x^2+d^2*e*x+d^3)*b*c^ (7/2)-64/7*x*(2/5*e^3*x^3+3/2*d*e^2*x^2+2*d^2*e*x+d^3)*c^(9/2)+((8/15*x^2* e^2+20/7*d*e*x+48/7*d^2)*c^(5/2)+((-2/3*e*x-30/7*d)*c^(3/2)+c^(1/2)*b*e)*e *b)*e*b^2)*(x*(c*x+b))^(1/2))/c^(9/2)
Time = 0.32 (sec) , antiderivative size = 497, normalized size of antiderivative = 2.37 \[ \int (d+e x)^3 \sqrt {b x+c x^2} \, dx=\left [-\frac {15 \, {\left (32 \, b^{2} c^{3} d^{3} - 48 \, b^{3} c^{2} d^{2} e + 30 \, b^{4} c d e^{2} - 7 \, b^{5} e^{3}\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (384 \, c^{5} e^{3} x^{4} + 480 \, b c^{4} d^{3} - 720 \, b^{2} c^{3} d^{2} e + 450 \, b^{3} c^{2} d e^{2} - 105 \, b^{4} c e^{3} + 48 \, {\left (30 \, c^{5} d e^{2} + b c^{4} e^{3}\right )} x^{3} + 8 \, {\left (240 \, c^{5} d^{2} e + 30 \, b c^{4} d e^{2} - 7 \, b^{2} c^{3} e^{3}\right )} x^{2} + 10 \, {\left (96 \, c^{5} d^{3} + 48 \, b c^{4} d^{2} e - 30 \, b^{2} c^{3} d e^{2} + 7 \, b^{3} c^{2} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{3840 \, c^{5}}, \frac {15 \, {\left (32 \, b^{2} c^{3} d^{3} - 48 \, b^{3} c^{2} d^{2} e + 30 \, b^{4} c d e^{2} - 7 \, b^{5} e^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (384 \, c^{5} e^{3} x^{4} + 480 \, b c^{4} d^{3} - 720 \, b^{2} c^{3} d^{2} e + 450 \, b^{3} c^{2} d e^{2} - 105 \, b^{4} c e^{3} + 48 \, {\left (30 \, c^{5} d e^{2} + b c^{4} e^{3}\right )} x^{3} + 8 \, {\left (240 \, c^{5} d^{2} e + 30 \, b c^{4} d e^{2} - 7 \, b^{2} c^{3} e^{3}\right )} x^{2} + 10 \, {\left (96 \, c^{5} d^{3} + 48 \, b c^{4} d^{2} e - 30 \, b^{2} c^{3} d e^{2} + 7 \, b^{3} c^{2} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{1920 \, c^{5}}\right ] \]
[-1/3840*(15*(32*b^2*c^3*d^3 - 48*b^3*c^2*d^2*e + 30*b^4*c*d*e^2 - 7*b^5*e ^3)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(384*c^5*e^3* x^4 + 480*b*c^4*d^3 - 720*b^2*c^3*d^2*e + 450*b^3*c^2*d*e^2 - 105*b^4*c*e^ 3 + 48*(30*c^5*d*e^2 + b*c^4*e^3)*x^3 + 8*(240*c^5*d^2*e + 30*b*c^4*d*e^2 - 7*b^2*c^3*e^3)*x^2 + 10*(96*c^5*d^3 + 48*b*c^4*d^2*e - 30*b^2*c^3*d*e^2 + 7*b^3*c^2*e^3)*x)*sqrt(c*x^2 + b*x))/c^5, 1/1920*(15*(32*b^2*c^3*d^3 - 4 8*b^3*c^2*d^2*e + 30*b^4*c*d*e^2 - 7*b^5*e^3)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (384*c^5*e^3*x^4 + 480*b*c^4*d^3 - 720*b^2*c^3*d^2 *e + 450*b^3*c^2*d*e^2 - 105*b^4*c*e^3 + 48*(30*c^5*d*e^2 + b*c^4*e^3)*x^3 + 8*(240*c^5*d^2*e + 30*b*c^4*d*e^2 - 7*b^2*c^3*e^3)*x^2 + 10*(96*c^5*d^3 + 48*b*c^4*d^2*e - 30*b^2*c^3*d*e^2 + 7*b^3*c^2*e^3)*x)*sqrt(c*x^2 + b*x) )/c^5]
Leaf count of result is larger than twice the leaf count of optimal. 434 vs. \(2 (204) = 408\).
Time = 0.47 (sec) , antiderivative size = 434, normalized size of antiderivative = 2.07 \[ \int (d+e x)^3 \sqrt {b x+c x^2} \, dx=\begin {cases} - \frac {b \left (b d^{3} - \frac {3 b \left (3 b d^{2} e - \frac {5 b \left (3 b d e^{2} - \frac {7 b \left (\frac {b e^{3}}{10} + 3 c d e^{2}\right )}{8 c} + 3 c d^{2} e\right )}{6 c} + c d^{3}\right )}{4 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{2 c} + \sqrt {b x + c x^{2}} \left (\frac {e^{3} x^{4}}{5} + \frac {x^{3} \left (\frac {b e^{3}}{10} + 3 c d e^{2}\right )}{4 c} + \frac {x^{2} \cdot \left (3 b d e^{2} - \frac {7 b \left (\frac {b e^{3}}{10} + 3 c d e^{2}\right )}{8 c} + 3 c d^{2} e\right )}{3 c} + \frac {x \left (3 b d^{2} e - \frac {5 b \left (3 b d e^{2} - \frac {7 b \left (\frac {b e^{3}}{10} + 3 c d e^{2}\right )}{8 c} + 3 c d^{2} e\right )}{6 c} + c d^{3}\right )}{2 c} + \frac {b d^{3} - \frac {3 b \left (3 b d^{2} e - \frac {5 b \left (3 b d e^{2} - \frac {7 b \left (\frac {b e^{3}}{10} + 3 c d e^{2}\right )}{8 c} + 3 c d^{2} e\right )}{6 c} + c d^{3}\right )}{4 c}}{c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {d^{3} \left (b x\right )^{\frac {3}{2}}}{3} + \frac {3 d^{2} e \left (b x\right )^{\frac {5}{2}}}{5 b} + \frac {3 d e^{2} \left (b x\right )^{\frac {7}{2}}}{7 b^{2}} + \frac {e^{3} \left (b x\right )^{\frac {9}{2}}}{9 b^{3}}\right )}{b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((-b*(b*d**3 - 3*b*(3*b*d**2*e - 5*b*(3*b*d*e**2 - 7*b*(b*e**3/10 + 3*c*d*e**2)/(8*c) + 3*c*d**2*e)/(6*c) + c*d**3)/(4*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))/(2*c) + sqrt(b*x + c*x**2)*(e**3*x**4/5 + x**3*(b*e**3/10 + 3*c*d*e**2)/(4*c) + x**2*(3*b*d *e**2 - 7*b*(b*e**3/10 + 3*c*d*e**2)/(8*c) + 3*c*d**2*e)/(3*c) + x*(3*b*d* *2*e - 5*b*(3*b*d*e**2 - 7*b*(b*e**3/10 + 3*c*d*e**2)/(8*c) + 3*c*d**2*e)/ (6*c) + c*d**3)/(2*c) + (b*d**3 - 3*b*(3*b*d**2*e - 5*b*(3*b*d*e**2 - 7*b* (b*e**3/10 + 3*c*d*e**2)/(8*c) + 3*c*d**2*e)/(6*c) + c*d**3)/(4*c))/c), Ne (c, 0)), (2*(d**3*(b*x)**(3/2)/3 + 3*d**2*e*(b*x)**(5/2)/(5*b) + 3*d*e**2* (b*x)**(7/2)/(7*b**2) + e**3*(b*x)**(9/2)/(9*b**3))/b, Ne(b, 0)), (0, True ))
Leaf count of result is larger than twice the leaf count of optimal. 439 vs. \(2 (190) = 380\).
Time = 0.21 (sec) , antiderivative size = 439, normalized size of antiderivative = 2.09 \[ \int (d+e x)^3 \sqrt {b x+c x^2} \, dx=\frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} e^{3} x^{2}}{5 \, c} + \frac {1}{2} \, \sqrt {c x^{2} + b x} d^{3} x - \frac {3 \, \sqrt {c x^{2} + b x} b d^{2} e x}{4 \, c} + \frac {15 \, \sqrt {c x^{2} + b x} b^{2} d e^{2} x}{32 \, c^{2}} + \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} d e^{2} x}{4 \, c} - \frac {7 \, \sqrt {c x^{2} + b x} b^{3} e^{3} x}{64 \, c^{3}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b e^{3} x}{40 \, c^{2}} - \frac {b^{2} d^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {3}{2}}} + \frac {3 \, b^{3} d^{2} e \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {5}{2}}} - \frac {15 \, b^{4} d e^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {7}{2}}} + \frac {7 \, b^{5} e^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {9}{2}}} + \frac {\sqrt {c x^{2} + b x} b d^{3}}{4 \, c} - \frac {3 \, \sqrt {c x^{2} + b x} b^{2} d^{2} e}{8 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} d^{2} e}{c} + \frac {15 \, \sqrt {c x^{2} + b x} b^{3} d e^{2}}{64 \, c^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b d e^{2}}{8 \, c^{2}} - \frac {7 \, \sqrt {c x^{2} + b x} b^{4} e^{3}}{128 \, c^{4}} + \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2} e^{3}}{48 \, c^{3}} \]
1/5*(c*x^2 + b*x)^(3/2)*e^3*x^2/c + 1/2*sqrt(c*x^2 + b*x)*d^3*x - 3/4*sqrt (c*x^2 + b*x)*b*d^2*e*x/c + 15/32*sqrt(c*x^2 + b*x)*b^2*d*e^2*x/c^2 + 3/4* (c*x^2 + b*x)^(3/2)*d*e^2*x/c - 7/64*sqrt(c*x^2 + b*x)*b^3*e^3*x/c^3 - 7/4 0*(c*x^2 + b*x)^(3/2)*b*e^3*x/c^2 - 1/8*b^2*d^3*log(2*c*x + b + 2*sqrt(c*x ^2 + b*x)*sqrt(c))/c^(3/2) + 3/16*b^3*d^2*e*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) - 15/128*b^4*d*e^2*log(2*c*x + b + 2*sqrt(c*x^2 + b *x)*sqrt(c))/c^(7/2) + 7/256*b^5*e^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*s qrt(c))/c^(9/2) + 1/4*sqrt(c*x^2 + b*x)*b*d^3/c - 3/8*sqrt(c*x^2 + b*x)*b^ 2*d^2*e/c^2 + (c*x^2 + b*x)^(3/2)*d^2*e/c + 15/64*sqrt(c*x^2 + b*x)*b^3*d* e^2/c^3 - 5/8*(c*x^2 + b*x)^(3/2)*b*d*e^2/c^2 - 7/128*sqrt(c*x^2 + b*x)*b^ 4*e^3/c^4 + 7/48*(c*x^2 + b*x)^(3/2)*b^2*e^3/c^3
Time = 0.28 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.21 \[ \int (d+e x)^3 \sqrt {b x+c x^2} \, dx=\frac {1}{1920} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, e^{3} x + \frac {30 \, c^{4} d e^{2} + b c^{3} e^{3}}{c^{4}}\right )} x + \frac {240 \, c^{4} d^{2} e + 30 \, b c^{3} d e^{2} - 7 \, b^{2} c^{2} e^{3}}{c^{4}}\right )} x + \frac {5 \, {\left (96 \, c^{4} d^{3} + 48 \, b c^{3} d^{2} e - 30 \, b^{2} c^{2} d e^{2} + 7 \, b^{3} c e^{3}\right )}}{c^{4}}\right )} x + \frac {15 \, {\left (32 \, b c^{3} d^{3} - 48 \, b^{2} c^{2} d^{2} e + 30 \, b^{3} c d e^{2} - 7 \, b^{4} e^{3}\right )}}{c^{4}}\right )} + \frac {{\left (32 \, b^{2} c^{3} d^{3} - 48 \, b^{3} c^{2} d^{2} e + 30 \, b^{4} c d e^{2} - 7 \, b^{5} e^{3}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{256 \, c^{\frac {9}{2}}} \]
1/1920*sqrt(c*x^2 + b*x)*(2*(4*(6*(8*e^3*x + (30*c^4*d*e^2 + b*c^3*e^3)/c^ 4)*x + (240*c^4*d^2*e + 30*b*c^3*d*e^2 - 7*b^2*c^2*e^3)/c^4)*x + 5*(96*c^4 *d^3 + 48*b*c^3*d^2*e - 30*b^2*c^2*d*e^2 + 7*b^3*c*e^3)/c^4)*x + 15*(32*b* c^3*d^3 - 48*b^2*c^2*d^2*e + 30*b^3*c*d*e^2 - 7*b^4*e^3)/c^4) + 1/256*(32* b^2*c^3*d^3 - 48*b^3*c^2*d^2*e + 30*b^4*c*d*e^2 - 7*b^5*e^3)*log(abs(2*(sq rt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(9/2)
Time = 10.24 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.72 \[ \int (d+e x)^3 \sqrt {b x+c x^2} \, dx=d^3\,\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )-\frac {7\,b\,e^3\,\left (\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,b\,\left (\frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}\right )}{8\,c}\right )}{10\,c}+\frac {e^3\,x^2\,{\left (c\,x^2+b\,x\right )}^{3/2}}{5\,c}-\frac {b^2\,d^3\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}+\frac {3\,d\,e^2\,x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {15\,b\,d\,e^2\,\left (\frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}\right )}{8\,c}+\frac {3\,b^3\,d^2\,e\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {d^2\,e\,\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{8\,c^2} \]
d^3*(b*x + c*x^2)^(1/2)*(x/2 + b/(4*c)) - (7*b*e^3*((x*(b*x + c*x^2)^(3/2) )/(4*c) - (5*b*((b^3*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2)^(1/2)))/(16 *c^(5/2)) + ((b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3*b^2 + 2*b*c*x))/(24*c^2))) /(8*c)))/(10*c) + (e^3*x^2*(b*x + c*x^2)^(3/2))/(5*c) - (b^2*d^3*log((b/2 + c*x)/c^(1/2) + (b*x + c*x^2)^(1/2)))/(8*c^(3/2)) + (3*d*e^2*x*(b*x + c*x ^2)^(3/2))/(4*c) - (15*b*d*e^2*((b^3*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c* x^2)^(1/2)))/(16*c^(5/2)) + ((b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3*b^2 + 2*b* c*x))/(24*c^2)))/(8*c) + (3*b^3*d^2*e*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c *x^2)^(1/2)))/(16*c^(5/2)) + (d^2*e*(b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3*b^2 + 2*b*c*x))/(8*c^2)