Integrand size = 36, antiderivative size = 227 \[ \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=-\frac {\left (4 b c d+8 a d e-5 b e^2\right ) (e+2 d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{64 d^3 (a+b x)}+\frac {(8 a d-5 b e+6 b d x) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{24 d^2 (a+b x)}-\frac {\left (4 c d-e^2\right ) \left (4 b c d+8 a d e-5 b e^2\right ) \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+e x+d x^2}}\right )}{128 d^{7/2} (a+b x)} \]
1/24*(6*b*d*x+8*a*d-5*b*e)*(d*x^2+e*x+c)^(3/2)*((b*x+a)^2)^(1/2)/d^2/(b*x+ a)-1/128*(4*c*d-e^2)*(8*a*d*e+4*b*c*d-5*b*e^2)*arctanh(1/2*(2*d*x+e)/d^(1/ 2)/(d*x^2+e*x+c)^(1/2))*((b*x+a)^2)^(1/2)/d^(7/2)/(b*x+a)-1/64*(8*a*d*e+4* b*c*d-5*b*e^2)*(2*d*x+e)*((b*x+a)^2)^(1/2)*(d*x^2+e*x+c)^(1/2)/d^3/(b*x+a)
Time = 0.42 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.78 \[ \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\frac {\sqrt {(a+b x)^2} \left (2 \sqrt {d} \sqrt {c+x (e+d x)} \left (8 a d \left (8 c d-3 e^2+2 d e x+8 d^2 x^2\right )+b \left (15 e^3-10 d e^2 x+8 d^2 e x^2+48 d^3 x^3+4 c d (-13 e+6 d x)\right )\right )+3 \left (4 c d-e^2\right ) \left (4 b c d+8 a d e-5 b e^2\right ) \log \left (e+2 d x-2 \sqrt {d} \sqrt {c+x (e+d x)}\right )\right )}{384 d^{7/2} (a+b x)} \]
(Sqrt[(a + b*x)^2]*(2*Sqrt[d]*Sqrt[c + x*(e + d*x)]*(8*a*d*(8*c*d - 3*e^2 + 2*d*e*x + 8*d^2*x^2) + b*(15*e^3 - 10*d*e^2*x + 8*d^2*e*x^2 + 48*d^3*x^3 + 4*c*d*(-13*e + 6*d*x))) + 3*(4*c*d - e^2)*(4*b*c*d + 8*a*d*e - 5*b*e^2) *Log[e + 2*d*x - 2*Sqrt[d]*Sqrt[c + x*(e + d*x)]]))/(384*d^(7/2)*(a + b*x) )
Time = 0.31 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.73, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1333, 27, 1225, 1087, 1092, 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.
| \(\displaystyle \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2+e x} \, dx\) |
| \(\Big \downarrow \) 1333 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int 2 b x (a+b x) \sqrt {d x^2+e x+c}dx}{2 b (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x (a+b x) \sqrt {d x^2+e x+c}dx}{a+b x}\) |
| \(\Big \downarrow \) 1225 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {\left (c+d x^2+e x\right )^{3/2} (8 a d+6 b d x-5 b e)}{24 d^2}-\frac {\left (8 a d e+4 b c d-5 b e^2\right ) \int \sqrt {d x^2+e x+c}dx}{16 d^2}\right )}{a+b x}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {\left (c+d x^2+e x\right )^{3/2} (8 a d+6 b d x-5 b e)}{24 d^2}-\frac {\left (8 a d e+4 b c d-5 b e^2\right ) \left (\frac {\left (4 c d-e^2\right ) \int \frac {1}{\sqrt {d x^2+e x+c}}dx}{8 d}+\frac {(2 d x+e) \sqrt {c+d x^2+e x}}{4 d}\right )}{16 d^2}\right )}{a+b x}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {\left (c+d x^2+e x\right )^{3/2} (8 a d+6 b d x-5 b e)}{24 d^2}-\frac {\left (8 a d e+4 b c d-5 b e^2\right ) \left (\frac {\left (4 c d-e^2\right ) \int \frac {1}{4 d-\frac {(e+2 d x)^2}{d x^2+e x+c}}d\frac {e+2 d x}{\sqrt {d x^2+e x+c}}}{4 d}+\frac {(2 d x+e) \sqrt {c+d x^2+e x}}{4 d}\right )}{16 d^2}\right )}{a+b x}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {\left (c+d x^2+e x\right )^{3/2} (8 a d+6 b d x-5 b e)}{24 d^2}-\frac {\left (8 a d e+4 b c d-5 b e^2\right ) \left (\frac {\left (4 c d-e^2\right ) \text {arctanh}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{8 d^{3/2}}+\frac {(2 d x+e) \sqrt {c+d x^2+e x}}{4 d}\right )}{16 d^2}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(((8*a*d - 5*b*e + 6*b*d*x)*(c + e*x + d*x^ 2)^(3/2))/(24*d^2) - ((4*b*c*d + 8*a*d*e - 5*b*e^2)*(((e + 2*d*x)*Sqrt[c + e*x + d*x^2])/(4*d) + ((4*c*d - e^2)*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[ c + e*x + d*x^2])])/(8*d^(3/2))))/(16*d^2)))/(a + b*x)
3.1.47.3.1 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[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, 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]
Int[((g_.) + (h_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_ ) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(a + b*x + c*x^2)^Fr acPart[p]/((4*c)^IntPart[p]*(b + 2*c*x)^(2*FracPart[p])) Int[(g + h*x)^m* (b + 2*c*x)^(2*p)*(d + e*x + f*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g , h, m, p, q}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Time = 0.67 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.88
| method | result | size |
| risch | \(\frac {\left (48 b \,x^{3} d^{3}+64 a \,d^{3} x^{2}+8 b \,d^{2} e \,x^{2}+16 a \,d^{2} e x +24 b c \,d^{2} x -10 b d \,e^{2} x +64 c \,d^{2} a -24 a d \,e^{2}-52 b c d e +15 b \,e^{3}\right ) \sqrt {d \,x^{2}+e x +c}\, \sqrt {\left (b x +a \right )^{2}}}{192 d^{3} \left (b x +a \right )}-\frac {\left (32 a c \,d^{2} e -8 a d \,e^{3}+16 b \,d^{2} c^{2}-24 b c d \,e^{2}+5 b \,e^{4}\right ) \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right ) \sqrt {\left (b x +a \right )^{2}}}{128 d^{\frac {7}{2}} \left (b x +a \right )}\) | \(199\) |
| default | \(\frac {\operatorname {csgn}\left (b x +a \right ) \left (96 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {7}{2}} b x +128 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {7}{2}} a -80 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {5}{2}} b e -96 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {7}{2}} a e x -48 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {7}{2}} b c x +60 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} b \,e^{2} x -48 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} a \,e^{2}-24 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} b c e +30 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {3}{2}} b \,e^{3}-96 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a c \,d^{3} e +24 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a \,d^{2} e^{3}-48 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b \,c^{2} d^{3}+72 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b c \,d^{2} e^{2}-15 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b d \,e^{4}\right )}{384 d^{\frac {9}{2}}}\) | \(381\) |
1/192*(48*b*d^3*x^3+64*a*d^3*x^2+8*b*d^2*e*x^2+16*a*d^2*e*x+24*b*c*d^2*x-1 0*b*d*e^2*x+64*a*c*d^2-24*a*d*e^2-52*b*c*d*e+15*b*e^3)*(d*x^2+e*x+c)^(1/2) /d^3*((b*x+a)^2)^(1/2)/(b*x+a)-1/128*(32*a*c*d^2*e-8*a*d*e^3+16*b*c^2*d^2- 24*b*c*d*e^2+5*b*e^4)/d^(7/2)*ln((1/2*e+d*x)/d^(1/2)+(d*x^2+e*x+c)^(1/2))* ((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.32 (sec) , antiderivative size = 391, normalized size of antiderivative = 1.72 \[ \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\left [\frac {3 \, {\left (16 \, b c^{2} d^{2} + 32 \, a c d^{2} e - 24 \, b c d e^{2} - 8 \, a d e^{3} + 5 \, b e^{4}\right )} \sqrt {d} \log \left (8 \, d^{2} x^{2} + 8 \, d e x - 4 \, \sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) + 4 \, {\left (48 \, b d^{4} x^{3} + 64 \, a c d^{3} - 52 \, b c d^{2} e - 24 \, a d^{2} e^{2} + 15 \, b d e^{3} + 8 \, {\left (8 \, a d^{4} + b d^{3} e\right )} x^{2} + 2 \, {\left (12 \, b c d^{3} + 8 \, a d^{3} e - 5 \, b d^{2} e^{2}\right )} x\right )} \sqrt {d x^{2} + e x + c}}{768 \, d^{4}}, \frac {3 \, {\left (16 \, b c^{2} d^{2} + 32 \, a c d^{2} e - 24 \, b c d e^{2} - 8 \, a d e^{3} + 5 \, b e^{4}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d e x + c d\right )}}\right ) + 2 \, {\left (48 \, b d^{4} x^{3} + 64 \, a c d^{3} - 52 \, b c d^{2} e - 24 \, a d^{2} e^{2} + 15 \, b d e^{3} + 8 \, {\left (8 \, a d^{4} + b d^{3} e\right )} x^{2} + 2 \, {\left (12 \, b c d^{3} + 8 \, a d^{3} e - 5 \, b d^{2} e^{2}\right )} x\right )} \sqrt {d x^{2} + e x + c}}{384 \, d^{4}}\right ] \]
[1/768*(3*(16*b*c^2*d^2 + 32*a*c*d^2*e - 24*b*c*d*e^2 - 8*a*d*e^3 + 5*b*e^ 4)*sqrt(d)*log(8*d^2*x^2 + 8*d*e*x - 4*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*s qrt(d) + 4*c*d + e^2) + 4*(48*b*d^4*x^3 + 64*a*c*d^3 - 52*b*c*d^2*e - 24*a *d^2*e^2 + 15*b*d*e^3 + 8*(8*a*d^4 + b*d^3*e)*x^2 + 2*(12*b*c*d^3 + 8*a*d^ 3*e - 5*b*d^2*e^2)*x)*sqrt(d*x^2 + e*x + c))/d^4, 1/384*(3*(16*b*c^2*d^2 + 32*a*c*d^2*e - 24*b*c*d*e^2 - 8*a*d*e^3 + 5*b*e^4)*sqrt(-d)*arctan(1/2*sq rt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(-d)/(d^2*x^2 + d*e*x + c*d)) + 2*(48* b*d^4*x^3 + 64*a*c*d^3 - 52*b*c*d^2*e - 24*a*d^2*e^2 + 15*b*d*e^3 + 8*(8*a *d^4 + b*d^3*e)*x^2 + 2*(12*b*c*d^3 + 8*a*d^3*e - 5*b*d^2*e^2)*x)*sqrt(d*x ^2 + e*x + c))/d^4]
\[ \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\int x \sqrt {c + d x^{2} + e x} \sqrt {\left (a + b x\right )^{2}}\, dx \]
\[ \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\int { \sqrt {d x^{2} + e x + c} \sqrt {{\left (b x + a\right )}^{2}} x \,d x } \]
Time = 0.29 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.17 \[ \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\frac {1}{192} \, \sqrt {d x^{2} + e x + c} {\left (2 \, {\left (4 \, {\left (6 \, b x \mathrm {sgn}\left (b x + a\right ) + \frac {8 \, a d^{3} \mathrm {sgn}\left (b x + a\right ) + b d^{2} e \mathrm {sgn}\left (b x + a\right )}{d^{3}}\right )} x + \frac {12 \, b c d^{2} \mathrm {sgn}\left (b x + a\right ) + 8 \, a d^{2} e \mathrm {sgn}\left (b x + a\right ) - 5 \, b d e^{2} \mathrm {sgn}\left (b x + a\right )}{d^{3}}\right )} x + \frac {64 \, a c d^{2} \mathrm {sgn}\left (b x + a\right ) - 52 \, b c d e \mathrm {sgn}\left (b x + a\right ) - 24 \, a d e^{2} \mathrm {sgn}\left (b x + a\right ) + 15 \, b e^{3} \mathrm {sgn}\left (b x + a\right )}{d^{3}}\right )} + \frac {{\left (16 \, b c^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) + 32 \, a c d^{2} e \mathrm {sgn}\left (b x + a\right ) - 24 \, b c d e^{2} \mathrm {sgn}\left (b x + a\right ) - 8 \, a d e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, b e^{4} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )} \sqrt {d} + e \right |}\right )}{128 \, d^{\frac {7}{2}}} \]
1/192*sqrt(d*x^2 + e*x + c)*(2*(4*(6*b*x*sgn(b*x + a) + (8*a*d^3*sgn(b*x + a) + b*d^2*e*sgn(b*x + a))/d^3)*x + (12*b*c*d^2*sgn(b*x + a) + 8*a*d^2*e* sgn(b*x + a) - 5*b*d*e^2*sgn(b*x + a))/d^3)*x + (64*a*c*d^2*sgn(b*x + a) - 52*b*c*d*e*sgn(b*x + a) - 24*a*d*e^2*sgn(b*x + a) + 15*b*e^3*sgn(b*x + a) )/d^3) + 1/128*(16*b*c^2*d^2*sgn(b*x + a) + 32*a*c*d^2*e*sgn(b*x + a) - 24 *b*c*d*e^2*sgn(b*x + a) - 8*a*d*e^3*sgn(b*x + a) + 5*b*e^4*sgn(b*x + a))*l og(abs(2*(sqrt(d)*x - sqrt(d*x^2 + e*x + c))*sqrt(d) + e))/d^(7/2)
Timed out. \[ \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\int x\,\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+e\,x+c} \,d x \]