Integrand size = 38, antiderivative size = 136 \[ \int \frac {(g+h x) \sqrt {a+b x+c x^2}}{\left (a d+b d x+c d x^2\right )^{3/2}} \, dx=-\frac {(2 c g-b h) \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c} d \sqrt {a d+b d x+c d x^2}}+\frac {h \sqrt {a+b x+c x^2} \log \left (a+b x+c x^2\right )}{2 c d \sqrt {a d+b d x+c d x^2}} \]
1/2*h*ln(c*x^2+b*x+a)*(c*x^2+b*x+a)^(1/2)/c/d/(c*d*x^2+b*d*x+a*d)^(1/2)-(- b*h+2*c*g)*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))*(c*x^2+b*x+a)^(1/2)/c/d/( -4*a*c+b^2)^(1/2)/(c*d*x^2+b*d*x+a*d)^(1/2)
Time = 0.23 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.79 \[ \int \frac {(g+h x) \sqrt {a+b x+c x^2}}{\left (a d+b d x+c d x^2\right )^{3/2}} \, dx=\frac {(a+x (b+c x))^{3/2} \left ((4 c g-2 b h) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )+\sqrt {-b^2+4 a c} h \log (a+x (b+c x))\right )}{2 c \sqrt {-b^2+4 a c} (d (a+x (b+c x)))^{3/2}} \]
((a + x*(b + c*x))^(3/2)*((4*c*g - 2*b*h)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4 *a*c]] + Sqrt[-b^2 + 4*a*c]*h*Log[a + x*(b + c*x)]))/(2*c*Sqrt[-b^2 + 4*a* c]*(d*(a + x*(b + c*x)))^(3/2))
Time = 0.29 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.82, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1330, 1142, 27, 1083, 219, 1103}
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.
\(\displaystyle \int \frac {(g+h x) \sqrt {a+b x+c x^2}}{\left (a d+b d x+c d x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1330 |
\(\displaystyle \frac {\sqrt {a+b x+c x^2} \int \frac {g+h x}{c d x^2+b d x+a d}dx}{\sqrt {a d+b d x+c d x^2}}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {\sqrt {a+b x+c x^2} \left (\frac {(2 c g-b h) \int \frac {1}{c d x^2+b d x+a d}dx}{2 c}+\frac {h \int \frac {d (b+2 c x)}{c d x^2+b d x+a d}dx}{2 c d}\right )}{\sqrt {a d+b d x+c d x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a+b x+c x^2} \left (\frac {(2 c g-b h) \int \frac {1}{c d x^2+b d x+a d}dx}{2 c}+\frac {h \int \frac {b+2 c x}{c d x^2+b d x+a d}dx}{2 c}\right )}{\sqrt {a d+b d x+c d x^2}}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {\sqrt {a+b x+c x^2} \left (\frac {h \int \frac {b+2 c x}{c d x^2+b d x+a d}dx}{2 c}-\frac {(2 c g-b h) \int \frac {1}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d(b d+2 c x d)}{c}\right )}{\sqrt {a d+b d x+c d x^2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {a+b x+c x^2} \left (\frac {h \int \frac {b+2 c x}{c d x^2+b d x+a d}dx}{2 c}-\frac {(2 c g-b h) \text {arctanh}\left (\frac {b d+2 c d x}{d \sqrt {b^2-4 a c}}\right )}{c d \sqrt {b^2-4 a c}}\right )}{\sqrt {a d+b d x+c d x^2}}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {\sqrt {a+b x+c x^2} \left (\frac {h \log \left (a+b x+c x^2\right )}{2 c d}-\frac {(2 c g-b h) \text {arctanh}\left (\frac {b d+2 c d x}{d \sqrt {b^2-4 a c}}\right )}{c d \sqrt {b^2-4 a c}}\right )}{\sqrt {a d+b d x+c d x^2}}\) |
(Sqrt[a + b*x + c*x^2]*(-(((2*c*g - b*h)*ArcTanh[(b*d + 2*c*d*x)/(Sqrt[b^2 - 4*a*c]*d)])/(c*Sqrt[b^2 - 4*a*c]*d)) + (h*Log[a + b*x + c*x^2])/(2*c*d) ))/Sqrt[a*d + b*d*x + c*d*x^2]
3.1.39.3.1 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)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[((g_.) + (h_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_ ) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b *x + c*x^2)^FracPart[p]/(d^IntPart[p]*(d + e*x + f*x^2)^FracPart[p])) Int [(g + h*x)^m*(d + e*x + f*x^2)^(p + q), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, p, q}, x] && EqQ[c*d - a*f, 0] && EqQ[b*d - a*e, 0] && !IntegerQ[p] && !IntegerQ[q] && !GtQ[c/f, 0]
Time = 0.90 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.89
method | result | size |
default | \(\frac {\sqrt {d \left (c \,x^{2}+b x +a \right )}\, \left (h \ln \left (c \,x^{2}+b x +a \right ) \sqrt {4 a c -b^{2}}-2 \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) b h +4 \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) c g \right )}{2 \sqrt {c \,x^{2}+b x +a}\, d^{2} c \sqrt {4 a c -b^{2}}}\) | \(121\) |
risch | \(\frac {\sqrt {c \,x^{2}+b x +a}\, \left (4 a c h -b^{2} h +\sqrt {-\left (b h -2 c g \right )^{2} \left (4 a c -b^{2}\right )}\right ) \ln \left (-4 a b c h +8 a \,c^{2} g +b^{3} h -2 b^{2} c g -2 \sqrt {-\left (b h -2 c g \right )^{2} \left (4 a c -b^{2}\right )}\, c x -\sqrt {-\left (b h -2 c g \right )^{2} \left (4 a c -b^{2}\right )}\, b \right )}{2 d \sqrt {d \left (c \,x^{2}+b x +a \right )}\, c \left (4 a c -b^{2}\right )}-\frac {\sqrt {c \,x^{2}+b x +a}\, \left (-4 a c h +b^{2} h +\sqrt {-\left (b h -2 c g \right )^{2} \left (4 a c -b^{2}\right )}\right ) \ln \left (-4 a b c h +8 a \,c^{2} g +b^{3} h -2 b^{2} c g +2 \sqrt {-\left (b h -2 c g \right )^{2} \left (4 a c -b^{2}\right )}\, c x +\sqrt {-\left (b h -2 c g \right )^{2} \left (4 a c -b^{2}\right )}\, b \right )}{2 d \sqrt {d \left (c \,x^{2}+b x +a \right )}\, c \left (4 a c -b^{2}\right )}\) | \(328\) |
1/2/(c*x^2+b*x+a)^(1/2)*(d*(c*x^2+b*x+a))^(1/2)*(h*ln(c*x^2+b*x+a)*(4*a*c- b^2)^(1/2)-2*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b*h+4*arctan((2*c*x+b)/(4 *a*c-b^2)^(1/2))*c*g)/d^2/c/(4*a*c-b^2)^(1/2)
\[ \int \frac {(g+h x) \sqrt {a+b x+c x^2}}{\left (a d+b d x+c d x^2\right )^{3/2}} \, dx=\int { \frac {\sqrt {c x^{2} + b x + a} {\left (h x + g\right )}}{{\left (c d x^{2} + b d x + a d\right )}^{\frac {3}{2}}} \,d x } \]
integral(sqrt(c*d*x^2 + b*d*x + a*d)*sqrt(c*x^2 + b*x + a)*(h*x + g)/(c^2* d^2*x^4 + 2*b*c*d^2*x^3 + 2*a*b*d^2*x + (b^2 + 2*a*c)*d^2*x^2 + a^2*d^2), x)
\[ \int \frac {(g+h x) \sqrt {a+b x+c x^2}}{\left (a d+b d x+c d x^2\right )^{3/2}} \, dx=\int \frac {\left (g + h x\right ) \sqrt {a + b x + c x^{2}}}{\left (d \left (a + b x + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {(g+h x) \sqrt {a+b x+c x^2}}{\left (a d+b d x+c d x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
\[ \int \frac {(g+h x) \sqrt {a+b x+c x^2}}{\left (a d+b d x+c d x^2\right )^{3/2}} \, dx=\int { \frac {\sqrt {c x^{2} + b x + a} {\left (h x + g\right )}}{{\left (c d x^{2} + b d x + a d\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(g+h x) \sqrt {a+b x+c x^2}}{\left (a d+b d x+c d x^2\right )^{3/2}} \, dx=\int \frac {\left (g+h\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{{\left (c\,d\,x^2+b\,d\,x+a\,d\right )}^{3/2}} \,d x \]