Integrand size = 22, antiderivative size = 103 \[ \int \frac {(d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx=\frac {e (8 c d-3 b e+2 c e x) \sqrt {a+b x+c x^2}}{4 c^2}+\frac {\left (\left (3 b^2-4 a c\right ) e^2+8 c d (c d-b e)\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{5/2}} \] Output:
1/4*e*(2*c*e*x-3*b*e+8*c*d)*(c*x^2+b*x+a)^(1/2)/c^2+1/8*((-4*a*c+3*b^2)*e^ 2+8*c*d*(-b*e+c*d))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^( 5/2)
Time = 0.45 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.03 \[ \int \frac {(d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx=\frac {2 \sqrt {c} e (8 c d-3 b e+2 c e x) \sqrt {a+x (b+c x)}+\left (-8 c^2 d^2-3 b^2 e^2+4 c e (2 b d+a e)\right ) \log \left (c^2 \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )}{8 c^{5/2}} \] Input:
Integrate[(d + e*x)^2/Sqrt[a + b*x + c*x^2],x]
Output:
(2*Sqrt[c]*e*(8*c*d - 3*b*e + 2*c*e*x)*Sqrt[a + x*(b + c*x)] + (-8*c^2*d^2 - 3*b^2*e^2 + 4*c*e*(2*b*d + a*e))*Log[c^2*(b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(8*c^(5/2))
Time = 0.29 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.29, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1166, 27, 1160, 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.
\(\displaystyle \int \frac {(d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx\) |
\(\Big \downarrow \) 1166 |
\(\displaystyle \frac {\int \frac {4 c d^2-e (b d+2 a e)+3 e (2 c d-b e) x}{2 \sqrt {c x^2+b x+a}}dx}{2 c}+\frac {e (d+e x) \sqrt {a+b x+c x^2}}{2 c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {4 c d^2-e (b d+2 a e)+3 e (2 c d-b e) x}{\sqrt {c x^2+b x+a}}dx}{4 c}+\frac {e (d+e x) \sqrt {a+b x+c x^2}}{2 c}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {\frac {\left (-4 c e (a e+2 b d)+3 b^2 e^2+8 c^2 d^2\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{2 c}+\frac {3 e \sqrt {a+b x+c x^2} (2 c d-b e)}{c}}{4 c}+\frac {e (d+e x) \sqrt {a+b x+c x^2}}{2 c}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {\frac {\left (-4 c e (a e+2 b d)+3 b^2 e^2+8 c^2 d^2\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{c}+\frac {3 e \sqrt {a+b x+c x^2} (2 c d-b e)}{c}}{4 c}+\frac {e (d+e x) \sqrt {a+b x+c x^2}}{2 c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-4 c e (a e+2 b d)+3 b^2 e^2+8 c^2 d^2\right )}{2 c^{3/2}}+\frac {3 e \sqrt {a+b x+c x^2} (2 c d-b e)}{c}}{4 c}+\frac {e (d+e x) \sqrt {a+b x+c x^2}}{2 c}\) |
Input:
Int[(d + e*x)^2/Sqrt[a + b*x + c*x^2],x]
Output:
(e*(d + e*x)*Sqrt[a + b*x + c*x^2])/(2*c) + ((3*e*(2*c*d - b*e)*Sqrt[a + b *x + c*x^2])/c + ((8*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(2*b*d + a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*c^(3/2)))/(4*c)
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[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_))*((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]
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]
Time = 1.02 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.91
method | result | size |
risch | \(-\frac {e \left (-2 c e x +3 b e -8 c d \right ) \sqrt {c \,x^{2}+b x +a}}{4 c^{2}}-\frac {\left (4 a c \,e^{2}-3 b^{2} e^{2}+8 b c d e -8 c^{2} d^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {5}{2}}}\) | \(94\) |
default | \(\frac {d^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+e^{2} \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )+2 d e \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )\) | \(194\) |
Input:
int((e*x+d)^2/(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/4*e*(-2*c*e*x+3*b*e-8*c*d)/c^2*(c*x^2+b*x+a)^(1/2)-1/8*(4*a*c*e^2-3*b^2 *e^2+8*b*c*d*e-8*c^2*d^2)/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/ 2))
Time = 0.10 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.40 \[ \int \frac {(d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx=\left [-\frac {{\left (8 \, c^{2} d^{2} - 8 \, b c d e + {\left (3 \, b^{2} - 4 \, a c\right )} e^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (2 \, c^{2} e^{2} x + 8 \, c^{2} d e - 3 \, b c e^{2}\right )} \sqrt {c x^{2} + b x + a}}{16 \, c^{3}}, -\frac {{\left (8 \, c^{2} d^{2} - 8 \, b c d e + {\left (3 \, b^{2} - 4 \, a c\right )} e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (2 \, c^{2} e^{2} x + 8 \, c^{2} d e - 3 \, b c e^{2}\right )} \sqrt {c x^{2} + b x + a}}{8 \, c^{3}}\right ] \] Input:
integrate((e*x+d)^2/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")
Output:
[-1/16*((8*c^2*d^2 - 8*b*c*d*e + (3*b^2 - 4*a*c)*e^2)*sqrt(c)*log(-8*c^2*x ^2 - 8*b*c*x - b^2 + 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(2*c^2*e^2*x + 8*c^2*d*e - 3*b*c*e^2)*sqrt(c*x^2 + b*x + a))/c^3, -1/8 *((8*c^2*d^2 - 8*b*c*d*e + (3*b^2 - 4*a*c)*e^2)*sqrt(-c)*arctan(1/2*sqrt(c *x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 2*(2*c^2*e ^2*x + 8*c^2*d*e - 3*b*c*e^2)*sqrt(c*x^2 + b*x + a))/c^3]
Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (94) = 188\).
Time = 0.81 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.36 \[ \int \frac {(d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx=\begin {cases} \left (\frac {e^{2} x}{2 c} + \frac {- \frac {3 b e^{2}}{4 c} + 2 d e}{c}\right ) \sqrt {a + b x + c x^{2}} + \left (- \frac {a e^{2}}{2 c} - \frac {b \left (- \frac {3 b e^{2}}{4 c} + 2 d e\right )}{2 c} + d^{2}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {e^{2} \left (a + b x\right )^{\frac {5}{2}}}{5 b^{2}} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (- 2 a e^{2} + 2 b d e\right )}{3 b^{2}} + \frac {\sqrt {a + b x} \left (a^{2} e^{2} - 2 a b d e + b^{2} d^{2}\right )}{b^{2}}\right )}{b} & \text {for}\: b \neq 0 \\\frac {\begin {cases} d^{2} x & \text {for}\: e = 0 \\\frac {\left (d + e x\right )^{3}}{3 e} & \text {otherwise} \end {cases}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:
integrate((e*x+d)**2/(c*x**2+b*x+a)**(1/2),x)
Output:
Piecewise(((e**2*x/(2*c) + (-3*b*e**2/(4*c) + 2*d*e)/c)*sqrt(a + b*x + c*x **2) + (-a*e**2/(2*c) - b*(-3*b*e**2/(4*c) + 2*d*e)/(2*c) + d**2)*Piecewis e((log(b + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(a - b**2/ (4*c), 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True )), Ne(c, 0)), (2*(e**2*(a + b*x)**(5/2)/(5*b**2) + (a + b*x)**(3/2)*(-2*a *e**2 + 2*b*d*e)/(3*b**2) + sqrt(a + b*x)*(a**2*e**2 - 2*a*b*d*e + b**2*d* *2)/b**2)/b, Ne(b, 0)), (Piecewise((d**2*x, Eq(e, 0)), ((d + e*x)**3/(3*e) , True))/sqrt(a), True))
Exception generated. \[ \int \frac {(d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((e*x+d)^2/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")
Output:
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
Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.02 \[ \int \frac {(d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx=\frac {1}{4} \, \sqrt {c x^{2} + b x + a} {\left (\frac {2 \, e^{2} x}{c} + \frac {8 \, c d e - 3 \, b e^{2}}{c^{2}}\right )} - \frac {{\left (8 \, c^{2} d^{2} - 8 \, b c d e + 3 \, b^{2} e^{2} - 4 \, a c e^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{8 \, c^{\frac {5}{2}}} \] Input:
integrate((e*x+d)^2/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")
Output:
1/4*sqrt(c*x^2 + b*x + a)*(2*e^2*x/c + (8*c*d*e - 3*b*e^2)/c^2) - 1/8*(8*c ^2*d^2 - 8*b*c*d*e + 3*b^2*e^2 - 4*a*c*e^2)*log(abs(2*(sqrt(c)*x - sqrt(c* x^2 + b*x + a))*sqrt(c) + b))/c^(5/2)
Timed out. \[ \int \frac {(d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx=\int \frac {{\left (d+e\,x\right )}^2}{\sqrt {c\,x^2+b\,x+a}} \,d x \] Input:
int((d + e*x)^2/(a + b*x + c*x^2)^(1/2),x)
Output:
int((d + e*x)^2/(a + b*x + c*x^2)^(1/2), x)
Time = 0.56 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.36 \[ \int \frac {(d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx=\frac {-6 \sqrt {c \,x^{2}+b x +a}\, b c \,e^{2}+16 \sqrt {c \,x^{2}+b x +a}\, c^{2} d e +4 \sqrt {c \,x^{2}+b x +a}\, c^{2} e^{2} x -4 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a c \,e^{2}+3 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{2} e^{2}-8 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b c d e +8 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) c^{2} d^{2}}{8 c^{3}} \] Input:
int((e*x+d)^2/(c*x^2+b*x+a)^(1/2),x)
Output:
( - 6*sqrt(a + b*x + c*x**2)*b*c*e**2 + 16*sqrt(a + b*x + c*x**2)*c**2*d*e + 4*sqrt(a + b*x + c*x**2)*c**2*e**2*x - 4*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a*c*e**2 + 3*sqrt(c)*log( (2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b**2*e* *2 - 8*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a *c - b**2))*b*c*d*e + 8*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*c**2*d**2)/(8*c**3)