Integrand size = 26, antiderivative size = 66 \[ \int \frac {(b d+2 c d x)^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 d^2 (b+2 c x)}{\sqrt {a+b x+c x^2}}+4 \sqrt {c} d^2 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \] Output:
-2*d^2*(2*c*x+b)/(c*x^2+b*x+a)^(1/2)+4*c^(1/2)*d^2*arctanh(1/2*(2*c*x+b)/c ^(1/2)/(c*x^2+b*x+a)^(1/2))
Time = 0.42 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.14 \[ \int \frac {(b d+2 c d x)^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 d^2 \left (b+2 c x-4 \sqrt {c} \sqrt {a+x (b+c x)} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )\right )}{\sqrt {a+x (b+c x)}} \] Input:
Integrate[(b*d + 2*c*d*x)^2/(a + b*x + c*x^2)^(3/2),x]
Output:
(-2*d^2*(b + 2*c*x - 4*Sqrt[c]*Sqrt[a + x*(b + c*x)]*ArcTanh[(Sqrt[c]*x)/( -Sqrt[a] + Sqrt[a + x*(b + c*x)])]))/Sqrt[a + x*(b + c*x)]
Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1110, 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.
\(\displaystyle \int \frac {(b d+2 c d x)^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1110 |
\(\displaystyle 4 c d^2 \int \frac {1}{\sqrt {c x^2+b x+a}}dx-\frac {2 d^2 (b+2 c x)}{\sqrt {a+b x+c x^2}}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle 8 c d^2 \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}-\frac {2 d^2 (b+2 c x)}{\sqrt {a+b x+c x^2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 4 \sqrt {c} d^2 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )-\frac {2 d^2 (b+2 c x)}{\sqrt {a+b x+c x^2}}\) |
Input:
Int[(b*d + 2*c*d*x)^2/(a + b*x + c*x^2)^(3/2),x]
Output:
(-2*d^2*(b + 2*c*x))/Sqrt[a + b*x + c*x^2] + 4*Sqrt[c]*d^2*ArcTanh[(b + 2* c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])]
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_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy mbol] :> Simp[d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), x] - Simp[d*e*((m - 1)/(b*(p + 1))) Int[(d + e*x)^(m - 2)*(a + b*x + c*x^ 2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0] && N eQ[m + 2*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]
Leaf count of result is larger than twice the leaf count of optimal. \(211\) vs. \(2(56)=112\).
Time = 1.06 (sec) , antiderivative size = 212, normalized size of antiderivative = 3.21
method | result | size |
default | \(d^{2} \left (\frac {2 b^{2} \left (2 c x +b \right )}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+4 c^{2} \left (-\frac {x}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}\right )+4 b c \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )\right )\) | \(212\) |
Input:
int((2*c*d*x+b*d)^2/(c*x^2+b*x+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
d^2*(2*b^2*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+4*c^2*(-x/c/(c*x^2+b* x+a)^(1/2)-1/2*b/c*(-1/c/(c*x^2+b*x+a)^(1/2)-b/c*(2*c*x+b)/(4*a*c-b^2)/(c* x^2+b*x+a)^(1/2))+1/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)))+4 *b*c*(-1/c/(c*x^2+b*x+a)^(1/2)-b/c*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/ 2)))
Time = 0.13 (sec) , antiderivative size = 225, normalized size of antiderivative = 3.41 \[ \int \frac {(b d+2 c d x)^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\left [\frac {2 \, {\left ({\left (c d^{2} x^{2} + b d^{2} x + a d^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - {\left (2 \, c d^{2} x + b d^{2}\right )} \sqrt {c x^{2} + b x + a}\right )}}{c x^{2} + b x + a}, -\frac {2 \, {\left (2 \, {\left (c d^{2} x^{2} + b d^{2} x + a d^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + {\left (2 \, c d^{2} x + b d^{2}\right )} \sqrt {c x^{2} + b x + a}\right )}}{c x^{2} + b x + a}\right ] \] Input:
integrate((2*c*d*x+b*d)^2/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")
Output:
[2*((c*d^2*x^2 + b*d^2*x + a*d^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) - (2*c*d^2*x + b*d^2 )*sqrt(c*x^2 + b*x + a))/(c*x^2 + b*x + a), -2*(2*(c*d^2*x^2 + b*d^2*x + a *d^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*c*d^2*x + b*d^2)*sqrt(c*x^2 + b*x + a))/(c*x^2 + b*x + a)]
\[ \int \frac {(b d+2 c d x)^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx=d^{2} \left (\int \frac {b^{2}}{a \sqrt {a + b x + c x^{2}} + b x \sqrt {a + b x + c x^{2}} + c x^{2} \sqrt {a + b x + c x^{2}}}\, dx + \int \frac {4 c^{2} x^{2}}{a \sqrt {a + b x + c x^{2}} + b x \sqrt {a + b x + c x^{2}} + c x^{2} \sqrt {a + b x + c x^{2}}}\, dx + \int \frac {4 b c x}{a \sqrt {a + b x + c x^{2}} + b x \sqrt {a + b x + c x^{2}} + c x^{2} \sqrt {a + b x + c x^{2}}}\, dx\right ) \] Input:
integrate((2*c*d*x+b*d)**2/(c*x**2+b*x+a)**(3/2),x)
Output:
d**2*(Integral(b**2/(a*sqrt(a + b*x + c*x**2) + b*x*sqrt(a + b*x + c*x**2) + c*x**2*sqrt(a + b*x + c*x**2)), x) + Integral(4*c**2*x**2/(a*sqrt(a + b *x + c*x**2) + b*x*sqrt(a + b*x + c*x**2) + c*x**2*sqrt(a + b*x + c*x**2)) , x) + Integral(4*b*c*x/(a*sqrt(a + b*x + c*x**2) + b*x*sqrt(a + b*x + c*x **2) + c*x**2*sqrt(a + b*x + c*x**2)), x))
Exception generated. \[ \int \frac {(b d+2 c d x)^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((2*c*d*x+b*d)^2/(c*x^2+b*x+a)^(3/2),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 = 0.42 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.68 \[ \int \frac {(b d+2 c d x)^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-4 \, \sqrt {c} d^{2} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right ) - \frac {2 \, {\left (\frac {2 \, {\left (b^{2} c d^{2} - 4 \, a c^{2} d^{2}\right )} x}{b^{2} - 4 \, a c} + \frac {b^{3} d^{2} - 4 \, a b c d^{2}}{b^{2} - 4 \, a c}\right )}}{\sqrt {c x^{2} + b x + a}} \] Input:
integrate((2*c*d*x+b*d)^2/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")
Output:
-4*sqrt(c)*d^2*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b)) - 2*(2*(b^2*c*d^2 - 4*a*c^2*d^2)*x/(b^2 - 4*a*c) + (b^3*d^2 - 4*a*b*c*d^2 )/(b^2 - 4*a*c))/sqrt(c*x^2 + b*x + a)
Time = 5.78 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.36 \[ \int \frac {(b d+2 c d x)^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx=4\,\sqrt {c}\,d^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )+\frac {b^2\,d^2\,\left (\frac {b}{2}+c\,x\right )}{\left (a\,c-\frac {b^2}{4}\right )\,\sqrt {c\,x^2+b\,x+a}}+\frac {4\,c\,d^2\,\left (\frac {a\,b}{2}-x\,\left (a\,c-\frac {b^2}{2}\right )\right )}{\left (a\,c-\frac {b^2}{4}\right )\,\sqrt {c\,x^2+b\,x+a}}-\frac {4\,b\,c\,d^2\,\left (4\,a+2\,b\,x\right )}{\left (4\,a\,c-b^2\right )\,\sqrt {c\,x^2+b\,x+a}} \] Input:
int((b*d + 2*c*d*x)^2/(a + b*x + c*x^2)^(3/2),x)
Output:
4*c^(1/2)*d^2*log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2)) + (b^2*d^ 2*(b/2 + c*x))/((a*c - b^2/4)*(a + b*x + c*x^2)^(1/2)) + (4*c*d^2*((a*b)/2 - x*(a*c - b^2/2)))/((a*c - b^2/4)*(a + b*x + c*x^2)^(1/2)) - (4*b*c*d^2* (4*a + 2*b*x))/((4*a*c - b^2)*(a + b*x + c*x^2)^(1/2))
Time = 0.19 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.92 \[ \int \frac {(b d+2 c d x)^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {2 d^{2} \left (-\sqrt {c \,x^{2}+b x +a}\, b -2 \sqrt {c \,x^{2}+b x +a}\, c x +2 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a +2 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b x +2 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) c \,x^{2}-2 \sqrt {c}\, a -2 \sqrt {c}\, b x -2 \sqrt {c}\, c \,x^{2}\right )}{c \,x^{2}+b x +a} \] Input:
int((2*c*d*x+b*d)^2/(c*x^2+b*x+a)^(3/2),x)
Output:
(2*d**2*( - sqrt(a + b*x + c*x**2)*b - 2*sqrt(a + b*x + c*x**2)*c*x + 2*sq rt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2 ))*a + 2*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4 *a*c - b**2))*b*x + 2*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*c*x**2 - 2*sqrt(c)*a - 2*sqrt(c)*b*x - 2*sqrt(c )*c*x**2))/(a + b*x + c*x**2)