Integrand size = 32, antiderivative size = 199 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+x} \sqrt {1+x} (d+e x)^3} \, dx=-\frac {\left (c d^2-b d e+a e^2\right ) \sqrt {-1+x} \sqrt {1+x}}{2 e \left (d^2-e^2\right ) (d+e x)^2}+\frac {\left (c d^3+b d^2 e-(3 a+4 c) d e^2+2 b e^3\right ) \sqrt {-1+x} \sqrt {1+x}}{2 e \left (d^2-e^2\right )^2 (d+e x)}+\frac {\left ((2 a+c) d^2-3 b d e+(a+2 c) e^2\right ) \text {arctanh}\left (\frac {\sqrt {d+e} \sqrt {1+x}}{\sqrt {d-e} \sqrt {-1+x}}\right )}{(d-e)^{5/2} (d+e)^{5/2}} \]
((2*a+c)*d^2-3*b*d*e+(a+2*c)*e^2)*arctanh((d+e)^(1/2)*(1+x)^(1/2)/(d-e)^(1 /2)/(-1+x)^(1/2))/(d-e)^(5/2)/(d+e)^(5/2)-1/2*(a*e^2-b*d*e+c*d^2)*(-1+x)^( 1/2)*(1+x)^(1/2)/e/(d^2-e^2)/(e*x+d)^2+1/2*(c*d^3+b*d^2*e-(3*a+4*c)*d*e^2+ 2*b*e^3)*(-1+x)^(1/2)*(1+x)^(1/2)/e/(d^2-e^2)^2/(e*x+d)
Time = 0.60 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.93 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+x} \sqrt {1+x} (d+e x)^3} \, dx=\frac {\sqrt {-1+x} \sqrt {1+x} \left (a e \left (-4 d^2+e^2-3 d e x\right )+c d \left (-3 d e+d^2 x-4 e^2 x\right )+b \left (2 d^3+d e^2+d^2 e x+2 e^3 x\right )\right )}{2 (d-e)^2 (d+e)^2 (d+e x)^2}-\frac {\left (-3 b d e+a \left (2 d^2+e^2\right )+c \left (d^2+2 e^2\right )\right ) \arctan \left (\frac {\sqrt {d-e} \sqrt {\frac {-1+x}{1+x}}}{\sqrt {-d-e}}\right )}{(-d-e)^{5/2} (d-e)^{5/2}} \]
(Sqrt[-1 + x]*Sqrt[1 + x]*(a*e*(-4*d^2 + e^2 - 3*d*e*x) + c*d*(-3*d*e + d^ 2*x - 4*e^2*x) + b*(2*d^3 + d*e^2 + d^2*e*x + 2*e^3*x)))/(2*(d - e)^2*(d + e)^2*(d + e*x)^2) - ((-3*b*d*e + a*(2*d^2 + e^2) + c*(d^2 + 2*e^2))*ArcTa n[(Sqrt[d - e]*Sqrt[(-1 + x)/(1 + x)])/Sqrt[-d - e]])/((-d - e)^(5/2)*(d - e)^(5/2))
Time = 0.50 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2113, 2182, 25, 679, 488, 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.
\(\displaystyle \int \frac {a+b x+c x^2}{\sqrt {x-1} \sqrt {x+1} (d+e x)^3} \, dx\) |
\(\Big \downarrow \) 2113 |
\(\displaystyle \frac {\sqrt {x^2-1} \int \frac {c x^2+b x+a}{(d+e x)^3 \sqrt {x^2-1}}dx}{\sqrt {x-1} \sqrt {x+1}}\) |
\(\Big \downarrow \) 2182 |
\(\displaystyle \frac {\sqrt {x^2-1} \left (-\frac {\int -\frac {2 (a d+c d-b e)+\left (\frac {c d^2}{e}+b d-a e-2 c e\right ) x}{(d+e x)^2 \sqrt {x^2-1}}dx}{2 \left (d^2-e^2\right )}-\frac {\sqrt {x^2-1} \left (a e^2-b d e+c d^2\right )}{2 e \left (d^2-e^2\right ) (d+e x)^2}\right )}{\sqrt {x-1} \sqrt {x+1}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sqrt {x^2-1} \left (\frac {\int \frac {2 (a d+c d-b e)+\left (\frac {c d^2}{e}+b d-a e-2 c e\right ) x}{(d+e x)^2 \sqrt {x^2-1}}dx}{2 \left (d^2-e^2\right )}-\frac {\sqrt {x^2-1} \left (a e^2-b d e+c d^2\right )}{2 e \left (d^2-e^2\right ) (d+e x)^2}\right )}{\sqrt {x-1} \sqrt {x+1}}\) |
\(\Big \downarrow \) 679 |
\(\displaystyle \frac {\sqrt {x^2-1} \left (\frac {\frac {\sqrt {x^2-1} \left (-3 a d e^2+b d^2 e+2 b e^3+c d^3-4 c d e^2\right )}{e \left (d^2-e^2\right ) (d+e x)}-\frac {\left (-a \left (2 d^2+e^2\right )+3 b d e-c \left (d^2+2 e^2\right )\right ) \int \frac {1}{(d+e x) \sqrt {x^2-1}}dx}{d^2-e^2}}{2 \left (d^2-e^2\right )}-\frac {\sqrt {x^2-1} \left (a e^2-b d e+c d^2\right )}{2 e \left (d^2-e^2\right ) (d+e x)^2}\right )}{\sqrt {x-1} \sqrt {x+1}}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle \frac {\sqrt {x^2-1} \left (\frac {\frac {\left (-a \left (2 d^2+e^2\right )+3 b d e-c \left (d^2+2 e^2\right )\right ) \int \frac {1}{d^2-e^2-\frac {(-e-d x)^2}{x^2-1}}d\frac {-e-d x}{\sqrt {x^2-1}}}{d^2-e^2}+\frac {\sqrt {x^2-1} \left (-3 a d e^2+b d^2 e+2 b e^3+c d^3-4 c d e^2\right )}{e \left (d^2-e^2\right ) (d+e x)}}{2 \left (d^2-e^2\right )}-\frac {\sqrt {x^2-1} \left (a e^2-b d e+c d^2\right )}{2 e \left (d^2-e^2\right ) (d+e x)^2}\right )}{\sqrt {x-1} \sqrt {x+1}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {x^2-1} \left (\frac {\frac {\text {arctanh}\left (\frac {-d x-e}{\sqrt {x^2-1} \sqrt {d^2-e^2}}\right ) \left (-a \left (2 d^2+e^2\right )+3 b d e-c \left (d^2+2 e^2\right )\right )}{\left (d^2-e^2\right )^{3/2}}+\frac {\sqrt {x^2-1} \left (-3 a d e^2+b d^2 e+2 b e^3+c d^3-4 c d e^2\right )}{e \left (d^2-e^2\right ) (d+e x)}}{2 \left (d^2-e^2\right )}-\frac {\sqrt {x^2-1} \left (a e^2-b d e+c d^2\right )}{2 e \left (d^2-e^2\right ) (d+e x)^2}\right )}{\sqrt {x-1} \sqrt {x+1}}\) |
(Sqrt[-1 + x^2]*(-1/2*((c*d^2 - b*d*e + a*e^2)*Sqrt[-1 + x^2])/(e*(d^2 - e ^2)*(d + e*x)^2) + (((c*d^3 + b*d^2*e - 3*a*d*e^2 - 4*c*d*e^2 + 2*b*e^3)*S qrt[-1 + x^2])/(e*(d^2 - e^2)*(d + e*x)) + ((3*b*d*e - a*(2*d^2 + e^2) - c *(d^2 + 2*e^2))*ArcTanh[(-e - d*x)/(Sqrt[d^2 - e^2]*Sqrt[-1 + x^2])])/(d^2 - e^2)^(3/2))/(2*(d^2 - e^2))))/(Sqrt[-1 + x]*Sqrt[1 + x])
3.1.40.3.1 Defintions of rubi rules used
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/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1 )/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[(c*d*f + a*e*g)/(c*d^2 + a*e^2) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[Simplify[m + 2*p + 3], 0]
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_. )*(x_))^(p_.), x_Symbol] :> Simp[(a + b*x)^FracPart[m]*((c + d*x)^FracPart[ m]/(a*c + b*d*x^2)^FracPart[m]) Int[Px*(a*c + b*d*x^2)^m*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && EqQ[b*c + a *d, 0] && EqQ[m, n] && !IntegerQ[m]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Qx = PolynomialQuotient[Pq, d + e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[e*R*(d + e*x)^(m + 1)*((a + b*x^2)^(p + 1)/((m + 1)*(b* d^2 + a*e^2))), x] + Simp[1/((m + 1)*(b*d^2 + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[(m + 1)*(b*d^2 + a*e^2)*Qx + b*d*R*(m + 1) - b *e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] && LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(1094\) vs. \(2(175)=350\).
Time = 1.69 (sec) , antiderivative size = 1095, normalized size of antiderivative = 5.50
-1/2*(ln(-2*(-(x^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2)*e+d*x+e)/(e*x+d))*c*d^2* e^2*x^2+4*ln(-2*(-(x^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2)*e+d*x+e)/(e*x+d))*a* d^3*e*x+2*ln(-2*(-(x^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2)*e+d*x+e)/(e*x+d))*a* d*e^3*x-6*ln(-2*(-(x^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2)*e+d*x+e)/(e*x+d))*b* d^2*e^2*x+2*ln(-2*(-(x^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2)*e+d*x+e)/(e*x+d))* c*d^3*e*x+4*ln(-2*(-(x^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2)*e+d*x+e)/(e*x+d))* c*d*e^3*x+4*a*d^2*e^2*(x^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2)-2*b*d^3*e*(x^2-1 )^(1/2)*((d^2-e^2)/e^2)^(1/2)-b*d*e^3*(x^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2)+ 3*c*d^2*e^2*(x^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2)+2*ln(-2*(-(x^2-1)^(1/2)*(( d^2-e^2)/e^2)^(1/2)*e+d*x+e)/(e*x+d))*a*d^2*e^2*x^2-3*ln(-2*(-(x^2-1)^(1/2 )*((d^2-e^2)/e^2)^(1/2)*e+d*x+e)/(e*x+d))*b*d*e^3*x^2+3*a*d*e^3*x*(x^2-1)^ (1/2)*((d^2-e^2)/e^2)^(1/2)-b*d^2*e^2*x*(x^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2 )-c*d^3*e*x*(x^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2)+4*c*d*e^3*x*(x^2-1)^(1/2)* ((d^2-e^2)/e^2)^(1/2)-2*b*e^4*x*(x^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2)+ln(-2* (-(x^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2)*e+d*x+e)/(e*x+d))*a*e^4*x^2+2*ln(-2* (-(x^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2)*e+d*x+e)/(e*x+d))*c*e^4*x^2+ln(-2*(- (x^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2)*e+d*x+e)/(e*x+d))*a*d^2*e^2-3*ln(-2*(- (x^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2)*e+d*x+e)/(e*x+d))*b*d^3*e+2*ln(-2*(-(x ^2-1)^(1/2)*((d^2-e^2)/e^2)^(1/2)*e+d*x+e)/(e*x+d))*c*d^2*e^2-a*e^4*(x^2-1 )^(1/2)*((d^2-e^2)/e^2)^(1/2)+2*ln(-2*(-(x^2-1)^(1/2)*((d^2-e^2)/e^2)^(...
Leaf count of result is larger than twice the leaf count of optimal. 587 vs. \(2 (175) = 350\).
Time = 0.32 (sec) , antiderivative size = 1186, normalized size of antiderivative = 5.96 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+x} \sqrt {1+x} (d+e x)^3} \, dx=\text {Too large to display} \]
[1/2*(c*d^7 + b*d^6*e - (3*a + 5*c)*d^5*e^2 + b*d^4*e^3 + (3*a + 4*c)*d^3* e^4 - 2*b*d^2*e^5 + (c*d^5*e^2 + b*d^4*e^3 - (3*a + 5*c)*d^3*e^4 + b*d^2*e ^5 + (3*a + 4*c)*d*e^6 - 2*b*e^7)*x^2 + ((2*a + c)*d^4*e^2 - 3*b*d^3*e^3 + (a + 2*c)*d^2*e^4 + ((2*a + c)*d^2*e^4 - 3*b*d*e^5 + (a + 2*c)*e^6)*x^2 + 2*((2*a + c)*d^3*e^3 - 3*b*d^2*e^4 + (a + 2*c)*d*e^5)*x)*sqrt(d^2 - e^2)* log((d^2*x + d*e + (d^2 - e^2 + sqrt(d^2 - e^2)*d)*sqrt(x + 1)*sqrt(x - 1) + sqrt(d^2 - e^2)*(d*x + e))/(e*x + d)) + (2*b*d^5*e^2 - (4*a + 3*c)*d^4* e^3 - b*d^3*e^4 + (5*a + 3*c)*d^2*e^5 - b*d*e^6 - a*e^7 + (c*d^5*e^2 + b*d ^4*e^3 - (3*a + 5*c)*d^3*e^4 + b*d^2*e^5 + (3*a + 4*c)*d*e^6 - 2*b*e^7)*x) *sqrt(x + 1)*sqrt(x - 1) + 2*(c*d^6*e + b*d^5*e^2 - (3*a + 5*c)*d^4*e^3 + b*d^3*e^4 + (3*a + 4*c)*d^2*e^5 - 2*b*d*e^6)*x)/(d^8*e^2 - 3*d^6*e^4 + 3*d ^4*e^6 - d^2*e^8 + (d^6*e^4 - 3*d^4*e^6 + 3*d^2*e^8 - e^10)*x^2 + 2*(d^7*e ^3 - 3*d^5*e^5 + 3*d^3*e^7 - d*e^9)*x), 1/2*(c*d^7 + b*d^6*e - (3*a + 5*c) *d^5*e^2 + b*d^4*e^3 + (3*a + 4*c)*d^3*e^4 - 2*b*d^2*e^5 + (c*d^5*e^2 + b* d^4*e^3 - (3*a + 5*c)*d^3*e^4 + b*d^2*e^5 + (3*a + 4*c)*d*e^6 - 2*b*e^7)*x ^2 - 2*((2*a + c)*d^4*e^2 - 3*b*d^3*e^3 + (a + 2*c)*d^2*e^4 + ((2*a + c)*d ^2*e^4 - 3*b*d*e^5 + (a + 2*c)*e^6)*x^2 + 2*((2*a + c)*d^3*e^3 - 3*b*d^2*e ^4 + (a + 2*c)*d*e^5)*x)*sqrt(-d^2 + e^2)*arctan(-(sqrt(-d^2 + e^2)*e*sqrt (x + 1)*sqrt(x - 1) - sqrt(-d^2 + e^2)*(e*x + d))/(d^2 - e^2)) + (2*b*d^5* e^2 - (4*a + 3*c)*d^4*e^3 - b*d^3*e^4 + (5*a + 3*c)*d^2*e^5 - b*d*e^6 -...
Timed out. \[ \int \frac {a+b x+c x^2}{\sqrt {-1+x} \sqrt {1+x} (d+e x)^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {a+b x+c x^2}{\sqrt {-1+x} \sqrt {1+x} (d+e x)^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume((e-d)*(e+d)>0)', see `assume?` f or more de
\[ \int \frac {a+b x+c x^2}{\sqrt {-1+x} \sqrt {1+x} (d+e x)^3} \, dx=\int { \frac {c x^{2} + b x + a}{{\left (e x + d\right )}^{3} \sqrt {x + 1} \sqrt {x - 1}} \,d x } \]
Time = 72.89 (sec) , antiderivative size = 7235, normalized size of antiderivative = 36.36 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+x} \sqrt {1+x} (d+e x)^3} \, dx=\text {Too large to display} \]
((((x - 1)^(1/2) - 1i)^2*(2*c*e^3 + c*d^2*e)*12i)/(d^2*((x + 1)^(1/2) - 1) ^2*(d^4 + e^4 - 2*d^2*e^2)) - (2*(7*c*d^4 + 14*c*d^2*e^2)*((x - 1)^(1/2) - 1i))/(7*d^3*((x + 1)^(1/2) - 1)*(d^4 + e^4 - 2*d^2*e^2)) + (((x - 1)^(1/2 ) - 1i)^4*(2*c*e^3 - c*d^2*e)*24i)/(d^2*((x + 1)^(1/2) - 1)^4*(d^4 + e^4 - 2*d^2*e^2)) - (2*(21*c*d^4 - 102*c*d^2*e^2)*((x - 1)^(1/2) - 1i)^5)/(3*d^ 3*((x + 1)^(1/2) - 1)^5*(d^4 + e^4 - 2*d^2*e^2)) - (2*(35*c*d^4 - 170*c*d^ 2*e^2)*((x - 1)^(1/2) - 1i)^3)/(5*d^3*((x + 1)^(1/2) - 1)^3*(d^4 + e^4 - 2 *d^2*e^2)) + (c*((x - 1)^(1/2) - 1i)^7*(d^2*1i + e^2*2i)*2i)/(d*((x + 1)^( 1/2) - 1)^7*(d^4 + e^4 - 2*d^2*e^2)) + (12*c*e*((x - 1)^(1/2) - 1i)^6*(d^2 *1i + e^2*2i))/(d^2*((x + 1)^(1/2) - 1)^6*(d^4 + e^4 - 2*d^2*e^2)))/(((x - 1)^(1/2) - 1i)^8/((x + 1)^(1/2) - 1)^8 - (e*((x - 1)^(1/2) - 1i)*8i)/(d*( (x + 1)^(1/2) - 1)) + (e*((x - 1)^(1/2) - 1i)^3*8i)/(d*((x + 1)^(1/2) - 1) ^3) + (e*((x - 1)^(1/2) - 1i)^5*8i)/(d*((x + 1)^(1/2) - 1)^5) - (e*((x - 1 )^(1/2) - 1i)^7*8i)/(d*((x + 1)^(1/2) - 1)^7) - (((x - 1)^(1/2) - 1i)^2*(4 *d^2 + 16*e^2))/(d^2*((x + 1)^(1/2) - 1)^2) - (((x - 1)^(1/2) - 1i)^6*(4*d ^2 + 16*e^2))/(d^2*((x + 1)^(1/2) - 1)^6) + (((x - 1)^(1/2) - 1i)^4*(6*d^2 - 32*e^2))/(d^2*((x + 1)^(1/2) - 1)^4) + 1) - ((2*((x - 1)^(1/2) - 1i)^3* (16*b*e^3 + 11*b*d^2*e))/(d^2*((x + 1)^(1/2) - 1)^3*(d^4 + e^4 - 2*d^2*e^2 )) - (6*b*e*((x - 1)^(1/2) - 1i)^7)/(((x + 1)^(1/2) - 1)^7*(d^4 + e^4 - 2* d^2*e^2)) - (6*b*e*((x - 1)^(1/2) - 1i))/(((x + 1)^(1/2) - 1)*(d^4 + e^...