Integrand size = 28, antiderivative size = 194 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^4} \, dx=\frac {(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^4 (a+b x) (d+e x)^3}-\frac {3 b (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^4 (a+b x) (d+e x)^2}+\frac {3 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x) (d+e x)}+\frac {b^3 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^4 (a+b x)} \] Output:
1/3*(-a*e+b*d)^3*((b*x+a)^2)^(1/2)/e^4/(b*x+a)/(e*x+d)^3-3/2*b*(-a*e+b*d)^ 2*((b*x+a)^2)^(1/2)/e^4/(b*x+a)/(e*x+d)^2+3*b^2*(-a*e+b*d)*((b*x+a)^2)^(1/ 2)/e^4/(b*x+a)/(e*x+d)+b^3*((b*x+a)^2)^(1/2)*ln(e*x+d)/e^4/(b*x+a)
Time = 1.06 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.54 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^4} \, dx=\frac {\sqrt {(a+b x)^2} \left ((b d-a e) \left (2 a^2 e^2+a b e (5 d+9 e x)+b^2 \left (11 d^2+27 d e x+18 e^2 x^2\right )\right )+6 b^3 (d+e x)^3 \log (d+e x)\right )}{6 e^4 (a+b x) (d+e x)^3} \] Input:
Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^4,x]
Output:
(Sqrt[(a + b*x)^2]*((b*d - a*e)*(2*a^2*e^2 + a*b*e*(5*d + 9*e*x) + b^2*(11 *d^2 + 27*d*e*x + 18*e^2*x^2)) + 6*b^3*(d + e*x)^3*Log[d + e*x]))/(6*e^4*( a + b*x)*(d + e*x)^3)
Time = 0.47 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.59, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1102, 27, 49, 2009}
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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^4} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3}{(d+e x)^4}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3}{(d+e x)^4}dx}{a+b x}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b^3}{e^3 (d+e x)}-\frac {3 (b d-a e) b^2}{e^3 (d+e x)^2}+\frac {3 (b d-a e)^2 b}{e^3 (d+e x)^3}+\frac {(a e-b d)^3}{e^3 (d+e x)^4}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {3 b^2 (b d-a e)}{e^4 (d+e x)}-\frac {3 b (b d-a e)^2}{2 e^4 (d+e x)^2}+\frac {(b d-a e)^3}{3 e^4 (d+e x)^3}+\frac {b^3 \log (d+e x)}{e^4}\right )}{a+b x}\) |
Input:
Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^4,x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((b*d - a*e)^3/(3*e^4*(d + e*x)^3) - (3*b*( b*d - a*e)^2)/(2*e^4*(d + e*x)^2) + (3*b^2*(b*d - a*e))/(e^4*(d + e*x)) + (b^3*Log[d + e*x])/e^4))/(a + b*x)
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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 1.45 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.75
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {3 b^{2} \left (a e -b d \right ) x^{2}}{e^{2}}-\frac {3 b \left (e^{2} a^{2}+2 a b d e -3 b^{2} d^{2}\right ) x}{2 e^{3}}-\frac {2 e^{3} a^{3}+3 a^{2} b d \,e^{2}+6 a \,b^{2} d^{2} e -11 b^{3} d^{3}}{6 e^{4}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{3}}+\frac {b^{3} \sqrt {\left (b x +a \right )^{2}}\, \ln \left (e x +d \right )}{e^{4} \left (b x +a \right )}\) | \(146\) |
default | \(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (6 \ln \left (e x +d \right ) b^{3} e^{3} x^{3}+18 \ln \left (e x +d \right ) b^{3} d \,e^{2} x^{2}+18 \ln \left (e x +d \right ) b^{3} d^{2} e x -18 x^{2} a \,b^{2} e^{3}+18 x^{2} b^{3} d \,e^{2}+6 \ln \left (e x +d \right ) b^{3} d^{3}-9 a^{2} b \,e^{3} x -18 x a \,b^{2} d \,e^{2}+27 b^{3} d^{2} e x -2 e^{3} a^{3}-3 a^{2} b d \,e^{2}-6 a \,b^{2} d^{2} e +11 b^{3} d^{3}\right )}{6 \left (b x +a \right )^{3} e^{4} \left (e x +d \right )^{3}}\) | \(186\) |
Input:
int((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^4,x,method=_RETURNVERBOSE)
Output:
((b*x+a)^2)^(1/2)/(b*x+a)*(-3*b^2*(a*e-b*d)/e^2*x^2-3/2*b*(a^2*e^2+2*a*b*d *e-3*b^2*d^2)/e^3*x-1/6*(2*a^3*e^3+3*a^2*b*d*e^2+6*a*b^2*d^2*e-11*b^3*d^3) /e^4)/(e*x+d)^3+b^3*((b*x+a)^2)^(1/2)*ln(e*x+d)/e^4/(b*x+a)
Time = 0.08 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^4} \, dx=\frac {11 \, b^{3} d^{3} - 6 \, a b^{2} d^{2} e - 3 \, a^{2} b d e^{2} - 2 \, a^{3} e^{3} + 18 \, {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 9 \, {\left (3 \, b^{3} d^{2} e - 2 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x + 6 \, {\left (b^{3} e^{3} x^{3} + 3 \, b^{3} d e^{2} x^{2} + 3 \, b^{3} d^{2} e x + b^{3} d^{3}\right )} \log \left (e x + d\right )}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \] Input:
integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^4,x, algorithm="fricas")
Output:
1/6*(11*b^3*d^3 - 6*a*b^2*d^2*e - 3*a^2*b*d*e^2 - 2*a^3*e^3 + 18*(b^3*d*e^ 2 - a*b^2*e^3)*x^2 + 9*(3*b^3*d^2*e - 2*a*b^2*d*e^2 - a^2*b*e^3)*x + 6*(b^ 3*e^3*x^3 + 3*b^3*d*e^2*x^2 + 3*b^3*d^2*e*x + b^3*d^3)*log(e*x + d))/(e^7* x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)
\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^4} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{4}}\, dx \] Input:
integrate((b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**4,x)
Output:
Integral(((a + b*x)**2)**(3/2)/(d + e*x)**4, x)
Exception generated. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^4} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^4,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(a*e-b*d>0)', see `assume?` for m ore detail
Time = 0.15 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^4} \, dx=\frac {b^{3} \log \left ({\left | e x + d \right |}\right ) \mathrm {sgn}\left (b x + a\right )}{e^{4}} + \frac {18 \, {\left (b^{3} d e \mathrm {sgn}\left (b x + a\right ) - a b^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 9 \, {\left (3 \, b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b^{2} d e \mathrm {sgn}\left (b x + a\right ) - a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right )\right )} x + \frac {11 \, b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 6 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )}{e}}{6 \, {\left (e x + d\right )}^{3} e^{3}} \] Input:
integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^4,x, algorithm="giac")
Output:
b^3*log(abs(e*x + d))*sgn(b*x + a)/e^4 + 1/6*(18*(b^3*d*e*sgn(b*x + a) - a *b^2*e^2*sgn(b*x + a))*x^2 + 9*(3*b^3*d^2*sgn(b*x + a) - 2*a*b^2*d*e*sgn(b *x + a) - a^2*b*e^2*sgn(b*x + a))*x + (11*b^3*d^3*sgn(b*x + a) - 6*a*b^2*d ^2*e*sgn(b*x + a) - 3*a^2*b*d*e^2*sgn(b*x + a) - 2*a^3*e^3*sgn(b*x + a))/e )/((e*x + d)^3*e^3)
Timed out. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^4} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{{\left (d+e\,x\right )}^4} \,d x \] Input:
int((a^2 + b^2*x^2 + 2*a*b*x)^(3/2)/(d + e*x)^4,x)
Output:
int((a^2 + b^2*x^2 + 2*a*b*x)^(3/2)/(d + e*x)^4, x)
Time = 0.22 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^4} \, dx=\frac {6 \,\mathrm {log}\left (e x +d \right ) b^{3} d^{4}+18 \,\mathrm {log}\left (e x +d \right ) b^{3} d^{3} e x +18 \,\mathrm {log}\left (e x +d \right ) b^{3} d^{2} e^{2} x^{2}+6 \,\mathrm {log}\left (e x +d \right ) b^{3} d \,e^{3} x^{3}-2 a^{3} d \,e^{3}-3 a^{2} b \,d^{2} e^{2}-9 a^{2} b d \,e^{3} x +6 a \,b^{2} e^{4} x^{3}+5 b^{3} d^{4}+9 b^{3} d^{3} e x -6 b^{3} d \,e^{3} x^{3}}{6 d \,e^{4} \left (e^{3} x^{3}+3 d \,e^{2} x^{2}+3 d^{2} e x +d^{3}\right )} \] Input:
int((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^4,x)
Output:
(6*log(d + e*x)*b**3*d**4 + 18*log(d + e*x)*b**3*d**3*e*x + 18*log(d + e*x )*b**3*d**2*e**2*x**2 + 6*log(d + e*x)*b**3*d*e**3*x**3 - 2*a**3*d*e**3 - 3*a**2*b*d**2*e**2 - 9*a**2*b*d*e**3*x + 6*a*b**2*e**4*x**3 + 5*b**3*d**4 + 9*b**3*d**3*e*x - 6*b**3*d*e**3*x**3)/(6*d*e**4*(d**3 + 3*d**2*e*x + 3*d *e**2*x**2 + e**3*x**3))