Integrand size = 35, antiderivative size = 150 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x) (d+e x)^{5/2}}+\frac {4 b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^{3/2}}-\frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt {d+e x}} \] Output:
-2/5*(-a*e+b*d)^2*((b*x+a)^2)^(1/2)/e^3/(b*x+a)/(e*x+d)^(5/2)+4/3*b*(-a*e+ b*d)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)/(e*x+d)^(3/2)-2*b^2*((b*x+a)^2)^(1/2)/e ^3/(b*x+a)/(e*x+d)^(1/2)
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.53 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (3 a^2 e^2+2 a b e (2 d+5 e x)+b^2 \left (8 d^2+20 d e x+15 e^2 x^2\right )\right )}{15 e^3 (a+b x) (d+e x)^{5/2}} \] Input:
Integrate[((a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(d + e*x)^(7/2),x]
Output:
(-2*Sqrt[(a + b*x)^2]*(3*a^2*e^2 + 2*a*b*e*(2*d + 5*e*x) + b^2*(8*d^2 + 20 *d*e*x + 15*e^2*x^2)))/(15*e^3*(a + b*x)*(d + e*x)^(5/2))
Time = 0.41 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.65, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {1187, 27, 53, 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.
\(\displaystyle \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b (a+b x)^2}{(d+e x)^{7/2}}dx}{b (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^2}{(d+e x)^{7/2}}dx}{a+b x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b^2}{e^2 (d+e x)^{3/2}}-\frac {2 (b d-a e) b}{e^2 (d+e x)^{5/2}}+\frac {(a e-b d)^2}{e^2 (d+e x)^{7/2}}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {4 b (b d-a e)}{3 e^3 (d+e x)^{3/2}}-\frac {2 (b d-a e)^2}{5 e^3 (d+e x)^{5/2}}-\frac {2 b^2}{e^3 \sqrt {d+e x}}\right )}{a+b x}\) |
Input:
Int[((a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(d + e*x)^(7/2),x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-2*(b*d - a*e)^2)/(5*e^3*(d + e*x)^(5/2)) + (4*b*(b*d - a*e))/(3*e^3*(d + e*x)^(3/2)) - (2*b^2)/(e^3*Sqrt[d + e*x]) ))/(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, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^ IntPart[p]*(b/2 + c*x)^(2*FracPart[p])) Int[(d + e*x)^m*(f + g*x)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 1.30 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.46
method | result | size |
default | \(-\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (15 b^{2} e^{2} x^{2}+10 x a b \,e^{2}+20 b^{2} d e x +3 e^{2} a^{2}+4 a b d e +8 b^{2} d^{2}\right )}{15 e^{3} \left (e x +d \right )^{\frac {5}{2}}}\) | \(69\) |
gosper | \(-\frac {2 \left (15 b^{2} e^{2} x^{2}+10 x a b \,e^{2}+20 b^{2} d e x +3 e^{2} a^{2}+4 a b d e +8 b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 \left (e x +d \right )^{\frac {5}{2}} e^{3} \left (b x +a \right )}\) | \(79\) |
orering | \(-\frac {2 \left (15 b^{2} e^{2} x^{2}+10 x a b \,e^{2}+20 b^{2} d e x +3 e^{2} a^{2}+4 a b d e +8 b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 \left (e x +d \right )^{\frac {5}{2}} e^{3} \left (b x +a \right )}\) | \(79\) |
Input:
int((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^(7/2),x,method=_RETURNVERBOSE)
Output:
-2/15*csgn(b*x+a)*(15*b^2*e^2*x^2+10*a*b*e^2*x+20*b^2*d*e*x+3*a^2*e^2+4*a* b*d*e+8*b^2*d^2)/e^3/(e*x+d)^(5/2)
Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.63 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (15 \, b^{2} e^{2} x^{2} + 8 \, b^{2} d^{2} + 4 \, a b d e + 3 \, a^{2} e^{2} + 10 \, {\left (2 \, b^{2} d e + a b e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \] Input:
integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^(7/2),x, algorithm="fricas")
Output:
-2/15*(15*b^2*e^2*x^2 + 8*b^2*d^2 + 4*a*b*d*e + 3*a^2*e^2 + 10*(2*b^2*d*e + a*b*e^2)*x)*sqrt(e*x + d)/(e^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^4*x + d^3*e^3 )
Timed out. \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=\text {Timed out} \] Input:
integrate((b*x+a)*((b*x+a)**2)**(1/2)/(e*x+d)**(7/2),x)
Output:
Timed out
Time = 0.06 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.79 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (5 \, b e x + 2 \, b d + 3 \, a e\right )} a}{15 \, {\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )} \sqrt {e x + d}} - \frac {2 \, {\left (15 \, b e^{2} x^{2} + 8 \, b d^{2} + 2 \, a d e + 5 \, {\left (4 \, b d e + a e^{2}\right )} x\right )} b}{15 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )} \sqrt {e x + d}} \] Input:
integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^(7/2),x, algorithm="maxima")
Output:
-2/15*(5*b*e*x + 2*b*d + 3*a*e)*a/((e^4*x^2 + 2*d*e^3*x + d^2*e^2)*sqrt(e* x + d)) - 2/15*(15*b*e^2*x^2 + 8*b*d^2 + 2*a*d*e + 5*(4*b*d*e + a*e^2)*x)* b/((e^5*x^2 + 2*d*e^4*x + d^2*e^3)*sqrt(e*x + d))
Time = 0.19 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.69 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} b^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, {\left (e x + d\right )} b^{2} d \mathrm {sgn}\left (b x + a\right ) + 3 \, b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, {\left (e x + d\right )} a b e \mathrm {sgn}\left (b x + a\right ) - 6 \, a b d e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{3}} \] Input:
integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^(7/2),x, algorithm="giac")
Output:
-2/15*(15*(e*x + d)^2*b^2*sgn(b*x + a) - 10*(e*x + d)*b^2*d*sgn(b*x + a) + 3*b^2*d^2*sgn(b*x + a) + 10*(e*x + d)*a*b*e*sgn(b*x + a) - 6*a*b*d*e*sgn( b*x + a) + 3*a^2*e^2*sgn(b*x + a))/((e*x + d)^(5/2)*e^3)
Time = 11.87 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.01 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {4\,x\,\left (a\,e+2\,b\,d\right )}{3\,e^4}+\frac {2\,b\,x^2}{e^3}+\frac {\frac {2\,a^2\,e^2}{5}+\frac {8\,a\,b\,d\,e}{15}+\frac {16\,b^2\,d^2}{15}}{b\,e^5}\right )}{x^3\,\sqrt {d+e\,x}+\frac {a\,d^2\,\sqrt {d+e\,x}}{b\,e^2}+\frac {x^2\,\left (a\,e^5+2\,b\,d\,e^4\right )\,\sqrt {d+e\,x}}{b\,e^5}+\frac {d\,x\,\left (2\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}} \] Input:
int((((a + b*x)^2)^(1/2)*(a + b*x))/(d + e*x)^(7/2),x)
Output:
-(((a + b*x)^2)^(1/2)*((4*x*(a*e + 2*b*d))/(3*e^4) + (2*b*x^2)/e^3 + ((2*a ^2*e^2)/5 + (16*b^2*d^2)/15 + (8*a*b*d*e)/15)/(b*e^5)))/(x^3*(d + e*x)^(1/ 2) + (a*d^2*(d + e*x)^(1/2))/(b*e^2) + (x^2*(a*e^5 + 2*b*d*e^4)*(d + e*x)^ (1/2))/(b*e^5) + (d*x*(2*a*e + b*d)*(d + e*x)^(1/2))/(b*e^2))
Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.54 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=\frac {-2 b^{2} e^{2} x^{2}-\frac {4}{3} a b \,e^{2} x -\frac {8}{3} b^{2} d e x -\frac {2}{5} a^{2} e^{2}-\frac {8}{15} a b d e -\frac {16}{15} b^{2} d^{2}}{\sqrt {e x +d}\, e^{3} \left (e^{2} x^{2}+2 d e x +d^{2}\right )} \] Input:
int((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^(7/2),x)
Output:
(2*( - 3*a**2*e**2 - 4*a*b*d*e - 10*a*b*e**2*x - 8*b**2*d**2 - 20*b**2*d*e *x - 15*b**2*e**2*x**2))/(15*sqrt(d + e*x)*e**3*(d**2 + 2*d*e*x + e**2*x** 2))