Integrand size = 35, antiderivative size = 160 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=-\frac {2 (b d-a e) (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^{3/2}}+\frac {2 (2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt {d+e x}}+\frac {2 b B \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)} \] Output:
-2/3*(-a*e+b*d)*(-A*e+B*d)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)/(e*x+d)^(3/2)+2*( -A*b*e-B*a*e+2*B*b*d)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)/(e*x+d)^(1/2)+2*b*B*(e *x+d)^(1/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)
Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.54 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (A b e (2 d+3 e x)+a e (2 B d+A e+3 B e x)-b B \left (8 d^2+12 d e x+3 e^2 x^2\right )\right )}{3 e^3 (a+b x) (d+e x)^{3/2}} \] Input:
Integrate[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(d + e*x)^(5/2),x]
Output:
(-2*Sqrt[(a + b*x)^2]*(A*b*e*(2*d + 3*e*x) + a*e*(2*B*d + A*e + 3*B*e*x) - b*B*(8*d^2 + 12*d*e*x + 3*e^2*x^2)))/(3*e^3*(a + b*x)*(d + e*x)^(3/2))
Time = 0.43 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.67, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {1187, 27, 86, 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.
\(\displaystyle \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} (A+B x)}{(d+e x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b (a+b x) (A+B x)}{(d+e x)^{5/2}}dx}{b (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x) (A+B x)}{(d+e x)^{5/2}}dx}{a+b x}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b B}{e^2 \sqrt {d+e x}}+\frac {-2 b B d+A b e+a B e}{e^2 (d+e x)^{3/2}}+\frac {(a e-b d) (A e-B d)}{e^2 (d+e x)^{5/2}}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {2 (-a B e-A b e+2 b B d)}{e^3 \sqrt {d+e x}}-\frac {2 (b d-a e) (B d-A e)}{3 e^3 (d+e x)^{3/2}}+\frac {2 b B \sqrt {d+e x}}{e^3}\right )}{a+b x}\) |
Input:
Int[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(d + e*x)^(5/2),x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-2*(b*d - a*e)*(B*d - A*e))/(3*e^3*(d + e *x)^(3/2)) + (2*(2*b*B*d - A*b*e - a*B*e))/(e^3*Sqrt[d + e*x]) + (2*b*B*Sq rt[d + e*x])/e^3))/(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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 1.61 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.49
method | result | size |
default | \(-\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (-3 B b \,e^{2} x^{2}+3 A b \,e^{2} x +3 B a \,e^{2} x -12 B b d e x +A a \,e^{2}+2 A b d e +2 B a d e -8 B b \,d^{2}\right )}{3 e^{3} \left (e x +d \right )^{\frac {3}{2}}}\) | \(78\) |
gosper | \(-\frac {2 \left (-3 B b \,e^{2} x^{2}+3 A b \,e^{2} x +3 B a \,e^{2} x -12 B b d e x +A a \,e^{2}+2 A b d e +2 B a d e -8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{3 \left (e x +d \right )^{\frac {3}{2}} e^{3} \left (b x +a \right )}\) | \(88\) |
orering | \(-\frac {2 \left (-3 B b \,e^{2} x^{2}+3 A b \,e^{2} x +3 B a \,e^{2} x -12 B b d e x +A a \,e^{2}+2 A b d e +2 B a d e -8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{3 \left (e x +d \right )^{\frac {3}{2}} e^{3} \left (b x +a \right )}\) | \(88\) |
risch | \(\frac {2 b B \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{e^{3} \left (b x +a \right )}-\frac {2 \left (3 A b \,e^{2} x +3 B a \,e^{2} x -6 B b d e x +A a \,e^{2}+2 A b d e +2 B a d e -5 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{3 e^{3} \left (e x +d \right )^{\frac {3}{2}} \left (b x +a \right )}\) | \(109\) |
Input:
int((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3*csgn(b*x+a)*(-3*B*b*e^2*x^2+3*A*b*e^2*x+3*B*a*e^2*x-12*B*b*d*e*x+A*a* e^2+2*A*b*d*e+2*B*a*d*e-8*B*b*d^2)/e^3/(e*x+d)^(3/2)
Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.57 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, B b e^{2} x^{2} + 8 \, B b d^{2} - A a e^{2} - 2 \, {\left (B a + A b\right )} d e + 3 \, {\left (4 \, B b d e - {\left (B a + A b\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \] Input:
integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(5/2),x, algorithm="fricas")
Output:
2/3*(3*B*b*e^2*x^2 + 8*B*b*d^2 - A*a*e^2 - 2*(B*a + A*b)*d*e + 3*(4*B*b*d* e - (B*a + A*b)*e^2)*x)*sqrt(e*x + d)/(e^5*x^2 + 2*d*e^4*x + d^2*e^3)
\[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=\int \frac {\left (A + B x\right ) \sqrt {\left (a + b x\right )^{2}}}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((B*x+A)*((b*x+a)**2)**(1/2)/(e*x+d)**(5/2),x)
Output:
Integral((A + B*x)*sqrt((a + b*x)**2)/(d + e*x)**(5/2), x)
Time = 0.06 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.60 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=-\frac {2 \, {\left (3 \, b e x + 2 \, b d + a e\right )} A}{3 \, {\left (e^{3} x + d e^{2}\right )} \sqrt {e x + d}} + \frac {2 \, {\left (3 \, b e^{2} x^{2} + 8 \, b d^{2} - 2 \, a d e + 3 \, {\left (4 \, b d e - a e^{2}\right )} x\right )} B}{3 \, {\left (e^{4} x + d e^{3}\right )} \sqrt {e x + d}} \] Input:
integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(5/2),x, algorithm="maxima")
Output:
-2/3*(3*b*e*x + 2*b*d + a*e)*A/((e^3*x + d*e^2)*sqrt(e*x + d)) + 2/3*(3*b* e^2*x^2 + 8*b*d^2 - 2*a*d*e + 3*(4*b*d*e - a*e^2)*x)*B/((e^4*x + d*e^3)*sq rt(e*x + d))
Time = 1.02 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.81 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, \sqrt {e x + d} B b \mathrm {sgn}\left (b x + a\right )}{e^{3}} + \frac {2 \, {\left (6 \, {\left (e x + d\right )} B b d \mathrm {sgn}\left (b x + a\right ) - B b d^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, {\left (e x + d\right )} B a e \mathrm {sgn}\left (b x + a\right ) - 3 \, {\left (e x + d\right )} A b e \mathrm {sgn}\left (b x + a\right ) + B a d e \mathrm {sgn}\left (b x + a\right ) + A b d e \mathrm {sgn}\left (b x + a\right ) - A a e^{2} \mathrm {sgn}\left (b x + a\right )\right )}}{3 \, {\left (e x + d\right )}^{\frac {3}{2}} e^{3}} \] Input:
integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(5/2),x, algorithm="giac")
Output:
2*sqrt(e*x + d)*B*b*sgn(b*x + a)/e^3 + 2/3*(6*(e*x + d)*B*b*d*sgn(b*x + a) - B*b*d^2*sgn(b*x + a) - 3*(e*x + d)*B*a*e*sgn(b*x + a) - 3*(e*x + d)*A*b *e*sgn(b*x + a) + B*a*d*e*sgn(b*x + a) + A*b*d*e*sgn(b*x + a) - A*a*e^2*sg n(b*x + a))/((e*x + d)^(3/2)*e^3)
Time = 12.16 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.91 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,A\,a\,e^2-16\,B\,b\,d^2+4\,A\,b\,d\,e+4\,B\,a\,d\,e}{3\,b\,e^4}-\frac {2\,B\,x^2}{e^2}+\frac {x\,\left (6\,A\,b\,e^2+6\,B\,a\,e^2-24\,B\,b\,d\,e\right )}{3\,b\,e^4}\right )}{x^2\,\sqrt {d+e\,x}+\frac {a\,d\,\sqrt {d+e\,x}}{b\,e}+\frac {x\,\left (3\,a\,e^4+3\,b\,d\,e^3\right )\,\sqrt {d+e\,x}}{3\,b\,e^4}} \] Input:
int((((a + b*x)^2)^(1/2)*(A + B*x))/(d + e*x)^(5/2),x)
Output:
-(((a + b*x)^2)^(1/2)*((2*A*a*e^2 - 16*B*b*d^2 + 4*A*b*d*e + 4*B*a*d*e)/(3 *b*e^4) - (2*B*x^2)/e^2 + (x*(6*A*b*e^2 + 6*B*a*e^2 - 24*B*b*d*e))/(3*b*e^ 4)))/(x^2*(d + e*x)^(1/2) + (a*d*(d + e*x)^(1/2))/(b*e) + (x*(3*a*e^4 + 3* b*d*e^3)*(d + e*x)^(1/2))/(3*b*e^4))
Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.44 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=\frac {2 b^{2} e^{2} x^{2}-4 a b \,e^{2} x +8 b^{2} d e x -\frac {2}{3} a^{2} e^{2}-\frac {8}{3} a b d e +\frac {16}{3} b^{2} d^{2}}{\sqrt {e x +d}\, e^{3} \left (e x +d \right )} \] Input:
int((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(5/2),x)
Output:
(2*( - a**2*e**2 - 4*a*b*d*e - 6*a*b*e**2*x + 8*b**2*d**2 + 12*b**2*d*e*x + 3*b**2*e**2*x**2))/(3*sqrt(d + e*x)*e**3*(d + e*x))