Integrand size = 35, antiderivative size = 114 \[ \int \frac {a+b x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {\sqrt {d+e x}}{(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:
-(e*x+d)^(1/2)/(-a*e+b*d)/((b*x+a)^2)^(1/2)+e*(b*x+a)*arctanh(b^(1/2)*(e*x +d)^(1/2)/(-a*e+b*d)^(1/2))/b^(1/2)/(-a*e+b*d)^(3/2)/((b*x+a)^2)^(1/2)
Time = 0.33 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.75 \[ \int \frac {a+b x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\frac {\sqrt {d+e x}}{-b d+a e}+\frac {e (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{\sqrt {b} (-b d+a e)^{3/2}}}{\sqrt {(a+b x)^2}} \] Input:
Integrate[(a + b*x)/(Sqrt[d + e*x]*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]
Output:
(Sqrt[d + e*x]/(-(b*d) + a*e) + (e*(a + b*x)*ArcTan[(Sqrt[b]*Sqrt[d + e*x] )/Sqrt[-(b*d) + a*e]])/(Sqrt[b]*(-(b*d) + a*e)^(3/2)))/Sqrt[(a + b*x)^2]
Time = 0.42 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1187, 27, 52, 73, 221}
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.
\(\displaystyle \int \frac {a+b x}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2} \sqrt {d+e x}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b^3 (a+b x) \int \frac {1}{b^3 (a+b x)^2 \sqrt {d+e x}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {1}{(a+b x)^2 \sqrt {d+e x}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {(a+b x) \left (-\frac {e \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{2 (b d-a e)}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(a+b x) \left (-\frac {\int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b d-a e}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(a+b x) \left (\frac {e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{3/2}}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
Input:
Int[(a + b*x)/(Sqrt[d + e*x]*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]
Output:
((a + b*x)*(-(Sqrt[d + e*x]/((b*d - a*e)*(a + b*x))) + (e*ArcTanh[(Sqrt[b] *Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(Sqrt[b]*(b*d - a*e)^(3/2))))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
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] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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]
Time = 1.37 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.98
method | result | size |
default | \(\frac {\left (\arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) b e x +\arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a e +\sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\right ) \left (b x +a \right )^{2}}{\sqrt {b \left (a e -b d \right )}\, \left (a e -b d \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(112\) |
Input:
int((b*x+a)/(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERB OSE)
Output:
(arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*b*e*x+arctan(b*(e*x+d)^(1/2)/ (b*(a*e-b*d))^(1/2))*a*e+(e*x+d)^(1/2)*(b*(a*e-b*d))^(1/2))*(b*x+a)^2/(b*( a*e-b*d))^(1/2)/(a*e-b*d)/((b*x+a)^2)^(3/2)
Time = 0.10 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.46 \[ \int \frac {a+b x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [-\frac {\sqrt {b^{2} d - a b e} {\left (b e x + a e\right )} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) + 2 \, {\left (b^{2} d - a b e\right )} \sqrt {e x + d}}{2 \, {\left (a b^{3} d^{2} - 2 \, a^{2} b^{2} d e + a^{3} b e^{2} + {\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} x\right )}}, -\frac {\sqrt {-b^{2} d + a b e} {\left (b e x + a e\right )} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) + {\left (b^{2} d - a b e\right )} \sqrt {e x + d}}{a b^{3} d^{2} - 2 \, a^{2} b^{2} d e + a^{3} b e^{2} + {\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} x}\right ] \] Input:
integrate((b*x+a)/(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm=" fricas")
Output:
[-1/2*(sqrt(b^2*d - a*b*e)*(b*e*x + a*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt (b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) + 2*(b^2*d - a*b*e)*sqrt(e*x + d ))/(a*b^3*d^2 - 2*a^2*b^2*d*e + a^3*b*e^2 + (b^4*d^2 - 2*a*b^3*d*e + a^2*b ^2*e^2)*x), -(sqrt(-b^2*d + a*b*e)*(b*e*x + a*e)*arctan(sqrt(-b^2*d + a*b* e)*sqrt(e*x + d)/(b*e*x + b*d)) + (b^2*d - a*b*e)*sqrt(e*x + d))/(a*b^3*d^ 2 - 2*a^2*b^2*d*e + a^3*b*e^2 + (b^4*d^2 - 2*a*b^3*d*e + a^2*b^2*e^2)*x)]
\[ \int \frac {a+b x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {a + b x}{\sqrt {d + e x} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((b*x+a)/(e*x+d)**(1/2)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)
Output:
Integral((a + b*x)/(sqrt(d + e*x)*((a + b*x)**2)**(3/2)), x)
\[ \int \frac {a+b x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {b x + a}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} \sqrt {e x + d}} \,d x } \] Input:
integrate((b*x+a)/(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm=" maxima")
Output:
integrate((b*x + a)/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*sqrt(e*x + d)), x)
Time = 0.16 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.97 \[ \int \frac {a+b x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {e \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} {\left (b d \mathrm {sgn}\left (b x + a\right ) - a e \mathrm {sgn}\left (b x + a\right )\right )}} - \frac {\sqrt {e x + d} e}{{\left (b d \mathrm {sgn}\left (b x + a\right ) - a e \mathrm {sgn}\left (b x + a\right )\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}} \] Input:
integrate((b*x+a)/(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm=" giac")
Output:
-e*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*(b*d *sgn(b*x + a) - a*e*sgn(b*x + a))) - sqrt(e*x + d)*e/((b*d*sgn(b*x + a) - a*e*sgn(b*x + a))*((e*x + d)*b - b*d + a*e))
Timed out. \[ \int \frac {a+b x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {a+b\,x}{\sqrt {d+e\,x}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \] Input:
int((a + b*x)/((d + e*x)^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)),x)
Output:
int((a + b*x)/((d + e*x)^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)), x)
Time = 0.26 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.37 \[ \int \frac {a+b x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a e +\sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) b e x +\sqrt {e x +d}\, a b e -\sqrt {e x +d}\, b^{2} d}{b \left (a^{2} b \,e^{2} x -2 a \,b^{2} d e x +b^{3} d^{2} x +a^{3} e^{2}-2 a^{2} b d e +a \,b^{2} d^{2}\right )} \] Input:
int((b*x+a)/(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)
Output:
(sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d))) *a*e + sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*b*e*x + sqrt(d + e*x)*a*b*e - sqrt(d + e*x)*b**2*d)/(b*(a**3*e**2 - 2*a**2*b*d*e + a**2*b*e**2*x + a*b**2*d**2 - 2*a*b**2*d*e*x + b**3*d**2*x ))