Integrand size = 17, antiderivative size = 95 \[ \int \frac {x}{\left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {x}{b^2 d}-\frac {4 a \sqrt {c+d x}}{b^3 d^2}+\frac {2 a \left (a^2-b^2 c\right )}{b^4 d^2 \left (a+b \sqrt {c+d x}\right )}+\frac {2 \left (3 a^2-b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{b^4 d^2} \] Output:
x/b^2/d-4*a*(d*x+c)^(1/2)/b^3/d^2+2*a*(-b^2*c+a^2)/b^4/d^2/(a+b*(d*x+c)^(1 /2))+2*(-b^2*c+3*a^2)*ln(a+b*(d*x+c)^(1/2))/b^4/d^2
Time = 0.09 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.14 \[ \int \frac {x}{\left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {2 a^3-2 a b^2 c-4 a^2 b \sqrt {c+d x}-3 a b^2 (c+d x)+b^3 (c+d x)^{3/2}}{b^4 d^2 \left (a+b \sqrt {c+d x}\right )}-\frac {2 \left (-3 a^2+b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{b^4 d^2} \] Input:
Integrate[x/(a + b*Sqrt[c + d*x])^2,x]
Output:
(2*a^3 - 2*a*b^2*c - 4*a^2*b*Sqrt[c + d*x] - 3*a*b^2*(c + d*x) + b^3*(c + d*x)^(3/2))/(b^4*d^2*(a + b*Sqrt[c + d*x])) - (2*(-3*a^2 + b^2*c)*Log[a + b*Sqrt[c + d*x]])/(b^4*d^2)
Time = 0.45 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {896, 25, 1732, 522, 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.
\(\displaystyle \int \frac {x}{\left (a+b \sqrt {c+d x}\right )^2} \, dx\) |
\(\Big \downarrow \) 896 |
\(\displaystyle \frac {\int \frac {d x}{\left (a+b \sqrt {c+d x}\right )^2}d(c+d x)}{d^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int -\frac {d x}{\left (a+b \sqrt {c+d x}\right )^2}d(c+d x)}{d^2}\) |
\(\Big \downarrow \) 1732 |
\(\displaystyle -\frac {2 \int -\frac {d x \sqrt {c+d x}}{\left (a+b \sqrt {c+d x}\right )^2}d\sqrt {c+d x}}{d^2}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle -\frac {2 \int \left (\frac {2 a}{b^3}-\frac {\sqrt {c+d x}}{b^2}+\frac {b^2 c-3 a^2}{b^3 \left (a+b \sqrt {c+d x}\right )}+\frac {a^3-a b^2 c}{b^3 \left (a+b \sqrt {c+d x}\right )^2}\right )d\sqrt {c+d x}}{d^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (-\frac {a \left (a^2-b^2 c\right )}{b^4 \left (a+b \sqrt {c+d x}\right )}-\frac {\left (3 a^2-b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{b^4}+\frac {2 a \sqrt {c+d x}}{b^3}-\frac {c+d x}{2 b^2}\right )}{d^2}\) |
Input:
Int[x/(a + b*Sqrt[c + d*x])^2,x]
Output:
(-2*((2*a*Sqrt[c + d*x])/b^3 - (c + d*x)/(2*b^2) - (a*(a^2 - b^2*c))/(b^4* (a + b*Sqrt[c + d*x])) - ((3*a^2 - b^2*c)*Log[a + b*Sqrt[c + d*x]])/b^4))/ d^2
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1) Subst[Int[Si mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symb ol] :> With[{g = Denominator[n]}, Simp[g Subst[Int[x^(g - 1)*(d + e*x^(g* n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p, q} , x] && EqQ[n2, 2*n] && FractionQ[n]
Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(\frac {-\frac {2 \left (-\frac {b \left (d x +c \right )}{2}+2 a \sqrt {d x +c}\right )}{b^{3}}+\frac {2 \left (-b^{2} c +3 a^{2}\right ) \ln \left (a +b \sqrt {d x +c}\right )}{b^{4}}+\frac {2 a \left (-b^{2} c +a^{2}\right )}{b^{4} \left (a +b \sqrt {d x +c}\right )}}{d^{2}}\) | \(87\) |
default | \(\frac {-\frac {2 \left (-\frac {b \left (d x +c \right )}{2}+2 a \sqrt {d x +c}\right )}{b^{3}}+\frac {2 \left (-b^{2} c +3 a^{2}\right ) \ln \left (a +b \sqrt {d x +c}\right )}{b^{4}}+\frac {2 a \left (-b^{2} c +a^{2}\right )}{b^{4} \left (a +b \sqrt {d x +c}\right )}}{d^{2}}\) | \(87\) |
Input:
int(x/(a+b*(d*x+c)^(1/2))^2,x,method=_RETURNVERBOSE)
Output:
2/d^2*(-1/b^3*(-1/2*b*(d*x+c)+2*a*(d*x+c)^(1/2))+1/b^4*(-b^2*c+3*a^2)*ln(a +b*(d*x+c)^(1/2))+a*(-b^2*c+a^2)/b^4/(a+b*(d*x+c)^(1/2)))
Time = 0.11 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.72 \[ \int \frac {x}{\left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {b^{4} d^{2} x^{2} + b^{4} c^{2} + a^{2} b^{2} c - 2 \, a^{4} + {\left (2 \, b^{4} c - a^{2} b^{2}\right )} d x - 2 \, {\left (b^{4} c^{2} - 4 \, a^{2} b^{2} c + 3 \, a^{4} + {\left (b^{4} c - 3 \, a^{2} b^{2}\right )} d x\right )} \log \left (\sqrt {d x + c} b + a\right ) - 2 \, {\left (2 \, a b^{3} d x + 3 \, a b^{3} c - 3 \, a^{3} b\right )} \sqrt {d x + c}}{b^{6} d^{3} x + {\left (b^{6} c - a^{2} b^{4}\right )} d^{2}} \] Input:
integrate(x/(a+b*(d*x+c)^(1/2))^2,x, algorithm="fricas")
Output:
(b^4*d^2*x^2 + b^4*c^2 + a^2*b^2*c - 2*a^4 + (2*b^4*c - a^2*b^2)*d*x - 2*( b^4*c^2 - 4*a^2*b^2*c + 3*a^4 + (b^4*c - 3*a^2*b^2)*d*x)*log(sqrt(d*x + c) *b + a) - 2*(2*a*b^3*d*x + 3*a*b^3*c - 3*a^3*b)*sqrt(d*x + c))/(b^6*d^3*x + (b^6*c - a^2*b^4)*d^2)
Time = 4.10 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.33 \[ \int \frac {x}{\left (a+b \sqrt {c+d x}\right )^2} \, dx=\begin {cases} \frac {2 \left (- \frac {a \left (a^{2} - b^{2} c\right ) \left (\begin {cases} \frac {\sqrt {c + d x}}{a^{2}} & \text {for}\: b = 0 \\- \frac {1}{b \left (a + b \sqrt {c + d x}\right )} & \text {otherwise} \end {cases}\right )}{b^{3}} - \frac {2 a \sqrt {c + d x}}{b^{3}} + \frac {c + d x}{2 b^{2}} + \frac {\left (3 a^{2} - b^{2} c\right ) \left (\begin {cases} \frac {\sqrt {c + d x}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b \sqrt {c + d x} \right )}}{b} & \text {otherwise} \end {cases}\right )}{b^{3}}\right )}{d^{2}} & \text {for}\: d \neq 0 \\\frac {x^{2}}{2 \left (a + b \sqrt {c}\right )^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(x/(a+b*(d*x+c)**(1/2))**2,x)
Output:
Piecewise((2*(-a*(a**2 - b**2*c)*Piecewise((sqrt(c + d*x)/a**2, Eq(b, 0)), (-1/(b*(a + b*sqrt(c + d*x))), True))/b**3 - 2*a*sqrt(c + d*x)/b**3 + (c + d*x)/(2*b**2) + (3*a**2 - b**2*c)*Piecewise((sqrt(c + d*x)/a, Eq(b, 0)), (log(a + b*sqrt(c + d*x))/b, True))/b**3)/d**2, Ne(d, 0)), (x**2/(2*(a + b*sqrt(c))**2), True))
Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.95 \[ \int \frac {x}{\left (a+b \sqrt {c+d x}\right )^2} \, dx=-\frac {\frac {2 \, {\left (a b^{2} c - a^{3}\right )}}{\sqrt {d x + c} b^{5} + a b^{4}} - \frac {{\left (d x + c\right )} b - 4 \, \sqrt {d x + c} a}{b^{3}} + \frac {2 \, {\left (b^{2} c - 3 \, a^{2}\right )} \log \left (\sqrt {d x + c} b + a\right )}{b^{4}}}{d^{2}} \] Input:
integrate(x/(a+b*(d*x+c)^(1/2))^2,x, algorithm="maxima")
Output:
-(2*(a*b^2*c - a^3)/(sqrt(d*x + c)*b^5 + a*b^4) - ((d*x + c)*b - 4*sqrt(d* x + c)*a)/b^3 + 2*(b^2*c - 3*a^2)*log(sqrt(d*x + c)*b + a)/b^4)/d^2
Time = 0.12 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.07 \[ \int \frac {x}{\left (a+b \sqrt {c+d x}\right )^2} \, dx=-\frac {\frac {2 \, {\left (b^{2} c - 3 \, a^{2}\right )} \log \left ({\left | \sqrt {d x + c} b + a \right |}\right )}{b^{4} d} - \frac {{\left (d x + c\right )} b^{2} d - 4 \, \sqrt {d x + c} a b d}{b^{4} d^{2}} + \frac {2 \, {\left (a b^{2} c - a^{3}\right )}}{{\left (\sqrt {d x + c} b + a\right )} b^{4} d}}{d} \] Input:
integrate(x/(a+b*(d*x+c)^(1/2))^2,x, algorithm="giac")
Output:
-(2*(b^2*c - 3*a^2)*log(abs(sqrt(d*x + c)*b + a))/(b^4*d) - ((d*x + c)*b^2 *d - 4*sqrt(d*x + c)*a*b*d)/(b^4*d^2) + 2*(a*b^2*c - a^3)/((sqrt(d*x + c)* b + a)*b^4*d))/d
Time = 0.06 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.03 \[ \int \frac {x}{\left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {x}{b^2\,d}+\frac {2\,\left (a^3-a\,b^2\,c\right )}{b\,\left (b^4\,d^2\,\sqrt {c+d\,x}+a\,b^3\,d^2\right )}-\frac {\ln \left (a+b\,\sqrt {c+d\,x}\right )\,\left (2\,b^2\,c-6\,a^2\right )}{b^4\,d^2}-\frac {4\,a\,\sqrt {c+d\,x}}{b^3\,d^2} \] Input:
int(x/(a + b*(c + d*x)^(1/2))^2,x)
Output:
x/(b^2*d) + (2*(a^3 - a*b^2*c))/(b*(b^4*d^2*(c + d*x)^(1/2) + a*b^3*d^2)) - (log(a + b*(c + d*x)^(1/2))*(2*b^2*c - 6*a^2))/(b^4*d^2) - (4*a*(c + d*x )^(1/2))/(b^3*d^2)
Time = 0.23 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.59 \[ \int \frac {x}{\left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {6 \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) a^{2} b -2 \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) b^{3} c -6 \sqrt {d x +c}\, a^{2} b +3 \sqrt {d x +c}\, b^{3} c +\sqrt {d x +c}\, b^{3} d x +6 \,\mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) a^{3}-2 \,\mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) a \,b^{2} c -3 a \,b^{2} c -3 a \,b^{2} d x}{b^{4} d^{2} \left (\sqrt {d x +c}\, b +a \right )} \] Input:
int(x/(a+b*(d*x+c)^(1/2))^2,x)
Output:
(6*sqrt(c + d*x)*log(sqrt(c + d*x)*b + a)*a**2*b - 2*sqrt(c + d*x)*log(sqr t(c + d*x)*b + a)*b**3*c - 6*sqrt(c + d*x)*a**2*b + 3*sqrt(c + d*x)*b**3*c + sqrt(c + d*x)*b**3*d*x + 6*log(sqrt(c + d*x)*b + a)*a**3 - 2*log(sqrt(c + d*x)*b + a)*a*b**2*c - 3*a*b**2*c - 3*a*b**2*d*x)/(b**4*d**2*(sqrt(c + d*x)*b + a))