3.7.34 \(\int \frac {x}{a+b \sqrt {c+d x}} \, dx\) [634]

Optimal. Leaf size=90 \[ -\frac {a x}{b^2 d}+\frac {2 \left (a^2-b^2 c\right ) \sqrt {c+d x}}{b^3 d^2}+\frac {2 (c+d x)^{3/2}}{3 b d^2}-\frac {2 a \left (a^2-b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{b^4 d^2} \]

[Out]

-a*x/b^2/d+2/3*(d*x+c)^(3/2)/b/d^2-2*a*(-b^2*c+a^2)*ln(a+b*(d*x+c)^(1/2))/b^4/d^2+2*(-b^2*c+a^2)*(d*x+c)^(1/2)
/b^3/d^2

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Rubi [A]
time = 0.06, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {378, 1412, 786} \begin {gather*} -\frac {2 a \left (a^2-b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{b^4 d^2}+\frac {2 \left (a^2-b^2 c\right ) \sqrt {c+d x}}{b^3 d^2}-\frac {a x}{b^2 d}+\frac {2 (c+d x)^{3/2}}{3 b d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Sqrt[c + d*x]),x]

[Out]

-((a*x)/(b^2*d)) + (2*(a^2 - b^2*c)*Sqrt[c + d*x])/(b^3*d^2) + (2*(c + d*x)^(3/2))/(3*b*d^2) - (2*a*(a^2 - b^2
*c)*Log[a + b*Sqrt[c + d*x]])/(b^4*d^2)

Rule 378

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(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]

Rule 786

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rule 1412

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, D
ist[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]

Rubi steps

\begin {align*} \int \frac {x}{a+b \sqrt {c+d x}} \, dx &=\frac {\text {Subst}\left (\int \frac {-c+x}{a+b \sqrt {x}} \, dx,x,c+d x\right )}{d^2}\\ &=\frac {2 \text {Subst}\left (\int \frac {x \left (-c+x^2\right )}{a+b x} \, dx,x,\sqrt {c+d x}\right )}{d^2}\\ &=\frac {2 \text {Subst}\left (\int \left (\frac {a^2-b^2 c}{b^3}-\frac {a x}{b^2}+\frac {x^2}{b}+\frac {-a^3+a b^2 c}{b^3 (a+b x)}\right ) \, dx,x,\sqrt {c+d x}\right )}{d^2}\\ &=-\frac {a x}{b^2 d}+\frac {2 \left (a^2-b^2 c\right ) \sqrt {c+d x}}{b^3 d^2}+\frac {2 (c+d x)^{3/2}}{3 b d^2}-\frac {2 a \left (a^2-b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{b^4 d^2}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 85, normalized size = 0.94 \begin {gather*} \frac {b \left (6 a^2 \sqrt {c+d x}+2 b^2 (-2 c+d x) \sqrt {c+d x}-3 a b (c+d x)\right )-6 \left (a^3-a b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{3 b^4 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*Sqrt[c + d*x]),x]

[Out]

(b*(6*a^2*Sqrt[c + d*x] + 2*b^2*(-2*c + d*x)*Sqrt[c + d*x] - 3*a*b*(c + d*x)) - 6*(a^3 - a*b^2*c)*Log[a + b*Sq
rt[c + d*x]])/(3*b^4*d^2)

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Maple [A]
time = 0.02, size = 85, normalized size = 0.94

method	result	size
derivativedivides	\(\frac {\frac {2 \left (\frac {\left (d x +c \right )^{\frac {3}{2}} b^{2}}{3}-\frac {a \left (d x +c \right ) b}{2}-b^{2} c \sqrt {d x +c}+a^{2} \sqrt {d x +c}\right )}{b^{3}}-\frac {2 a \left (-b^{2} c +a^{2}\right ) \ln \left (a +b \sqrt {d x +c}\right )}{b^{4}}}{d^{2}}\)	\(85\)
default	\(\frac {\frac {2 \left (\frac {\left (d x +c \right )^{\frac {3}{2}} b^{2}}{3}-\frac {a \left (d x +c \right ) b}{2}-b^{2} c \sqrt {d x +c}+a^{2} \sqrt {d x +c}\right )}{b^{3}}-\frac {2 a \left (-b^{2} c +a^{2}\right ) \ln \left (a +b \sqrt {d x +c}\right )}{b^{4}}}{d^{2}}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a+b*(d*x+c)^(1/2)),x,method=_RETURNVERBOSE)

[Out]

2/d^2*(1/b^3*(1/3*(d*x+c)^(3/2)*b^2-1/2*a*(d*x+c)*b-b^2*c*(d*x+c)^(1/2)+a^2*(d*x+c)^(1/2))-a*(-b^2*c+a^2)/b^4*
ln(a+b*(d*x+c)^(1/2)))

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Maxima [A]
time = 0.29, size = 81, normalized size = 0.90 \begin {gather*} \frac {\frac {2 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} - 3 \, {\left (d x + c\right )} a b - 6 \, {\left (b^{2} c - a^{2}\right )} \sqrt {d x + c}}{b^{3}} + \frac {6 \, {\left (a b^{2} c - a^{3}\right )} \log \left (\sqrt {d x + c} b + a\right )}{b^{4}}}{3 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*(d*x+c)^(1/2)),x, algorithm="maxima")

[Out]

1/3*((2*(d*x + c)^(3/2)*b^2 - 3*(d*x + c)*a*b - 6*(b^2*c - a^2)*sqrt(d*x + c))/b^3 + 6*(a*b^2*c - a^3)*log(sqr
t(d*x + c)*b + a)/b^4)/d^2

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Fricas [A]
time = 0.36, size = 71, normalized size = 0.79 \begin {gather*} -\frac {3 \, a b^{2} d x - 6 \, {\left (a b^{2} c - a^{3}\right )} \log \left (\sqrt {d x + c} b + a\right ) - 2 \, {\left (b^{3} d x - 2 \, b^{3} c + 3 \, a^{2} b\right )} \sqrt {d x + c}}{3 \, b^{4} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*(d*x+c)^(1/2)),x, algorithm="fricas")

[Out]

-1/3*(3*a*b^2*d*x - 6*(a*b^2*c - a^3)*log(sqrt(d*x + c)*b + a) - 2*(b^3*d*x - 2*b^3*c + 3*a^2*b)*sqrt(d*x + c)
)/(b^4*d^2)

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Sympy [A]
time = 2.25, size = 109, normalized size = 1.21 \begin {gather*} \begin {cases} \frac {2 \left (- \frac {a \left (c + d x\right )}{2 b^{2} d} - \frac {a \left (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} d} + \frac {\left (c + d x\right )^{\frac {3}{2}}}{3 b d} + \frac {\left (a^{2} - b^{2} c\right ) \sqrt {c + d x}}{b^{3} d}\right )}{d} & \text {for}\: d \neq 0 \\\frac {x^{2}}{2 \left (a + b \sqrt {c}\right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*(d*x+c)**(1/2)),x)

[Out]

Piecewise((2*(-a*(c + d*x)/(2*b**2*d) - a*(a**2 - b**2*c)*Piecewise((sqrt(c + d*x)/a, Eq(b, 0)), (log(a + b*sq
rt(c + d*x))/b, True))/(b**3*d) + (c + d*x)**(3/2)/(3*b*d) + (a**2 - b**2*c)*sqrt(c + d*x)/(b**3*d))/d, Ne(d,
0)), (x**2/(2*(a + b*sqrt(c))), True))

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Giac [A]
time = 4.04, size = 105, normalized size = 1.17 \begin {gather*} \frac {\frac {6 \, {\left (a b^{2} c - a^{3}\right )} \log \left ({\left | \sqrt {d x + c} b + a \right |}\right )}{b^{4} d} + \frac {2 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} d^{2} - 6 \, \sqrt {d x + c} b^{2} c d^{2} - 3 \, {\left (d x + c\right )} a b d^{2} + 6 \, \sqrt {d x + c} a^{2} d^{2}}{b^{3} d^{3}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*(d*x+c)^(1/2)),x, algorithm="giac")

[Out]

1/3*(6*(a*b^2*c - a^3)*log(abs(sqrt(d*x + c)*b + a))/(b^4*d) + (2*(d*x + c)^(3/2)*b^2*d^2 - 6*sqrt(d*x + c)*b^
2*c*d^2 - 3*(d*x + c)*a*b*d^2 + 6*sqrt(d*x + c)*a^2*d^2)/(b^3*d^3))/d

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Mupad [B]
time = 0.05, size = 89, normalized size = 0.99 \begin {gather*} \frac {2\,{\left (c+d\,x\right )}^{3/2}}{3\,b\,d^2}-\left (\frac {2\,c}{b\,d^2}-\frac {2\,a^2}{b^3\,d^2}\right )\,\sqrt {c+d\,x}-\frac {\ln \left (a+b\,\sqrt {c+d\,x}\right )\,\left (2\,a^3-2\,a\,b^2\,c\right )}{b^4\,d^2}-\frac {a\,x}{b^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*(c + d*x)^(1/2)),x)

[Out]

(2*(c + d*x)^(3/2))/(3*b*d^2) - ((2*c)/(b*d^2) - (2*a^2)/(b^3*d^2))*(c + d*x)^(1/2) - (log(a + b*(c + d*x)^(1/
2))*(2*a^3 - 2*a*b^2*c))/(b^4*d^2) - (a*x)/(b^2*d)

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