3.7.33 \(\int \frac {x^2}{a+b \sqrt {c+d x}} \, dx\) [633]

Optimal. Leaf size=151 \[ -\frac {a \left (a^2-2 b^2 c\right ) x}{b^4 d^2}+\frac {2 \left (a^2-b^2 c\right )^2 \sqrt {c+d x}}{b^5 d^3}+\frac {2 \left (a^2-2 b^2 c\right ) (c+d x)^{3/2}}{3 b^3 d^3}-\frac {a (c+d x)^2}{2 b^2 d^3}+\frac {2 (c+d x)^{5/2}}{5 b d^3}-\frac {2 a \left (a^2-b^2 c\right )^2 \log \left (a+b \sqrt {c+d x}\right )}{b^6 d^3} \]

[Out]

-a*(-2*b^2*c+a^2)*x/b^4/d^2+2/3*(-2*b^2*c+a^2)*(d*x+c)^(3/2)/b^3/d^3-1/2*a*(d*x+c)^2/b^2/d^3+2/5*(d*x+c)^(5/2)
/b/d^3-2*a*(-b^2*c+a^2)^2*ln(a+b*(d*x+c)^(1/2))/b^6/d^3+2*(-b^2*c+a^2)^2*(d*x+c)^(1/2)/b^5/d^3

________________________________________________________________________________________

Rubi [A]
time = 0.12, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {378, 1412, 786} \begin {gather*} -\frac {2 a \left (a^2-b^2 c\right )^2 \log \left (a+b \sqrt {c+d x}\right )}{b^6 d^3}+\frac {2 \left (a^2-b^2 c\right )^2 \sqrt {c+d x}}{b^5 d^3}-\frac {a x \left (a^2-2 b^2 c\right )}{b^4 d^2}+\frac {2 \left (a^2-2 b^2 c\right ) (c+d x)^{3/2}}{3 b^3 d^3}-\frac {a (c+d x)^2}{2 b^2 d^3}+\frac {2 (c+d x)^{5/2}}{5 b d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*Sqrt[c + d*x]),x]

[Out]

-((a*(a^2 - 2*b^2*c)*x)/(b^4*d^2)) + (2*(a^2 - b^2*c)^2*Sqrt[c + d*x])/(b^5*d^3) + (2*(a^2 - 2*b^2*c)*(c + d*x
)^(3/2))/(3*b^3*d^3) - (a*(c + d*x)^2)/(2*b^2*d^3) + (2*(c + d*x)^(5/2))/(5*b*d^3) - (2*a*(a^2 - b^2*c)^2*Log[
a + b*Sqrt[c + d*x]])/(b^6*d^3)

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^2}{a+b \sqrt {c+d x}} \, dx &=\frac {\text {Subst}\left (\int \frac {(-c+x)^2}{a+b \sqrt {x}} \, dx,x,c+d x\right )}{d^3}\\ &=\frac {2 \text {Subst}\left (\int \frac {x \left (-c+x^2\right )^2}{a+b x} \, dx,x,\sqrt {c+d x}\right )}{d^3}\\ &=\frac {2 \text {Subst}\left (\int \left (\frac {\left (-a^2+b^2 c\right )^2}{b^5}-\frac {a \left (a^2-2 b^2 c\right ) x}{b^4}-\frac {\left (-a^2+2 b^2 c\right ) x^2}{b^3}-\frac {a x^3}{b^2}+\frac {x^4}{b}-\frac {a \left (a^2-b^2 c\right )^2}{b^5 (a+b x)}\right ) \, dx,x,\sqrt {c+d x}\right )}{d^3}\\ &=-\frac {a \left (a^2-2 b^2 c\right ) x}{b^4 d^2}+\frac {2 \left (a^2-b^2 c\right )^2 \sqrt {c+d x}}{b^5 d^3}+\frac {2 \left (a^2-2 b^2 c\right ) (c+d x)^{3/2}}{3 b^3 d^3}-\frac {a (c+d x)^2}{2 b^2 d^3}+\frac {2 (c+d x)^{5/2}}{5 b d^3}-\frac {2 a \left (a^2-b^2 c\right )^2 \log \left (a+b \sqrt {c+d x}\right )}{b^6 d^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.11, size = 151, normalized size = 1.00 \begin {gather*} \frac {b \left (60 a^4 \sqrt {c+d x}-20 a^2 b^2 (5 c-d x) \sqrt {c+d x}-30 a^3 b (c+d x)+15 a b^3 \left (3 c^2+2 c d x-d^2 x^2\right )+4 b^4 \sqrt {c+d x} \left (8 c^2-4 c d x+3 d^2 x^2\right )\right )-60 a \left (a^2-b^2 c\right )^2 \log \left (a+b \sqrt {c+d x}\right )}{30 b^6 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*Sqrt[c + d*x]),x]

[Out]

(b*(60*a^4*Sqrt[c + d*x] - 20*a^2*b^2*(5*c - d*x)*Sqrt[c + d*x] - 30*a^3*b*(c + d*x) + 15*a*b^3*(3*c^2 + 2*c*d
*x - d^2*x^2) + 4*b^4*Sqrt[c + d*x]*(8*c^2 - 4*c*d*x + 3*d^2*x^2)) - 60*a*(a^2 - b^2*c)^2*Log[a + b*Sqrt[c + d
*x]])/(30*b^6*d^3)

________________________________________________________________________________________

Maple [A]
time = 0.03, size = 166, normalized size = 1.10

method	result	size
derivativedivides	\(\frac {\frac {2 \left (\frac {\left (d x +c \right )^{\frac {5}{2}} b^{4}}{5}-\frac {a \left (d x +c \right )^{2} b^{3}}{4}-\frac {2 b^{4} c \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {a^{2} b^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+a \,b^{3} c \left (d x +c \right )+b^{4} c^{2} \sqrt {d x +c}-\frac {a^{3} b \left (d x +c \right )}{2}-2 a^{2} b^{2} c \sqrt {d x +c}+a^{4} \sqrt {d x +c}\right )}{b^{5}}-\frac {2 a \left (b^{4} c^{2}-2 a^{2} b^{2} c +a^{4}\right ) \ln \left (a +b \sqrt {d x +c}\right )}{b^{6}}}{d^{3}}\)	\(166\)
default	\(\frac {\frac {2 \left (\frac {\left (d x +c \right )^{\frac {5}{2}} b^{4}}{5}-\frac {a \left (d x +c \right )^{2} b^{3}}{4}-\frac {2 b^{4} c \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {a^{2} b^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+a \,b^{3} c \left (d x +c \right )+b^{4} c^{2} \sqrt {d x +c}-\frac {a^{3} b \left (d x +c \right )}{2}-2 a^{2} b^{2} c \sqrt {d x +c}+a^{4} \sqrt {d x +c}\right )}{b^{5}}-\frac {2 a \left (b^{4} c^{2}-2 a^{2} b^{2} c +a^{4}\right ) \ln \left (a +b \sqrt {d x +c}\right )}{b^{6}}}{d^{3}}\)	\(166\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d^3*(1/b^5*(1/5*(d*x+c)^(5/2)*b^4-1/4*a*(d*x+c)^2*b^3-2/3*b^4*c*(d*x+c)^(3/2)+1/3*a^2*b^2*(d*x+c)^(3/2)+a*b^
3*c*(d*x+c)+b^4*c^2*(d*x+c)^(1/2)-1/2*a^3*b*(d*x+c)-2*a^2*b^2*c*(d*x+c)^(1/2)+a^4*(d*x+c)^(1/2))-a*(b^4*c^2-2*
a^2*b^2*c+a^4)/b^6*ln(a+b*(d*x+c)^(1/2)))

________________________________________________________________________________________

Maxima [A]
time = 0.28, size = 148, normalized size = 0.98 \begin {gather*} \frac {\frac {12 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4} - 15 \, {\left (d x + c\right )}^{2} a b^{3} - 20 \, {\left (2 \, b^{4} c - a^{2} b^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} + 30 \, {\left (2 \, a b^{3} c - a^{3} b\right )} {\left (d x + c\right )} + 60 \, {\left (b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}\right )} \sqrt {d x + c}}{b^{5}} - \frac {60 \, {\left (a b^{4} c^{2} - 2 \, a^{3} b^{2} c + a^{5}\right )} \log \left (\sqrt {d x + c} b + a\right )}{b^{6}}}{30 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*((12*(d*x + c)^(5/2)*b^4 - 15*(d*x + c)^2*a*b^3 - 20*(2*b^4*c - a^2*b^2)*(d*x + c)^(3/2) + 30*(2*a*b^3*c
- a^3*b)*(d*x + c) + 60*(b^4*c^2 - 2*a^2*b^2*c + a^4)*sqrt(d*x + c))/b^5 - 60*(a*b^4*c^2 - 2*a^3*b^2*c + a^5)*
log(sqrt(d*x + c)*b + a)/b^6)/d^3

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 138, normalized size = 0.91 \begin {gather*} -\frac {15 \, a b^{4} d^{2} x^{2} - 30 \, {\left (a b^{4} c - a^{3} b^{2}\right )} d x + 60 \, {\left (a b^{4} c^{2} - 2 \, a^{3} b^{2} c + a^{5}\right )} \log \left (\sqrt {d x + c} b + a\right ) - 4 \, {\left (3 \, b^{5} d^{2} x^{2} + 8 \, b^{5} c^{2} - 25 \, a^{2} b^{3} c + 15 \, a^{4} b - {\left (4 \, b^{5} c - 5 \, a^{2} b^{3}\right )} d x\right )} \sqrt {d x + c}}{30 \, b^{6} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(15*a*b^4*d^2*x^2 - 30*(a*b^4*c - a^3*b^2)*d*x + 60*(a*b^4*c^2 - 2*a^3*b^2*c + a^5)*log(sqrt(d*x + c)*b
+ a) - 4*(3*b^5*d^2*x^2 + 8*b^5*c^2 - 25*a^2*b^3*c + 15*a^4*b - (4*b^5*c - 5*a^2*b^3)*d*x)*sqrt(d*x + c))/(b^6
*d^3)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{a + b \sqrt {c + d x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/(a + b*sqrt(c + d*x)), x)

________________________________________________________________________________________

Giac [A]
time = 3.90, size = 198, normalized size = 1.31 \begin {gather*} -\frac {2 \, {\left (a b^{4} c^{2} - 2 \, a^{3} b^{2} c + a^{5}\right )} \log \left ({\left | \sqrt {d x + c} b + a \right |}\right )}{b^{6} d^{3}} + \frac {12 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4} d^{12} - 40 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{4} c d^{12} + 60 \, \sqrt {d x + c} b^{4} c^{2} d^{12} - 15 \, {\left (d x + c\right )}^{2} a b^{3} d^{12} + 60 \, {\left (d x + c\right )} a b^{3} c d^{12} + 20 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} b^{2} d^{12} - 120 \, \sqrt {d x + c} a^{2} b^{2} c d^{12} - 30 \, {\left (d x + c\right )} a^{3} b d^{12} + 60 \, \sqrt {d x + c} a^{4} d^{12}}{30 \, b^{5} d^{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(a*b^4*c^2 - 2*a^3*b^2*c + a^5)*log(abs(sqrt(d*x + c)*b + a))/(b^6*d^3) + 1/30*(12*(d*x + c)^(5/2)*b^4*d^12
- 40*(d*x + c)^(3/2)*b^4*c*d^12 + 60*sqrt(d*x + c)*b^4*c^2*d^12 - 15*(d*x + c)^2*a*b^3*d^12 + 60*(d*x + c)*a*
b^3*c*d^12 + 20*(d*x + c)^(3/2)*a^2*b^2*d^12 - 120*sqrt(d*x + c)*a^2*b^2*c*d^12 - 30*(d*x + c)*a^3*b*d^12 + 60
*sqrt(d*x + c)*a^4*d^12)/(b^5*d^15)

________________________________________________________________________________________

Mupad [B]
time = 3.21, size = 184, normalized size = 1.22 \begin {gather*} \frac {2\,{\left (c+d\,x\right )}^{5/2}}{5\,b\,d^3}-\left (\frac {a^2\,\left (\frac {4\,c}{b\,d^3}-\frac {2\,a^2}{b^3\,d^3}\right )}{b^2}-\frac {2\,c^2}{b\,d^3}\right )\,\sqrt {c+d\,x}-\left (\frac {4\,c}{3\,b\,d^3}-\frac {2\,a^2}{3\,b^3\,d^3}\right )\,{\left (c+d\,x\right )}^{3/2}-\frac {\ln \left (a+b\,\sqrt {c+d\,x}\right )\,\left (2\,a^5-4\,a^3\,b^2\,c+2\,a\,b^4\,c^2\right )}{b^6\,d^3}-\frac {a\,{\left (c+d\,x\right )}^2}{2\,b^2\,d^3}+\frac {a\,d\,x\,\left (\frac {4\,c}{b\,d^3}-\frac {2\,a^2}{b^3\,d^3}\right )}{2\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*(c + d*x)^(1/2)),x)

[Out]

(2*(c + d*x)^(5/2))/(5*b*d^3) - ((a^2*((4*c)/(b*d^3) - (2*a^2)/(b^3*d^3)))/b^2 - (2*c^2)/(b*d^3))*(c + d*x)^(1
/2) - ((4*c)/(3*b*d^3) - (2*a^2)/(3*b^3*d^3))*(c + d*x)^(3/2) - (log(a + b*(c + d*x)^(1/2))*(2*a^5 - 4*a^3*b^2
*c + 2*a*b^4*c^2))/(b^6*d^3) - (a*(c + d*x)^2)/(2*b^2*d^3) + (a*d*x*((4*c)/(b*d^3) - (2*a^2)/(b^3*d^3)))/(2*b)

________________________________________________________________________________________