3.640 \(\int \frac {x^2}{(a+b \sqrt {c+d x})^2} \, dx\)

Optimal. Leaf size=166 \[ \frac {2 a \left (a^2-b^2 c\right )^2}{b^6 d^3 \left (a+b \sqrt {c+d x}\right )}-\frac {8 a \left (a^2-b^2 c\right ) \sqrt {c+d x}}{b^5 d^3}+\frac {x \left (3 a^2-2 b^2 c\right )}{b^4 d^2}+\frac {2 \left (5 a^4-6 a^2 b^2 c+b^4 c^2\right ) \log \left (a+b \sqrt {c+d x}\right )}{b^6 d^3}-\frac {4 a (c+d x)^{3/2}}{3 b^3 d^3}+\frac {(c+d x)^2}{2 b^2 d^3} \]

[Out]

(-2*b^2*c+3*a^2)*x/b^4/d^2-4/3*a*(d*x+c)^(3/2)/b^3/d^3+1/2*(d*x+c)^2/b^2/d^3+2*(b^4*c^2-6*a^2*b^2*c+5*a^4)*ln(
a+b*(d*x+c)^(1/2))/b^6/d^3-8*a*(-b^2*c+a^2)*(d*x+c)^(1/2)/b^5/d^3+2*a*(-b^2*c+a^2)^2/b^6/d^3/(a+b*(d*x+c)^(1/2
))

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Rubi [A]  time = 0.17, antiderivative size = 166, 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 = {371, 1398, 772} \[ \frac {2 \left (-6 a^2 b^2 c+5 a^4+b^4 c^2\right ) \log \left (a+b \sqrt {c+d x}\right )}{b^6 d^3}+\frac {2 a \left (a^2-b^2 c\right )^2}{b^6 d^3 \left (a+b \sqrt {c+d x}\right )}-\frac {8 a \left (a^2-b^2 c\right ) \sqrt {c+d x}}{b^5 d^3}+\frac {x \left (3 a^2-2 b^2 c\right )}{b^4 d^2}-\frac {4 a (c+d x)^{3/2}}{3 b^3 d^3}+\frac {(c+d x)^2}{2 b^2 d^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*Sqrt[c + d*x])^2,x]

[Out]

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

Rule 371

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 772

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 1398

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}{\left (a+b \sqrt {c+d x}\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(-c+x)^2}{\left (a+b \sqrt {x}\right )^2} \, dx,x,c+d x\right )}{d^3}\\ &=\frac {2 \operatorname {Subst}\left (\int \frac {x \left (-c+x^2\right )^2}{(a+b x)^2} \, dx,x,\sqrt {c+d x}\right )}{d^3}\\ &=\frac {2 \operatorname {Subst}\left (\int \left (-\frac {4 a \left (a^2-b^2 c\right )}{b^5}-\frac {\left (-3 a^2+2 b^2 c\right ) x}{b^4}-\frac {2 a x^2}{b^3}+\frac {x^3}{b^2}-\frac {a \left (a^2-b^2 c\right )^2}{b^5 (a+b x)^2}+\frac {5 a^4-6 a^2 b^2 c+b^4 c^2}{b^5 (a+b x)}\right ) \, dx,x,\sqrt {c+d x}\right )}{d^3}\\ &=\frac {\left (3 a^2-2 b^2 c\right ) x}{b^4 d^2}-\frac {8 a \left (a^2-b^2 c\right ) \sqrt {c+d x}}{b^5 d^3}-\frac {4 a (c+d x)^{3/2}}{3 b^3 d^3}+\frac {(c+d x)^2}{2 b^2 d^3}+\frac {2 a \left (a^2-b^2 c\right )^2}{b^6 d^3 \left (a+b \sqrt {c+d x}\right )}+\frac {2 \left (5 a^4-6 a^2 b^2 c+b^4 c^2\right ) \log \left (a+b \sqrt {c+d x}\right )}{b^6 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 185, normalized size = 1.11 \[ \frac {16 a^5-44 a^4 b \sqrt {c+d x}-2 a^3 b^2 (38 c+15 d x)+2 a^2 b^3 \sqrt {c+d x} (18 c+5 d x)+12 \left (5 a^4-6 a^2 b^2 c+b^4 c^2\right ) \left (a+b \sqrt {c+d x}\right ) \log \left (a+b \sqrt {c+d x}\right )+a b^4 \left (52 c^2+26 c d x-5 d^2 x^2\right )+3 b^5 d x (d x-2 c) \sqrt {c+d x}}{6 b^6 d^3 \left (a+b \sqrt {c+d x}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*a^5 - 44*a^4*b*Sqrt[c + d*x] + 3*b^5*d*x*(-2*c + d*x)*Sqrt[c + d*x] + 2*a^2*b^3*Sqrt[c + d*x]*(18*c + 5*d*
x) - 2*a^3*b^2*(38*c + 15*d*x) + a*b^4*(52*c^2 + 26*c*d*x - 5*d^2*x^2) + 12*(5*a^4 - 6*a^2*b^2*c + b^4*c^2)*(a
+ b*Sqrt[c + d*x])*Log[a + b*Sqrt[c + d*x]])/(6*b^6*d^3*(a + b*Sqrt[c + d*x]))

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fricas [A]  time = 0.47, size = 269, normalized size = 1.62 \[ \frac {3 \, b^{6} d^{3} x^{3} - 9 \, b^{6} c^{3} + 15 \, a^{2} b^{4} c^{2} + 6 \, a^{4} b^{2} c - 12 \, a^{6} - 3 \, {\left (b^{6} c - 5 \, a^{2} b^{4}\right )} d^{2} x^{2} - 3 \, {\left (5 \, b^{6} c^{2} - 14 \, a^{2} b^{4} c + 6 \, a^{4} b^{2}\right )} d x + 12 \, {\left (b^{6} c^{3} - 7 \, a^{2} b^{4} c^{2} + 11 \, a^{4} b^{2} c - 5 \, a^{6} + {\left (b^{6} c^{2} - 6 \, a^{2} b^{4} c + 5 \, a^{4} b^{2}\right )} d x\right )} \log \left (\sqrt {d x + c} b + a\right ) - 4 \, {\left (2 \, a b^{5} d^{2} x^{2} - 13 \, a b^{5} c^{2} + 28 \, a^{3} b^{3} c - 15 \, a^{5} b - 2 \, {\left (4 \, a b^{5} c - 5 \, a^{3} b^{3}\right )} d x\right )} \sqrt {d x + c}}{6 \, {\left (b^{8} d^{4} x + {\left (b^{8} c - a^{2} b^{6}\right )} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*b^6*d^3*x^3 - 9*b^6*c^3 + 15*a^2*b^4*c^2 + 6*a^4*b^2*c - 12*a^6 - 3*(b^6*c - 5*a^2*b^4)*d^2*x^2 - 3*(5*
b^6*c^2 - 14*a^2*b^4*c + 6*a^4*b^2)*d*x + 12*(b^6*c^3 - 7*a^2*b^4*c^2 + 11*a^4*b^2*c - 5*a^6 + (b^6*c^2 - 6*a^
2*b^4*c + 5*a^4*b^2)*d*x)*log(sqrt(d*x + c)*b + a) - 4*(2*a*b^5*d^2*x^2 - 13*a*b^5*c^2 + 28*a^3*b^3*c - 15*a^5
*b - 2*(4*a*b^5*c - 5*a^3*b^3)*d*x)*sqrt(d*x + c))/(b^8*d^4*x + (b^8*c - a^2*b^6)*d^3)

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giac [A]  time = 0.37, size = 191, normalized size = 1.15 \[ \frac {2 \, {\left (b^{4} c^{2} - 6 \, a^{2} b^{2} c + 5 \, a^{4}\right )} \log \left ({\left | \sqrt {d x + c} b + a \right |}\right )}{b^{6} d^{3}} + \frac {2 \, {\left (a b^{4} c^{2} - 2 \, a^{3} b^{2} c + a^{5}\right )}}{{\left (\sqrt {d x + c} b + a\right )} b^{6} d^{3}} + \frac {3 \, {\left (d x + c\right )}^{2} b^{6} d^{9} - 12 \, {\left (d x + c\right )} b^{6} c d^{9} - 8 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{5} d^{9} + 48 \, \sqrt {d x + c} a b^{5} c d^{9} + 18 \, {\left (d x + c\right )} a^{2} b^{4} d^{9} - 48 \, \sqrt {d x + c} a^{3} b^{3} d^{9}}{6 \, b^{8} d^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b^4*c^2 - 6*a^2*b^2*c + 5*a^4)*log(abs(sqrt(d*x + c)*b + a))/(b^6*d^3) + 2*(a*b^4*c^2 - 2*a^3*b^2*c + a^5)/
((sqrt(d*x + c)*b + a)*b^6*d^3) + 1/6*(3*(d*x + c)^2*b^6*d^9 - 12*(d*x + c)*b^6*c*d^9 - 8*(d*x + c)^(3/2)*a*b^
5*d^9 + 48*sqrt(d*x + c)*a*b^5*c*d^9 + 18*(d*x + c)*a^2*b^4*d^9 - 48*sqrt(d*x + c)*a^3*b^3*d^9)/(b^8*d^12)

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maple [A]  time = 0.01, size = 253, normalized size = 1.52 \[ \frac {x^{2}}{2 b^{2} d}+\frac {2 a \,c^{2}}{\left (a +\sqrt {d x +c}\, b \right ) b^{2} d^{3}}+\frac {2 c^{2} \ln \left (a +\sqrt {d x +c}\, b \right )}{b^{2} d^{3}}-\frac {c x}{b^{2} d^{2}}-\frac {4 a^{3} c}{\left (a +\sqrt {d x +c}\, b \right ) b^{4} d^{3}}-\frac {12 a^{2} c \ln \left (a +\sqrt {d x +c}\, b \right )}{b^{4} d^{3}}+\frac {3 a^{2} x}{b^{4} d^{2}}-\frac {3 c^{2}}{2 b^{2} d^{3}}+\frac {2 a^{5}}{\left (a +\sqrt {d x +c}\, b \right ) b^{6} d^{3}}+\frac {10 a^{4} \ln \left (a +\sqrt {d x +c}\, b \right )}{b^{6} d^{3}}+\frac {3 a^{2} c}{b^{4} d^{3}}+\frac {8 \sqrt {d x +c}\, a c}{b^{3} d^{3}}-\frac {8 \sqrt {d x +c}\, a^{3}}{b^{5} d^{3}}-\frac {4 \left (d x +c \right )^{\frac {3}{2}} a}{3 b^{3} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/b^2*x^2-1/d^2/b^2*x*c-3/2/d^3/b^2*c^2-4/3*a*(d*x+c)^(3/2)/b^3/d^3+3/d^2/b^4*x*a^2+3/d^3/b^4*a^2*c+8/d^3/
b^3*a*c*(d*x+c)^(1/2)-8/d^3/b^5*a^3*(d*x+c)^(1/2)+2/d^3*a/b^2/(a+(d*x+c)^(1/2)*b)*c^2-4/d^3*a^3/b^4/(a+(d*x+c)
^(1/2)*b)*c+2/d^3*a^5/b^6/(a+(d*x+c)^(1/2)*b)+2/d^3/b^2*ln(a+(d*x+c)^(1/2)*b)*c^2-12/d^3/b^4*ln(a+(d*x+c)^(1/2
)*b)*a^2*c+10/d^3/b^6*ln(a+(d*x+c)^(1/2)*b)*a^4

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maxima [A]  time = 0.93, size = 158, normalized size = 0.95 \[ \frac {\frac {12 \, {\left (a b^{4} c^{2} - 2 \, a^{3} b^{2} c + a^{5}\right )}}{\sqrt {d x + c} b^{7} + a b^{6}} + \frac {3 \, {\left (d x + c\right )}^{2} b^{3} - 8 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{2} - 6 \, {\left (2 \, b^{3} c - 3 \, a^{2} b\right )} {\left (d x + c\right )} + 48 \, {\left (a b^{2} c - a^{3}\right )} \sqrt {d x + c}}{b^{5}} + \frac {12 \, {\left (b^{4} c^{2} - 6 \, a^{2} b^{2} c + 5 \, a^{4}\right )} \log \left (\sqrt {d x + c} b + a\right )}{b^{6}}}{6 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(12*(a*b^4*c^2 - 2*a^3*b^2*c + a^5)/(sqrt(d*x + c)*b^7 + a*b^6) + (3*(d*x + c)^2*b^3 - 8*(d*x + c)^(3/2)*a
*b^2 - 6*(2*b^3*c - 3*a^2*b)*(d*x + c) + 48*(a*b^2*c - a^3)*sqrt(d*x + c))/b^5 + 12*(b^4*c^2 - 6*a^2*b^2*c + 5
*a^4)*log(sqrt(d*x + c)*b + a)/b^6)/d^3

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mupad [B]  time = 3.20, size = 197, normalized size = 1.19 \[ \left (\frac {4\,a^3}{b^5\,d^3}+\frac {2\,a\,\left (\frac {4\,c}{b^2\,d^3}-\frac {6\,a^2}{b^4\,d^3}\right )}{b}\right )\,\sqrt {c+d\,x}+\frac {2\,\left (a^5-2\,a^3\,b^2\,c+a\,b^4\,c^2\right )}{b\,\left (b^6\,d^3\,\sqrt {c+d\,x}+a\,b^5\,d^3\right )}+\frac {{\left (c+d\,x\right )}^2}{2\,b^2\,d^3}-d\,x\,\left (\frac {2\,c}{b^2\,d^3}-\frac {3\,a^2}{b^4\,d^3}\right )-\frac {4\,a\,{\left (c+d\,x\right )}^{3/2}}{3\,b^3\,d^3}+\frac {\ln \left (a+b\,\sqrt {c+d\,x}\right )\,\left (10\,a^4-12\,a^2\,b^2\,c+2\,b^4\,c^2\right )}{b^6\,d^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*(c + d*x)^(1/2))^2,x)

[Out]

((4*a^3)/(b^5*d^3) + (2*a*((4*c)/(b^2*d^3) - (6*a^2)/(b^4*d^3)))/b)*(c + d*x)^(1/2) + (2*(a^5 - 2*a^3*b^2*c +
a*b^4*c^2))/(b*(b^6*d^3*(c + d*x)^(1/2) + a*b^5*d^3)) + (c + d*x)^2/(2*b^2*d^3) - d*x*((2*c)/(b^2*d^3) - (3*a^
2)/(b^4*d^3)) - (4*a*(c + d*x)^(3/2))/(3*b^3*d^3) + (log(a + b*(c + d*x)^(1/2))*(10*a^4 + 2*b^4*c^2 - 12*a^2*b
^2*c))/(b^6*d^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (a + b \sqrt {c + d x}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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