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

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

[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)

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Rubi [A]  time = 0.157475, antiderivative size = 151, normalized size of antiderivative = 1., 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} \[ \frac{2 \left (a^2-2 b^2 c\right ) (c+d x)^{3/2}}{3 b^3 d^3}+\frac{2 \left (a^2-b^2 c\right )^2 \sqrt{c+d x}}{b^5 d^3}-\frac{a x \left (a^2-2 b^2 c\right )}{b^4 d^2}-\frac{2 a \left (a^2-b^2 c\right )^2 \log \left (a+b \sqrt{c+d x}\right )}{b^6 d^3}-\frac{a (c+d x)^2}{2 b^2 d^3}+\frac{2 (c+d x)^{5/2}}{5 b d^3} \]

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 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 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]

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]

Rubi steps

\begin{align*} \int \frac{x^2}{a+b \sqrt{c+d x}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(-c+x)^2}{a+b \sqrt{x}} \, 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} \, dx,x,\sqrt{c+d x}\right )}{d^3}\\ &=\frac{2 \operatorname{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 = 0.134633, size = 138, normalized size = 0.91 \[ \frac{b \left (-20 a^2 b^2 (5 c-d x) \sqrt{c+d x}-30 a^3 b d x+60 a^4 \sqrt{c+d x}-15 a b^3 d x (d x-2 c)+4 b^4 \sqrt{c+d x} \left (8 c^2-4 c d x+3 d^2 x^2\right )\right )-60 a \left (a^2-b^2 c\right )^2 \log \left (a+b \sqrt{c+d x}\right )}{30 b^6 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 235, normalized size = 1.6 \begin{align*}{\frac{2}{5\,b{d}^{3}} \left ( dx+c \right ) ^{{\frac{5}{2}}}}-{\frac{a{x}^{2}}{2\,{b}^{2}d}}+{\frac{axc}{{b}^{2}{d}^{2}}}+{\frac{3\,{c}^{2}a}{2\,{d}^{3}{b}^{2}}}-{\frac{4\,c}{3\,b{d}^{3}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+{\frac{2\,{a}^{2}}{3\,{b}^{3}{d}^{3}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+2\,{\frac{{c}^{2}\sqrt{dx+c}}{b{d}^{3}}}-{\frac{{a}^{3}x}{{b}^{4}{d}^{2}}}-{\frac{{a}^{3}c}{{b}^{4}{d}^{3}}}-4\,{\frac{{a}^{2}c\sqrt{dx+c}}{{b}^{3}{d}^{3}}}+2\,{\frac{{a}^{4}\sqrt{dx+c}}{{d}^{3}{b}^{5}}}-2\,{\frac{a\ln \left ( a+b\sqrt{dx+c} \right ){c}^{2}}{{d}^{3}{b}^{2}}}+4\,{\frac{{a}^{3}\ln \left ( a+b\sqrt{dx+c} \right ) c}{{b}^{4}{d}^{3}}}-2\,{\frac{{a}^{5}\ln \left ( a+b\sqrt{dx+c} \right ) }{{d}^{3}{b}^{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.2224, size = 200, normalized size = 1.32 \begin{align*} \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{align*}

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

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Fricas [A]  time = 1.74785, size = 306, normalized size = 2.03 \begin{align*} -\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{align*}

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)

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

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)

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Giac [A]  time = 1.2403, size = 320, normalized size = 2.12 \begin{align*} -\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{2 \,{\left (a b^{4} c^{2} \log \left ({\left | a \right |}\right ) - 2 \, a^{3} b^{2} c \log \left ({\left | a \right |}\right ) + a^{5} \log \left ({\left | a \right |}\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{align*}

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) + 2*(a*b^4*c^2*log(abs(a)) - 2*a^3*
b^2*c*log(abs(a)) + a^5*log(abs(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)