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

Optimal. Leaf size=138 \[ \frac{\left (a^2-2 b^2 c\right ) (c+d x)^3}{3 d^3}-\frac{c \left (2 a^2-b^2 c\right ) (c+d x)^2}{2 d^3}+\frac{a^2 c^2 x}{d^2}+\frac{4 a b c^2 (c+d x)^{3/2}}{3 d^3}+\frac{4 a b (c+d x)^{7/2}}{7 d^3}-\frac{8 a b c (c+d x)^{5/2}}{5 d^3}+\frac{b^2 (c+d x)^4}{4 d^3} \]

[Out]

(a^2*c^2*x)/d^2 + (4*a*b*c^2*(c + d*x)^(3/2))/(3*d^3) - (c*(2*a^2 - b^2*c)*(c + d*x)^2)/(2*d^3) - (8*a*b*c*(c
+ d*x)^(5/2))/(5*d^3) + ((a^2 - 2*b^2*c)*(c + d*x)^3)/(3*d^3) + (4*a*b*(c + d*x)^(7/2))/(7*d^3) + (b^2*(c + d*
x)^4)/(4*d^3)

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Rubi [A]  time = 0.16633, antiderivative size = 138, 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} \[ \frac{\left (a^2-2 b^2 c\right ) (c+d x)^3}{3 d^3}-\frac{c \left (2 a^2-b^2 c\right ) (c+d x)^2}{2 d^3}+\frac{a^2 c^2 x}{d^2}+\frac{4 a b c^2 (c+d x)^{3/2}}{3 d^3}+\frac{4 a b (c+d x)^{7/2}}{7 d^3}-\frac{8 a b c (c+d x)^{5/2}}{5 d^3}+\frac{b^2 (c+d x)^4}{4 d^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*Sqrt[c + d*x])^2,x]

[Out]

(a^2*c^2*x)/d^2 + (4*a*b*c^2*(c + d*x)^(3/2))/(3*d^3) - (c*(2*a^2 - b^2*c)*(c + d*x)^2)/(2*d^3) - (8*a*b*c*(c
+ d*x)^(5/2))/(5*d^3) + ((a^2 - 2*b^2*c)*(c + d*x)^3)/(3*d^3) + (4*a*b*(c + d*x)^(7/2))/(7*d^3) + (b^2*(c + d*
x)^4)/(4*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 x^2 \left (a+b \sqrt{c+d x}\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (a+b \sqrt{x}\right )^2 (-c+x)^2 \, dx,x,c+d x\right )}{d^3}\\ &=\frac{2 \operatorname{Subst}\left (\int x (a+b x)^2 \left (-c+x^2\right )^2 \, dx,x,\sqrt{c+d x}\right )}{d^3}\\ &=\frac{2 \operatorname{Subst}\left (\int \left (a^2 c^2 x+2 a b c^2 x^2+c \left (-2 a^2+b^2 c\right ) x^3-4 a b c x^4+\left (a^2-2 b^2 c\right ) x^5+2 a b x^6+b^2 x^7\right ) \, dx,x,\sqrt{c+d x}\right )}{d^3}\\ &=\frac{a^2 c^2 x}{d^2}+\frac{4 a b c^2 (c+d x)^{3/2}}{3 d^3}-\frac{c \left (2 a^2-b^2 c\right ) (c+d x)^2}{2 d^3}-\frac{8 a b c (c+d x)^{5/2}}{5 d^3}+\frac{\left (a^2-2 b^2 c\right ) (c+d x)^3}{3 d^3}+\frac{4 a b (c+d x)^{7/2}}{7 d^3}+\frac{b^2 (c+d x)^4}{4 d^3}\\ \end{align*}

Mathematica [A]  time = 0.189345, size = 77, normalized size = 0.56 \[ \frac{a^2 x^3}{3}+\frac{4 a b \sqrt{c+d x} \left (-4 c^2 d x+8 c^3+3 c d^2 x^2+15 d^3 x^3\right )}{105 d^3}+\frac{1}{12} b^2 x^3 (4 c+3 d x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x^3)/3 + (b^2*x^3*(4*c + 3*d*x))/12 + (4*a*b*Sqrt[c + d*x]*(8*c^3 - 4*c^2*d*x + 3*c*d^2*x^2 + 15*d^3*x^3)
)/(105*d^3)

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Maple [A]  time = 0.003, size = 66, normalized size = 0.5 \begin{align*}{b}^{2} \left ({\frac{d{x}^{4}}{4}}+{\frac{c{x}^{3}}{3}} \right ) +4\,{\frac{ab \left ( 1/7\, \left ( dx+c \right ) ^{7/2}-2/5\,c \left ( dx+c \right ) ^{5/2}+1/3\,{c}^{2} \left ( dx+c \right ) ^{3/2} \right ) }{{d}^{3}}}+{\frac{{a}^{2}{x}^{3}}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*(1/4*d*x^4+1/3*c*x^3)+4*a*b/d^3*(1/7*(d*x+c)^(7/2)-2/5*c*(d*x+c)^(5/2)+1/3*c^2*(d*x+c)^(3/2))+1/3*a^2*x^3

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Maxima [A]  time = 1.14879, size = 151, normalized size = 1.09 \begin{align*} \frac{105 \,{\left (d x + c\right )}^{4} b^{2} + 240 \,{\left (d x + c\right )}^{\frac{7}{2}} a b - 672 \,{\left (d x + c\right )}^{\frac{5}{2}} a b c + 560 \,{\left (d x + c\right )}^{\frac{3}{2}} a b c^{2} + 420 \,{\left (d x + c\right )} a^{2} c^{2} - 140 \,{\left (2 \, b^{2} c - a^{2}\right )}{\left (d x + c\right )}^{3} + 210 \,{\left (b^{2} c^{2} - 2 \, a^{2} c\right )}{\left (d x + c\right )}^{2}}{420 \, 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))^2,x, algorithm="maxima")

[Out]

1/420*(105*(d*x + c)^4*b^2 + 240*(d*x + c)^(7/2)*a*b - 672*(d*x + c)^(5/2)*a*b*c + 560*(d*x + c)^(3/2)*a*b*c^2
+ 420*(d*x + c)*a^2*c^2 - 140*(2*b^2*c - a^2)*(d*x + c)^3 + 210*(b^2*c^2 - 2*a^2*c)*(d*x + c)^2)/d^3

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Fricas [A]  time = 1.92185, size = 188, normalized size = 1.36 \begin{align*} \frac{105 \, b^{2} d^{4} x^{4} + 140 \,{\left (b^{2} c + a^{2}\right )} d^{3} x^{3} + 16 \,{\left (15 \, a b d^{3} x^{3} + 3 \, a b c d^{2} x^{2} - 4 \, a b c^{2} d x + 8 \, a b c^{3}\right )} \sqrt{d x + c}}{420 \, 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))^2,x, algorithm="fricas")

[Out]

1/420*(105*b^2*d^4*x^4 + 140*(b^2*c + a^2)*d^3*x^3 + 16*(15*a*b*d^3*x^3 + 3*a*b*c*d^2*x^2 - 4*a*b*c^2*d*x + 8*
a*b*c^3)*sqrt(d*x + c))/d^3

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Sympy [A]  time = 4.49434, size = 110, normalized size = 0.8 \begin{align*} \begin{cases} \frac{\frac{a^{2} d x^{3}}{3} + \frac{4 a b \left (\frac{c^{2} \left (c + d x\right )^{\frac{3}{2}}}{3} - \frac{2 c \left (c + d x\right )^{\frac{5}{2}}}{5} + \frac{\left (c + d x\right )^{\frac{7}{2}}}{7}\right )}{d^{2}} + \frac{2 b^{2} \left (\frac{c^{2} \left (c + d x\right )^{2}}{4} - \frac{c \left (c + d x\right )^{3}}{3} + \frac{\left (c + d x\right )^{4}}{8}\right )}{d^{2}}}{d} & \text{for}\: d \neq 0 \\\frac{x^{3} \left (a + b \sqrt{c}\right )^{2}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((a**2*d*x**3/3 + 4*a*b*(c**2*(c + d*x)**(3/2)/3 - 2*c*(c + d*x)**(5/2)/5 + (c + d*x)**(7/2)/7)/d**2
+ 2*b**2*(c**2*(c + d*x)**2/4 - c*(c + d*x)**3/3 + (c + d*x)**4/8)/d**2)/d, Ne(d, 0)), (x**3*(a + b*sqrt(c))*
*2/3, True))

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Giac [A]  time = 1.28828, size = 139, normalized size = 1.01 \begin{align*} \frac{140 \,{\left (d x^{3} + \frac{c^{3}}{d^{2}}\right )} a^{2} + \frac{16 \,{\left (15 \,{\left (d x + c\right )}^{\frac{7}{2}} - 42 \,{\left (d x + c\right )}^{\frac{5}{2}} c + 35 \,{\left (d x + c\right )}^{\frac{3}{2}} c^{2}\right )} a b}{d^{2}} + \frac{35 \,{\left (3 \,{\left (d x + c\right )}^{4} - 8 \,{\left (d x + c\right )}^{3} c + 6 \,{\left (d x + c\right )}^{2} c^{2}\right )} b^{2}}{d^{2}}}{420 \, d} \end{align*}

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]

1/420*(140*(d*x^3 + c^3/d^2)*a^2 + 16*(15*(d*x + c)^(7/2) - 42*(d*x + c)^(5/2)*c + 35*(d*x + c)^(3/2)*c^2)*a*b
/d^2 + 35*(3*(d*x + c)^4 - 8*(d*x + c)^3*c + 6*(d*x + c)^2*c^2)*b^2/d^2)/d