3.618 \(\int x^3 (a+b \sqrt{c+d x})^2 \, dx\)

Optimal. Leaf size=185 \[ \frac{c^2 \left (3 a^2-b^2 c\right ) (c+d x)^2}{2 d^4}+\frac{\left (a^2-3 b^2 c\right ) (c+d x)^4}{4 d^4}-\frac{c \left (a^2-b^2 c\right ) (c+d x)^3}{d^4}-\frac{a^2 c^3 x}{d^3}+\frac{12 a b c^2 (c+d x)^{5/2}}{5 d^4}-\frac{4 a b c^3 (c+d x)^{3/2}}{3 d^4}+\frac{4 a b (c+d x)^{9/2}}{9 d^4}-\frac{12 a b c (c+d x)^{7/2}}{7 d^4}+\frac{b^2 (c+d x)^5}{5 d^4} \]

[Out]

-((a^2*c^3*x)/d^3) - (4*a*b*c^3*(c + d*x)^(3/2))/(3*d^4) + (c^2*(3*a^2 - b^2*c)*(c + d*x)^2)/(2*d^4) + (12*a*b
*c^2*(c + d*x)^(5/2))/(5*d^4) - (c*(a^2 - b^2*c)*(c + d*x)^3)/d^4 - (12*a*b*c*(c + d*x)^(7/2))/(7*d^4) + ((a^2
- 3*b^2*c)*(c + d*x)^4)/(4*d^4) + (4*a*b*(c + d*x)^(9/2))/(9*d^4) + (b^2*(c + d*x)^5)/(5*d^4)

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Rubi [A]  time = 0.25477, antiderivative size = 185, 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} \[ \frac{c^2 \left (3 a^2-b^2 c\right ) (c+d x)^2}{2 d^4}+\frac{\left (a^2-3 b^2 c\right ) (c+d x)^4}{4 d^4}-\frac{c \left (a^2-b^2 c\right ) (c+d x)^3}{d^4}-\frac{a^2 c^3 x}{d^3}+\frac{12 a b c^2 (c+d x)^{5/2}}{5 d^4}-\frac{4 a b c^3 (c+d x)^{3/2}}{3 d^4}+\frac{4 a b (c+d x)^{9/2}}{9 d^4}-\frac{12 a b c (c+d x)^{7/2}}{7 d^4}+\frac{b^2 (c+d x)^5}{5 d^4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Sqrt[c + d*x])^2,x]

[Out]

-((a^2*c^3*x)/d^3) - (4*a*b*c^3*(c + d*x)^(3/2))/(3*d^4) + (c^2*(3*a^2 - b^2*c)*(c + d*x)^2)/(2*d^4) + (12*a*b
*c^2*(c + d*x)^(5/2))/(5*d^4) - (c*(a^2 - b^2*c)*(c + d*x)^3)/d^4 - (12*a*b*c*(c + d*x)^(7/2))/(7*d^4) + ((a^2
- 3*b^2*c)*(c + d*x)^4)/(4*d^4) + (4*a*b*(c + d*x)^(9/2))/(9*d^4) + (b^2*(c + d*x)^5)/(5*d^4)

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^3 \left (a+b \sqrt{c+d x}\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (a+b \sqrt{x}\right )^2 (-c+x)^3 \, dx,x,c+d x\right )}{d^4}\\ &=\frac{2 \operatorname{Subst}\left (\int x (a+b x)^2 \left (-c+x^2\right )^3 \, dx,x,\sqrt{c+d x}\right )}{d^4}\\ &=\frac{2 \operatorname{Subst}\left (\int \left (-a^2 c^3 x-2 a b c^3 x^2-c^2 \left (-3 a^2+b^2 c\right ) x^3+6 a b c^2 x^4+3 c \left (-a^2+b^2 c\right ) x^5-6 a b c x^6+\left (a^2-3 b^2 c\right ) x^7+2 a b x^8+b^2 x^9\right ) \, dx,x,\sqrt{c+d x}\right )}{d^4}\\ &=-\frac{a^2 c^3 x}{d^3}-\frac{4 a b c^3 (c+d x)^{3/2}}{3 d^4}+\frac{c^2 \left (3 a^2-b^2 c\right ) (c+d x)^2}{2 d^4}+\frac{12 a b c^2 (c+d x)^{5/2}}{5 d^4}-\frac{c \left (a^2-b^2 c\right ) (c+d x)^3}{d^4}-\frac{12 a b c (c+d x)^{7/2}}{7 d^4}+\frac{\left (a^2-3 b^2 c\right ) (c+d x)^4}{4 d^4}+\frac{4 a b (c+d x)^{9/2}}{9 d^4}+\frac{b^2 (c+d x)^5}{5 d^4}\\ \end{align*}

Mathematica [A]  time = 0.313332, size = 88, normalized size = 0.48 \[ \frac{a^2 x^4}{4}+\frac{4 a b \sqrt{c+d x} \left (-6 c^2 d^2 x^2+8 c^3 d x-16 c^4+5 c d^3 x^3+35 d^4 x^4\right )}{315 d^4}+\frac{1}{20} b^2 x^4 (5 c+4 d x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*Sqrt[c + d*x])^2,x]

[Out]

(a^2*x^4)/4 + (b^2*x^4*(5*c + 4*d*x))/20 + (4*a*b*Sqrt[c + d*x]*(-16*c^4 + 8*c^3*d*x - 6*c^2*d^2*x^2 + 5*c*d^3
*x^3 + 35*d^4*x^4))/(315*d^4)

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Maple [A]  time = 0.003, size = 78, normalized size = 0.4 \begin{align*}{b}^{2} \left ({\frac{d{x}^{5}}{5}}+{\frac{c{x}^{4}}{4}} \right ) +4\,{\frac{ab \left ( 1/9\, \left ( dx+c \right ) ^{9/2}-3/7\,c \left ( dx+c \right ) ^{7/2}+3/5\,{c}^{2} \left ( dx+c \right ) ^{5/2}-1/3\,{c}^{3} \left ( dx+c \right ) ^{3/2} \right ) }{{d}^{4}}}+{\frac{{a}^{2}{x}^{4}}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*(d*x+c)^(1/2))^2,x)

[Out]

b^2*(1/5*d*x^5+1/4*c*x^4)+4*a*b/d^4*(1/9*(d*x+c)^(9/2)-3/7*c*(d*x+c)^(7/2)+3/5*c^2*(d*x+c)^(5/2)-1/3*c^3*(d*x+
c)^(3/2))+1/4*a^2*x^4

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Maxima [A]  time = 1.10767, size = 204, normalized size = 1.1 \begin{align*} \frac{252 \,{\left (d x + c\right )}^{5} b^{2} + 560 \,{\left (d x + c\right )}^{\frac{9}{2}} a b - 2160 \,{\left (d x + c\right )}^{\frac{7}{2}} a b c + 3024 \,{\left (d x + c\right )}^{\frac{5}{2}} a b c^{2} - 1680 \,{\left (d x + c\right )}^{\frac{3}{2}} a b c^{3} - 1260 \,{\left (d x + c\right )} a^{2} c^{3} - 315 \,{\left (3 \, b^{2} c - a^{2}\right )}{\left (d x + c\right )}^{4} + 1260 \,{\left (b^{2} c^{2} - a^{2} c\right )}{\left (d x + c\right )}^{3} - 630 \,{\left (b^{2} c^{3} - 3 \, a^{2} c^{2}\right )}{\left (d x + c\right )}^{2}}{1260 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*(d*x+c)^(1/2))^2,x, algorithm="maxima")

[Out]

1/1260*(252*(d*x + c)^5*b^2 + 560*(d*x + c)^(9/2)*a*b - 2160*(d*x + c)^(7/2)*a*b*c + 3024*(d*x + c)^(5/2)*a*b*
c^2 - 1680*(d*x + c)^(3/2)*a*b*c^3 - 1260*(d*x + c)*a^2*c^3 - 315*(3*b^2*c - a^2)*(d*x + c)^4 + 1260*(b^2*c^2
- a^2*c)*(d*x + c)^3 - 630*(b^2*c^3 - 3*a^2*c^2)*(d*x + c)^2)/d^4

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Fricas [A]  time = 1.98351, size = 217, normalized size = 1.17 \begin{align*} \frac{252 \, b^{2} d^{5} x^{5} + 315 \,{\left (b^{2} c + a^{2}\right )} d^{4} x^{4} + 16 \,{\left (35 \, a b d^{4} x^{4} + 5 \, a b c d^{3} x^{3} - 6 \, a b c^{2} d^{2} x^{2} + 8 \, a b c^{3} d x - 16 \, a b c^{4}\right )} \sqrt{d x + c}}{1260 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*(d*x+c)^(1/2))^2,x, algorithm="fricas")

[Out]

1/1260*(252*b^2*d^5*x^5 + 315*(b^2*c + a^2)*d^4*x^4 + 16*(35*a*b*d^4*x^4 + 5*a*b*c*d^3*x^3 - 6*a*b*c^2*d^2*x^2
+ 8*a*b*c^3*d*x - 16*a*b*c^4)*sqrt(d*x + c))/d^4

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Sympy [A]  time = 5.39549, size = 139, normalized size = 0.75 \begin{align*} \begin{cases} \frac{\frac{a^{2} d x^{4}}{4} + \frac{4 a b \left (- \frac{c^{3} \left (c + d x\right )^{\frac{3}{2}}}{3} + \frac{3 c^{2} \left (c + d x\right )^{\frac{5}{2}}}{5} - \frac{3 c \left (c + d x\right )^{\frac{7}{2}}}{7} + \frac{\left (c + d x\right )^{\frac{9}{2}}}{9}\right )}{d^{3}} + \frac{2 b^{2} \left (- \frac{c^{3} \left (c + d x\right )^{2}}{4} + \frac{c^{2} \left (c + d x\right )^{3}}{2} - \frac{3 c \left (c + d x\right )^{4}}{8} + \frac{\left (c + d x\right )^{5}}{10}\right )}{d^{3}}}{d} & \text{for}\: d \neq 0 \\\frac{x^{4} \left (a + b \sqrt{c}\right )^{2}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*(d*x+c)**(1/2))**2,x)

[Out]

Piecewise(((a**2*d*x**4/4 + 4*a*b*(-c**3*(c + d*x)**(3/2)/3 + 3*c**2*(c + d*x)**(5/2)/5 - 3*c*(c + d*x)**(7/2)
/7 + (c + d*x)**(9/2)/9)/d**3 + 2*b**2*(-c**3*(c + d*x)**2/4 + c**2*(c + d*x)**3/2 - 3*c*(c + d*x)**4/8 + (c +
d*x)**5/10)/d**3)/d, Ne(d, 0)), (x**4*(a + b*sqrt(c))**2/4, True))

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Giac [A]  time = 1.24226, size = 173, normalized size = 0.94 \begin{align*} \frac{315 \,{\left (d x^{4} - \frac{c^{4}}{d^{3}}\right )} a^{2} + \frac{16 \,{\left (35 \,{\left (d x + c\right )}^{\frac{9}{2}} - 135 \,{\left (d x + c\right )}^{\frac{7}{2}} c + 189 \,{\left (d x + c\right )}^{\frac{5}{2}} c^{2} - 105 \,{\left (d x + c\right )}^{\frac{3}{2}} c^{3}\right )} a b}{d^{3}} + \frac{63 \,{\left (4 \,{\left (d x + c\right )}^{5} - 15 \,{\left (d x + c\right )}^{4} c + 20 \,{\left (d x + c\right )}^{3} c^{2} - 10 \,{\left (d x + c\right )}^{2} c^{3}\right )} b^{2}}{d^{3}}}{1260 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*(d*x+c)^(1/2))^2,x, algorithm="giac")

[Out]

1/1260*(315*(d*x^4 - c^4/d^3)*a^2 + 16*(35*(d*x + c)^(9/2) - 135*(d*x + c)^(7/2)*c + 189*(d*x + c)^(5/2)*c^2 -
105*(d*x + c)^(3/2)*c^3)*a*b/d^3 + 63*(4*(d*x + c)^5 - 15*(d*x + c)^4*c + 20*(d*x + c)^3*c^2 - 10*(d*x + c)^2
*c^3)*b^2/d^3)/d