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

Optimal. Leaf size=89 \[ \frac{\left (a^2-b^2 c\right ) (c+d x)^2}{2 d^2}-\frac{a^2 c x}{d}+\frac{4 a b (c+d x)^{5/2}}{5 d^2}-\frac{4 a b c (c+d x)^{3/2}}{3 d^2}+\frac{b^2 (c+d x)^3}{3 d^2} \]

[Out]

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

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Rubi [A]  time = 0.090381, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {371, 1398, 772} \[ \frac{\left (a^2-b^2 c\right ) (c+d x)^2}{2 d^2}-\frac{a^2 c x}{d}+\frac{4 a b (c+d x)^{5/2}}{5 d^2}-\frac{4 a b c (c+d x)^{3/2}}{3 d^2}+\frac{b^2 (c+d x)^3}{3 d^2} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*Sqrt[c + d*x])^2,x]

[Out]

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

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

Mathematica [A]  time = 0.102714, size = 63, normalized size = 0.71 \[ \frac{1}{30} \left (15 a^2 x^2+\frac{8 a b \sqrt{c+d x} \left (-2 c^2+c d x+3 d^2 x^2\right )}{d^2}+5 b^2 x^2 (3 c+2 d x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.002, size = 54, normalized size = 0.6 \begin{align*}{b}^{2} \left ({\frac{d{x}^{3}}{3}}+{\frac{c{x}^{2}}{2}} \right ) +4\,{\frac{ab \left ( 1/5\, \left ( dx+c \right ) ^{5/2}-1/3\,c \left ( dx+c \right ) ^{3/2} \right ) }{{d}^{2}}}+{\frac{{a}^{2}{x}^{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.10944, size = 97, normalized size = 1.09 \begin{align*} \frac{10 \,{\left (d x + c\right )}^{3} b^{2} + 24 \,{\left (d x + c\right )}^{\frac{5}{2}} a b - 40 \,{\left (d x + c\right )}^{\frac{3}{2}} a b c - 30 \,{\left (d x + c\right )} a^{2} c - 15 \,{\left (b^{2} c - a^{2}\right )}{\left (d x + c\right )}^{2}}{30 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(10*(d*x + c)^3*b^2 + 24*(d*x + c)^(5/2)*a*b - 40*(d*x + c)^(3/2)*a*b*c - 30*(d*x + c)*a^2*c - 15*(b^2*c
- a^2)*(d*x + c)^2)/d^2

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Fricas [A]  time = 1.91882, size = 151, normalized size = 1.7 \begin{align*} \frac{10 \, b^{2} d^{3} x^{3} + 15 \,{\left (b^{2} c + a^{2}\right )} d^{2} x^{2} + 8 \,{\left (3 \, a b d^{2} x^{2} + a b c d x - 2 \, a b c^{2}\right )} \sqrt{d x + c}}{30 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(10*b^2*d^3*x^3 + 15*(b^2*c + a^2)*d^2*x^2 + 8*(3*a*b*d^2*x^2 + a*b*c*d*x - 2*a*b*c^2)*sqrt(d*x + c))/d^2

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Sympy [A]  time = 3.60007, size = 94, normalized size = 1.06 \begin{align*} \begin{cases} \frac{\frac{2 a^{2} \left (- \frac{c \left (c + d x\right )}{2} + \frac{\left (c + d x\right )^{2}}{4}\right )}{d} + \frac{4 a b \left (- \frac{c \left (c + d x\right )^{\frac{3}{2}}}{3} + \frac{\left (c + d x\right )^{\frac{5}{2}}}{5}\right )}{d} + \frac{2 b^{2} \left (- \frac{c \left (c + d x\right )^{2}}{4} + \frac{\left (c + d x\right )^{3}}{6}\right )}{d}}{d} & \text{for}\: d \neq 0 \\\frac{x^{2} \left (a + b \sqrt{c}\right )^{2}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((2*a**2*(-c*(c + d*x)/2 + (c + d*x)**2/4)/d + 4*a*b*(-c*(c + d*x)**(3/2)/3 + (c + d*x)**(5/2)/5)/d
+ 2*b**2*(-c*(c + d*x)**2/4 + (c + d*x)**3/6)/d)/d, Ne(d, 0)), (x**2*(a + b*sqrt(c))**2/2, True))

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Giac [A]  time = 1.21314, size = 115, normalized size = 1.29 \begin{align*} \frac{\frac{15 \,{\left ({\left (d x + c\right )}^{2} - 2 \,{\left (d x + c\right )} c\right )} a^{2}}{d} + \frac{8 \,{\left (3 \,{\left (d x + c\right )}^{\frac{5}{2}} - 5 \,{\left (d x + c\right )}^{\frac{3}{2}} c\right )} a b}{d} + \frac{5 \,{\left (2 \,{\left (d x + c\right )}^{3} - 3 \,{\left (d x + c\right )}^{2} c\right )} b^{2}}{d}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(15*((d*x + c)^2 - 2*(d*x + c)*c)*a^2/d + 8*(3*(d*x + c)^(5/2) - 5*(d*x + c)^(3/2)*c)*a*b/d + 5*(2*(d*x +
c)^3 - 3*(d*x + c)^2*c)*b^2/d)/d