3.7.20 \(\int x (a+b \sqrt {c+d x})^2 \, dx\) [620]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 89, normalized size of antiderivative = 1.00, 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 = {378, 1412, 786} \begin {gather*} \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} \end {gather*}

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 378

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 786

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 1412

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 x \left (a+b \sqrt {c+d x}\right )^2 \, dx &=\frac {\text {Subst}\left (\int \left (a+b \sqrt {x}\right )^2 (-c+x) \, dx,x,c+d x\right )}{d^2}\\ &=\frac {2 \text {Subst}\left (\int x (a+b x)^2 \left (-c+x^2\right ) \, dx,x,\sqrt {c+d x}\right )}{d^2}\\ &=\frac {2 \text {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*}

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Mathematica [A]
time = 0.04, size = 68, normalized size = 0.76 \begin {gather*} \frac {(c+d x) \left (-15 a^2 (c-d x)-8 a b (2 c-3 d x) \sqrt {c+d x}+5 b^2 \left (-c^2+c d x+2 d^2 x^2\right )\right )}{30 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.29, size = 54, normalized size = 0.61

method	result	size
default	\(b^{2} \left (\frac {1}{3} d \,x^{3}+\frac {1}{2} c \,x^{2}\right )+\frac {4 a b \left (\frac {\left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {c \left (d x +c \right )^{\frac {3}{2}}}{3}\right )}{d^{2}}+\frac {a^{2} x^{2}}{2}\)	\(54\)
trager	\(\frac {\left (2 b^{2} d x +3 b^{2} c +3 a^{2}\right ) x^{2}}{6}-\frac {4 a b \left (-3 d^{2} x^{2}-c d x +2 c^{2}\right ) \sqrt {d x +c}}{15 d^{2}}\)	\(59\)
derivativedivides	\(\frac {\frac {b^{2} \left (d x +c \right )^{3}}{3}+\frac {4 a b \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {\left (-b^{2} c +a^{2}\right ) \left (d x +c \right )^{2}}{2}-\frac {4 c a b \left (d x +c \right )^{\frac {3}{2}}}{3}-c \,a^{2} \left (d x +c \right )}{d^{2}}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.28, size = 72, normalized size = 0.81 \begin {gather*} \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 {gather*}

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 = 0.51, size = 67, normalized size = 0.75 \begin {gather*} \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 {gather*}

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 = 2.49, size = 94, normalized size = 1.06 \begin {gather*} \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 {gather*}

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 = 4.24, size = 131, normalized size = 1.47 \begin {gather*} \frac {10 \, b^{2} d^{2} x^{3} + \frac {40 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a b c}{d} + \frac {15 \, {\left ({\left (d x + c\right )}^{2} - 2 \, {\left (d x + c\right )} c\right )} b^{2} c}{d} + \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}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} a b}{d}}{30 \, d} \end {gather*}

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*(10*b^2*d^2*x^3 + 40*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*a*b*c/d + 15*((d*x + c)^2 - 2*(d*x + c)*c)*b^2
*c/d + 15*((d*x + c)^2 - 2*(d*x + c)*c)*a^2/d + 8*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)
*c^2)*a*b/d)/d

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Mupad [B]
time = 0.03, size = 79, normalized size = 0.89 \begin {gather*} \frac {b^2\,{\left (c+d\,x\right )}^3}{3\,d^2}-\frac {\left (2\,b^2\,c-2\,a^2\right )\,{\left (c+d\,x\right )}^2}{4\,d^2}+\frac {4\,a\,b\,{\left (c+d\,x\right )}^{5/2}}{5\,d^2}-\frac {a^2\,c\,x}{d}-\frac {4\,a\,b\,c\,{\left (c+d\,x\right )}^{3/2}}{3\,d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*(c + d*x)^(1/2))^2,x)

[Out]

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

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