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+4/3*a*b*c^2*(d*x+c)^(3/2)/d^3-1/2*c*(-b^2*c+2*a^2)*(d*x+c)^2/d^3-8/5*a*b*c*(d*x+c)^(5/2)/d^3+1/3
*(-2*b^2*c+a^2)*(d*x+c)^3/d^3+4/7*a*b*(d*x+c)^(7/2)/d^3+1/4*b^2*(d*x+c)^4/d^3

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Rubi [A]  time = 0.17, antiderivative size = 138, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.19, size = 77, normalized size = 0.56 \[ \frac {a^2 x^3}{3}+\frac {4 a b \sqrt {c+d x} \left (8 c^3-4 c^2 d x+3 c d^2 x^2+15 d^3 x^3\right )}{105 d^3}+\frac {1}{12} b^2 x^3 (4 c+3 d x) \]

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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fricas [A]  time = 0.85, size = 81, normalized size = 0.59 \[ \frac {105 \, b^{2} d^{4} x^{4} + 140 \, {\left (b^{2} c + a^{2}\right )} d^{3} x^{3} + 16 \, {\left (15 \, a b d^{3} x^{3} + 3 \, a b c d^{2} x^{2} - 4 \, a b c^{2} d x + 8 \, a b c^{3}\right )} \sqrt {d x + c}}{420 \, d^{3}} \]

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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giac [A]  time = 0.43, size = 127, normalized size = 0.92 \[ \frac {105 \, b^{2} d^{2} x^{4} + 140 \, b^{2} c d x^{3} + 140 \, a^{2} d x^{3} + \frac {112 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} a b c}{d^{2}} + \frac {48 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} c + 35 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} - 35 \, \sqrt {d x + c} c^{3}\right )} a b}{d^{2}}}{420 \, d} \]

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*(105*b^2*d^2*x^4 + 140*b^2*c*d*x^3 + 140*a^2*d*x^3 + 112*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*
sqrt(d*x + c)*c^2)*a*b*c/d^2 + 48*(5*(d*x + c)^(7/2) - 21*(d*x + c)^(5/2)*c + 35*(d*x + c)^(3/2)*c^2 - 35*sqrt
(d*x + c)*c^3)*a*b/d^2)/d

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maple [A]  time = 0.00, size = 66, normalized size = 0.48 \[ \frac {a^{2} x^{3}}{3}+\left (\frac {1}{4} d \,x^{4}+\frac {1}{3} c \,x^{3}\right ) b^{2}+\frac {4 \left (\frac {\left (d x +c \right )^{\frac {3}{2}} c^{2}}{3}-\frac {2 \left (d x +c \right )^{\frac {5}{2}} c}{5}+\frac {\left (d x +c \right )^{\frac {7}{2}}}{7}\right ) a b}{d^{3}} \]

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

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maxima [A]  time = 0.88, size = 112, normalized size = 0.81 \[ \frac {105 \, {\left (d x + c\right )}^{4} b^{2} + 240 \, {\left (d x + c\right )}^{\frac {7}{2}} a b - 672 \, {\left (d x + c\right )}^{\frac {5}{2}} a b c + 560 \, {\left (d x + c\right )}^{\frac {3}{2}} a b c^{2} + 420 \, {\left (d x + c\right )} a^{2} c^{2} - 140 \, {\left (2 \, b^{2} c - a^{2}\right )} {\left (d x + c\right )}^{3} + 210 \, {\left (b^{2} c^{2} - 2 \, a^{2} c\right )} {\left (d x + c\right )}^{2}}{420 \, d^{3}} \]

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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mupad [B]  time = 0.06, size = 124, normalized size = 0.90 \[ \frac {b^2\,{\left (c+d\,x\right )}^4}{4\,d^3}-\frac {\left (4\,a^2\,c-2\,b^2\,c^2\right )\,{\left (c+d\,x\right )}^2}{4\,d^3}-\frac {\left (4\,b^2\,c-2\,a^2\right )\,{\left (c+d\,x\right )}^3}{6\,d^3}+\frac {a^2\,c^2\,x}{d^2}+\frac {4\,a\,b\,{\left (c+d\,x\right )}^{7/2}}{7\,d^3}+\frac {4\,a\,b\,c^2\,{\left (c+d\,x\right )}^{3/2}}{3\,d^3}-\frac {8\,a\,b\,c\,{\left (c+d\,x\right )}^{5/2}}{5\,d^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*(c + d*x)^(1/2))^2,x)

[Out]

(b^2*(c + d*x)^4)/(4*d^3) - ((4*a^2*c - 2*b^2*c^2)*(c + d*x)^2)/(4*d^3) - ((4*b^2*c - 2*a^2)*(c + d*x)^3)/(6*d
^3) + (a^2*c^2*x)/d^2 + (4*a*b*(c + d*x)^(7/2))/(7*d^3) + (4*a*b*c^2*(c + d*x)^(3/2))/(3*d^3) - (8*a*b*c*(c +
d*x)^(5/2))/(5*d^3)

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sympy [A]  time = 5.97, size = 110, normalized size = 0.80 \[ \begin {cases} \frac {\frac {a^{2} d x^{3}}{3} + \frac {4 a b \left (\frac {c^{2} \left (c + d x\right )^{\frac {3}{2}}}{3} - \frac {2 c \left (c + d x\right )^{\frac {5}{2}}}{5} + \frac {\left (c + d x\right )^{\frac {7}{2}}}{7}\right )}{d^{2}} + \frac {2 b^{2} \left (\frac {c^{2} \left (c + d x\right )^{2}}{4} - \frac {c \left (c + d x\right )^{3}}{3} + \frac {\left (c + d x\right )^{4}}{8}\right )}{d^{2}}}{d} & \text {for}\: d \neq 0 \\\frac {x^{3} \left (a + b \sqrt {c}\right )^{2}}{3} & \text {otherwise} \end {cases} \]

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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