3.627 \(\int x \sqrt {a+b \sqrt {c+d x}} \, dx\)

Optimal. Leaf size=133 \[ \frac {4 \left (3 a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{5/2}}{5 b^4 d^2}-\frac {4 a \left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{3/2}}{3 b^4 d^2}+\frac {4 \left (a+b \sqrt {c+d x}\right )^{9/2}}{9 b^4 d^2}-\frac {12 a \left (a+b \sqrt {c+d x}\right )^{7/2}}{7 b^4 d^2} \]

[Out]

-4/3*a*(-b^2*c+a^2)*(a+b*(d*x+c)^(1/2))^(3/2)/b^4/d^2+4/5*(-b^2*c+3*a^2)*(a+b*(d*x+c)^(1/2))^(5/2)/b^4/d^2-12/
7*a*(a+b*(d*x+c)^(1/2))^(7/2)/b^4/d^2+4/9*(a+b*(d*x+c)^(1/2))^(9/2)/b^4/d^2

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Rubi [A]  time = 0.10, antiderivative size = 133, 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} \[ \frac {4 \left (3 a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{5/2}}{5 b^4 d^2}-\frac {4 a \left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{3/2}}{3 b^4 d^2}+\frac {4 \left (a+b \sqrt {c+d x}\right )^{9/2}}{9 b^4 d^2}-\frac {12 a \left (a+b \sqrt {c+d x}\right )^{7/2}}{7 b^4 d^2} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a + b*Sqrt[c + d*x]],x]

[Out]

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

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Mathematica [A]  time = 0.14, size = 84, normalized size = 0.63 \[ \frac {4 \left (a+b \sqrt {c+d x}\right )^{3/2} \left (-16 a^3+24 a^2 b \sqrt {c+d x}+6 a b^2 (2 c-5 d x)+7 b^3 \sqrt {c+d x} (5 d x-4 c)\right )}{315 b^4 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[a + b*Sqrt[c + d*x]],x]

[Out]

(4*(a + b*Sqrt[c + d*x])^(3/2)*(-16*a^3 + 6*a*b^2*(2*c - 5*d*x) + 24*a^2*b*Sqrt[c + d*x] + 7*b^3*Sqrt[c + d*x]
*(-4*c + 5*d*x)))/(315*b^4*d^2)

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fricas [A]  time = 0.56, size = 103, normalized size = 0.77 \[ \frac {4 \, {\left (35 \, b^{4} d^{2} x^{2} - 28 \, b^{4} c^{2} + 36 \, a^{2} b^{2} c - 16 \, a^{4} + {\left (7 \, b^{4} c - 6 \, a^{2} b^{2}\right )} d x + {\left (5 \, a b^{3} d x - 16 \, a b^{3} c + 8 \, a^{3} b\right )} \sqrt {d x + c}\right )} \sqrt {\sqrt {d x + c} b + a}}{315 \, b^{4} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/315*(35*b^4*d^2*x^2 - 28*b^4*c^2 + 36*a^2*b^2*c - 16*a^4 + (7*b^4*c - 6*a^2*b^2)*d*x + (5*a*b^3*d*x - 16*a*b
^3*c + 8*a^3*b)*sqrt(d*x + c))*sqrt(sqrt(d*x + c)*b + a)/(b^4*d^2)

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giac [B]  time = 0.37, size = 279, normalized size = 2.10 \[ -\frac {4 \, {\left (\frac {3 \, {\left (35 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} b^{2} c - 105 \, \sqrt {\sqrt {d x + c} b + a} a b^{2} c - 15 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {7}{2}} + 63 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {5}{2}} a - 105 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} a^{2} + 105 \, \sqrt {\sqrt {d x + c} b + a} a^{3}\right )} a}{b^{3} d} + \frac {63 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {5}{2}} b^{2} c - 210 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} a b^{2} c + 315 \, \sqrt {\sqrt {d x + c} b + a} a^{2} b^{2} c - 35 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {9}{2}} + 180 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {7}{2}} a - 378 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {5}{2}} a^{2} + 420 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} a^{3} - 315 \, \sqrt {\sqrt {d x + c} b + a} a^{4}}{b^{3} d}\right )}}{315 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/315*(3*(35*(sqrt(d*x + c)*b + a)^(3/2)*b^2*c - 105*sqrt(sqrt(d*x + c)*b + a)*a*b^2*c - 15*(sqrt(d*x + c)*b
+ a)^(7/2) + 63*(sqrt(d*x + c)*b + a)^(5/2)*a - 105*(sqrt(d*x + c)*b + a)^(3/2)*a^2 + 105*sqrt(sqrt(d*x + c)*b
+ a)*a^3)*a/(b^3*d) + (63*(sqrt(d*x + c)*b + a)^(5/2)*b^2*c - 210*(sqrt(d*x + c)*b + a)^(3/2)*a*b^2*c + 315*s
qrt(sqrt(d*x + c)*b + a)*a^2*b^2*c - 35*(sqrt(d*x + c)*b + a)^(9/2) + 180*(sqrt(d*x + c)*b + a)^(7/2)*a - 378*
(sqrt(d*x + c)*b + a)^(5/2)*a^2 + 420*(sqrt(d*x + c)*b + a)^(3/2)*a^3 - 315*sqrt(sqrt(d*x + c)*b + a)*a^4)/(b^
3*d))/(b*d)

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maple [A]  time = 0.00, size = 94, normalized size = 0.71 \[ \frac {-\frac {12 \left (a +\sqrt {d x +c}\, b \right )^{\frac {7}{2}} a}{7}-\frac {4 \left (-b^{2} c +a^{2}\right ) \left (a +\sqrt {d x +c}\, b \right )^{\frac {3}{2}} a}{3}+\frac {4 \left (a +\sqrt {d x +c}\, b \right )^{\frac {9}{2}}}{9}+\frac {4 \left (-b^{2} c +3 a^{2}\right ) \left (a +\sqrt {d x +c}\, b \right )^{\frac {5}{2}}}{5}}{b^{4} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.94, size = 93, normalized size = 0.70 \[ \frac {4 \, {\left (35 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {9}{2}} - 135 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {7}{2}} a - 63 \, {\left (b^{2} c - 3 \, a^{2}\right )} {\left (\sqrt {d x + c} b + a\right )}^{\frac {5}{2}} + 105 \, {\left (a b^{2} c - a^{3}\right )} {\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}}\right )}}{315 \, b^{4} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/315*(35*(sqrt(d*x + c)*b + a)^(9/2) - 135*(sqrt(d*x + c)*b + a)^(7/2)*a - 63*(b^2*c - 3*a^2)*(sqrt(d*x + c)*
b + a)^(5/2) + 105*(a*b^2*c - a^3)*(sqrt(d*x + c)*b + a)^(3/2))/(b^4*d^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\sqrt {a+b\,\sqrt {c+d\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sqrt {a + b \sqrt {c + d x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(a + b*sqrt(c + d*x)), x)

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