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

Optimal. Leaf size=131 \[ \frac {4 \left (3 a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{3/2}}{3 b^4 d^2}-\frac {4 a \left (a^2-b^2 c\right ) \sqrt {a+b \sqrt {c+d x}}}{b^4 d^2}+\frac {4 \left (a+b \sqrt {c+d x}\right )^{7/2}}{7 b^4 d^2}-\frac {12 a \left (a+b \sqrt {c+d x}\right )^{5/2}}{5 b^4 d^2} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 131, 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} \[ \frac {4 \left (3 a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{3/2}}{3 b^4 d^2}-\frac {4 a \left (a^2-b^2 c\right ) \sqrt {a+b \sqrt {c+d x}}}{b^4 d^2}+\frac {4 \left (a+b \sqrt {c+d x}\right )^{7/2}}{7 b^4 d^2}-\frac {12 a \left (a+b \sqrt {c+d x}\right )^{5/2}}{5 b^4 d^2} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a + b*Sqrt[c + d*x]],x]

[Out]

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

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Mathematica [A]  time = 0.10, size = 84, normalized size = 0.64 \[ \frac {4 \sqrt {a+b \sqrt {c+d x}} \left (-48 a^3+24 a^2 b \sqrt {c+d x}+2 a b^2 (26 c-9 d x)+5 b^3 \sqrt {c+d x} (3 d x-4 c)\right )}{105 b^4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[a + b*Sqrt[c + d*x]]*(-48*a^3 + 2*a*b^2*(26*c - 9*d*x) + 24*a^2*b*Sqrt[c + d*x] + 5*b^3*Sqrt[c + d*x]*
(-4*c + 3*d*x)))/(105*b^4*d^2)

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fricas [A]  time = 0.58, size = 71, normalized size = 0.54 \[ -\frac {4 \, {\left (18 \, a b^{2} d x - 52 \, a b^{2} c + 48 \, a^{3} - {\left (15 \, b^{3} d x - 20 \, b^{3} c + 24 \, a^{2} b\right )} \sqrt {d x + c}\right )} \sqrt {\sqrt {d x + c} b + a}}{105 \, b^{4} d^{2}} \]

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]

-4/105*(18*a*b^2*d*x - 52*a*b^2*c + 48*a^3 - (15*b^3*d*x - 20*b^3*c + 24*a^2*b)*sqrt(d*x + c))*sqrt(sqrt(d*x +
c)*b + a)/(b^4*d^2)

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giac [A]  time = 0.37, size = 115, normalized size = 0.88 \[ -\frac {4 \, {\left (35 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} b^{2} c - 105 \, \sqrt {\sqrt {d x + c} b + a} a b^{2} c - 15 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {7}{2}} + 63 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {5}{2}} a - 105 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} a^{2} + 105 \, \sqrt {\sqrt {d x + c} b + a} a^{3}\right )}}{105 \, b^{4} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/105*(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)/(b^4*d^2)

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maple [A]  time = 0.00, size = 94, normalized size = 0.72 \[ \frac {-\frac {12 \left (a +\sqrt {d x +c}\, b \right )^{\frac {5}{2}} a}{5}-4 \left (-b^{2} c +a^{2}\right ) \sqrt {a +\sqrt {d x +c}\, b}\, a +\frac {4 \left (a +\sqrt {d x +c}\, b \right )^{\frac {7}{2}}}{7}+\frac {4 \left (-b^{2} c +3 a^{2}\right ) \left (a +\sqrt {d x +c}\, b \right )^{\frac {3}{2}}}{3}}{b^{4} d^{2}} \]

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

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maxima [A]  time = 0.93, size = 93, normalized size = 0.71 \[ \frac {4 \, {\left (15 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {7}{2}} - 63 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {5}{2}} a - 35 \, {\left (b^{2} c - 3 \, a^{2}\right )} {\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} + 105 \, {\left (a b^{2} c - a^{3}\right )} \sqrt {\sqrt {d x + c} b + a}\right )}}{105 \, b^{4} d^{2}} \]

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]

4/105*(15*(sqrt(d*x + c)*b + a)^(7/2) - 63*(sqrt(d*x + c)*b + a)^(5/2)*a - 35*(b^2*c - 3*a^2)*(sqrt(d*x + c)*b
+ a)^(3/2) + 105*(a*b^2*c - a^3)*sqrt(sqrt(d*x + c)*b + a))/(b^4*d^2)

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

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 \[ \int \frac {x}{\sqrt {a + b \sqrt {c + d x}}}\, dx \]

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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