3.24.83 \(\int \frac {\sqrt {a+b \sqrt {c x^3}}}{x^4} \, dx\)

Optimal. Leaf size=97 \[ \frac {b^2 c \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c x^3}}}{\sqrt {a}}\right )}{6 a^{3/2}}-\frac {b c \sqrt {a+b \sqrt {c x^3}}}{6 a \sqrt {c x^3}}-\frac {\sqrt {a+b \sqrt {c x^3}}}{3 x^3} \]

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Rubi [A]  time = 0.05, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {369, 266, 47, 51, 63, 208} \begin {gather*} \frac {b^2 c \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c x^3}}}{\sqrt {a}}\right )}{6 a^{3/2}}-\frac {b c \sqrt {a+b \sqrt {c x^3}}}{6 a \sqrt {c x^3}}-\frac {\sqrt {a+b \sqrt {c x^3}}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[c*x^3]]/x^4,x]

[Out]

-Sqrt[a + b*Sqrt[c*x^3]]/(3*x^3) - (b*c*Sqrt[a + b*Sqrt[c*x^3]])/(6*a*Sqrt[c*x^3]) + (b^2*c*ArcTanh[Sqrt[a + b
*Sqrt[c*x^3]]/Sqrt[a]])/(6*a^(3/2))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> With[{k = Denominator[n]}, Su
bst[Int[(d*x)^m*(a + b*c^n*x^(n*q))^p, x], x^(1/k), (c*x^q)^(1/k)/(c^(1/k)*(x^(1/k))^(q - 1))]] /; FreeQ[{a, b
, c, d, m, p, q}, x] && FractionQ[n]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b \sqrt {c x^3}}}{x^4} \, dx &=\operatorname {Subst}\left (\int \frac {\sqrt {a+b \sqrt {c} x^{3/2}}}{x^4} \, dx,\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=\operatorname {Subst}\left (\frac {2}{3} \operatorname {Subst}\left (\int \frac {\sqrt {a+b \sqrt {c} x}}{x^3} \, dx,x,x^{3/2}\right ),\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=-\frac {\sqrt {a+b \sqrt {c x^3}}}{3 x^3}+\operatorname {Subst}\left (\frac {1}{6} \left (b \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b \sqrt {c} x}} \, dx,x,x^{3/2}\right ),\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=-\frac {\sqrt {a+b \sqrt {c x^3}}}{3 x^3}-\frac {b c \sqrt {a+b \sqrt {c x^3}}}{6 a \sqrt {c x^3}}-\operatorname {Subst}\left (\frac {\left (b^2 c\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b \sqrt {c} x}} \, dx,x,x^{3/2}\right )}{12 a},\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=-\frac {\sqrt {a+b \sqrt {c x^3}}}{3 x^3}-\frac {b c \sqrt {a+b \sqrt {c x^3}}}{6 a \sqrt {c x^3}}-\operatorname {Subst}\left (\frac {\left (b \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b \sqrt {c}}+\frac {x^2}{b \sqrt {c}}} \, dx,x,\sqrt {a+b \sqrt {c} x^{3/2}}\right )}{6 a},\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=-\frac {\sqrt {a+b \sqrt {c x^3}}}{3 x^3}-\frac {b c \sqrt {a+b \sqrt {c x^3}}}{6 a \sqrt {c x^3}}+\frac {b^2 c \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c x^3}}}{\sqrt {a}}\right )}{6 a^{3/2}}\\ \end {align*}

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Mathematica [F]  time = 0.05, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a+b \sqrt {c x^3}}}{x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[Sqrt[a + b*Sqrt[c*x^3]]/x^4,x]

[Out]

Integrate[Sqrt[a + b*Sqrt[c*x^3]]/x^4, x]

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IntegrateAlgebraic [A]  time = 1.27, size = 80, normalized size = 0.82 \begin {gather*} \frac {b^2 c \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c x^3}}}{\sqrt {a}}\right )}{6 a^{3/2}}-\frac {\sqrt {a+b \sqrt {c x^3}} \left (2 a+b \sqrt {c x^3}\right )}{6 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[a + b*Sqrt[c*x^3]]/x^4,x]

[Out]

-1/6*(Sqrt[a + b*Sqrt[c*x^3]]*(2*a + b*Sqrt[c*x^3]))/(a*x^3) + (b^2*c*ArcTanh[Sqrt[a + b*Sqrt[c*x^3]]/Sqrt[a]]
)/(6*a^(3/2))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*x^3)^(1/2))^(1/2)/x^4,x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.21, size = 122, normalized size = 1.26 \begin {gather*} -\frac {{\left (\frac {b^{3} c^{3} \arctan \left (\frac {\sqrt {\sqrt {c x} b c^{2} x + a c^{2}}}{\sqrt {-a} c}\right )}{\sqrt {-a} a} + \frac {\sqrt {\sqrt {c x} b c^{2} x + a c^{2}} a b^{3} c^{6} + {\left (\sqrt {c x} b c^{2} x + a c^{2}\right )}^{\frac {3}{2}} b^{3} c^{4}}{a b^{2} c^{5} x^{3}}\right )} {\left | c \right |}}{6 \, b c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*x^3)^(1/2))^(1/2)/x^4,x, algorithm="giac")

[Out]

-1/6*(b^3*c^3*arctan(sqrt(sqrt(c*x)*b*c^2*x + a*c^2)/(sqrt(-a)*c))/(sqrt(-a)*a) + (sqrt(sqrt(c*x)*b*c^2*x + a*
c^2)*a*b^3*c^6 + (sqrt(c*x)*b*c^2*x + a*c^2)^(3/2)*b^3*c^4)/(a*b^2*c^5*x^3))*abs(c)/(b*c^3)

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maple [A]  time = 0.30, size = 81, normalized size = 0.84 \begin {gather*} -\frac {-a \,b^{2} c \,x^{3} \arctanh \left (\frac {\sqrt {a +\sqrt {c \,x^{3}}\, b}}{\sqrt {a}}\right )+2 \sqrt {a +\sqrt {c \,x^{3}}\, b}\, a^{\frac {5}{2}}+\sqrt {c \,x^{3}}\, \sqrt {a +\sqrt {c \,x^{3}}\, b}\, a^{\frac {3}{2}} b}{6 a^{\frac {5}{2}} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+(c*x^3)^(1/2)*b)^(1/2)/x^4,x)

[Out]

-1/6*(-b^2*arctanh((a+(c*x^3)^(1/2)*b)^(1/2)/a^(1/2))*c*x^3*a+(c*x^3)^(1/2)*b*(a+(c*x^3)^(1/2)*b)^(1/2)*a^(3/2
)+2*(a+(c*x^3)^(1/2)*b)^(1/2)*a^(5/2))/x^3/a^(5/2)

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maxima [A]  time = 1.16, size = 126, normalized size = 1.30 \begin {gather*} -\frac {1}{12} \, {\left (\frac {b^{2} \log \left (\frac {\sqrt {\sqrt {c x^{3}} b + a} - \sqrt {a}}{\sqrt {\sqrt {c x^{3}} b + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2 \, {\left ({\left (\sqrt {c x^{3}} b + a\right )}^{\frac {3}{2}} b^{2} + \sqrt {\sqrt {c x^{3}} b + a} a b^{2}\right )}}{{\left (\sqrt {c x^{3}} b + a\right )}^{2} a - 2 \, {\left (\sqrt {c x^{3}} b + a\right )} a^{2} + a^{3}}\right )} c \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*x^3)^(1/2))^(1/2)/x^4,x, algorithm="maxima")

[Out]

-1/12*(b^2*log((sqrt(sqrt(c*x^3)*b + a) - sqrt(a))/(sqrt(sqrt(c*x^3)*b + a) + sqrt(a)))/a^(3/2) + 2*((sqrt(c*x
^3)*b + a)^(3/2)*b^2 + sqrt(sqrt(c*x^3)*b + a)*a*b^2)/((sqrt(c*x^3)*b + a)^2*a - 2*(sqrt(c*x^3)*b + a)*a^2 + a
^3))*c

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {a+b\,\sqrt {c\,x^3}}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*(c*x^3)^(1/2))^(1/2)/x^4,x)

[Out]

int((a + b*(c*x^3)^(1/2))^(1/2)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b \sqrt {c x^{3}}}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*x**3)**(1/2))**(1/2)/x**4,x)

[Out]

Integral(sqrt(a + b*sqrt(c*x**3))/x**4, x)

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