3.2965 \(\int \frac {\sqrt {a+b \sqrt {c x^3}}}{x^2} \, dx\)

Optimal. Leaf size=355 \[ \frac {3^{3/4} \sqrt {2+\sqrt {3}} b^{2/3} \sqrt [3]{c} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}+b^{2/3} \sqrt [3]{c} x}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}\right )^2}} F\left (\sin ^{-1}\left (\frac {\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{\sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}\right )^2}} \sqrt {a+b \sqrt {c x^3}}}-\frac {\sqrt {a+b \sqrt {c x^3}}}{x} \]

[Out]

3^(3/4)*b^(2/3)*c^(1/3)*EllipticF((a^(1/3)*(1-3^(1/2))+b^(1/3)*c^(2/3)*x^2/(c*x^3)^(1/2))/(a^(1/3)*(1+3^(1/2))
+b^(1/3)*c^(2/3)*x^2/(c*x^3)^(1/2)),I*3^(1/2)+2*I)*(a^(1/3)+b^(1/3)*c^(2/3)*x^2/(c*x^3)^(1/2))*(1/2*6^(1/2)+1/
2*2^(1/2))*((a^(2/3)+b^(2/3)*c^(1/3)*x-a^(1/3)*b^(1/3)*c^(2/3)*x^2/(c*x^3)^(1/2))/(a^(1/3)*(1+3^(1/2))+b^(1/3)
*c^(2/3)*x^2/(c*x^3)^(1/2))^2)^(1/2)/(a^(1/3)*(a^(1/3)+b^(1/3)*c^(2/3)*x^2/(c*x^3)^(1/2))/(a^(1/3)*(1+3^(1/2))
+b^(1/3)*c^(2/3)*x^2/(c*x^3)^(1/2))^2)^(1/2)/(a+b*(c*x^3)^(1/2))^(1/2)-(a+b*(c*x^3)^(1/2))^(1/2)/x

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Rubi [A]  time = 0.15, antiderivative size = 355, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {369, 341, 277, 218} \[ \frac {3^{3/4} \sqrt {2+\sqrt {3}} b^{2/3} \sqrt [3]{c} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}+b^{2/3} \sqrt [3]{c} x}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}\right )^2}} F\left (\sin ^{-1}\left (\frac {\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{\sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}\right )^2}} \sqrt {a+b \sqrt {c x^3}}}-\frac {\sqrt {a+b \sqrt {c x^3}}}{x} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[c*x^3]]/x^2,x]

[Out]

-(Sqrt[a + b*Sqrt[c*x^3]]/x) + (3^(3/4)*Sqrt[2 + Sqrt[3]]*b^(2/3)*c^(1/3)*(a^(1/3) + (b^(1/3)*c^(2/3)*x^2)/Sqr
t[c*x^3])*Sqrt[(a^(2/3) + b^(2/3)*c^(1/3)*x - (a^(1/3)*b^(1/3)*c^(2/3)*x^2)/Sqrt[c*x^3])/((1 + Sqrt[3])*a^(1/3
) + (b^(1/3)*c^(2/3)*x^2)/Sqrt[c*x^3])^2]*EllipticF[ArcSin[((1 - Sqrt[3])*a^(1/3) + (b^(1/3)*c^(2/3)*x^2)/Sqrt
[c*x^3])/((1 + Sqrt[3])*a^(1/3) + (b^(1/3)*c^(2/3)*x^2)/Sqrt[c*x^3])], -7 - 4*Sqrt[3]])/(Sqrt[(a^(1/3)*(a^(1/3
) + (b^(1/3)*c^(2/3)*x^2)/Sqrt[c*x^3]))/((1 + Sqrt[3])*a^(1/3) + (b^(1/3)*c^(2/3)*x^2)/Sqrt[c*x^3])^2]*Sqrt[a
+ b*Sqrt[c*x^3]])

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 341

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k*(
m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[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^2} \, dx &=\operatorname {Subst}\left (\int \frac {\sqrt {a+b \sqrt {c} x^{3/2}}}{x^2} \, dx,\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=\operatorname {Subst}\left (2 \operatorname {Subst}\left (\int \frac {\sqrt {a+b \sqrt {c} x^3}}{x^3} \, dx,x,\sqrt {x}\right ),\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=-\frac {\sqrt {a+b \sqrt {c x^3}}}{x}+\operatorname {Subst}\left (\frac {1}{2} \left (3 b \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b \sqrt {c} x^3}} \, dx,x,\sqrt {x}\right ),\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=-\frac {\sqrt {a+b \sqrt {c x^3}}}{x}+\frac {3^{3/4} \sqrt {2+\sqrt {3}} b^{2/3} \sqrt [3]{c} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}\right ) \sqrt {\frac {a^{2/3}+b^{2/3} \sqrt [3]{c} x-\frac {\sqrt [3]{a} \sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}}\right )|-7-4 \sqrt {3}\right )}{\sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b} c^{2/3} x^2}{\sqrt {c x^3}}\right )^2}} \sqrt {a+b \sqrt {c x^3}}}\\ \end {align*}

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Mathematica [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a+b \sqrt {c x^3}}}{x^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sqrt[a + b*Sqrt[c*x^3]]/x^2,x]

[Out]

Integrate[Sqrt[a + b*Sqrt[c*x^3]]/x^2, x]

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fricas [F]  time = 6.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\sqrt {c x^{3}} b + a}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(sqrt(c*x^3)*b + a)/x^2, x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(x)]sym2poly
/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument ValueEvaluation time: 22.1-2*abs
(c)/c^2/c*4*c^2/8*sqrt(a*c^2/(c*x)^2+b*c/sqrt(c*x))+integrate(2*abs(c)/c^2/c*6*c^3*b/8/2*c*c/(sqrt(c*x)*sqrt(b
*c^2*x*sqrt(c*x)+a*c^2)*abs(c)*sign(x)),x)

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maple [A]  time = 0.29, size = 306, normalized size = 0.86 \[ -\frac {i \sqrt {3}\, \left (-a \,b^{2} c \right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {-\frac {i \left (-2 \sqrt {c \,x^{3}}\, b +i \sqrt {3}\, \left (-a \,b^{2} c \right )^{\frac {1}{3}} x -\left (-a \,b^{2} c \right )^{\frac {1}{3}} x \right ) \sqrt {3}}{\left (-a \,b^{2} c \right )^{\frac {1}{3}} x}}\, \sqrt {\frac {\sqrt {c \,x^{3}}\, b -\left (-a \,b^{2} c \right )^{\frac {1}{3}} x}{\left (-a \,b^{2} c \right )^{\frac {1}{3}} \left (i \sqrt {3}-3\right ) x}}\, \sqrt {-\frac {i \left (2 \sqrt {c \,x^{3}}\, b +i \sqrt {3}\, \left (-a \,b^{2} c \right )^{\frac {1}{3}} x +\left (-a \,b^{2} c \right )^{\frac {1}{3}} x \right ) \sqrt {3}}{\left (-a \,b^{2} c \right )^{\frac {1}{3}} x}}\, x \EllipticF \left (\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {-\frac {i \left (-2 \sqrt {c \,x^{3}}\, b +i \sqrt {3}\, \left (-a \,b^{2} c \right )^{\frac {1}{3}} x -\left (-a \,b^{2} c \right )^{\frac {1}{3}} x \right ) \sqrt {3}}{\left (-a \,b^{2} c \right )^{\frac {1}{3}} x}}}{6}, \sqrt {2}\, \sqrt {\frac {i \sqrt {3}}{i \sqrt {3}-3}}\right )+2 a +2 \sqrt {c \,x^{3}}\, b}{2 \sqrt {a +\sqrt {c \,x^{3}}\, b}\, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(I*3^(1/2)*(-a*b^2*c)^(1/3)*2^(1/2)*(-I*(-2*(c*x^3)^(1/2)*b+I*3^(1/2)*(-a*b^2*c)^(1/3)*x-(-a*b^2*c)^(1/3)
*x)*3^(1/2)/(-a*b^2*c)^(1/3)/x)^(1/2)*(((c*x^3)^(1/2)*b-(-a*b^2*c)^(1/3)*x)/(-a*b^2*c)^(1/3)/(I*3^(1/2)-3)/x)^
(1/2)*(-I*(2*(c*x^3)^(1/2)*b+I*3^(1/2)*(-a*b^2*c)^(1/3)*x+(-a*b^2*c)^(1/3)*x)*3^(1/2)/(-a*b^2*c)^(1/3)/x)^(1/2
)*EllipticF(1/6*3^(1/2)*2^(1/2)*(-I*(-2*(c*x^3)^(1/2)*b+I*3^(1/2)*(-a*b^2*c)^(1/3)*x-(-a*b^2*c)^(1/3)*x)*3^(1/
2)/(-a*b^2*c)^(1/3)/x)^(1/2),2^(1/2)*(I*3^(1/2)/(I*3^(1/2)-3))^(1/2))*x+2*a+2*(c*x^3)^(1/2)*b)/x/(a+(c*x^3)^(1
/2)*b)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\sqrt {c x^{3}} b + a}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(sqrt(c*x^3)*b + a)/x^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {a+b\,\sqrt {c\,x^3}}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + b \sqrt {c x^{3}}}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*sqrt(c*x**3))/x**2, x)

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