3.366 \(\int \frac {\sqrt {c+d x^3}}{x^2 (a+b x^3)} \, dx\)
Optimal. Leaf size=62 \[ -\frac {\sqrt {c+d x^3} F_1\left (-\frac {1}{3};1,-\frac {1}{2};\frac {2}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a x \sqrt {\frac {d x^3}{c}+1}} \]
[Out]
-AppellF1(-1/3,1,-1/2,2/3,-b*x^3/a,-d*x^3/c)*(d*x^3+c)^(1/2)/a/x/(1+d*x^3/c)^(1/2)
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Rubi [A] time = 0.05, antiderivative size = 62, normalized size of antiderivative = 1.00,
number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used =
{511, 510} \[ -\frac {\sqrt {c+d x^3} F_1\left (-\frac {1}{3};1,-\frac {1}{2};\frac {2}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a x \sqrt {\frac {d x^3}{c}+1}} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[c + d*x^3]/(x^2*(a + b*x^3)),x]
[Out]
-((Sqrt[c + d*x^3]*AppellF1[-1/3, 1, -1/2, 2/3, -((b*x^3)/a), -((d*x^3)/c)])/(a*x*Sqrt[1 + (d*x^3)/c]))
Rule 510
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])
Rule 511
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && !(IntegerQ[
p] || GtQ[a, 0])
Rubi steps
\begin {align*} \int \frac {\sqrt {c+d x^3}}{x^2 \left (a+b x^3\right )} \, dx &=\frac {\sqrt {c+d x^3} \int \frac {\sqrt {1+\frac {d x^3}{c}}}{x^2 \left (a+b x^3\right )} \, dx}{\sqrt {1+\frac {d x^3}{c}}}\\ &=-\frac {\sqrt {c+d x^3} F_1\left (-\frac {1}{3};1,-\frac {1}{2};\frac {2}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a x \sqrt {1+\frac {d x^3}{c}}}\\ \end {align*}
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Mathematica [B] time = 0.12, size = 139, normalized size = 2.24 \[ \frac {5 x^3 \sqrt {\frac {d x^3}{c}+1} (3 a d-2 b c) F_1\left (\frac {2}{3};\frac {1}{2},1;\frac {5}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+2 b d x^6 \sqrt {\frac {d x^3}{c}+1} F_1\left (\frac {5}{3};\frac {1}{2},1;\frac {8}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )-20 a \left (c+d x^3\right )}{20 a^2 x \sqrt {c+d x^3}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sqrt[c + d*x^3]/(x^2*(a + b*x^3)),x]
[Out]
(-20*a*(c + d*x^3) + 5*(-2*b*c + 3*a*d)*x^3*Sqrt[1 + (d*x^3)/c]*AppellF1[2/3, 1/2, 1, 5/3, -((d*x^3)/c), -((b*
x^3)/a)] + 2*b*d*x^6*Sqrt[1 + (d*x^3)/c]*AppellF1[5/3, 1/2, 1, 8/3, -((d*x^3)/c), -((b*x^3)/a)])/(20*a^2*x*Sqr
t[c + d*x^3])
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^3+c)^(1/2)/x^2/(b*x^3+a),x, algorithm="fricas")
[Out]
Timed out
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d x^{3} + c}}{{\left (b x^{3} + a\right )} x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^3+c)^(1/2)/x^2/(b*x^3+a),x, algorithm="giac")
[Out]
integrate(sqrt(d*x^3 + c)/((b*x^3 + a)*x^2), x)
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maple [C] time = 0.27, size = 1314, normalized size = 21.19 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*x^3+c)^(1/2)/x^2/(b*x^3+a),x)
[Out]
-1/a*b*(-2/3*I/b*3^(1/2)*(-c*d^2)^(1/3)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c
*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d))^(1/2)*(-I*(
x+1/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*((-3/2*
(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*EllipticE(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2
)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(
1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2))+(-c*d^2)^(1/3)/d*EllipticF(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/
2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^
(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2)))+1/3*I/b/d^2*2^(1/2)*sum(1/_alpha*(-c*d^2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c
*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*
d^2)^(1/3))*d)^(1/2)*(-1/2*I*(2*x+(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+
c)^(1/2)*(2*_alpha^2*d^2+I*(-c*d^2)^(1/3)*3^(1/2)*_alpha*d-(-c*d^2)^(1/3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-
c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^
2)^(1/3)*d)^(1/2),1/2*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d+I*3^(1/2)*c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alp
ha-3*(-c*d^2)^(2/3)*_alpha)/(a*d-b*c)*b/d,(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d
^2)^(1/3)/d)/d)^(1/2)),_alpha=RootOf(_Z^3*b+a)))+1/a*(-(d*x^3+c)^(1/2)/x-I*3^(1/2)*(-c*d^2)^(1/3)*(I*(x+1/2*(-
c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3/2*(-c
*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d))^(1/2)*(-I*(x+1/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/
d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*((-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*Ell
ipticE(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),
(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2))+(-c*d^2)^(1/3)/d*El
lipticF(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2)
,(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2))))
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d x^{3} + c}}{{\left (b x^{3} + a\right )} x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^3+c)^(1/2)/x^2/(b*x^3+a),x, algorithm="maxima")
[Out]
integrate(sqrt(d*x^3 + c)/((b*x^3 + a)*x^2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\sqrt {d\,x^3+c}}{x^2\,\left (b\,x^3+a\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c + d*x^3)^(1/2)/(x^2*(a + b*x^3)),x)
[Out]
int((c + d*x^3)^(1/2)/(x^2*(a + b*x^3)), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c + d x^{3}}}{x^{2} \left (a + b x^{3}\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x**3+c)**(1/2)/x**2/(b*x**3+a),x)
[Out]
Integral(sqrt(c + d*x**3)/(x**2*(a + b*x**3)), x)
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