3.363 \(\int \frac {x^3 \sqrt {c+d x^3}}{a+b x^3} \, dx\)
Optimal. Leaf size=64 \[ \frac {x^4 \sqrt {c+d x^3} F_1\left (\frac {4}{3};1,-\frac {1}{2};\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a \sqrt {\frac {d x^3}{c}+1}} \]
[Out]
1/4*x^4*AppellF1(4/3,1,-1/2,7/3,-b*x^3/a,-d*x^3/c)*(d*x^3+c)^(1/2)/a/(1+d*x^3/c)^(1/2)
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Rubi [A] time = 0.05, antiderivative size = 64, 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} \[ \frac {x^4 \sqrt {c+d x^3} F_1\left (\frac {4}{3};1,-\frac {1}{2};\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a \sqrt {\frac {d x^3}{c}+1}} \]
Antiderivative was successfully verified.
[In]
Int[(x^3*Sqrt[c + d*x^3])/(a + b*x^3),x]
[Out]
(x^4*Sqrt[c + d*x^3]*AppellF1[4/3, 1, -1/2, 7/3, -((b*x^3)/a), -((d*x^3)/c)])/(4*a*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 {x^3 \sqrt {c+d x^3}}{a+b x^3} \, dx &=\frac {\sqrt {c+d x^3} \int \frac {x^3 \sqrt {1+\frac {d x^3}{c}}}{a+b x^3} \, dx}{\sqrt {1+\frac {d x^3}{c}}}\\ &=\frac {x^4 \sqrt {c+d x^3} F_1\left (\frac {4}{3};1,-\frac {1}{2};\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a \sqrt {1+\frac {d x^3}{c}}}\\ \end {align*}
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Mathematica [B] time = 0.62, size = 241, normalized size = 3.77 \[ \frac {x \left (8 \left (\frac {8 a^2 c^2 F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{\left (a+b x^3\right ) \left (3 x^3 \left (2 b c F_1\left (\frac {4}{3};\frac {1}{2},2;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d F_1\left (\frac {4}{3};\frac {3}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )-8 a c F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )}+c+d x^3\right )+\frac {x^3 \sqrt {\frac {d x^3}{c}+1} (3 b c-5 a d) F_1\left (\frac {4}{3};\frac {1}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{a}\right )}{20 b \sqrt {c+d x^3}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(x^3*Sqrt[c + d*x^3])/(a + b*x^3),x]
[Out]
(x*(((3*b*c - 5*a*d)*x^3*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), -((b*x^3)/a)])/a + 8*(c
+ d*x^3 + (8*a^2*c^2*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3)/a)])/((a + b*x^3)*(-8*a*c*AppellF1[1/3
, 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3)/a)] + 3*x^3*(2*b*c*AppellF1[4/3, 1/2, 2, 7/3, -((d*x^3)/c), -((b*x^3)/a
)] + a*d*AppellF1[4/3, 3/2, 1, 7/3, -((d*x^3)/c), -((b*x^3)/a)]))))))/(20*b*Sqrt[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(x^3*(d*x^3+c)^(1/2)/(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 \[ \int \frac {\sqrt {d x^{3} + c} x^{3}}{b x^{3} + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(d*x^3+c)^(1/2)/(b*x^3+a),x, algorithm="giac")
[Out]
integrate(sqrt(d*x^3 + c)*x^3/(b*x^3 + a), x)
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maple [C] time = 0.35, size = 1012, normalized size = 15.81 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^3*(d*x^3+c)^(1/2)/(b*x^3+a),x)
[Out]
1/b*(2/5*(d*x^3+c)^(1/2)*x-2/5*I*c*3^(1/2)*(-c*d^2)^(1/3)/d*(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)*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)))-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)*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/_a
lpha^2*(-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)*_alpha-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)))
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d x^{3} + c} x^{3}}{b x^{3} + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(d*x^3+c)^(1/2)/(b*x^3+a),x, algorithm="maxima")
[Out]
integrate(sqrt(d*x^3 + c)*x^3/(b*x^3 + a), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^3\,\sqrt {d\,x^3+c}}{b\,x^3+a} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((x^3*(c + d*x^3)^(1/2))/(a + b*x^3),x)
[Out]
int((x^3*(c + d*x^3)^(1/2))/(a + b*x^3), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \sqrt {c + d x^{3}}}{a + b x^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**3*(d*x**3+c)**(1/2)/(b*x**3+a),x)
[Out]
Integral(x**3*sqrt(c + d*x**3)/(a + b*x**3), x)
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