3.373 \(\int \frac {x^3 (c+d x^3)^{3/2}}{a+b x^3} \, dx\)
Optimal. Leaf size=65 \[ \frac {c x^4 \sqrt {c+d x^3} F_1\left (\frac {4}{3};1,-\frac {3}{2};\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a \sqrt {\frac {d x^3}{c}+1}} \]
[Out]
1/4*c*x^4*AppellF1(4/3,1,-3/2,7/3,-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 = 0.06, antiderivative size = 65, 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} \[ \frac {c x^4 \sqrt {c+d x^3} F_1\left (\frac {4}{3};1,-\frac {3}{2};\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a \sqrt {\frac {d x^3}{c}+1}} \]
Antiderivative was successfully verified.
[In]
Int[(x^3*(c + d*x^3)^(3/2))/(a + b*x^3),x]
[Out]
(c*x^4*Sqrt[c + d*x^3]*AppellF1[4/3, 1, -3/2, 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 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx &=\frac {\left (c \sqrt {c+d x^3}\right ) \int \frac {x^3 \left (1+\frac {d x^3}{c}\right )^{3/2}}{a+b x^3} \, dx}{\sqrt {1+\frac {d x^3}{c}}}\\ &=\frac {c x^4 \sqrt {c+d x^3} F_1\left (\frac {4}{3};1,-\frac {3}{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 = 0.57, size = 280, normalized size = 4.31 \[ \frac {x \left (\frac {x^3 \sqrt {\frac {d x^3}{c}+1} \left (55 a^2 d^2-88 a b c d+27 b^2 c^2\right ) F_1\left (\frac {4}{3};\frac {1}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{a}-\frac {64 a^2 c^2 (11 a d-14 b c) F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{\left (a+b x^3\right ) \left (3 x^3 \left (2 b c F_1\left (\frac {4}{3};\frac {1}{2},2;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d F_1\left (\frac {4}{3};\frac {3}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )-8 a c F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )}+8 \left (c+d x^3\right ) \left (-11 a d+14 b c+5 b d x^3\right )\right )}{220 b^2 \sqrt {c+d x^3}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(x^3*(c + d*x^3)^(3/2))/(a + b*x^3),x]
[Out]
(x*(8*(c + d*x^3)*(14*b*c - 11*a*d + 5*b*d*x^3) + ((27*b^2*c^2 - 88*a*b*c*d + 55*a^2*d^2)*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 - (64*a^2*c^2*(-14*b*c + 11*a*d)*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)])))))/(220*b^2*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)^(3/2)/(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 \[ \int \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}} x^{3}}{b x^{3} + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(d*x^3+c)^(3/2)/(b*x^3+a),x, algorithm="giac")
[Out]
integrate((d*x^3 + c)^(3/2)*x^3/(b*x^3 + a), x)
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maple [C] time = 0.38, size = 1101, normalized size = 16.94 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^3*(d*x^3+c)^(3/2)/(b*x^3+a),x)
[Out]
1/b*(2/11*d*x^4*(d*x^3+c)^(1/2)+28/55*c*x*(d*x^3+c)^(1/2)-18/55*I*c^2*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*(2/5*d/b*x*(d*x^3+c)^(1/2)-2/3*I*(-(a*d-2*b*c)/b^2*d-2/5/b*c*d)*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))+1/3*I/b^2/d^2*2^(1/2)*sum((-a^2*d^2+2*a*b*c*d-b
^2*c^2)/_alpha^2/(a*d-b*c)*(-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 \[ \int \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}} x^{3}}{b x^{3} + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(d*x^3+c)^(3/2)/(b*x^3+a),x, algorithm="maxima")
[Out]
integrate((d*x^3 + c)^(3/2)*x^3/(b*x^3 + a), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^3\,{\left (d\,x^3+c\right )}^{3/2}}{b\,x^3+a} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((x^3*(c + d*x^3)^(3/2))/(a + b*x^3),x)
[Out]
int((x^3*(c + d*x^3)^(3/2))/(a + b*x^3), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \left (c + d x^{3}\right )^{\frac {3}{2}}}{a + b x^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**3*(d*x**3+c)**(3/2)/(b*x**3+a),x)
[Out]
Integral(x**3*(c + d*x**3)**(3/2)/(a + b*x**3), x)
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