3.4.73 \(\int \frac {x^3 (c+d x^3)^{3/2}}{a+b x^3} \, dx\) [373]

Optimal. Leaf size=65 \[ \frac {c x^4 \sqrt {c+d x^3} F_1\left (\frac {4}{3};1,-\frac {3}{2};\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a \sqrt {1+\frac {d x^3}{c}}} \]

[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 = 0.04, 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 = {525, 524} \begin {gather*} \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}} \end {gather*}

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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[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] Leaf count is larger than twice the leaf count of optimal. \(280\) vs. \(2(65)=130\).
time = 6.61, size = 280, normalized size = 4.31 \begin {gather*} \frac {x \left (8 \left (c+d x^3\right ) \left (14 b c-11 a d+5 b d x^3\right )+\frac {\left (27 b^2 c^2-88 a b c d+55 a^2 d^2\right ) x^3 \sqrt {1+\frac {d x^3}{c}} 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 (-14 b c+11 a d) 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 (-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 )+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 )\right )}\right )}{220 b^2 \sqrt {c+d x^3}} \end {gather*}

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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 6.
time = 0.38, size = 1101, normalized size = 16.94 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,method=_RETURNVERBOSE)

[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)/d*(-c*d^2)^(1/3)*(I*(x+1/2/d*(-c
*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2)*((x-1/d*(-c*d^2)^(1/3))/(-3/2/d*(-
c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)*(-I*(x+1/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)
)*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/
2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*
3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)))-a/b*(2/5*d/b*x*(d*x^3+c)^(1/2)-2/3*I*(-d*(a*d-2*b*c)/b^2-2/5*c*d/b)*3^(1/2)
/d*(-c*d^2)^(1/3)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2)*(
(x-1/d*(-c*d^2)^(1/3))/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)*(-I*(x+1/2/d*(-c*d^2)^(1/
3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x
+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),(I*3^(1/2)/d*(-c*d^2)^(1
/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(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*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-
c*d^2)^(1/3))^(1/2)*(d*(x-1/d*(-c*d^2)^(1/3))/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3)))^(1/2)*(-1/2*I*d*(2
*x+1/d*(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-c*d^2)^(1/3)*_alp
ha*3^(1/2)*d-I*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3)*_alpha*d-(-c*d^2)^(2/3))*EllipticPi(1/3*3^
(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),1/2*b/d*(2*I*
(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-c*d^2)^(2/3)*_alpha-3*c*d)
/(a*d-b*c),(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)),_alpha=R
ootOf(_Z^3*b+a)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [F(-1)] Timed out
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(x^3*(d*x^3+c)^(3/2)/(b*x^3+a),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \left (c + d x^{3}\right )^{\frac {3}{2}}}{a + b x^{3}}\, dx \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^3\,{\left (d\,x^3+c\right )}^{3/2}}{b\,x^3+a} \,d x \end {gather*}

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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