3.4.65 \(\int \frac {\sqrt {c+d x^3}}{a+b x^3} \, dx\) [365]

Optimal. Leaf size=59 \[ \frac {x \sqrt {c+d x^3} F_1\left (\frac {1}{3};1,-\frac {1}{2};\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a \sqrt {1+\frac {d x^3}{c}}} \]

[Out]

x*AppellF1(1/3,1,-1/2,4/3,-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 = 0.02, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {441, 440} \begin {gather*} \frac {x \sqrt {c+d x^3} F_1\left (\frac {1}{3};1,-\frac {1}{2};\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a \sqrt {\frac {d x^3}{c}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x^3]/(a + b*x^3),x]

[Out]

(x*Sqrt[c + d*x^3]*AppellF1[1/3, 1, -1/2, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/(a*Sqrt[1 + (d*x^3)/c])

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rubi steps

\begin {align*} \int \frac {\sqrt {c+d x^3}}{a+b x^3} \, dx &=\frac {\sqrt {c+d x^3} \int \frac {\sqrt {1+\frac {d x^3}{c}}}{a+b x^3} \, dx}{\sqrt {1+\frac {d x^3}{c}}}\\ &=\frac {x \sqrt {c+d x^3} F_1\left (\frac {1}{3};1,-\frac {1}{2};\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{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. \(161\) vs. \(2(59)=118\).
time = 9.99, size = 161, normalized size = 2.73 \begin {gather*} \frac {8 a c x \sqrt {c+d x^3} 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 {1}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[c + d*x^3]/(a + b*x^3),x]

[Out]

(8*a*c*x*Sqrt[c + d*x^3]*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, 1/2, 1, 7/3, -((d*x^3)/c), -((b*x^3)/a)])))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 6.
time = 0.30, size = 705, normalized size = 11.95 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)^(1/2)/(b*x^3+a),x,method=_RETURNVERBOSE)

[Out]

-2/3*I/b*3^(1/2)*(-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/d^2*2^(1/2)*sum(1/_al
pha^2*(-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)*_alpha*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)*_a
lpha^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=RootOf(_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((d*x^3+c)^(1/2)/(b*x^3+a),x, algorithm="maxima")

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   Not integrable (provided residues have n
o relations)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^3)^(1/2)/(a + b*x^3),x)

[Out]

int((c + d*x^3)^(1/2)/(a + b*x^3), x)

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