3.268 \(\int \frac{\sqrt{c+d x^3}}{4 c+d x^3} \, dx\)

Optimal. Leaf size=64 \[ \frac{x \sqrt{c+d x^3} F_1\left (\frac{1}{3};1,-\frac{1}{2};\frac{4}{3};-\frac{d x^3}{4 c},-\frac{d x^3}{c}\right )}{4 c \sqrt{\frac{d x^3}{c}+1}} \]

[Out]

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

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Rubi [A]  time = 0.0280723, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {430, 429} \[ \frac{x \sqrt{c+d x^3} F_1\left (\frac{1}{3};1,-\frac{1}{2};\frac{4}{3};-\frac{d x^3}{4 c},-\frac{d x^3}{c}\right )}{4 c \sqrt{\frac{d x^3}{c}+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x^3]/(4*c + d*x^3),x]

[Out]

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

Rule 430

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])

Rule 429

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])

Rubi steps

\begin{align*} \int \frac{\sqrt{c+d x^3}}{4 c+d x^3} \, dx &=\frac{\sqrt{c+d x^3} \int \frac{\sqrt{1+\frac{d x^3}{c}}}{4 c+d 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{d x^3}{4 c},-\frac{d x^3}{c}\right )}{4 c \sqrt{1+\frac{d x^3}{c}}}\\ \end{align*}

Mathematica [B]  time = 0.141994, size = 165, normalized size = 2.58 \[ \frac{16 c x \sqrt{c+d x^3} F_1\left (\frac{1}{3};-\frac{1}{2},1;\frac{4}{3};-\frac{d x^3}{c},-\frac{d x^3}{4 c}\right )}{\left (4 c+d x^3\right ) \left (16 c F_1\left (\frac{1}{3};-\frac{1}{2},1;\frac{4}{3};-\frac{d x^3}{c},-\frac{d x^3}{4 c}\right )-3 d x^3 \left (F_1\left (\frac{4}{3};-\frac{1}{2},2;\frac{7}{3};-\frac{d x^3}{c},-\frac{d x^3}{4 c}\right )-2 F_1\left (\frac{4}{3};\frac{1}{2},1;\frac{7}{3};-\frac{d x^3}{c},-\frac{d x^3}{4 c}\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[c + d*x^3]/(4*c + d*x^3),x]

[Out]

(16*c*x*Sqrt[c + d*x^3]*AppellF1[1/3, -1/2, 1, 4/3, -((d*x^3)/c), -(d*x^3)/(4*c)])/((4*c + d*x^3)*(16*c*Appell
F1[1/3, -1/2, 1, 4/3, -((d*x^3)/c), -(d*x^3)/(4*c)] - 3*d*x^3*(AppellF1[4/3, -1/2, 2, 7/3, -((d*x^3)/c), -(d*x
^3)/(4*c)] - 2*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), -(d*x^3)/(4*c)])))

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Maple [C]  time = 0.006, size = 696, normalized size = 10.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)^(1/2)/(d*x^3+4*c),x)

[Out]

-2/3*I*3^(1/2)/d*(-d^2*c)^(1/3)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^
(1/3))^(1/2)*((x-1/d*(-d^2*c)^(1/3))/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)*(-I*(x+1/2/
d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*EllipticF(1/3
*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),(I*3^(1/2)
/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2))+1/3*I/d^3*2^(1/2)*sum(1/_alph
a^2*(-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1
/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^
(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^
2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^
(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),1/6/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha
^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-d^2*c)^(1/3
)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d+4*c))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d x^{3} + c}}{d x^{3} + 4 \, c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="maxima")

[Out]

integrate(sqrt(d*x^3 + c)/(d*x^3 + 4*c), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d x^{3} + c}}{d x^{3} + 4 \, c}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="fricas")

[Out]

integral(sqrt(d*x^3 + c)/(d*x^3 + 4*c), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c + d x^{3}}}{4 c + d x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**3+c)**(1/2)/(d*x**3+4*c),x)

[Out]

Integral(sqrt(c + d*x**3)/(4*c + d*x**3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d x^{3} + c}}{d x^{3} + 4 \, c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="giac")

[Out]

integrate(sqrt(d*x^3 + c)/(d*x^3 + 4*c), x)