3.279 \(\int \frac {1}{\sqrt {c+d x^3} (4 c+d x^3)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 64, normalized size of antiderivative = 1.00, 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} \[ \frac {x \sqrt {\frac {d x^3}{c}+1} F_1\left (\frac {1}{3};1,\frac {1}{2};\frac {4}{3};-\frac {d x^3}{4 c},-\frac {d x^3}{c}\right )}{4 c \sqrt {c+d x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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])

Rubi steps

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

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Mathematica [B]  time = 0.07, size = 165, normalized size = 2.58 \[ \frac {16 c x F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )}{\sqrt {c+d x^3} \left (4 c+d x^3\right ) \left (16 c F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )-3 d x^3 \left (F_1\left (\frac {4}{3};\frac {1}{2},2;\frac {7}{3};-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )+2 F_1\left (\frac {4}{3};\frac {3}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 1.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d x^{3} + c}}{d^{2} x^{6} + 5 \, c d x^{3} + 4 \, c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*x^3 + c)/(d^2*x^6 + 5*c*d*x^3 + 4*c^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (d x^{3} + 4 \, c\right )} \sqrt {d x^{3} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.15, size = 416, normalized size = 6.50 \[ -\frac {i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {\left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (2 \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )^{2} d^{2}+i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right ) d -\left (-c \,d^{2}\right )^{\frac {1}{3}} \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right ) d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}-\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, \frac {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )^{2} d +i \sqrt {3}\, c d -3 c d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )-3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )}{6 c d}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) d}}\right )}{9 c \,d^{3} \sqrt {d \,x^{3}+c}\, \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*I/d^3/c*2^(1/2)*sum(1/_alpha^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),1/6*(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)/c/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*
d+4*c))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (d x^{3} + 4 \, c\right )} \sqrt {d x^{3} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sqrt {d\,x^3+c}\,\left (d\,x^3+4\,c\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c + d x^{3}} \left (4 c + d x^{3}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(c + d*x**3)*(4*c + d*x**3)), x)

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