3.69 \(\int \frac {c+d x^3}{(a+b x^3)^{8/3}} \, dx\)

Optimal. Leaf size=94 \[ \frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} (a d+4 b c) \, _2F_1\left (\frac {1}{3},\frac {5}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{5 a^2 b \left (a+b x^3\right )^{2/3}}+\frac {x (b c-a d)}{5 a b \left (a+b x^3\right )^{5/3}} \]

[Out]

1/5*(-a*d+b*c)*x/a/b/(b*x^3+a)^(5/3)+1/5*(a*d+4*b*c)*x*(1+b*x^3/a)^(2/3)*hypergeom([1/3, 5/3],[4/3],-b*x^3/a)/
a^2/b/(b*x^3+a)^(2/3)

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Rubi [A]  time = 0.03, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {385, 246, 245} \[ \frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} (a d+4 b c) \, _2F_1\left (\frac {1}{3},\frac {5}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{5 a^2 b \left (a+b x^3\right )^{2/3}}+\frac {x (b c-a d)}{5 a b \left (a+b x^3\right )^{5/3}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)/(a + b*x^3)^(8/3),x]

[Out]

((b*c - a*d)*x)/(5*a*b*(a + b*x^3)^(5/3)) + ((4*b*c + a*d)*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 5/3,
4/3, -((b*x^3)/a)])/(5*a^2*b*(a + b*x^3)^(2/3))

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 385

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

Rubi steps

\begin {align*} \int \frac {c+d x^3}{\left (a+b x^3\right )^{8/3}} \, dx &=\frac {(b c-a d) x}{5 a b \left (a+b x^3\right )^{5/3}}+\frac {(4 b c+a d) \int \frac {1}{\left (a+b x^3\right )^{5/3}} \, dx}{5 a b}\\ &=\frac {(b c-a d) x}{5 a b \left (a+b x^3\right )^{5/3}}+\frac {\left ((4 b c+a d) \left (1+\frac {b x^3}{a}\right )^{2/3}\right ) \int \frac {1}{\left (1+\frac {b x^3}{a}\right )^{5/3}} \, dx}{5 a^2 b \left (a+b x^3\right )^{2/3}}\\ &=\frac {(b c-a d) x}{5 a b \left (a+b x^3\right )^{5/3}}+\frac {(4 b c+a d) x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {5}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{5 a^2 b \left (a+b x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 75, normalized size = 0.80 \[ \frac {x \left (\frac {\left (a+b x^3\right ) \left (\frac {b x^3}{a}+1\right )^{2/3} (a d+4 b c) \, _2F_1\left (\frac {1}{3},\frac {8}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{a^2}-d\right )}{4 b \left (a+b x^3\right )^{5/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^3)/(a + b*x^3)^(8/3),x]

[Out]

(x*(-d + ((4*b*c + a*d)*(a + b*x^3)*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 8/3, 4/3, -((b*x^3)/a)])/a^2)
)/(4*b*(a + b*x^3)^(5/3))

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (d x^{3} + c\right )}}{b^{3} x^{9} + 3 \, a b^{2} x^{6} + 3 \, a^{2} b x^{3} + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)/(b*x^3+a)^(8/3),x, algorithm="fricas")

[Out]

integral((b*x^3 + a)^(1/3)*(d*x^3 + c)/(b^3*x^9 + 3*a*b^2*x^6 + 3*a^2*b*x^3 + a^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x^{3} + c}{{\left (b x^{3} + a\right )}^{\frac {8}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)/(b*x^3+a)^(8/3),x, algorithm="giac")

[Out]

integrate((d*x^3 + c)/(b*x^3 + a)^(8/3), x)

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maple [F]  time = 0.40, size = 0, normalized size = 0.00 \[ \int \frac {d \,x^{3}+c}{\left (b \,x^{3}+a \right )^{\frac {8}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)/(b*x^3+a)^(8/3),x)

[Out]

int((d*x^3+c)/(b*x^3+a)^(8/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x^{3} + c}{{\left (b x^{3} + a\right )}^{\frac {8}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)/(b*x^3+a)^(8/3),x, algorithm="maxima")

[Out]

integrate((d*x^3 + c)/(b*x^3 + a)^(8/3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {d\,x^3+c}{{\left (b\,x^3+a\right )}^{8/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^3)/(a + b*x^3)^(8/3),x)

[Out]

int((c + d*x^3)/(a + b*x^3)^(8/3), x)

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sympy [C]  time = 117.70, size = 78, normalized size = 0.83 \[ \frac {c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {8}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {8}{3}} \Gamma \left (\frac {4}{3}\right )} + \frac {d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {8}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {8}{3}} \Gamma \left (\frac {7}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**3+c)/(b*x**3+a)**(8/3),x)

[Out]

c*x*gamma(1/3)*hyper((1/3, 8/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(8/3)*gamma(4/3)) + d*x**4*gamma(4/3)
*hyper((4/3, 8/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(8/3)*gamma(7/3))

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