3.68 \(\int \frac{c+d x^3}{(a+b x^3)^{5/3}} \, dx\)

Optimal. Leaf size=93 \[ \frac{x \left (\frac{b x^3}{a}+1\right )^{2/3} (a d+b c) \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{2 a b \left (a+b x^3\right )^{2/3}}+\frac{x (b c-a d)}{2 a b \left (a+b x^3\right )^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.0252296, antiderivative size = 93, normalized size of antiderivative = 1., 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} \[ \frac{x \left (\frac{b x^3}{a}+1\right )^{2/3} (a d+b c) \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{2 a b \left (a+b x^3\right )^{2/3}}+\frac{x (b c-a d)}{2 a b \left (a+b x^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)/(a + b*x^3)^(5/3),x]

[Out]

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

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])

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 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])

Rubi steps

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

Mathematica [A]  time = 0.0342796, size = 66, normalized size = 0.71 \[ \frac{x \left (\frac{b x^3}{a}+1\right )^{2/3} (a d+b c) \, _2F_1\left (\frac{1}{3},\frac{5}{3};\frac{4}{3};-\frac{b x^3}{a}\right )-a d x}{a b \left (a+b x^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^3)/(a + b*x^3)^(5/3),x]

[Out]

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

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Maple [F]  time = 0.224, size = 0, normalized size = 0. \begin{align*} \int{(d{x}^{3}+c) \left ( b{x}^{3}+a \right ) ^{-{\frac{5}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)/(b*x^3+a)^(5/3),x)

[Out]

int((d*x^3+c)/(b*x^3+a)^(5/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d x^{3} + c}{{\left (b x^{3} + a\right )}^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)/(b*x^3+a)^(5/3),x, algorithm="maxima")

[Out]

integrate((d*x^3 + c)/(b*x^3 + a)^(5/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (d x^{3} + c\right )}}{b^{2} x^{6} + 2 \, a b x^{3} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)/(b*x^3+a)^(5/3),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 14.1314, size = 78, normalized size = 0.84 \begin{align*} \frac{c x \Gamma \left (\frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{5}{3} \\ \frac{4}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac{5}{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{5}{3} \\ \frac{7}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac{5}{3}} \Gamma \left (\frac{7}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**3+c)/(b*x**3+a)**(5/3),x)

[Out]

c*x*gamma(1/3)*hyper((1/3, 5/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(5/3)*gamma(4/3)) + d*x**4*gamma(4/3)
*hyper((4/3, 5/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(5/3)*gamma(7/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d x^{3} + c}{{\left (b x^{3} + a\right )}^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)/(b*x^3+a)^(5/3),x, algorithm="giac")

[Out]

integrate((d*x^3 + c)/(b*x^3 + a)^(5/3), x)