3.82 \(\int \frac {(c+d x^3)^2}{(a+b x^3)^{2/3}} \, dx\)

Optimal. Leaf size=132 \[ \frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} \left (2 a^2 d^2-5 a b c d+5 b^2 c^2\right ) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{5 b^2 \left (a+b x^3\right )^{2/3}}+\frac {2 d x \sqrt [3]{a+b x^3} (2 b c-a d)}{5 b^2}+\frac {d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {416, 388, 246, 245} \[ \frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} \left (2 a^2 d^2-5 a b c d+5 b^2 c^2\right ) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{5 b^2 \left (a+b x^3\right )^{2/3}}+\frac {2 d x \sqrt [3]{a+b x^3} (2 b c-a d)}{5 b^2}+\frac {d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)^2/(a + b*x^3)^(2/3),x]

[Out]

(2*d*(2*b*c - a*d)*x*(a + b*x^3)^(1/3))/(5*b^2) + (d*x*(a + b*x^3)^(1/3)*(c + d*x^3))/(5*b) + ((5*b^2*c^2 - 5*
a*b*c*d + 2*a^2*d^2)*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(5*b^2*(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 388

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

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rubi steps

\begin {align*} \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx &=\frac {d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b}+\frac {\int \frac {c (5 b c-a d)+4 d (2 b c-a d) x^3}{\left (a+b x^3\right )^{2/3}} \, dx}{5 b}\\ &=\frac {2 d (2 b c-a d) x \sqrt [3]{a+b x^3}}{5 b^2}+\frac {d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b}-\frac {(4 a d (2 b c-a d)-2 b c (5 b c-a d)) \int \frac {1}{\left (a+b x^3\right )^{2/3}} \, dx}{10 b^2}\\ &=\frac {2 d (2 b c-a d) x \sqrt [3]{a+b x^3}}{5 b^2}+\frac {d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b}-\frac {\left ((4 a d (2 b c-a d)-2 b c (5 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}{10 b^2 \left (a+b x^3\right )^{2/3}}\\ &=\frac {2 d (2 b c-a d) x \sqrt [3]{a+b x^3}}{5 b^2}+\frac {d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b}+\frac {\left (5 b^2 c^2-5 a b c d+2 a^2 d^2\right ) 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 )}{5 b^2 \left (a+b x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [A]  time = 5.14, size = 104, normalized size = 0.79 \[ \frac {x \left (\left (\frac {b x^3}{a}+1\right )^{2/3} \left (2 a^2 d^2-5 a b c d+5 b^2 c^2\right ) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )-d \left (a+b x^3\right ) \left (2 a d-b \left (5 c+d x^3\right )\right )\right )}{5 b^2 \left (a+b x^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^3)^2/(a + b*x^3)^(2/3),x]

[Out]

(x*(-(d*(a + b*x^3)*(2*a*d - b*(5*c + d*x^3))) + (5*b^2*c^2 - 5*a*b*c*d + 2*a^2*d^2)*(1 + (b*x^3)/a)^(2/3)*Hyp
ergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)]))/(5*b^2*(a + b*x^3)^(2/3))

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d^{2} x^{6} + 2 \, c d x^{3} + c^{2}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{3} + c\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.39, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \,x^{3}+c \right )^{2}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)^2/(b*x^3+a)^(2/3),x)

[Out]

int((d*x^3+c)^2/(b*x^3+a)^(2/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{3} + c\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d\,x^3+c\right )}^2}{{\left (b\,x^3+a\right )}^{2/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^3)^2/(a + b*x^3)^(2/3),x)

[Out]

int((c + d*x^3)^2/(a + b*x^3)^(2/3), x)

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sympy [C]  time = 4.27, size = 126, normalized size = 0.95 \[ \frac {c^{2} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {4}{3}\right )} + \frac {2 c d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {7}{3}\right )} + \frac {d^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {10}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**2*x*gamma(1/3)*hyper((1/3, 2/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(4/3)) + 2*c*d*x**4*gam
ma(4/3)*hyper((2/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(7/3)) + d**2*x**7*gamma(7/3)*hyp
er((2/3, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(10/3))

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