3.84 \(\int \frac {(c+d x^3)^2}{(a+b x^3)^{8/3}} \, dx\)

Optimal. Leaf size=147 \[ \frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} \left (2 a^2 d^2+a b c d+2 b^2 c^2\right ) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{5 a^2 b^2 \left (a+b x^3\right )^{2/3}}+\frac {2 x \left (\frac {c^2}{a^2}-\frac {d^2}{b^2}\right )}{5 \left (a+b x^3\right )^{2/3}}+\frac {x \left (c+d x^3\right ) (b c-a d)}{5 a b \left (a+b x^3\right )^{5/3}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 147, 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 = {413, 385, 246, 245} \[ \frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} \left (2 a^2 d^2+a b c d+2 b^2 c^2\right ) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{5 a^2 b^2 \left (a+b x^3\right )^{2/3}}+\frac {2 x \left (\frac {c^2}{a^2}-\frac {d^2}{b^2}\right )}{5 \left (a+b x^3\right )^{2/3}}+\frac {x \left (c+d x^3\right ) (b c-a d)}{5 a b \left (a+b x^3\right )^{5/3}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)^2/(a + b*x^3)^(8/3),x]

[Out]

(2*(c^2/a^2 - d^2/b^2)*x)/(5*(a + b*x^3)^(2/3)) + ((b*c - a*d)*x*(c + d*x^3))/(5*a*b*(a + b*x^3)^(5/3)) + ((2*
b^2*c^2 + 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*a^2*
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 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 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[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 )^{8/3}} \, dx &=\frac {(b c-a d) x \left (c+d x^3\right )}{5 a b \left (a+b x^3\right )^{5/3}}+\frac {\int \frac {c (4 b c+a d)+d (b c+4 a d) x^3}{\left (a+b x^3\right )^{5/3}} \, dx}{5 a b}\\ &=\frac {2 \left (\frac {c^2}{a^2}-\frac {d^2}{b^2}\right ) x}{5 \left (a+b x^3\right )^{2/3}}+\frac {(b c-a d) x \left (c+d x^3\right )}{5 a b \left (a+b x^3\right )^{5/3}}+\frac {\left (2 b^2 c^2+a b c d+2 a^2 d^2\right ) \int \frac {1}{\left (a+b x^3\right )^{2/3}} \, dx}{5 a^2 b^2}\\ &=\frac {2 \left (\frac {c^2}{a^2}-\frac {d^2}{b^2}\right ) x}{5 \left (a+b x^3\right )^{2/3}}+\frac {(b c-a d) x \left (c+d x^3\right )}{5 a b \left (a+b x^3\right )^{5/3}}+\frac {\left (\left (2 b^2 c^2+a b c d+2 a^2 d^2\right ) \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}{5 a^2 b^2 \left (a+b x^3\right )^{2/3}}\\ &=\frac {2 \left (\frac {c^2}{a^2}-\frac {d^2}{b^2}\right ) x}{5 \left (a+b x^3\right )^{2/3}}+\frac {(b c-a d) x \left (c+d x^3\right )}{5 a b \left (a+b x^3\right )^{5/3}}+\frac {\left (2 b^2 c^2+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 a^2 b^2 \left (a+b x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [A]  time = 5.16, size = 128, normalized size = 0.87 \[ \frac {x \left (\left (a+b x^3\right ) \left (\frac {b x^3}{a}+1\right )^{2/3} \left (2 a^2 d^2+a b c d+2 b^2 c^2\right ) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )+\left (a+b x^3\right ) \left (-3 a^2 d^2+a b c d+2 b^2 c^2\right )+a (b c-a d)^2\right )}{5 a^2 b^2 \left (a+b x^3\right )^{5/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^3)^2/(a + b*x^3)^(8/3),x]

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d^{2} x^{6} + 2 \, c d x^{3} + c^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{b^{3} x^{9} + 3 \, a b^{2} x^{6} + 3 \, a^{2} b x^{3} + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^2/(b*x^3+a)^(8/3),x, algorithm="fricas")

[Out]

integral((d^2*x^6 + 2*c*d*x^3 + c^2)*(b*x^3 + a)^(1/3)/(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 \[ \int \frac {{\left (d x^{3} + c\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {8}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^2/(b*x^3+a)^(8/3),x, algorithm="giac")

[Out]

integrate((d*x^3 + c)^2/(b*x^3 + a)^(8/3), x)

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maple [F]  time = 0.60, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \,x^{3}+c \right )^{2}}{\left (b \,x^{3}+a \right )^{\frac {8}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)^2/(b*x^3+a)^(8/3),x)

[Out]

int((d*x^3+c)^2/(b*x^3+a)^(8/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{3} + c\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {8}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^2/(b*x^3+a)^(8/3),x, algorithm="maxima")

[Out]

integrate((d*x^3 + c)^2/(b*x^3 + a)^(8/3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d\,x^3+c\right )}^2}{{\left (b\,x^3+a\right )}^{8/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^3)^2/(a + b*x^3)^(8/3),x)

[Out]

int((c + d*x^3)^2/(a + b*x^3)^(8/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c + d x^{3}\right )^{2}}{\left (a + b x^{3}\right )^{\frac {8}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**3+c)**2/(b*x**3+a)**(8/3),x)

[Out]

Integral((c + d*x**3)**2/(a + b*x**3)**(8/3), x)

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