3.1.84 \(\int \frac {(c+d x^3)^2}{(a+b x^3)^{8/3}} \, dx\) [84]

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

[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 = 0.06, 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 = {424, 393, 252, 251} \begin {gather*} \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}} \end {gather*}

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 251

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 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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 393

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 424

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 = 13.13, size = 171, normalized size = 1.16 \begin {gather*} \frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} \Gamma \left (\frac {2}{3}\right ) \left (5 a \left (14 c^2+7 c d x^3+2 d^2 x^6\right ) \, _2F_1\left (\frac {1}{3},\frac {8}{3};\frac {10}{3};-\frac {b x^3}{a}\right )-2 b x^3 \left (11 c^2+16 c d x^3+5 d^2 x^6\right ) \, _2F_1\left (\frac {4}{3},\frac {11}{3};\frac {13}{3};-\frac {b x^3}{a}\right )-6 b x^3 \left (c+d x^3\right )^2 \, _3F_2\left (\frac {4}{3},2,\frac {11}{3};1,\frac {13}{3};-\frac {b x^3}{a}\right )\right )}{63 a^3 \left (a+b x^3\right )^{2/3} \Gamma \left (\frac {8}{3}\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(1 + (b*x^3)/a)^(2/3)*Gamma[2/3]*(5*a*(14*c^2 + 7*c*d*x^3 + 2*d^2*x^6)*Hypergeometric2F1[1/3, 8/3, 10/3, -(
(b*x^3)/a)] - 2*b*x^3*(11*c^2 + 16*c*d*x^3 + 5*d^2*x^6)*Hypergeometric2F1[4/3, 11/3, 13/3, -((b*x^3)/a)] - 6*b
*x^3*(c + d*x^3)^2*HypergeometricPFQ[{4/3, 2, 11/3}, {1, 13/3}, -((b*x^3)/a)]))/(63*a^3*(a + b*x^3)^(2/3)*Gamm
a[8/3])

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Maple [F]
time = 0.05, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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