3.1.36 \(\int \frac {(c+d x)^4}{(a+b x^3)^{2/3}} \, dx\) [36]

Optimal. Leaf size=306 \[ \frac {6 c^2 d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^4 x^2 \sqrt [3]{a+b x^3}}{3 b}-\frac {4 c^3 d \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}+\frac {2 a d^4 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{5/3}}+\frac {c^4 x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}+\frac {c d^3 x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}-\frac {2 c^3 d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{b^{2/3}}+\frac {a d^4 \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{3 b^{5/3}} \]

[Out]

6*c^2*d^2*(b*x^3+a)^(1/3)/b+1/3*d^4*x^2*(b*x^3+a)^(1/3)/b+c^4*x*(1+b*x^3/a)^(2/3)*hypergeom([1/3, 2/3],[4/3],-
b*x^3/a)/(b*x^3+a)^(2/3)+c*d^3*x^4*(1+b*x^3/a)^(2/3)*hypergeom([2/3, 4/3],[7/3],-b*x^3/a)/(b*x^3+a)^(2/3)-2*c^
3*d*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(2/3)+1/3*a*d^4*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(5/3)-4/3*c^3*d*arctan(1/3
*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/b^(2/3)*3^(1/2)+2/9*a*d^4*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))
*3^(1/2))/b^(5/3)*3^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 306, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {1907, 252, 251, 337, 267, 372, 371, 327} \begin {gather*} -\frac {4 c^3 d \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}+\frac {2 a d^4 \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{5/3}}-\frac {2 c^3 d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{b^{2/3}}+\frac {a d^4 \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{3 b^{5/3}}+\frac {c^4 x \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}+\frac {6 c^2 d^2 \sqrt [3]{a+b x^3}}{b}+\frac {c d^3 x^4 \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}+\frac {d^4 x^2 \sqrt [3]{a+b x^3}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^4/(a + b*x^3)^(2/3),x]

[Out]

(6*c^2*d^2*(a + b*x^3)^(1/3))/b + (d^4*x^2*(a + b*x^3)^(1/3))/(3*b) - (4*c^3*d*ArcTan[(1 + (2*b^(1/3)*x)/(a +
b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(2/3)) + (2*a*d^4*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(3
*Sqrt[3]*b^(5/3)) + (c^4*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(a + b*x^3)^(
2/3) + (c*d^3*x^4*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[2/3, 4/3, 7/3, -((b*x^3)/a)])/(a + b*x^3)^(2/3) - (2
*c^3*d*Log[b^(1/3)*x - (a + b*x^3)^(1/3)])/b^(2/3) + (a*d^4*Log[b^(1/3)*x - (a + b*x^3)^(1/3)])/(3*b^(5/3))

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 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 327

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

Rule 337

Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Simp[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(
1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 1907

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n])

Rubi steps

\begin {align*} \int \frac {(c+d x)^4}{\left (a+b x^3\right )^{2/3}} \, dx &=\int \left (\frac {c^4}{\left (a+b x^3\right )^{2/3}}+\frac {4 c^3 d x}{\left (a+b x^3\right )^{2/3}}+\frac {6 c^2 d^2 x^2}{\left (a+b x^3\right )^{2/3}}+\frac {4 c d^3 x^3}{\left (a+b x^3\right )^{2/3}}+\frac {d^4 x^4}{\left (a+b x^3\right )^{2/3}}\right ) \, dx\\ &=c^4 \int \frac {1}{\left (a+b x^3\right )^{2/3}} \, dx+\left (4 c^3 d\right ) \int \frac {x}{\left (a+b x^3\right )^{2/3}} \, dx+\left (6 c^2 d^2\right ) \int \frac {x^2}{\left (a+b x^3\right )^{2/3}} \, dx+\left (4 c d^3\right ) \int \frac {x^3}{\left (a+b x^3\right )^{2/3}} \, dx+d^4 \int \frac {x^4}{\left (a+b x^3\right )^{2/3}} \, dx\\ &=\frac {6 c^2 d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^4 x^2 \sqrt [3]{a+b x^3}}{3 b}+\left (4 c^3 d\right ) \text {Subst}\left (\int \frac {x}{1-b x^3} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )-\frac {\left (2 a d^4\right ) \int \frac {x}{\left (a+b x^3\right )^{2/3}} \, dx}{3 b}+\frac {\left (c^4 \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}{\left (a+b x^3\right )^{2/3}}+\frac {\left (4 c d^3 \left (1+\frac {b x^3}{a}\right )^{2/3}\right ) \int \frac {x^3}{\left (1+\frac {b x^3}{a}\right )^{2/3}} \, dx}{\left (a+b x^3\right )^{2/3}}\\ &=\frac {6 c^2 d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^4 x^2 \sqrt [3]{a+b x^3}}{3 b}+\frac {c^4 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 )}{\left (a+b x^3\right )^{2/3}}+\frac {c d^3 x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}+\frac {\left (4 c^3 d\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt [3]{b} x} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{3 \sqrt [3]{b}}-\frac {\left (4 c^3 d\right ) \text {Subst}\left (\int \frac {1-\sqrt [3]{b} x}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{3 \sqrt [3]{b}}-\frac {\left (2 a d^4\right ) \text {Subst}\left (\int \frac {x}{1-b x^3} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{3 b}\\ &=\frac {6 c^2 d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^4 x^2 \sqrt [3]{a+b x^3}}{3 b}+\frac {c^4 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 )}{\left (a+b x^3\right )^{2/3}}+\frac {c d^3 x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}-\frac {4 c^3 d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}+\frac {\left (2 c^3 d\right ) \text {Subst}\left (\int \frac {\sqrt [3]{b}+2 b^{2/3} x}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}-\frac {\left (2 c^3 d\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{\sqrt [3]{b}}-\frac {\left (2 a d^4\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt [3]{b} x} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{4/3}}+\frac {\left (2 a d^4\right ) \text {Subst}\left (\int \frac {1-\sqrt [3]{b} x}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{4/3}}\\ &=\frac {6 c^2 d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^4 x^2 \sqrt [3]{a+b x^3}}{3 b}+\frac {c^4 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 )}{\left (a+b x^3\right )^{2/3}}+\frac {c d^3 x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}-\frac {4 c^3 d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}+\frac {2 a d^4 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{5/3}}+\frac {2 c^3 d \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}+\frac {\left (4 c^3 d\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{b^{2/3}}-\frac {\left (a d^4\right ) \text {Subst}\left (\int \frac {\sqrt [3]{b}+2 b^{2/3} x}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{5/3}}+\frac {\left (a d^4\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{4/3}}\\ &=\frac {6 c^2 d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^4 x^2 \sqrt [3]{a+b x^3}}{3 b}-\frac {4 c^3 d \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}+\frac {c^4 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 )}{\left (a+b x^3\right )^{2/3}}+\frac {c d^3 x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}-\frac {4 c^3 d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}+\frac {2 a d^4 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{5/3}}+\frac {2 c^3 d \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}-\frac {a d^4 \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{5/3}}-\frac {\left (2 a d^4\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{5/3}}\\ &=\frac {6 c^2 d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^4 x^2 \sqrt [3]{a+b x^3}}{3 b}-\frac {4 c^3 d \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}+\frac {2 a d^4 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{5/3}}+\frac {c^4 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 )}{\left (a+b x^3\right )^{2/3}}+\frac {c d^3 x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}-\frac {4 c^3 d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}+\frac {2 a d^4 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{5/3}}+\frac {2 c^3 d \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}-\frac {a d^4 \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{5/3}}\\ \end {align*}

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Mathematica [A]
time = 10.13, size = 166, normalized size = 0.54 \begin {gather*} \frac {3 b c^4 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 )+d \left (\left (6 b c^3-a d^3\right ) x^2 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {b x^3}{a+b x^3}\right )+d \left (\left (18 c^2+d^2 x^2\right ) \left (a+b x^3\right )+3 b c d x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^3}{a}\right )\right )\right )}{3 b \left (a+b x^3\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*c^4*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)] + d*((6*b*c^3 - a*d^3)*x^2*Hyp
ergeometric2F1[2/3, 1, 5/3, (b*x^3)/(a + b*x^3)] + d*((18*c^2 + d^2*x^2)*(a + b*x^3) + 3*b*c*d*x^4*(1 + (b*x^3
)/a)^(2/3)*Hypergeometric2F1[2/3, 4/3, 7/3, -((b*x^3)/a)])))/(3*b*(a + b*x^3)^(2/3))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (d x +c \right )^{4}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^4/(b*x^3+a)^(2/3),x)

[Out]

int((d*x+c)^4/(b*x^3+a)^(2/3),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+c)^4/(b*x^3+a)^(2/3),x, algorithm="maxima")

[Out]

integrate((d*x + c)^4/(b*x^3 + a)^(2/3), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^4/(b*x^3+a)^(2/3),x, algorithm="fricas")

[Out]

Timed out

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Sympy [A]
time = 2.73, size = 204, normalized size = 0.67 \begin {gather*} 6 c^{2} d^{2} \left (\begin {cases} \frac {x^{3}}{3 a^{\frac {2}{3}}} & \text {for}\: b = 0 \\\frac {\sqrt [3]{a + b x^{3}}}{b} & \text {otherwise} \end {cases}\right ) + \frac {c^{4} 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 {4 c^{3} d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {5}{3}\right )} + \frac {4 c d^{3} 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^{4} x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {8}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**4/(b*x**3+a)**(2/3),x)

[Out]

6*c**2*d**2*Piecewise((x**3/(3*a**(2/3)), Eq(b, 0)), ((a + b*x**3)**(1/3)/b, True)) + c**4*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)) + 4*c**3*d*x**2*gamma(2/3)*hyper((2/3, 2
/3), (5/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(5/3)) + 4*c*d**3*x**4*gamma(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**4*x**5*gamma(5/3)*hyper((2/3, 5/3), (8/3,), b*x**
3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(8/3))

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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+c)^4/(b*x^3+a)^(2/3),x, algorithm="giac")

[Out]

integrate((d*x + c)^4/(b*x^3 + a)^(2/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c+d\,x\right )}^4}{{\left (b\,x^3+a\right )}^{2/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^4/(a + b*x^3)^(2/3),x)

[Out]

int((c + d*x)^4/(a + b*x^3)^(2/3), x)

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