3.1.37 \(\int \frac {(c+d x)^3}{(a+b x^3)^{2/3}} \, dx\) [37]

Optimal. Leaf size=187 \[ \frac {3 c d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b}-\frac {\sqrt {3} c^2 d \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{b^{2/3}}+\frac {\left (2 b c^3-a d^3\right ) x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 b \left (a+b x^3\right )^{2/3}}-\frac {3 c^2 d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3}} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {1902, 1900, 267, 1907, 252, 251, 337} \begin {gather*} -\frac {\sqrt {3} c^2 d \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{b^{2/3}}-\frac {3 c^2 d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3}}+\frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} \left (2 b c^3-a d^3\right ) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 b \left (a+b x^3\right )^{2/3}}+\frac {3 c d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3/(a + b*x^3)^(2/3),x]

[Out]

(3*c*d^2*(a + b*x^3)^(1/3))/b + (d^3*x*(a + b*x^3)^(1/3))/(2*b) - (Sqrt[3]*c^2*d*ArcTan[(1 + (2*b^(1/3)*x)/(a
+ b*x^3)^(1/3))/Sqrt[3]])/b^(2/3) + ((2*b*c^3 - a*d^3)*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3
, -((b*x^3)/a)])/(2*b*(a + b*x^3)^(2/3)) - (3*c^2*d*Log[b^(1/3)*x - (a + b*x^3)^(1/3)])/(2*b^(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 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 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 1900

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[Coeff[Pq, x, n - 1], Int[x^(n - 1)*(a + b*x^n)^p, x
], x] + Int[ExpandToSum[Pq - Coeff[Pq, x, n - 1]*x^(n - 1), x]*(a + b*x^n)^p, x] /; FreeQ[{a, b, p}, x] && Pol
yQ[Pq, x] && IGtQ[n, 0] && Expon[Pq, x] == n - 1

Rule 1902

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, With[{Pqq = Coeff[Pq, x, q]}, D
ist[1/(b*(q + n*p + 1)), Int[ExpandToSum[b*(q + n*p + 1)*(Pq - Pqq*x^q) - a*Pqq*(q - n + 1)*x^(q - n), x]*(a +
b*x^n)^p, x], x] + Simp[Pqq*x^(q - n + 1)*((a + b*x^n)^(p + 1)/(b*(q + n*p + 1))), x]] /; NeQ[q + n*p + 1, 0]
&& q - n >= 0 && (IntegerQ[2*p] || IntegerQ[p + (q + 1)/(2*n)])] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && IG
tQ[n, 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)^3}{\left (a+b x^3\right )^{2/3}} \, dx &=\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b}+\frac {\int \frac {2 b c^3-a d^3+6 b c^2 d x+6 b c d^2 x^2}{\left (a+b x^3\right )^{2/3}} \, dx}{2 b}\\ &=\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b}+\frac {\int \frac {2 b c^3-a d^3+6 b c^2 d x}{\left (a+b x^3\right )^{2/3}} \, dx}{2 b}+\left (3 c d^2\right ) \int \frac {x^2}{\left (a+b x^3\right )^{2/3}} \, dx\\ &=\frac {3 c d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b}+\frac {\int \left (\frac {2 b c^3 \left (1-\frac {a d^3}{2 b c^3}\right )}{\left (a+b x^3\right )^{2/3}}+\frac {6 b c^2 d x}{\left (a+b x^3\right )^{2/3}}\right ) \, dx}{2 b}\\ &=\frac {3 c d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b}+\left (3 c^2 d\right ) \int \frac {x}{\left (a+b x^3\right )^{2/3}} \, dx+\frac {\left (2 b c^3-a d^3\right ) \int \frac {1}{\left (a+b x^3\right )^{2/3}} \, dx}{2 b}\\ &=\frac {3 c d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b}+\left (3 c^2 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 (\left (2 b c^3-a d^3\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}{2 b \left (a+b x^3\right )^{2/3}}\\ &=\frac {3 c d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b}+\frac {\left (2 b c^3-a d^3\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 )}{2 b \left (a+b x^3\right )^{2/3}}+\frac {\left (c^2 d\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt [3]{b} x} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{\sqrt [3]{b}}-\frac {\left (c^2 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 )}{\sqrt [3]{b}}\\ &=\frac {3 c d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b}+\frac {\left (2 b c^3-a d^3\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 )}{2 b \left (a+b x^3\right )^{2/3}}-\frac {c^2 d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{b^{2/3}}+\frac {\left (c^2 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 )}{2 b^{2/3}}-\frac {\left (3 c^2 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 )}{2 \sqrt [3]{b}}\\ &=\frac {3 c d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b}+\frac {\left (2 b c^3-a d^3\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 )}{2 b \left (a+b x^3\right )^{2/3}}-\frac {c^2 d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{b^{2/3}}+\frac {c^2 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 )}{2 b^{2/3}}+\frac {\left (3 c^2 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 {3 c d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b}-\frac {\sqrt {3} c^2 d \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{b^{2/3}}+\frac {\left (2 b c^3-a d^3\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 )}{2 b \left (a+b x^3\right )^{2/3}}-\frac {c^2 d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{b^{2/3}}+\frac {c^2 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 )}{2 b^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 10.08, size = 145, normalized size = 0.78 \begin {gather*} \frac {4 b c^3 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 (6 b c^2 x^2 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {b x^3}{a+b x^3}\right )+d \left (12 c \left (a+b x^3\right )+b 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 )}{4 b \left (a+b x^3\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*c^3*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)] + d*(6*b*c^2*x^2*Hypergeometri
c2F1[2/3, 1, 5/3, (b*x^3)/(a + b*x^3)] + d*(12*c*(a + b*x^3) + b*d*x^4*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1
[2/3, 4/3, 7/3, -((b*x^3)/a)])))/(4*b*(a + b*x^3)^(2/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [A]
time = 2.12, size = 153, normalized size = 0.82 \begin {gather*} 3 c 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^{3} 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 {c^{2} 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 )}}{a^{\frac {2}{3}} \Gamma \left (\frac {5}{3}\right )} + \frac {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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

[Out]

int((c + d*x)^3/(a + b*x^3)^(2/3), x)

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