Integrand size = 19, antiderivative size = 187 \[ \int \frac {(c+d x)^3}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {3 c d^2 \sqrt [3]{a+b x^3}}{b}+\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b}-\frac {\sqrt {3} c^2 d \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{b^{2/3}}+\frac {\left (2 b c^3-a d^3\right ) x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{2 b \left (a+b x^3\right )^{2/3}}-\frac {3 c^2 d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3}} \]
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)
Time = 10.08 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.78 \[ \int \frac {(c+d x)^3}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {4 b c^3 x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )+d \left (6 b c^2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},\frac {b x^3}{a+b x^3}\right )+d \left (12 c \left (a+b x^3\right )+b d x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {7}{3},-\frac {b x^3}{a}\right )\right )\right )}{4 b \left (a+b x^3\right )^{2/3}} \]
(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*Hypergeometric2F1[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))
Time = 0.45 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2427, 2425, 793, 2432, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(c+d x)^3}{\left (a+b x^3\right )^{2/3}} \, dx\) |
| \(\Big \downarrow \) 2427 |
| \(\displaystyle \frac {\int \frac {2 b c^3+6 b d x c^2+6 b d^2 x^2 c-a d^3}{\left (b x^3+a\right )^{2/3}}dx}{2 b}+\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b}\) |
| \(\Big \downarrow \) 2425 |
| \(\displaystyle \frac {\int \frac {2 b c^3+6 b d x c^2-a d^3}{\left (b x^3+a\right )^{2/3}}dx+6 b c d^2 \int \frac {x^2}{\left (b x^3+a\right )^{2/3}}dx}{2 b}+\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b}\) |
| \(\Big \downarrow \) 793 |
| \(\displaystyle \frac {\int \frac {2 b c^3+6 b d x c^2-a d^3}{\left (b x^3+a\right )^{2/3}}dx+6 c d^2 \sqrt [3]{a+b x^3}}{2 b}+\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b}\) |
| \(\Big \downarrow \) 2432 |
| \(\displaystyle \frac {\int \left (\frac {2 b \left (1-\frac {a d^3}{2 b c^3}\right ) c^3}{\left (b x^3+a\right )^{2/3}}+\frac {6 b d x c^2}{\left (b x^3+a\right )^{2/3}}\right )dx+6 c d^2 \sqrt [3]{a+b x^3}}{2 b}+\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-2 \sqrt {3} \sqrt [3]{b} c^2 d \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )+\frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} \left (2 b c^3-a d^3\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}-3 \sqrt [3]{b} c^2 d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )+6 c d^2 \sqrt [3]{a+b x^3}}{2 b}+\frac {d^3 x \sqrt [3]{a+b x^3}}{2 b}\) |
(d^3*x*(a + b*x^3)^(1/3))/(2*b) + (6*c*d^2*(a + b*x^3)^(1/3) - 2*Sqrt[3]*b ^(1/3)*c^2*d*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[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)])/(a + b*x^3)^(2/3) - 3*b^(1/3)*c^2*d*Log[b^(1/3)*x - (a + b*x^3 )^(1/3)])/(2*b)
3.1.37.3.1 Defintions of rubi rules used
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]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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] && PolyQ[P q, x] && IGtQ[n, 0] && Expon[Pq, x] == n - 1
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{q = Expon[Pq, x ]}, With[{Pqq = Coeff[Pq, x, q]}, Simp[Pqq*x^(q - n + 1)*((a + b*x^n)^(p + 1)/(b*(q + n*p + 1))), x] + Simp[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]] /; 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] && IGtQ[n, 0]
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] || Poly Q[Pq, x^n])
\[\int \frac {\left (d x +c \right )^{3}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x\]
\[ \int \frac {(c+d x)^3}{\left (a+b x^3\right )^{2/3}} \, dx=\int { \frac {{\left (d x + c\right )}^{3}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}} \,d x } \]
Time = 1.72 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.82 \[ \int \frac {(c+d x)^3}{\left (a+b x^3\right )^{2/3}} \, dx=3 c d^{2} \left (\begin {cases} \frac {x^{3}}{3 a^{\frac {2}{3}}} & \text {for}\: b = 0 \\\frac {\sqrt [3]{a + b x^{3}}}{b} & \text {otherwise} \end {cases}\right ) + \frac {c^{3} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {4}{3}\right )} + \frac {c^{2} d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{a^{\frac {2}{3}} \Gamma \left (\frac {5}{3}\right )} + \frac {d^{3} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {7}{3}\right )} \]
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))
\[ \int \frac {(c+d x)^3}{\left (a+b x^3\right )^{2/3}} \, dx=\int { \frac {{\left (d x + c\right )}^{3}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}} \,d x } \]
\[ \int \frac {(c+d x)^3}{\left (a+b x^3\right )^{2/3}} \, dx=\int { \frac {{\left (d x + c\right )}^{3}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}} \,d x } \]
Timed out. \[ \int \frac {(c+d x)^3}{\left (a+b x^3\right )^{2/3}} \, dx=\int \frac {{\left (c+d\,x\right )}^3}{{\left (b\,x^3+a\right )}^{2/3}} \,d x \]