Integrand size = 13, antiderivative size = 35 \[ \int \left (a x+b x^2\right )^{4/3} \, dx=\frac {3 \left (a x+b x^2\right )^{7/3} \operatorname {Hypergeometric2F1}\left (1,\frac {14}{3},\frac {10}{3},-\frac {b x}{a}\right )}{7 a} \] Output:
3/7*(b*x^2+a*x)^(7/3)*hypergeom([1, 14/3],[10/3],-b*x/a)/a
Time = 10.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.37 \[ \int \left (a x+b x^2\right )^{4/3} \, dx=\frac {3 a x^2 \sqrt [3]{x (a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {7}{3},\frac {10}{3},-\frac {b x}{a}\right )}{7 \sqrt [3]{1+\frac {b x}{a}}} \] Input:
Integrate[(a*x + b*x^2)^(4/3),x]
Output:
(3*a*x^2*(x*(a + b*x))^(1/3)*Hypergeometric2F1[-4/3, 7/3, 10/3, -((b*x)/a) ])/(7*(1 + (b*x)/a)^(1/3))
Leaf count is larger than twice the leaf count of optimal. \(350\) vs. \(2(35)=70\).
Time = 0.61 (sec) , antiderivative size = 350, normalized size of antiderivative = 10.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {1087, 1087, 1093, 1090, 234, 760}
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.
\(\displaystyle \int \left (a x+b x^2\right )^{4/3} \, dx\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {3 (a+2 b x) \left (a x+b x^2\right )^{4/3}}{22 b}-\frac {2 a^2 \int \sqrt [3]{b x^2+a x}dx}{11 b}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {3 (a+2 b x) \left (a x+b x^2\right )^{4/3}}{22 b}-\frac {2 a^2 \left (\frac {3 (a+2 b x) \sqrt [3]{a x+b x^2}}{10 b}-\frac {a^2 \int \frac {1}{\left (b x^2+a x\right )^{2/3}}dx}{10 b}\right )}{11 b}\) |
\(\Big \downarrow \) 1093 |
\(\displaystyle \frac {3 (a+2 b x) \left (a x+b x^2\right )^{4/3}}{22 b}-\frac {2 a^2 \left (\frac {3 (a+2 b x) \sqrt [3]{a x+b x^2}}{10 b}-\frac {a^2 \left (-\frac {b \left (a x+b x^2\right )}{a^2}\right )^{2/3} \int \frac {1}{\left (-\frac {b^2 x^2}{a^2}-\frac {b x}{a}\right )^{2/3}}dx}{10 b \left (a x+b x^2\right )^{2/3}}\right )}{11 b}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {3 (a+2 b x) \left (a x+b x^2\right )^{4/3}}{22 b}-\frac {2 a^2 \left (\frac {a^4 \left (-\frac {b \left (a x+b x^2\right )}{a^2}\right )^{2/3} \int \frac {1}{\left (1-\frac {a^2 \left (-\frac {2 x b^2}{a^2}-\frac {b}{a}\right )^2}{b^2}\right )^{2/3}}d\left (-\frac {2 x b^2}{a^2}-\frac {b}{a}\right )}{5\ 2^{2/3} b^3 \left (a x+b x^2\right )^{2/3}}+\frac {3 (a+2 b x) \sqrt [3]{a x+b x^2}}{10 b}\right )}{11 b}\) |
\(\Big \downarrow \) 234 |
\(\displaystyle \frac {3 (a+2 b x) \left (a x+b x^2\right )^{4/3}}{22 b}-\frac {2 a^2 \left (\frac {3 (a+2 b x) \sqrt [3]{a x+b x^2}}{10 b}-\frac {3 a^2 \sqrt {-\frac {a^2 \left (-\frac {2 b^2 x}{a^2}-\frac {b}{a}\right )^2}{b^2}} \left (-\frac {b \left (a x+b x^2\right )}{a^2}\right )^{2/3} \int \frac {1}{\sqrt {-\frac {a^2 \left (-\frac {2 x b^2}{a^2}-\frac {b}{a}\right )^2}{b^2}}}d\sqrt [3]{1-\frac {a^2 \left (-\frac {2 x b^2}{a^2}-\frac {b}{a}\right )^2}{b^2}}}{10\ 2^{2/3} b \left (-\frac {2 b^2 x}{a^2}-\frac {b}{a}\right ) \left (a x+b x^2\right )^{2/3}}\right )}{11 b}\) |
\(\Big \downarrow \) 760 |
\(\displaystyle \frac {3 (a+2 b x) \left (a x+b x^2\right )^{4/3}}{22 b}-\frac {2 a^2 \left (\frac {3^{3/4} \sqrt {2-\sqrt {3}} a^2 \left (\frac {2 b^2 x}{a^2}+\frac {b}{a}+1\right ) \left (-\frac {b \left (a x+b x^2\right )}{a^2}\right )^{2/3} \sqrt {\frac {\left (-\frac {2 b^2 x}{a^2}-\frac {b}{a}\right )^2+\sqrt [3]{1-\frac {a^2 \left (-\frac {2 b^2 x}{a^2}-\frac {b}{a}\right )^2}{b^2}}+1}{\left (\frac {2 b^2 x}{a^2}+\frac {b}{a}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\frac {2 x b^2}{a^2}+\frac {b}{a}+\sqrt {3}+1}{\frac {2 x b^2}{a^2}+\frac {b}{a}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{5\ 2^{2/3} b \left (-\frac {2 b^2 x}{a^2}-\frac {b}{a}\right ) \sqrt {-\frac {\frac {2 b^2 x}{a^2}+\frac {b}{a}+1}{\left (\frac {2 b^2 x}{a^2}+\frac {b}{a}-\sqrt {3}+1\right )^2}} \left (a x+b x^2\right )^{2/3}}+\frac {3 (a+2 b x) \sqrt [3]{a x+b x^2}}{10 b}\right )}{11 b}\) |
Input:
Int[(a*x + b*x^2)^(4/3),x]
Output:
(3*(a + 2*b*x)*(a*x + b*x^2)^(4/3))/(22*b) - (2*a^2*((3*(a + 2*b*x)*(a*x + b*x^2)^(1/3))/(10*b) + (3^(3/4)*Sqrt[2 - Sqrt[3]]*a^2*(1 + b/a + (2*b^2*x )/a^2)*(-((b*(a*x + b*x^2))/a^2))^(2/3)*Sqrt[(1 + (-(b/a) - (2*b^2*x)/a^2) ^2 + (1 - (a^2*(-(b/a) - (2*b^2*x)/a^2)^2)/b^2)^(1/3))/(1 - Sqrt[3] + b/a + (2*b^2*x)/a^2)^2]*EllipticF[ArcSin[(1 + Sqrt[3] + b/a + (2*b^2*x)/a^2)/( 1 - Sqrt[3] + b/a + (2*b^2*x)/a^2)], -7 + 4*Sqrt[3]])/(5*2^(2/3)*b*(-(b/a) - (2*b^2*x)/a^2)*Sqrt[-((1 + b/a + (2*b^2*x)/a^2)/(1 - Sqrt[3] + b/a + (2 *b^2*x)/a^2)^2)]*(a*x + b*x^2)^(2/3))))/(11*b)
Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) Subst[Int[1/Sqrt[-a + x^3], x], x, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b }, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) *s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x ] && NegQ[a]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
Int[((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b*x + c*x^2)^p/((- c)*((b*x + c*x^2)/b^2))^p Int[((-c)*(x/b) - c^2*(x^2/b^2))^p, x], x] /; F reeQ[{b, c}, x] && (IntegerQ[4*p] || IntegerQ[3*p])
\[\int \left (b \,x^{2}+a x \right )^{\frac {4}{3}}d x\]
Input:
int((b*x^2+a*x)^(4/3),x)
Output:
int((b*x^2+a*x)^(4/3),x)
\[ \int \left (a x+b x^2\right )^{4/3} \, dx=\int { {\left (b x^{2} + a x\right )}^{\frac {4}{3}} \,d x } \] Input:
integrate((b*x^2+a*x)^(4/3),x, algorithm="fricas")
Output:
integral((b*x^2 + a*x)^(4/3), x)
\[ \int \left (a x+b x^2\right )^{4/3} \, dx=\int \left (a x + b x^{2}\right )^{\frac {4}{3}}\, dx \] Input:
integrate((b*x**2+a*x)**(4/3),x)
Output:
Integral((a*x + b*x**2)**(4/3), x)
\[ \int \left (a x+b x^2\right )^{4/3} \, dx=\int { {\left (b x^{2} + a x\right )}^{\frac {4}{3}} \,d x } \] Input:
integrate((b*x^2+a*x)^(4/3),x, algorithm="maxima")
Output:
integrate((b*x^2 + a*x)^(4/3), x)
\[ \int \left (a x+b x^2\right )^{4/3} \, dx=\int { {\left (b x^{2} + a x\right )}^{\frac {4}{3}} \,d x } \] Input:
integrate((b*x^2+a*x)^(4/3),x, algorithm="giac")
Output:
integrate((b*x^2 + a*x)^(4/3), x)
Time = 9.43 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \left (a x+b x^2\right )^{4/3} \, dx=\frac {3\,x\,{\left (b\,x^2+a\,x\right )}^{4/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {4}{3},\frac {7}{3};\ \frac {10}{3};\ -\frac {b\,x}{a}\right )}{7\,{\left (\frac {b\,x}{a}+1\right )}^{4/3}} \] Input:
int((a*x + b*x^2)^(4/3),x)
Output:
(3*x*(a*x + b*x^2)^(4/3)*hypergeom([-4/3, 7/3], 10/3, -(b*x)/a))/(7*((b*x) /a + 1)^(4/3))
\[ \int \left (a x+b x^2\right )^{4/3} \, dx=\frac {-6 \left (b x +a \right )^{\frac {1}{3}} a^{4}-6 \left (b x +a \right )^{\frac {1}{3}} a^{3} b x +3 \left (b x +a \right )^{\frac {1}{3}} a^{2} b^{2} x^{2}+45 \left (b x +a \right )^{\frac {1}{3}} a \,b^{3} x^{3}+30 \left (b x +a \right )^{\frac {1}{3}} b^{4} x^{4}-4 x^{\frac {2}{3}} \left (\int \frac {\left (b x +a \right )^{\frac {1}{3}}}{x^{\frac {5}{3}} a +x^{\frac {8}{3}} b}d x \right ) a^{5}}{110 x^{\frac {2}{3}} b^{3}} \] Input:
int((b*x^2+a*x)^(4/3),x)
Output:
( - 6*(a + b*x)**(1/3)*a**4 - 6*(a + b*x)**(1/3)*a**3*b*x + 3*(a + b*x)**( 1/3)*a**2*b**2*x**2 + 45*(a + b*x)**(1/3)*a*b**3*x**3 + 30*(a + b*x)**(1/3 )*b**4*x**4 - 4*x**(2/3)*int((a + b*x)**(1/3)/(x**(2/3)*a*x + x**(2/3)*b*x **2),x)*a**5)/(110*x**(2/3)*b**3)