Integrand size = 15, antiderivative size = 64 \[ \int x^m \left (a+b x^3\right )^{3/2} \, dx=\frac {a x^{1+m} \sqrt {a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )}{(1+m) \sqrt {1+\frac {b x^3}{a}}} \] Output:
a*x^(1+m)*(b*x^3+a)^(1/2)*hypergeom([-3/2, 1/3+1/3*m],[4/3+1/3*m],-b*x^3/a )/(1+m)/(1+b*x^3/a)^(1/2)
Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03 \[ \int x^m \left (a+b x^3\right )^{3/2} \, dx=\frac {a x^{1+m} \sqrt {a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{3},1+\frac {1+m}{3},-\frac {b x^3}{a}\right )}{(1+m) \sqrt {1+\frac {b x^3}{a}}} \] Input:
Integrate[x^m*(a + b*x^3)^(3/2),x]
Output:
(a*x^(1 + m)*Sqrt[a + b*x^3]*Hypergeometric2F1[-3/2, (1 + m)/3, 1 + (1 + m )/3, -((b*x^3)/a)])/((1 + m)*Sqrt[1 + (b*x^3)/a])
Time = 0.31 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {889, 888}
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.
\(\displaystyle \int x^m \left (a+b x^3\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 889 |
\(\displaystyle \frac {a \sqrt {a+b x^3} \int x^m \left (\frac {b x^3}{a}+1\right )^{3/2}dx}{\sqrt {\frac {b x^3}{a}+1}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {a x^{m+1} \sqrt {a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m+1}{3},\frac {m+4}{3},-\frac {b x^3}{a}\right )}{(m+1) \sqrt {\frac {b x^3}{a}+1}}\) |
Input:
Int[x^m*(a + b*x^3)^(3/2),x]
Output:
(a*x^(1 + m)*Sqrt[a + b*x^3]*Hypergeometric2F1[-3/2, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)])/((1 + m)*Sqrt[1 + (b*x^3)/a])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^I ntPart[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])
\[\int x^{m} \left (b \,x^{3}+a \right )^{\frac {3}{2}}d x\]
Input:
int(x^m*(b*x^3+a)^(3/2),x)
Output:
int(x^m*(b*x^3+a)^(3/2),x)
\[ \int x^m \left (a+b x^3\right )^{3/2} \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {3}{2}} x^{m} \,d x } \] Input:
integrate(x^m*(b*x^3+a)^(3/2),x, algorithm="fricas")
Output:
integral((b*x^3 + a)^(3/2)*x^m, x)
Result contains complex when optimal does not.
Time = 1.44 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.84 \[ \int x^m \left (a+b x^3\right )^{3/2} \, dx=\frac {a^{\frac {3}{2}} x^{m + 1} \Gamma \left (\frac {m}{3} + \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {m}{3} + \frac {1}{3} \\ \frac {m}{3} + \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {m}{3} + \frac {4}{3}\right )} \] Input:
integrate(x**m*(b*x**3+a)**(3/2),x)
Output:
a**(3/2)*x**(m + 1)*gamma(m/3 + 1/3)*hyper((-3/2, m/3 + 1/3), (m/3 + 4/3,) , b*x**3*exp_polar(I*pi)/a)/(3*gamma(m/3 + 4/3))
\[ \int x^m \left (a+b x^3\right )^{3/2} \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {3}{2}} x^{m} \,d x } \] Input:
integrate(x^m*(b*x^3+a)^(3/2),x, algorithm="maxima")
Output:
integrate((b*x^3 + a)^(3/2)*x^m, x)
\[ \int x^m \left (a+b x^3\right )^{3/2} \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {3}{2}} x^{m} \,d x } \] Input:
integrate(x^m*(b*x^3+a)^(3/2),x, algorithm="giac")
Output:
integrate((b*x^3 + a)^(3/2)*x^m, x)
Timed out. \[ \int x^m \left (a+b x^3\right )^{3/2} \, dx=\int x^m\,{\left (b\,x^3+a\right )}^{3/2} \,d x \] Input:
int(x^m*(a + b*x^3)^(3/2),x)
Output:
int(x^m*(a + b*x^3)^(3/2), x)
\[ \int x^m \left (a+b x^3\right )^{3/2} \, dx=\frac {4 x^{m} \sqrt {b \,x^{3}+a}\, a m x +28 x^{m} \sqrt {b \,x^{3}+a}\, a x +4 x^{m} \sqrt {b \,x^{3}+a}\, b m \,x^{4}+10 x^{m} \sqrt {b \,x^{3}+a}\, b \,x^{4}+108 \left (\int \frac {x^{m} \sqrt {b \,x^{3}+a}}{4 b \,m^{2} x^{3}+32 b m \,x^{3}+55 b \,x^{3}+4 a \,m^{2}+32 a m +55 a}d x \right ) a^{2} m^{2}+864 \left (\int \frac {x^{m} \sqrt {b \,x^{3}+a}}{4 b \,m^{2} x^{3}+32 b m \,x^{3}+55 b \,x^{3}+4 a \,m^{2}+32 a m +55 a}d x \right ) a^{2} m +1485 \left (\int \frac {x^{m} \sqrt {b \,x^{3}+a}}{4 b \,m^{2} x^{3}+32 b m \,x^{3}+55 b \,x^{3}+4 a \,m^{2}+32 a m +55 a}d x \right ) a^{2}}{4 m^{2}+32 m +55} \] Input:
int(x^m*(b*x^3+a)^(3/2),x)
Output:
(4*x**m*sqrt(a + b*x**3)*a*m*x + 28*x**m*sqrt(a + b*x**3)*a*x + 4*x**m*sqr t(a + b*x**3)*b*m*x**4 + 10*x**m*sqrt(a + b*x**3)*b*x**4 + 108*int((x**m*s qrt(a + b*x**3))/(4*a*m**2 + 32*a*m + 55*a + 4*b*m**2*x**3 + 32*b*m*x**3 + 55*b*x**3),x)*a**2*m**2 + 864*int((x**m*sqrt(a + b*x**3))/(4*a*m**2 + 32* a*m + 55*a + 4*b*m**2*x**3 + 32*b*m*x**3 + 55*b*x**3),x)*a**2*m + 1485*int ((x**m*sqrt(a + b*x**3))/(4*a*m**2 + 32*a*m + 55*a + 4*b*m**2*x**3 + 32*b* m*x**3 + 55*b*x**3),x)*a**2)/(4*m**2 + 32*m + 55)