Integrand size = 15, antiderivative size = 99 \[ \int \frac {\sqrt [4]{a+b x}}{x^{5/2}} \, dx=-\frac {2 \sqrt [4]{a+b x}}{3 x^{3/2}}-\frac {b \sqrt [4]{a+b x}}{3 a \sqrt {x}}-\frac {b^{3/2} \left (1+\frac {b x}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right ),2\right )}{3 \sqrt {a} (a+b x)^{3/4}} \] Output:
-2/3*(b*x+a)^(1/4)/x^(3/2)-1/3*b*(b*x+a)^(1/4)/a/x^(1/2)-1/3*b^(3/2)*(1+b* x/a)^(3/4)*InverseJacobiAM(1/2*arctan(b^(1/2)*x^(1/2)/a^(1/2)),2^(1/2))/a^ (1/2)/(b*x+a)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.47 \[ \int \frac {\sqrt [4]{a+b x}}{x^{5/2}} \, dx=-\frac {2 \sqrt [4]{a+b x} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{4},-\frac {1}{2},-\frac {b x}{a}\right )}{3 x^{3/2} \sqrt [4]{1+\frac {b x}{a}}} \] Input:
Integrate[(a + b*x)^(1/4)/x^(5/2),x]
Output:
(-2*(a + b*x)^(1/4)*Hypergeometric2F1[-3/2, -1/4, -1/2, -((b*x)/a)])/(3*x^ (3/2)*(1 + (b*x)/a)^(1/4))
Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {57, 61, 73, 765, 762}
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.
\(\displaystyle \int \frac {\sqrt [4]{a+b x}}{x^{5/2}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {1}{6} b \int \frac {1}{x^{3/2} (a+b x)^{3/4}}dx-\frac {2 \sqrt [4]{a+b x}}{3 x^{3/2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{6} b \left (-\frac {b \int \frac {1}{\sqrt {x} (a+b x)^{3/4}}dx}{2 a}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )-\frac {2 \sqrt [4]{a+b x}}{3 x^{3/2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{6} b \left (-\frac {2 \int \frac {1}{\sqrt {\frac {a+b x}{b}-\frac {a}{b}}}d\sqrt [4]{a+b x}}{a}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )-\frac {2 \sqrt [4]{a+b x}}{3 x^{3/2}}\) |
\(\Big \downarrow \) 765 |
\(\displaystyle \frac {1}{6} b \left (-\frac {2 \sqrt {1-\frac {a+b x}{a}} \int \frac {1}{\sqrt {1-\frac {a+b x}{a}}}d\sqrt [4]{a+b x}}{a \sqrt {\frac {a+b x}{b}-\frac {a}{b}}}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )-\frac {2 \sqrt [4]{a+b x}}{3 x^{3/2}}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {1}{6} b \left (-\frac {2 \sqrt {1-\frac {a+b x}{a}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{a+b x}}{\sqrt [4]{a}}\right ),-1\right )}{a^{3/4} \sqrt {\frac {a+b x}{b}-\frac {a}{b}}}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )-\frac {2 \sqrt [4]{a+b x}}{3 x^{3/2}}\) |
Input:
Int[(a + b*x)^(1/4)/x^(5/2),x]
Output:
(-2*(a + b*x)^(1/4))/(3*x^(3/2)) + (b*((-2*(a + b*x)^(1/4))/(a*Sqrt[x]) - (2*Sqrt[1 - (a + b*x)/a]*EllipticF[ArcSin[(a + b*x)^(1/4)/a^(1/4)], -1])/( a^(3/4)*Sqrt[-(a/b) + (a + b*x)/b])))/6
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) )*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[Sqrt[1 + b*(x^4/a)]/Sqrt [a + b*x^4] Int[1/Sqrt[1 + b*(x^4/a)], x], x] /; FreeQ[{a, b}, x] && NegQ [b/a] && !GtQ[a, 0]
\[\int \frac {\left (b x +a \right )^{\frac {1}{4}}}{x^{\frac {5}{2}}}d x\]
Input:
int((b*x+a)^(1/4)/x^(5/2),x)
Output:
int((b*x+a)^(1/4)/x^(5/2),x)
\[ \int \frac {\sqrt [4]{a+b x}}{x^{5/2}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{x^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*x+a)^(1/4)/x^(5/2),x, algorithm="fricas")
Output:
integral((b*x + a)^(1/4)/x^(5/2), x)
Result contains complex when optimal does not.
Time = 0.97 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.36 \[ \int \frac {\sqrt [4]{a+b x}}{x^{5/2}} \, dx=- \frac {2 \sqrt [4]{a} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {1}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{3 x^{\frac {3}{2}}} \] Input:
integrate((b*x+a)**(1/4)/x**(5/2),x)
Output:
-2*a**(1/4)*hyper((-3/2, -1/4), (-1/2,), b*x*exp_polar(I*pi)/a)/(3*x**(3/2 ))
\[ \int \frac {\sqrt [4]{a+b x}}{x^{5/2}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{x^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*x+a)^(1/4)/x^(5/2),x, algorithm="maxima")
Output:
integrate((b*x + a)^(1/4)/x^(5/2), x)
\[ \int \frac {\sqrt [4]{a+b x}}{x^{5/2}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{x^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*x+a)^(1/4)/x^(5/2),x, algorithm="giac")
Output:
integrate((b*x + a)^(1/4)/x^(5/2), x)
Timed out. \[ \int \frac {\sqrt [4]{a+b x}}{x^{5/2}} \, dx=\int \frac {{\left (a+b\,x\right )}^{1/4}}{x^{5/2}} \,d x \] Input:
int((a + b*x)^(1/4)/x^(5/2),x)
Output:
int((a + b*x)^(1/4)/x^(5/2), x)
\[ \int \frac {\sqrt [4]{a+b x}}{x^{5/2}} \, dx=\frac {-4 \left (b x +a \right )^{\frac {1}{4}}-\sqrt {x}\, \left (\int \frac {\sqrt {x}\, \left (b x +a \right )^{\frac {1}{4}}}{b \,x^{4}+a \,x^{3}}d x \right ) a x}{5 \sqrt {x}\, x} \] Input:
int((b*x+a)^(1/4)/x^(5/2),x)
Output:
( - 4*(a + b*x)**(1/4) - sqrt(x)*int((sqrt(x)*(a + b*x)**(1/4))/(a*x**3 + b*x**4),x)*a*x)/(5*sqrt(x)*x)