Integrand size = 15, antiderivative size = 77 \[ \int \frac {1}{x^{5/4} (a+b x)^{5/4}} \, dx=-\frac {4}{a \sqrt [4]{x} \sqrt [4]{a+b x}}+\frac {8 \sqrt [4]{\frac {b x}{a+b x}} \sqrt [4]{a+b x} E\left (\left .\frac {1}{2} \arcsin \left (\frac {\sqrt {a}}{\sqrt {a+b x}}\right )\right |2\right )}{a^{3/2} \sqrt [4]{x}} \] Output:
-4/a/x^(1/4)/(b*x+a)^(1/4)+8*(b*x/(b*x+a))^(1/4)*(b*x+a)^(1/4)*EllipticE(s in(1/2*arcsin(a^(1/2)/(b*x+a)^(1/2))),2^(1/2))/a^(3/2)/x^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x^{5/4} (a+b x)^{5/4}} \, dx=-\frac {4 \sqrt [4]{1+\frac {b x}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{4},\frac {3}{4},-\frac {b x}{a}\right )}{a \sqrt [4]{x} \sqrt [4]{a+b x}} \] Input:
Integrate[1/(x^(5/4)*(a + b*x)^(5/4)),x]
Output:
(-4*(1 + (b*x)/a)^(1/4)*Hypergeometric2F1[-1/4, 5/4, 3/4, -((b*x)/a)])/(a* x^(1/4)*(a + b*x)^(1/4))
Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.75, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {61, 61, 73, 839, 813, 858, 807, 212}
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.
\(\displaystyle \int \frac {1}{x^{5/4} (a+b x)^{5/4}} \, dx\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {2 \int \frac {1}{x^{5/4} \sqrt [4]{a+b x}}dx}{a}+\frac {4}{a \sqrt [4]{x} \sqrt [4]{a+b x}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {2 \left (\frac {2 b \int \frac {1}{\sqrt [4]{x} \sqrt [4]{a+b x}}dx}{a}-\frac {4 (a+b x)^{3/4}}{a \sqrt [4]{x}}\right )}{a}+\frac {4}{a \sqrt [4]{x} \sqrt [4]{a+b x}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 \left (\frac {8 b \int \frac {\sqrt {x}}{\sqrt [4]{a+b x}}d\sqrt [4]{x}}{a}-\frac {4 (a+b x)^{3/4}}{a \sqrt [4]{x}}\right )}{a}+\frac {4}{a \sqrt [4]{x} \sqrt [4]{a+b x}}\) |
\(\Big \downarrow \) 839 |
\(\displaystyle \frac {2 \left (\frac {8 b \left (\frac {x^{3/4}}{2 \sqrt [4]{a+b x}}-\frac {1}{2} a \int \frac {\sqrt {x}}{(a+b x)^{5/4}}d\sqrt [4]{x}\right )}{a}-\frac {4 (a+b x)^{3/4}}{a \sqrt [4]{x}}\right )}{a}+\frac {4}{a \sqrt [4]{x} \sqrt [4]{a+b x}}\) |
\(\Big \downarrow \) 813 |
\(\displaystyle \frac {2 \left (\frac {8 b \left (\frac {x^{3/4}}{2 \sqrt [4]{a+b x}}-\frac {a \sqrt [4]{x} \sqrt [4]{\frac {a}{b x}+1} \int \frac {1}{\left (\frac {a}{b x}+1\right )^{5/4} x^{3/4}}d\sqrt [4]{x}}{2 b \sqrt [4]{a+b x}}\right )}{a}-\frac {4 (a+b x)^{3/4}}{a \sqrt [4]{x}}\right )}{a}+\frac {4}{a \sqrt [4]{x} \sqrt [4]{a+b x}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {2 \left (\frac {8 b \left (\frac {a \sqrt [4]{x} \sqrt [4]{\frac {a}{b x}+1} \int \frac {1}{\sqrt [4]{x} \left (\frac {a x}{b}+1\right )^{5/4}}d\frac {1}{\sqrt [4]{x}}}{2 b \sqrt [4]{a+b x}}+\frac {x^{3/4}}{2 \sqrt [4]{a+b x}}\right )}{a}-\frac {4 (a+b x)^{3/4}}{a \sqrt [4]{x}}\right )}{a}+\frac {4}{a \sqrt [4]{x} \sqrt [4]{a+b x}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {2 \left (\frac {8 b \left (\frac {a \sqrt [4]{x} \sqrt [4]{\frac {a}{b x}+1} \int \frac {1}{\left (\frac {\sqrt {x} a}{b}+1\right )^{5/4}}d\sqrt {x}}{4 b \sqrt [4]{a+b x}}+\frac {x^{3/4}}{2 \sqrt [4]{a+b x}}\right )}{a}-\frac {4 (a+b x)^{3/4}}{a \sqrt [4]{x}}\right )}{a}+\frac {4}{a \sqrt [4]{x} \sqrt [4]{a+b x}}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {2 \left (\frac {8 b \left (\frac {\sqrt {a} \sqrt [4]{x} \sqrt [4]{\frac {a}{b x}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )\right |2\right )}{2 \sqrt {b} \sqrt [4]{a+b x}}+\frac {x^{3/4}}{2 \sqrt [4]{a+b x}}\right )}{a}-\frac {4 (a+b x)^{3/4}}{a \sqrt [4]{x}}\right )}{a}+\frac {4}{a \sqrt [4]{x} \sqrt [4]{a+b x}}\) |
Input:
Int[1/(x^(5/4)*(a + b*x)^(5/4)),x]
Output:
4/(a*x^(1/4)*(a + b*x)^(1/4)) + (2*((-4*(a + b*x)^(3/4))/(a*x^(1/4)) + (8* b*(x^(3/4)/(2*(a + b*x)^(1/4)) + (Sqrt[a]*(1 + a/(b*x))^(1/4)*x^(1/4)*Elli pticE[ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]]/2, 2])/(2*Sqrt[b]*(a + b*x)^(1/4)) ))/a))/a
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[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) )*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x*((1 + a/(b*x^4) )^(1/4)/(b*(a + b*x^4)^(1/4))) Int[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[x^3/(2*(a + b*x^4 )^(1/4)), x] - Simp[a/2 Int[x^2/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b} , x] && PosQ[b/a]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
\[\int \frac {1}{x^{\frac {5}{4}} \left (b x +a \right )^{\frac {5}{4}}}d x\]
Input:
int(1/x^(5/4)/(b*x+a)^(5/4),x)
Output:
int(1/x^(5/4)/(b*x+a)^(5/4),x)
\[ \int \frac {1}{x^{5/4} (a+b x)^{5/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{4}} x^{\frac {5}{4}}} \,d x } \] Input:
integrate(1/x^(5/4)/(b*x+a)^(5/4),x, algorithm="fricas")
Output:
integral((b*x + a)^(3/4)*x^(3/4)/(b^2*x^4 + 2*a*b*x^3 + a^2*x^2), x)
Result contains complex when optimal does not.
Time = 1.97 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.51 \[ \int \frac {1}{x^{5/4} (a+b x)^{5/4}} \, dx=\frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{a^{\frac {5}{4}} \sqrt [4]{x} \Gamma \left (\frac {3}{4}\right )} \] Input:
integrate(1/x**(5/4)/(b*x+a)**(5/4),x)
Output:
gamma(-1/4)*hyper((-1/4, 5/4), (3/4,), b*x*exp_polar(I*pi)/a)/(a**(5/4)*x* *(1/4)*gamma(3/4))
\[ \int \frac {1}{x^{5/4} (a+b x)^{5/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{4}} x^{\frac {5}{4}}} \,d x } \] Input:
integrate(1/x^(5/4)/(b*x+a)^(5/4),x, algorithm="maxima")
Output:
integrate(1/((b*x + a)^(5/4)*x^(5/4)), x)
\[ \int \frac {1}{x^{5/4} (a+b x)^{5/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{4}} x^{\frac {5}{4}}} \,d x } \] Input:
integrate(1/x^(5/4)/(b*x+a)^(5/4),x, algorithm="giac")
Output:
integrate(1/((b*x + a)^(5/4)*x^(5/4)), x)
Timed out. \[ \int \frac {1}{x^{5/4} (a+b x)^{5/4}} \, dx=\int \frac {1}{x^{5/4}\,{\left (a+b\,x\right )}^{5/4}} \,d x \] Input:
int(1/(x^(5/4)*(a + b*x)^(5/4)),x)
Output:
int(1/(x^(5/4)*(a + b*x)^(5/4)), x)
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.35 \[ \int \frac {1}{x^{5/4} (a+b x)^{5/4}} \, dx=\frac {4 x^{\frac {1}{4}} \left (b x +a \right )^{\frac {1}{4}}}{\sqrt {x}\, \sqrt {b x +a}\, a} \] Input:
int(1/x^(5/4)/(b*x+a)^(5/4),x)
Output:
(4*x**(1/4)*(a + b*x)**(1/4))/(sqrt(x)*sqrt(a + b*x)*a)