Integrand size = 19, antiderivative size = 43 \[ \int \frac {1}{(2-5 x)^{5/4} \sqrt [4]{3-5 x}} \, dx=\frac {4}{5 \sqrt [4]{2-5 x} \sqrt [4]{3-5 x}}-\frac {4}{5} E\left (\left .\frac {1}{2} \arcsin \left (\frac {1}{\sqrt {3-5 x}}\right )\right |2\right ) \] Output:
4/5/(2-5*x)^(1/4)/(3-5*x)^(1/4)-4/5*EllipticE(sin(1/2*arcsin(1/(3-5*x)^(1/ 2))),2^(1/2))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.65 \[ \int \frac {1}{(2-5 x)^{5/4} \sqrt [4]{3-5 x}} \, dx=\frac {4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4},\frac {3}{4},-2+5 x\right )}{5 \sqrt [4]{2-5 x}} \] Input:
Integrate[1/((2 - 5*x)^(5/4)*(3 - 5*x)^(1/4)),x]
Output:
(4*Hypergeometric2F1[-1/4, 1/4, 3/4, -2 + 5*x])/(5*(2 - 5*x)^(1/4))
Leaf count is larger than twice the leaf count of optimal. \(101\) vs. \(2(43)=86\).
Time = 0.22 (sec) , antiderivative size = 101, normalized size of antiderivative = 2.35, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {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.
\(\displaystyle \int \frac {1}{(2-5 x)^{5/4} \sqrt [4]{3-5 x}} \, dx\) |
\(\Big \downarrow \) 61 |
\(\displaystyle 2 \int \frac {1}{\sqrt [4]{2-5 x} \sqrt [4]{3-5 x}}dx+\frac {4 (3-5 x)^{3/4}}{5 \sqrt [4]{2-5 x}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {4 (3-5 x)^{3/4}}{5 \sqrt [4]{2-5 x}}-\frac {8}{5} \int \frac {\sqrt {2-5 x}}{\sqrt [4]{3-5 x}}d\sqrt [4]{2-5 x}\) |
\(\Big \downarrow \) 839 |
\(\displaystyle \frac {4 (3-5 x)^{3/4}}{5 \sqrt [4]{2-5 x}}-\frac {8}{5} \left (\frac {(2-5 x)^{3/4}}{2 \sqrt [4]{3-5 x}}-\frac {1}{2} \int \frac {\sqrt {2-5 x}}{(3-5 x)^{5/4}}d\sqrt [4]{2-5 x}\right )\) |
\(\Big \downarrow \) 813 |
\(\displaystyle \frac {4 (3-5 x)^{3/4}}{5 \sqrt [4]{2-5 x}}-\frac {8}{5} \left (\frac {(2-5 x)^{3/4}}{2 \sqrt [4]{3-5 x}}-\frac {\sqrt [4]{\frac {1}{2-5 x}+1} \sqrt [4]{2-5 x} \int \frac {1}{\left (1+\frac {1}{2-5 x}\right )^{5/4} (2-5 x)^{3/4}}d\sqrt [4]{2-5 x}}{2 \sqrt [4]{3-5 x}}\right )\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {4 (3-5 x)^{3/4}}{5 \sqrt [4]{2-5 x}}-\frac {8}{5} \left (\frac {\sqrt [4]{\frac {1}{2-5 x}+1} \sqrt [4]{2-5 x} \int \frac {1}{\sqrt [4]{2-5 x} (3-5 x)^{5/4}}d\frac {1}{\sqrt [4]{2-5 x}}}{2 \sqrt [4]{3-5 x}}+\frac {(2-5 x)^{3/4}}{2 \sqrt [4]{3-5 x}}\right )\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {4 (3-5 x)^{3/4}}{5 \sqrt [4]{2-5 x}}-\frac {8}{5} \left (\frac {\sqrt [4]{\frac {1}{2-5 x}+1} \sqrt [4]{2-5 x} \int \frac {1}{\left (\sqrt {2-5 x}+1\right )^{5/4}}d\sqrt {2-5 x}}{4 \sqrt [4]{3-5 x}}+\frac {(2-5 x)^{3/4}}{2 \sqrt [4]{3-5 x}}\right )\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {4 (3-5 x)^{3/4}}{5 \sqrt [4]{2-5 x}}-\frac {8}{5} \left (\frac {\sqrt [4]{\frac {1}{2-5 x}+1} \sqrt [4]{2-5 x} E\left (\left .\frac {1}{2} \arctan \left (\sqrt {2-5 x}\right )\right |2\right )}{2 \sqrt [4]{3-5 x}}+\frac {(2-5 x)^{3/4}}{2 \sqrt [4]{3-5 x}}\right )\) |
Input:
Int[1/((2 - 5*x)^(5/4)*(3 - 5*x)^(1/4)),x]
Output:
(4*(3 - 5*x)^(3/4))/(5*(2 - 5*x)^(1/4)) - (8*((2 - 5*x)^(3/4)/(2*(3 - 5*x) ^(1/4)) + ((1 + (2 - 5*x)^(-1))^(1/4)*(2 - 5*x)^(1/4)*EllipticE[ArcTan[Sqr t[2 - 5*x]]/2, 2])/(2*(3 - 5*x)^(1/4))))/5
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]
\[\int \frac {1}{\left (2-5 x \right )^{\frac {5}{4}} \left (3-5 x \right )^{\frac {1}{4}}}d x\]
Input:
int(1/(2-5*x)^(5/4)/(3-5*x)^(1/4),x)
Output:
int(1/(2-5*x)^(5/4)/(3-5*x)^(1/4),x)
\[ \int \frac {1}{(2-5 x)^{5/4} \sqrt [4]{3-5 x}} \, dx=\int { \frac {1}{{\left (-5 \, x + 3\right )}^{\frac {1}{4}} {\left (-5 \, x + 2\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(1/(2-5*x)^(5/4)/(3-5*x)^(1/4),x, algorithm="fricas")
Output:
integral(-(-5*x + 3)^(3/4)*(-5*x + 2)^(3/4)/(125*x^3 - 175*x^2 + 80*x - 12 ), x)
Result contains complex when optimal does not.
Time = 1.32 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(2-5 x)^{5/4} \sqrt [4]{3-5 x}} \, dx=- \frac {2 \sqrt [4]{-1} \sqrt {5} e^{\frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {1}{5 \left (x - \frac {2}{5}\right )}} \right )}}{25 \sqrt {x - \frac {2}{5}}} \] Input:
integrate(1/(2-5*x)**(5/4)/(3-5*x)**(1/4),x)
Output:
-2*(-1)**(1/4)*sqrt(5)*exp(I*pi/4)*hyper((1/4, 1/2), (3/2,), 1/(5*(x - 2/5 )))/(25*sqrt(x - 2/5))
\[ \int \frac {1}{(2-5 x)^{5/4} \sqrt [4]{3-5 x}} \, dx=\int { \frac {1}{{\left (-5 \, x + 3\right )}^{\frac {1}{4}} {\left (-5 \, x + 2\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(1/(2-5*x)^(5/4)/(3-5*x)^(1/4),x, algorithm="maxima")
Output:
integrate(1/((-5*x + 3)^(1/4)*(-5*x + 2)^(5/4)), x)
\[ \int \frac {1}{(2-5 x)^{5/4} \sqrt [4]{3-5 x}} \, dx=\int { \frac {1}{{\left (-5 \, x + 3\right )}^{\frac {1}{4}} {\left (-5 \, x + 2\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(1/(2-5*x)^(5/4)/(3-5*x)^(1/4),x, algorithm="giac")
Output:
integrate(1/((-5*x + 3)^(1/4)*(-5*x + 2)^(5/4)), x)
Timed out. \[ \int \frac {1}{(2-5 x)^{5/4} \sqrt [4]{3-5 x}} \, dx=\int \frac {1}{{\left (2-5\,x\right )}^{5/4}\,{\left (3-5\,x\right )}^{1/4}} \,d x \] Input:
int(1/((2 - 5*x)^(5/4)*(3 - 5*x)^(1/4)),x)
Output:
int(1/((2 - 5*x)^(5/4)*(3 - 5*x)^(1/4)), x)
\[ \int \frac {1}{(2-5 x)^{5/4} \sqrt [4]{3-5 x}} \, dx=-\left (\int \frac {1}{5 \left (-5 x +2\right )^{\frac {1}{4}} \left (-5 x +3\right )^{\frac {1}{4}} x -2 \left (-5 x +2\right )^{\frac {1}{4}} \left (-5 x +3\right )^{\frac {1}{4}}}d x \right ) \] Input:
int(1/(2-5*x)^(5/4)/(3-5*x)^(1/4),x)
Output:
- int(1/(5*( - 5*x + 2)**(1/4)*( - 5*x + 3)**(1/4)*x - 2*( - 5*x + 2)**(1 /4)*( - 5*x + 3)**(1/4)),x)