Integrand size = 21, antiderivative size = 22 \[ \int \frac {1}{\sqrt [4]{2-b x} (3-b x)^{5/4}} \, dx=\frac {4 E\left (\left .\frac {1}{2} \arcsin \left (\frac {1}{\sqrt {3-b x}}\right )\right |2\right )}{b} \] Output:
4*EllipticE(sin(1/2*arcsin(1/(-b*x+3)^(1/2))),2^(1/2))/b
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {1}{\sqrt [4]{2-b x} (3-b x)^{5/4}} \, dx=-\frac {4 (2-b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-2+b x\right )}{3 b} \] Input:
Integrate[1/((2 - b*x)^(1/4)*(3 - b*x)^(5/4)),x]
Output:
(-4*(2 - b*x)^(3/4)*Hypergeometric2F1[3/4, 5/4, 7/4, -2 + b*x])/(3*b)
Leaf count is larger than twice the leaf count of optimal. \(111\) vs. \(2(22)=44\).
Time = 0.22 (sec) , antiderivative size = 111, normalized size of antiderivative = 5.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, 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.
\(\displaystyle \int \frac {1}{\sqrt [4]{2-b x} (3-b x)^{5/4}} \, dx\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -2 \int \frac {1}{\sqrt [4]{2-b x} \sqrt [4]{3-b x}}dx-\frac {4 (2-b x)^{3/4}}{b \sqrt [4]{3-b x}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {8 \int \frac {\sqrt {2-b x}}{\sqrt [4]{3-b x}}d\sqrt [4]{2-b x}}{b}-\frac {4 (2-b x)^{3/4}}{b \sqrt [4]{3-b x}}\) |
\(\Big \downarrow \) 839 |
\(\displaystyle \frac {8 \left (\frac {(2-b x)^{3/4}}{2 \sqrt [4]{3-b x}}-\frac {1}{2} \int \frac {\sqrt {2-b x}}{(3-b x)^{5/4}}d\sqrt [4]{2-b x}\right )}{b}-\frac {4 (2-b x)^{3/4}}{b \sqrt [4]{3-b x}}\) |
\(\Big \downarrow \) 813 |
\(\displaystyle \frac {8 \left (\frac {(2-b x)^{3/4}}{2 \sqrt [4]{3-b x}}-\frac {\sqrt [4]{2-b x} \sqrt [4]{\frac {1}{2-b x}+1} \int \frac {1}{(2-b x)^{3/4} \left (1+\frac {1}{2-b x}\right )^{5/4}}d\sqrt [4]{2-b x}}{2 \sqrt [4]{3-b x}}\right )}{b}-\frac {4 (2-b x)^{3/4}}{b \sqrt [4]{3-b x}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {8 \left (\frac {\sqrt [4]{2-b x} \sqrt [4]{\frac {1}{2-b x}+1} \int \frac {1}{\sqrt [4]{2-b x} (3-b x)^{5/4}}d\frac {1}{\sqrt [4]{2-b x}}}{2 \sqrt [4]{3-b x}}+\frac {(2-b x)^{3/4}}{2 \sqrt [4]{3-b x}}\right )}{b}-\frac {4 (2-b x)^{3/4}}{b \sqrt [4]{3-b x}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {8 \left (\frac {\sqrt [4]{2-b x} \sqrt [4]{\frac {1}{2-b x}+1} \int \frac {1}{\left (\sqrt {2-b x}+1\right )^{5/4}}d\sqrt {2-b x}}{4 \sqrt [4]{3-b x}}+\frac {(2-b x)^{3/4}}{2 \sqrt [4]{3-b x}}\right )}{b}-\frac {4 (2-b x)^{3/4}}{b \sqrt [4]{3-b x}}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {8 \left (\frac {\sqrt [4]{2-b x} \sqrt [4]{\frac {1}{2-b x}+1} E\left (\left .\frac {1}{2} \arctan \left (\sqrt {2-b x}\right )\right |2\right )}{2 \sqrt [4]{3-b x}}+\frac {(2-b x)^{3/4}}{2 \sqrt [4]{3-b x}}\right )}{b}-\frac {4 (2-b x)^{3/4}}{b \sqrt [4]{3-b x}}\) |
Input:
Int[1/((2 - b*x)^(1/4)*(3 - b*x)^(5/4)),x]
Output:
(-4*(2 - b*x)^(3/4))/(b*(3 - b*x)^(1/4)) + (8*((2 - b*x)^(3/4)/(2*(3 - b*x )^(1/4)) + ((2 - b*x)^(1/4)*(1 + (2 - b*x)^(-1))^(1/4)*EllipticE[ArcTan[Sq rt[2 - b*x]]/2, 2])/(2*(3 - b*x)^(1/4))))/b
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]
\[\int \frac {1}{\left (-b x +2\right )^{\frac {1}{4}} \left (-b x +3\right )^{\frac {5}{4}}}d x\]
Input:
int(1/(-b*x+2)^(1/4)/(-b*x+3)^(5/4),x)
Output:
int(1/(-b*x+2)^(1/4)/(-b*x+3)^(5/4),x)
\[ \int \frac {1}{\sqrt [4]{2-b x} (3-b x)^{5/4}} \, dx=\int { \frac {1}{{\left (-b x + 3\right )}^{\frac {5}{4}} {\left (-b x + 2\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/(-b*x+2)^(1/4)/(-b*x+3)^(5/4),x, algorithm="fricas")
Output:
integral(-(-b*x + 3)^(3/4)*(-b*x + 2)^(3/4)/(b^3*x^3 - 8*b^2*x^2 + 21*b*x - 18), x)
\[ \int \frac {1}{\sqrt [4]{2-b x} (3-b x)^{5/4}} \, dx=\int \frac {1}{\sqrt [4]{- b x + 2} \left (- b x + 3\right )^{\frac {5}{4}}}\, dx \] Input:
integrate(1/(-b*x+2)**(1/4)/(-b*x+3)**(5/4),x)
Output:
Integral(1/((-b*x + 2)**(1/4)*(-b*x + 3)**(5/4)), x)
\[ \int \frac {1}{\sqrt [4]{2-b x} (3-b x)^{5/4}} \, dx=\int { \frac {1}{{\left (-b x + 3\right )}^{\frac {5}{4}} {\left (-b x + 2\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/(-b*x+2)^(1/4)/(-b*x+3)^(5/4),x, algorithm="maxima")
Output:
integrate(1/((-b*x + 3)^(5/4)*(-b*x + 2)^(1/4)), x)
\[ \int \frac {1}{\sqrt [4]{2-b x} (3-b x)^{5/4}} \, dx=\int { \frac {1}{{\left (-b x + 3\right )}^{\frac {5}{4}} {\left (-b x + 2\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/(-b*x+2)^(1/4)/(-b*x+3)^(5/4),x, algorithm="giac")
Output:
integrate(1/((-b*x + 3)^(5/4)*(-b*x + 2)^(1/4)), x)
Timed out. \[ \int \frac {1}{\sqrt [4]{2-b x} (3-b x)^{5/4}} \, dx=\int \frac {1}{{\left (2-b\,x\right )}^{1/4}\,{\left (3-b\,x\right )}^{5/4}} \,d x \] Input:
int(1/((2 - b*x)^(1/4)*(3 - b*x)^(5/4)),x)
Output:
int(1/((2 - b*x)^(1/4)*(3 - b*x)^(5/4)), x)
\[ \int \frac {1}{\sqrt [4]{2-b x} (3-b x)^{5/4}} \, dx=-\left (\int \frac {1}{\left (-b x +2\right )^{\frac {1}{4}} \left (-b x +3\right )^{\frac {1}{4}} b x -3 \left (-b x +2\right )^{\frac {1}{4}} \left (-b x +3\right )^{\frac {1}{4}}}d x \right ) \] Input:
int(1/(-b*x+2)^(1/4)/(-b*x+3)^(5/4),x)
Output:
- int(1/(( - b*x + 2)**(1/4)*( - b*x + 3)**(1/4)*b*x - 3*( - b*x + 2)**(1 /4)*( - b*x + 3)**(1/4)),x)