Integrand size = 15, antiderivative size = 68 \[ \int \frac {1}{x^{15/4} \sqrt [4]{a+b x}} \, dx=-\frac {4 (a+b x)^{3/4}}{11 a x^{11/4}}+\frac {32 b (a+b x)^{3/4}}{77 a^2 x^{7/4}}-\frac {128 b^2 (a+b x)^{3/4}}{231 a^3 x^{3/4}} \] Output:
-4/11*(b*x+a)^(3/4)/a/x^(11/4)+32/77*b*(b*x+a)^(3/4)/a^2/x^(7/4)-128/231*b ^2*(b*x+a)^(3/4)/a^3/x^(3/4)
Time = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x^{15/4} \sqrt [4]{a+b x}} \, dx=-\frac {4 (a+b x)^{3/4} \left (21 a^2-24 a b x+32 b^2 x^2\right )}{231 a^3 x^{11/4}} \] Input:
Integrate[1/(x^(15/4)*(a + b*x)^(1/4)),x]
Output:
(-4*(a + b*x)^(3/4)*(21*a^2 - 24*a*b*x + 32*b^2*x^2))/(231*a^3*x^(11/4))
Time = 0.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {55, 55, 48}
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.
\(\displaystyle \int \frac {1}{x^{15/4} \sqrt [4]{a+b x}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {8 b \int \frac {1}{x^{11/4} \sqrt [4]{a+b x}}dx}{11 a}-\frac {4 (a+b x)^{3/4}}{11 a x^{11/4}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {8 b \left (-\frac {4 b \int \frac {1}{x^{7/4} \sqrt [4]{a+b x}}dx}{7 a}-\frac {4 (a+b x)^{3/4}}{7 a x^{7/4}}\right )}{11 a}-\frac {4 (a+b x)^{3/4}}{11 a x^{11/4}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {8 b \left (\frac {16 b (a+b x)^{3/4}}{21 a^2 x^{3/4}}-\frac {4 (a+b x)^{3/4}}{7 a x^{7/4}}\right )}{11 a}-\frac {4 (a+b x)^{3/4}}{11 a x^{11/4}}\) |
Input:
Int[1/(x^(15/4)*(a + b*x)^(1/4)),x]
Output:
(-4*(a + b*x)^(3/4))/(11*a*x^(11/4)) - (8*b*((-4*(a + b*x)^(3/4))/(7*a*x^( 7/4)) + (16*b*(a + b*x)^(3/4))/(21*a^2*x^(3/4))))/(11*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] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
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*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.51
method | result | size |
gosper | \(-\frac {4 \left (b x +a \right )^{\frac {3}{4}} \left (32 b^{2} x^{2}-24 a b x +21 a^{2}\right )}{231 x^{\frac {11}{4}} a^{3}}\) | \(35\) |
risch | \(-\frac {4 \left (b x +a \right )^{\frac {3}{4}} \left (32 b^{2} x^{2}-24 a b x +21 a^{2}\right )}{231 x^{\frac {11}{4}} a^{3}}\) | \(35\) |
orering | \(-\frac {4 \left (b x +a \right )^{\frac {3}{4}} \left (32 b^{2} x^{2}-24 a b x +21 a^{2}\right )}{231 x^{\frac {11}{4}} a^{3}}\) | \(35\) |
Input:
int(1/x^(15/4)/(b*x+a)^(1/4),x,method=_RETURNVERBOSE)
Output:
-4/231*(b*x+a)^(3/4)*(32*b^2*x^2-24*a*b*x+21*a^2)/x^(11/4)/a^3
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x^{15/4} \sqrt [4]{a+b x}} \, dx=-\frac {4 \, {\left (32 \, b^{2} x^{2} - 24 \, a b x + 21 \, a^{2}\right )} {\left (b x + a\right )}^{\frac {3}{4}}}{231 \, a^{3} x^{\frac {11}{4}}} \] Input:
integrate(1/x^(15/4)/(b*x+a)^(1/4),x, algorithm="fricas")
Output:
-4/231*(32*b^2*x^2 - 24*a*b*x + 21*a^2)*(b*x + a)^(3/4)/(a^3*x^(11/4))
Leaf count of result is larger than twice the leaf count of optimal. 396 vs. \(2 (63) = 126\).
Time = 92.33 (sec) , antiderivative size = 396, normalized size of antiderivative = 5.82 \[ \int \frac {1}{x^{15/4} \sqrt [4]{a+b x}} \, dx=\frac {21 a^{4} b^{\frac {19}{4}} \left (\frac {a}{b x} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {11}{4}\right )}{16 a^{5} b^{4} x^{2} \Gamma \left (\frac {1}{4}\right ) + 32 a^{4} b^{5} x^{3} \Gamma \left (\frac {1}{4}\right ) + 16 a^{3} b^{6} x^{4} \Gamma \left (\frac {1}{4}\right )} + \frac {18 a^{3} b^{\frac {23}{4}} x \left (\frac {a}{b x} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {11}{4}\right )}{16 a^{5} b^{4} x^{2} \Gamma \left (\frac {1}{4}\right ) + 32 a^{4} b^{5} x^{3} \Gamma \left (\frac {1}{4}\right ) + 16 a^{3} b^{6} x^{4} \Gamma \left (\frac {1}{4}\right )} + \frac {5 a^{2} b^{\frac {27}{4}} x^{2} \left (\frac {a}{b x} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {11}{4}\right )}{16 a^{5} b^{4} x^{2} \Gamma \left (\frac {1}{4}\right ) + 32 a^{4} b^{5} x^{3} \Gamma \left (\frac {1}{4}\right ) + 16 a^{3} b^{6} x^{4} \Gamma \left (\frac {1}{4}\right )} + \frac {40 a b^{\frac {31}{4}} x^{3} \left (\frac {a}{b x} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {11}{4}\right )}{16 a^{5} b^{4} x^{2} \Gamma \left (\frac {1}{4}\right ) + 32 a^{4} b^{5} x^{3} \Gamma \left (\frac {1}{4}\right ) + 16 a^{3} b^{6} x^{4} \Gamma \left (\frac {1}{4}\right )} + \frac {32 b^{\frac {35}{4}} x^{4} \left (\frac {a}{b x} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {11}{4}\right )}{16 a^{5} b^{4} x^{2} \Gamma \left (\frac {1}{4}\right ) + 32 a^{4} b^{5} x^{3} \Gamma \left (\frac {1}{4}\right ) + 16 a^{3} b^{6} x^{4} \Gamma \left (\frac {1}{4}\right )} \] Input:
integrate(1/x**(15/4)/(b*x+a)**(1/4),x)
Output:
21*a**4*b**(19/4)*(a/(b*x) + 1)**(3/4)*gamma(-11/4)/(16*a**5*b**4*x**2*gam ma(1/4) + 32*a**4*b**5*x**3*gamma(1/4) + 16*a**3*b**6*x**4*gamma(1/4)) + 1 8*a**3*b**(23/4)*x*(a/(b*x) + 1)**(3/4)*gamma(-11/4)/(16*a**5*b**4*x**2*ga mma(1/4) + 32*a**4*b**5*x**3*gamma(1/4) + 16*a**3*b**6*x**4*gamma(1/4)) + 5*a**2*b**(27/4)*x**2*(a/(b*x) + 1)**(3/4)*gamma(-11/4)/(16*a**5*b**4*x**2 *gamma(1/4) + 32*a**4*b**5*x**3*gamma(1/4) + 16*a**3*b**6*x**4*gamma(1/4)) + 40*a*b**(31/4)*x**3*(a/(b*x) + 1)**(3/4)*gamma(-11/4)/(16*a**5*b**4*x** 2*gamma(1/4) + 32*a**4*b**5*x**3*gamma(1/4) + 16*a**3*b**6*x**4*gamma(1/4) ) + 32*b**(35/4)*x**4*(a/(b*x) + 1)**(3/4)*gamma(-11/4)/(16*a**5*b**4*x**2 *gamma(1/4) + 32*a**4*b**5*x**3*gamma(1/4) + 16*a**3*b**6*x**4*gamma(1/4))
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.68 \[ \int \frac {1}{x^{15/4} \sqrt [4]{a+b x}} \, dx=-\frac {4 \, {\left (\frac {77 \, {\left (b x + a\right )}^{\frac {3}{4}} b^{2}}{x^{\frac {3}{4}}} - \frac {66 \, {\left (b x + a\right )}^{\frac {7}{4}} b}{x^{\frac {7}{4}}} + \frac {21 \, {\left (b x + a\right )}^{\frac {11}{4}}}{x^{\frac {11}{4}}}\right )}}{231 \, a^{3}} \] Input:
integrate(1/x^(15/4)/(b*x+a)^(1/4),x, algorithm="maxima")
Output:
-4/231*(77*(b*x + a)^(3/4)*b^2/x^(3/4) - 66*(b*x + a)^(7/4)*b/x^(7/4) + 21 *(b*x + a)^(11/4)/x^(11/4))/a^3
\[ \int \frac {1}{x^{15/4} \sqrt [4]{a+b x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {1}{4}} x^{\frac {15}{4}}} \,d x } \] Input:
integrate(1/x^(15/4)/(b*x+a)^(1/4),x, algorithm="giac")
Output:
integrate(1/((b*x + a)^(1/4)*x^(15/4)), x)
Timed out. \[ \int \frac {1}{x^{15/4} \sqrt [4]{a+b x}} \, dx=\int \frac {1}{x^{15/4}\,{\left (a+b\,x\right )}^{1/4}} \,d x \] Input:
int(1/(x^(15/4)*(a + b*x)^(1/4)),x)
Output:
int(1/(x^(15/4)*(a + b*x)^(1/4)), x)
\[ \int \frac {1}{x^{15/4} \sqrt [4]{a+b x}} \, dx=\int \frac {1}{x^{\frac {15}{4}} \left (b x +a \right )^{\frac {1}{4}}}d x \] Input:
int(1/x^(15/4)/(b*x+a)^(1/4),x)
Output:
int(1/(x**(3/4)*(a + b*x)**(1/4)*x**3),x)