Integrand size = 19, antiderivative size = 101 \[ \int \frac {1}{(a+b x)^{15/4} \sqrt [4]{c+d x}} \, dx=-\frac {4 (c+d x)^{3/4}}{11 (b c-a d) (a+b x)^{11/4}}+\frac {32 d (c+d x)^{3/4}}{77 (b c-a d)^2 (a+b x)^{7/4}}-\frac {128 d^2 (c+d x)^{3/4}}{231 (b c-a d)^3 (a+b x)^{3/4}} \] Output:
-4/11*(d*x+c)^(3/4)/(-a*d+b*c)/(b*x+a)^(11/4)+32/77*d*(d*x+c)^(3/4)/(-a*d+ b*c)^2/(b*x+a)^(7/4)-128/231*d^2*(d*x+c)^(3/4)/(-a*d+b*c)^3/(b*x+a)^(3/4)
Time = 0.35 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(a+b x)^{15/4} \sqrt [4]{c+d x}} \, dx=-\frac {4 (c+d x)^{3/4} \left (77 a^2 d^2+22 a b d (-3 c+4 d x)+b^2 \left (21 c^2-24 c d x+32 d^2 x^2\right )\right )}{231 (b c-a d)^3 (a+b x)^{11/4}} \] Input:
Integrate[1/((a + b*x)^(15/4)*(c + d*x)^(1/4)),x]
Output:
(-4*(c + d*x)^(3/4)*(77*a^2*d^2 + 22*a*b*d*(-3*c + 4*d*x) + b^2*(21*c^2 - 24*c*d*x + 32*d^2*x^2)))/(231*(b*c - a*d)^3*(a + b*x)^(11/4))
Time = 0.18 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.13, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, 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.
\(\displaystyle \int \frac {1}{(a+b x)^{15/4} \sqrt [4]{c+d x}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {8 d \int \frac {1}{(a+b x)^{11/4} \sqrt [4]{c+d x}}dx}{11 (b c-a d)}-\frac {4 (c+d x)^{3/4}}{11 (a+b x)^{11/4} (b c-a d)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {8 d \left (-\frac {4 d \int \frac {1}{(a+b x)^{7/4} \sqrt [4]{c+d x}}dx}{7 (b c-a d)}-\frac {4 (c+d x)^{3/4}}{7 (a+b x)^{7/4} (b c-a d)}\right )}{11 (b c-a d)}-\frac {4 (c+d x)^{3/4}}{11 (a+b x)^{11/4} (b c-a d)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {8 d \left (\frac {16 d (c+d x)^{3/4}}{21 (a+b x)^{3/4} (b c-a d)^2}-\frac {4 (c+d x)^{3/4}}{7 (a+b x)^{7/4} (b c-a d)}\right )}{11 (b c-a d)}-\frac {4 (c+d x)^{3/4}}{11 (a+b x)^{11/4} (b c-a d)}\) |
Input:
Int[1/((a + b*x)^(15/4)*(c + d*x)^(1/4)),x]
Output:
(-4*(c + d*x)^(3/4))/(11*(b*c - a*d)*(a + b*x)^(11/4)) - (8*d*((-4*(c + d* x)^(3/4))/(7*(b*c - a*d)*(a + b*x)^(7/4)) + (16*d*(c + d*x)^(3/4))/(21*(b* c - a*d)^2*(a + b*x)^(3/4))))/(11*(b*c - a*d))
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 = 0.17 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(\frac {4 \left (x d +c \right )^{\frac {3}{4}} \left (32 d^{2} x^{2} b^{2}+88 x a b \,d^{2}-24 x \,b^{2} c d +77 a^{2} d^{2}-66 a b c d +21 b^{2} c^{2}\right )}{231 \left (b x +a \right )^{\frac {11}{4}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
orering | \(\frac {4 \left (x d +c \right )^{\frac {3}{4}} \left (32 d^{2} x^{2} b^{2}+88 x a b \,d^{2}-24 x \,b^{2} c d +77 a^{2} d^{2}-66 a b c d +21 b^{2} c^{2}\right )}{231 \left (b x +a \right )^{\frac {11}{4}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
Input:
int(1/(b*x+a)^(15/4)/(d*x+c)^(1/4),x,method=_RETURNVERBOSE)
Output:
4/231*(d*x+c)^(3/4)*(32*b^2*d^2*x^2+88*a*b*d^2*x-24*b^2*c*d*x+77*a^2*d^2-6 6*a*b*c*d+21*b^2*c^2)/(b*x+a)^(11/4)/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d- b^3*c^3)
Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (83) = 166\).
Time = 0.41 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.50 \[ \int \frac {1}{(a+b x)^{15/4} \sqrt [4]{c+d x}} \, dx=-\frac {4 \, {\left (32 \, b^{2} d^{2} x^{2} + 21 \, b^{2} c^{2} - 66 \, a b c d + 77 \, a^{2} d^{2} - 8 \, {\left (3 \, b^{2} c d - 11 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{231 \, {\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} + {\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{3} + 3 \, {\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x\right )}} \] Input:
integrate(1/(b*x+a)^(15/4)/(d*x+c)^(1/4),x, algorithm="fricas")
Output:
-4/231*(32*b^2*d^2*x^2 + 21*b^2*c^2 - 66*a*b*c*d + 77*a^2*d^2 - 8*(3*b^2*c *d - 11*a*b*d^2)*x)*(b*x + a)^(1/4)*(d*x + c)^(3/4)/(a^3*b^3*c^3 - 3*a^4*b ^2*c^2*d + 3*a^5*b*c*d^2 - a^6*d^3 + (b^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4* c*d^2 - a^3*b^3*d^3)*x^3 + 3*(a*b^5*c^3 - 3*a^2*b^4*c^2*d + 3*a^3*b^3*c*d^ 2 - a^4*b^2*d^3)*x^2 + 3*(a^2*b^4*c^3 - 3*a^3*b^3*c^2*d + 3*a^4*b^2*c*d^2 - a^5*b*d^3)*x)
Timed out. \[ \int \frac {1}{(a+b x)^{15/4} \sqrt [4]{c+d x}} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x+a)**(15/4)/(d*x+c)**(1/4),x)
Output:
Timed out
\[ \int \frac {1}{(a+b x)^{15/4} \sqrt [4]{c+d x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {15}{4}} {\left (d x + c\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/(b*x+a)^(15/4)/(d*x+c)^(1/4),x, algorithm="maxima")
Output:
integrate(1/((b*x + a)^(15/4)*(d*x + c)^(1/4)), x)
\[ \int \frac {1}{(a+b x)^{15/4} \sqrt [4]{c+d x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {15}{4}} {\left (d x + c\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/(b*x+a)^(15/4)/(d*x+c)^(1/4),x, algorithm="giac")
Output:
integrate(1/((b*x + a)^(15/4)*(d*x + c)^(1/4)), x)
Timed out. \[ \int \frac {1}{(a+b x)^{15/4} \sqrt [4]{c+d x}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{15/4}\,{\left (c+d\,x\right )}^{1/4}} \,d x \] Input:
int(1/((a + b*x)^(15/4)*(c + d*x)^(1/4)),x)
Output:
int(1/((a + b*x)^(15/4)*(c + d*x)^(1/4)), x)
\[ \int \frac {1}{(a+b x)^{15/4} \sqrt [4]{c+d x}} \, dx=\int \frac {1}{\left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} a^{3}+3 \left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} a^{2} b x +3 \left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} a \,b^{2} x^{2}+\left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} b^{3} x^{3}}d x \] Input:
int(1/(b*x+a)^(15/4)/(d*x+c)^(1/4),x)
Output:
int(1/((c + d*x)**(1/4)*(a + b*x)**(3/4)*a**3 + 3*(c + d*x)**(1/4)*(a + b* x)**(3/4)*a**2*b*x + 3*(c + d*x)**(1/4)*(a + b*x)**(3/4)*a*b**2*x**2 + (c + d*x)**(1/4)*(a + b*x)**(3/4)*b**3*x**3),x)