Integrand size = 19, antiderivative size = 101 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{21/4}} \, dx=-\frac {4 (c+d x)^{9/4}}{17 (b c-a d) (a+b x)^{17/4}}+\frac {32 d (c+d x)^{9/4}}{221 (b c-a d)^2 (a+b x)^{13/4}}-\frac {128 d^2 (c+d x)^{9/4}}{1989 (b c-a d)^3 (a+b x)^{9/4}} \] Output:
-4/17*(d*x+c)^(9/4)/(-a*d+b*c)/(b*x+a)^(17/4)+32/221*d*(d*x+c)^(9/4)/(-a*d +b*c)^2/(b*x+a)^(13/4)-128/1989*d^2*(d*x+c)^(9/4)/(-a*d+b*c)^3/(b*x+a)^(9/ 4)
Time = 0.85 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{21/4}} \, dx=-\frac {4 (c+d x)^{9/4} \left (221 a^2 d^2+34 a b d (-9 c+4 d x)+b^2 \left (117 c^2-72 c d x+32 d^2 x^2\right )\right )}{1989 (b c-a d)^3 (a+b x)^{17/4}} \] Input:
Integrate[(c + d*x)^(5/4)/(a + b*x)^(21/4),x]
Output:
(-4*(c + d*x)^(9/4)*(221*a^2*d^2 + 34*a*b*d*(-9*c + 4*d*x) + b^2*(117*c^2 - 72*c*d*x + 32*d^2*x^2)))/(1989*(b*c - a*d)^3*(a + b*x)^(17/4))
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.
| \(\displaystyle \int \frac {(c+d x)^{5/4}}{(a+b x)^{21/4}} \, dx\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle -\frac {8 d \int \frac {(c+d x)^{5/4}}{(a+b x)^{17/4}}dx}{17 (b c-a d)}-\frac {4 (c+d x)^{9/4}}{17 (a+b x)^{17/4} (b c-a d)}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle -\frac {8 d \left (-\frac {4 d \int \frac {(c+d x)^{5/4}}{(a+b x)^{13/4}}dx}{13 (b c-a d)}-\frac {4 (c+d x)^{9/4}}{13 (a+b x)^{13/4} (b c-a d)}\right )}{17 (b c-a d)}-\frac {4 (c+d x)^{9/4}}{17 (a+b x)^{17/4} (b c-a d)}\) |
| \(\Big \downarrow \) 48 |
| \(\displaystyle -\frac {4 (c+d x)^{9/4}}{17 (a+b x)^{17/4} (b c-a d)}-\frac {8 d \left (\frac {16 d (c+d x)^{9/4}}{117 (a+b x)^{9/4} (b c-a d)^2}-\frac {4 (c+d x)^{9/4}}{13 (a+b x)^{13/4} (b c-a d)}\right )}{17 (b c-a d)}\) |
Input:
Int[(c + d*x)^(5/4)/(a + b*x)^(21/4),x]
Output:
(-4*(c + d*x)^(9/4))/(17*(b*c - a*d)*(a + b*x)^(17/4)) - (8*d*((-4*(c + d* x)^(9/4))/(13*(b*c - a*d)*(a + b*x)^(13/4)) + (16*d*(c + d*x)^(9/4))/(117* (b*c - a*d)^2*(a + b*x)^(9/4))))/(17*(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 = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04
| method | result | size |
| gosper | \(\frac {4 \left (x d +c \right )^{\frac {9}{4}} \left (32 d^{2} x^{2} b^{2}+136 x a b \,d^{2}-72 x \,b^{2} c d +221 a^{2} d^{2}-306 a b c d +117 b^{2} c^{2}\right )}{1989 \left (b x +a \right )^{\frac {17}{4}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
| orering | \(\frac {4 \left (x d +c \right )^{\frac {9}{4}} \left (32 d^{2} x^{2} b^{2}+136 x a b \,d^{2}-72 x \,b^{2} c d +221 a^{2} d^{2}-306 a b c d +117 b^{2} c^{2}\right )}{1989 \left (b x +a \right )^{\frac {17}{4}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
Input:
int((d*x+c)^(5/4)/(b*x+a)^(21/4),x,method=_RETURNVERBOSE)
Output:
4/1989*(d*x+c)^(9/4)*(32*b^2*d^2*x^2+136*a*b*d^2*x-72*b^2*c*d*x+221*a^2*d^ 2-306*a*b*c*d+117*b^2*c^2)/(b*x+a)^(17/4)/(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. 426 vs. \(2 (83) = 166\).
Time = 0.16 (sec) , antiderivative size = 426, normalized size of antiderivative = 4.22 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{21/4}} \, dx=-\frac {4 \, {\left (32 \, b^{2} d^{4} x^{4} + 117 \, b^{2} c^{4} - 306 \, a b c^{3} d + 221 \, a^{2} c^{2} d^{2} - 8 \, {\left (b^{2} c d^{3} - 17 \, a b d^{4}\right )} x^{3} + {\left (5 \, b^{2} c^{2} d^{2} - 34 \, a b c d^{3} + 221 \, a^{2} d^{4}\right )} x^{2} + 2 \, {\left (81 \, b^{2} c^{3} d - 238 \, a b c^{2} d^{2} + 221 \, a^{2} c d^{3}\right )} x\right )} {\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}}}{1989 \, {\left (a^{5} b^{3} c^{3} - 3 \, a^{6} b^{2} c^{2} d + 3 \, a^{7} b c d^{2} - a^{8} d^{3} + {\left (b^{8} c^{3} - 3 \, a b^{7} c^{2} d + 3 \, a^{2} b^{6} c d^{2} - a^{3} b^{5} d^{3}\right )} x^{5} + 5 \, {\left (a b^{7} c^{3} - 3 \, a^{2} b^{6} c^{2} d + 3 \, a^{3} b^{5} c d^{2} - a^{4} b^{4} d^{3}\right )} x^{4} + 10 \, {\left (a^{2} b^{6} c^{3} - 3 \, a^{3} b^{5} c^{2} d + 3 \, a^{4} b^{4} c d^{2} - a^{5} b^{3} d^{3}\right )} x^{3} + 10 \, {\left (a^{3} b^{5} c^{3} - 3 \, a^{4} b^{4} c^{2} d + 3 \, a^{5} b^{3} c d^{2} - a^{6} b^{2} d^{3}\right )} x^{2} + 5 \, {\left (a^{4} b^{4} c^{3} - 3 \, a^{5} b^{3} c^{2} d + 3 \, a^{6} b^{2} c d^{2} - a^{7} b d^{3}\right )} x\right )}} \] Input:
integrate((d*x+c)^(5/4)/(b*x+a)^(21/4),x, algorithm="fricas")
Output:
-4/1989*(32*b^2*d^4*x^4 + 117*b^2*c^4 - 306*a*b*c^3*d + 221*a^2*c^2*d^2 - 8*(b^2*c*d^3 - 17*a*b*d^4)*x^3 + (5*b^2*c^2*d^2 - 34*a*b*c*d^3 + 221*a^2*d ^4)*x^2 + 2*(81*b^2*c^3*d - 238*a*b*c^2*d^2 + 221*a^2*c*d^3)*x)*(b*x + a)^ (3/4)*(d*x + c)^(1/4)/(a^5*b^3*c^3 - 3*a^6*b^2*c^2*d + 3*a^7*b*c*d^2 - a^8 *d^3 + (b^8*c^3 - 3*a*b^7*c^2*d + 3*a^2*b^6*c*d^2 - a^3*b^5*d^3)*x^5 + 5*( a*b^7*c^3 - 3*a^2*b^6*c^2*d + 3*a^3*b^5*c*d^2 - a^4*b^4*d^3)*x^4 + 10*(a^2 *b^6*c^3 - 3*a^3*b^5*c^2*d + 3*a^4*b^4*c*d^2 - a^5*b^3*d^3)*x^3 + 10*(a^3* b^5*c^3 - 3*a^4*b^4*c^2*d + 3*a^5*b^3*c*d^2 - a^6*b^2*d^3)*x^2 + 5*(a^4*b^ 4*c^3 - 3*a^5*b^3*c^2*d + 3*a^6*b^2*c*d^2 - a^7*b*d^3)*x)
Timed out. \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{21/4}} \, dx=\text {Timed out} \] Input:
integrate((d*x+c)**(5/4)/(b*x+a)**(21/4),x)
Output:
Timed out
\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{21/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {21}{4}}} \,d x } \] Input:
integrate((d*x+c)^(5/4)/(b*x+a)^(21/4),x, algorithm="maxima")
Output:
integrate((d*x + c)^(5/4)/(b*x + a)^(21/4), x)
\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{21/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {21}{4}}} \,d x } \] Input:
integrate((d*x+c)^(5/4)/(b*x+a)^(21/4),x, algorithm="giac")
Output:
integrate((d*x + c)^(5/4)/(b*x + a)^(21/4), x)
Time = 0.64 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.65 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{21/4}} \, dx=\frac {{\left (c+d\,x\right )}^{1/4}\,\left (\frac {884\,a^2\,c^2\,d^2-1224\,a\,b\,c^3\,d+468\,b^2\,c^4}{1989\,b^4\,{\left (a\,d-b\,c\right )}^3}+\frac {x^2\,\left (884\,a^2\,d^4-136\,a\,b\,c\,d^3+20\,b^2\,c^2\,d^2\right )}{1989\,b^4\,{\left (a\,d-b\,c\right )}^3}+\frac {128\,d^4\,x^4}{1989\,b^2\,{\left (a\,d-b\,c\right )}^3}+\frac {32\,d^3\,x^3\,\left (17\,a\,d-b\,c\right )}{1989\,b^3\,{\left (a\,d-b\,c\right )}^3}+\frac {8\,c\,d\,x\,\left (221\,a^2\,d^2-238\,a\,b\,c\,d+81\,b^2\,c^2\right )}{1989\,b^4\,{\left (a\,d-b\,c\right )}^3}\right )}{x^4\,{\left (a+b\,x\right )}^{1/4}+\frac {a^4\,{\left (a+b\,x\right )}^{1/4}}{b^4}+\frac {6\,a^2\,x^2\,{\left (a+b\,x\right )}^{1/4}}{b^2}+\frac {4\,a\,x^3\,{\left (a+b\,x\right )}^{1/4}}{b}+\frac {4\,a^3\,x\,{\left (a+b\,x\right )}^{1/4}}{b^3}} \] Input:
int((c + d*x)^(5/4)/(a + b*x)^(21/4),x)
Output:
((c + d*x)^(1/4)*((468*b^2*c^4 + 884*a^2*c^2*d^2 - 1224*a*b*c^3*d)/(1989*b ^4*(a*d - b*c)^3) + (x^2*(884*a^2*d^4 + 20*b^2*c^2*d^2 - 136*a*b*c*d^3))/( 1989*b^4*(a*d - b*c)^3) + (128*d^4*x^4)/(1989*b^2*(a*d - b*c)^3) + (32*d^3 *x^3*(17*a*d - b*c))/(1989*b^3*(a*d - b*c)^3) + (8*c*d*x*(221*a^2*d^2 + 81 *b^2*c^2 - 238*a*b*c*d))/(1989*b^4*(a*d - b*c)^3)))/(x^4*(a + b*x)^(1/4) + (a^4*(a + b*x)^(1/4))/b^4 + (6*a^2*x^2*(a + b*x)^(1/4))/b^2 + (4*a*x^3*(a + b*x)^(1/4))/b + (4*a^3*x*(a + b*x)^(1/4))/b^3)
Time = 0.27 (sec) , antiderivative size = 401, normalized size of antiderivative = 3.97 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{21/4}} \, dx=\frac {4 \left (d x +c \right )^{\frac {1}{4}} \left (32 b^{2} d^{4} x^{4}+136 a b \,d^{4} x^{3}-8 b^{2} c \,d^{3} x^{3}+221 a^{2} d^{4} x^{2}-34 a b c \,d^{3} x^{2}+5 b^{2} c^{2} d^{2} x^{2}+442 a^{2} c \,d^{3} x -476 a b \,c^{2} d^{2} x +162 b^{2} c^{3} d x +221 a^{2} c^{2} d^{2}-306 a b \,c^{3} d +117 b^{2} c^{4}\right )}{1989 \left (b x +a \right )^{\frac {1}{4}} \left (a^{3} b^{4} d^{3} x^{4}-3 a^{2} b^{5} c \,d^{2} x^{4}+3 a \,b^{6} c^{2} d \,x^{4}-b^{7} c^{3} x^{4}+4 a^{4} b^{3} d^{3} x^{3}-12 a^{3} b^{4} c \,d^{2} x^{3}+12 a^{2} b^{5} c^{2} d \,x^{3}-4 a \,b^{6} c^{3} x^{3}+6 a^{5} b^{2} d^{3} x^{2}-18 a^{4} b^{3} c \,d^{2} x^{2}+18 a^{3} b^{4} c^{2} d \,x^{2}-6 a^{2} b^{5} c^{3} x^{2}+4 a^{6} b \,d^{3} x -12 a^{5} b^{2} c \,d^{2} x +12 a^{4} b^{3} c^{2} d x -4 a^{3} b^{4} c^{3} x +a^{7} d^{3}-3 a^{6} b c \,d^{2}+3 a^{5} b^{2} c^{2} d -a^{4} b^{3} c^{3}\right )} \] Input:
int((d*x+c)^(5/4)/(b*x+a)^(21/4),x)
Output:
(4*(c + d*x)**(1/4)*(221*a**2*c**2*d**2 + 442*a**2*c*d**3*x + 221*a**2*d** 4*x**2 - 306*a*b*c**3*d - 476*a*b*c**2*d**2*x - 34*a*b*c*d**3*x**2 + 136*a *b*d**4*x**3 + 117*b**2*c**4 + 162*b**2*c**3*d*x + 5*b**2*c**2*d**2*x**2 - 8*b**2*c*d**3*x**3 + 32*b**2*d**4*x**4))/(1989*(a + b*x)**(1/4)*(a**7*d** 3 - 3*a**6*b*c*d**2 + 4*a**6*b*d**3*x + 3*a**5*b**2*c**2*d - 12*a**5*b**2* c*d**2*x + 6*a**5*b**2*d**3*x**2 - a**4*b**3*c**3 + 12*a**4*b**3*c**2*d*x - 18*a**4*b**3*c*d**2*x**2 + 4*a**4*b**3*d**3*x**3 - 4*a**3*b**4*c**3*x + 18*a**3*b**4*c**2*d*x**2 - 12*a**3*b**4*c*d**2*x**3 + a**3*b**4*d**3*x**4 - 6*a**2*b**5*c**3*x**2 + 12*a**2*b**5*c**2*d*x**3 - 3*a**2*b**5*c*d**2*x* *4 - 4*a*b**6*c**3*x**3 + 3*a*b**6*c**2*d*x**4 - b**7*c**3*x**4))