Integrand size = 17, antiderivative size = 123 \[ \int \frac {(a+b x)^4}{(c+d x)^{3/2}} \, dx=-\frac {2 (b c-a d)^4}{d^5 \sqrt {c+d x}}-\frac {8 b (b c-a d)^3 \sqrt {c+d x}}{d^5}+\frac {4 b^2 (b c-a d)^2 (c+d x)^{3/2}}{d^5}-\frac {8 b^3 (b c-a d) (c+d x)^{5/2}}{5 d^5}+\frac {2 b^4 (c+d x)^{7/2}}{7 d^5} \] Output:
-2*(-a*d+b*c)^4/d^5/(d*x+c)^(1/2)-8*b*(-a*d+b*c)^3*(d*x+c)^(1/2)/d^5+4*b^2 *(-a*d+b*c)^2*(d*x+c)^(3/2)/d^5-8/5*b^3*(-a*d+b*c)*(d*x+c)^(5/2)/d^5+2/7*b ^4*(d*x+c)^(7/2)/d^5
Time = 0.07 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b x)^4}{(c+d x)^{3/2}} \, dx=\frac {2 \left (-35 a^4 d^4+140 a^3 b d^3 (2 c+d x)+70 a^2 b^2 d^2 \left (-8 c^2-4 c d x+d^2 x^2\right )+28 a b^3 d \left (16 c^3+8 c^2 d x-2 c d^2 x^2+d^3 x^3\right )+b^4 \left (-128 c^4-64 c^3 d x+16 c^2 d^2 x^2-8 c d^3 x^3+5 d^4 x^4\right )\right )}{35 d^5 \sqrt {c+d x}} \] Input:
Integrate[(a + b*x)^4/(c + d*x)^(3/2),x]
Output:
(2*(-35*a^4*d^4 + 140*a^3*b*d^3*(2*c + d*x) + 70*a^2*b^2*d^2*(-8*c^2 - 4*c *d*x + d^2*x^2) + 28*a*b^3*d*(16*c^3 + 8*c^2*d*x - 2*c*d^2*x^2 + d^3*x^3) + b^4*(-128*c^4 - 64*c^3*d*x + 16*c^2*d^2*x^2 - 8*c*d^3*x^3 + 5*d^4*x^4))) /(35*d^5*Sqrt[c + d*x])
Time = 0.22 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {53, 2009}
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.
| \(\displaystyle \int \frac {(a+b x)^4}{(c+d x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle \int \left (-\frac {4 b^3 (c+d x)^{3/2} (b c-a d)}{d^4}+\frac {6 b^2 \sqrt {c+d x} (b c-a d)^2}{d^4}-\frac {4 b (b c-a d)^3}{d^4 \sqrt {c+d x}}+\frac {(a d-b c)^4}{d^4 (c+d x)^{3/2}}+\frac {b^4 (c+d x)^{5/2}}{d^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {8 b^3 (c+d x)^{5/2} (b c-a d)}{5 d^5}+\frac {4 b^2 (c+d x)^{3/2} (b c-a d)^2}{d^5}-\frac {8 b \sqrt {c+d x} (b c-a d)^3}{d^5}-\frac {2 (b c-a d)^4}{d^5 \sqrt {c+d x}}+\frac {2 b^4 (c+d x)^{7/2}}{7 d^5}\) |
Input:
Int[(a + b*x)^4/(c + d*x)^(3/2),x]
Output:
(-2*(b*c - a*d)^4)/(d^5*Sqrt[c + d*x]) - (8*b*(b*c - a*d)^3*Sqrt[c + d*x]) /d^5 + (4*b^2*(b*c - a*d)^2*(c + d*x)^(3/2))/d^5 - (8*b^3*(b*c - a*d)*(c + d*x)^(5/2))/(5*d^5) + (2*b^4*(c + d*x)^(7/2))/(7*d^5)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Time = 0.20 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.18
| method | result | size |
| pseudoelliptic | \(\frac {\frac {2 \left (5 b^{4} x^{4}+28 a \,x^{3} b^{3}+70 a^{2} b^{2} x^{2}+140 a^{3} b x -35 a^{4}\right ) d^{4}}{35}+16 c \left (-\frac {1}{35} b^{3} x^{3}-\frac {1}{5} a \,b^{2} x^{2}-a^{2} b x +a^{3}\right ) b \,d^{3}-32 c^{2} \left (-\frac {1}{35} b^{2} x^{2}-\frac {2}{5} a b x +a^{2}\right ) b^{2} d^{2}+\frac {128 c^{3} \left (-\frac {b x}{7}+a \right ) b^{3} d}{5}-\frac {256 c^{4} b^{4}}{35}}{\sqrt {x d +c}\, d^{5}}\) | \(145\) |
| risch | \(\frac {2 b \left (5 d^{3} x^{3} b^{3}+28 x^{2} a \,b^{2} d^{3}-13 x^{2} b^{3} c \,d^{2}+70 x \,a^{2} b \,d^{3}-84 x a \,b^{2} c \,d^{2}+29 x \,b^{3} c^{2} d +140 a^{3} d^{3}-350 a^{2} b c \,d^{2}+308 a \,b^{2} c^{2} d -93 b^{3} c^{3}\right ) \sqrt {x d +c}}{35 d^{5}}-\frac {2 \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}{d^{5} \sqrt {x d +c}}\) | \(179\) |
| gosper | \(-\frac {2 \left (-5 d^{4} x^{4} b^{4}-28 a \,b^{3} d^{4} x^{3}+8 b^{4} c \,d^{3} x^{3}-70 a^{2} b^{2} d^{4} x^{2}+56 a \,b^{3} c \,d^{3} x^{2}-16 b^{4} c^{2} d^{2} x^{2}-140 a^{3} b \,d^{4} x +280 a^{2} b^{2} c \,d^{3} x -224 a \,b^{3} c^{2} d^{2} x +64 b^{4} c^{3} d x +35 d^{4} a^{4}-280 a^{3} b c \,d^{3}+560 a^{2} b^{2} c^{2} d^{2}-448 a \,b^{3} c^{3} d +128 c^{4} b^{4}\right )}{35 \sqrt {x d +c}\, d^{5}}\) | \(186\) |
| trager | \(-\frac {2 \left (-5 d^{4} x^{4} b^{4}-28 a \,b^{3} d^{4} x^{3}+8 b^{4} c \,d^{3} x^{3}-70 a^{2} b^{2} d^{4} x^{2}+56 a \,b^{3} c \,d^{3} x^{2}-16 b^{4} c^{2} d^{2} x^{2}-140 a^{3} b \,d^{4} x +280 a^{2} b^{2} c \,d^{3} x -224 a \,b^{3} c^{2} d^{2} x +64 b^{4} c^{3} d x +35 d^{4} a^{4}-280 a^{3} b c \,d^{3}+560 a^{2} b^{2} c^{2} d^{2}-448 a \,b^{3} c^{3} d +128 c^{4} b^{4}\right )}{35 \sqrt {x d +c}\, d^{5}}\) | \(186\) |
| orering | \(-\frac {2 \left (-5 d^{4} x^{4} b^{4}-28 a \,b^{3} d^{4} x^{3}+8 b^{4} c \,d^{3} x^{3}-70 a^{2} b^{2} d^{4} x^{2}+56 a \,b^{3} c \,d^{3} x^{2}-16 b^{4} c^{2} d^{2} x^{2}-140 a^{3} b \,d^{4} x +280 a^{2} b^{2} c \,d^{3} x -224 a \,b^{3} c^{2} d^{2} x +64 b^{4} c^{3} d x +35 d^{4} a^{4}-280 a^{3} b c \,d^{3}+560 a^{2} b^{2} c^{2} d^{2}-448 a \,b^{3} c^{3} d +128 c^{4} b^{4}\right )}{35 \sqrt {x d +c}\, d^{5}}\) | \(186\) |
| derivativedivides | \(\frac {\frac {2 b^{4} \left (x d +c \right )^{\frac {7}{2}}}{7}+\frac {8 a \,b^{3} d \left (x d +c \right )^{\frac {5}{2}}}{5}-\frac {8 b^{4} c \left (x d +c \right )^{\frac {5}{2}}}{5}+4 a^{2} b^{2} d^{2} \left (x d +c \right )^{\frac {3}{2}}-8 a \,b^{3} c d \left (x d +c \right )^{\frac {3}{2}}+4 b^{4} c^{2} \left (x d +c \right )^{\frac {3}{2}}+8 a^{3} b \,d^{3} \sqrt {x d +c}-24 a^{2} b^{2} c \,d^{2} \sqrt {x d +c}+24 a \,b^{3} c^{2} d \sqrt {x d +c}-8 b^{4} c^{3} \sqrt {x d +c}-\frac {2 \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}{\sqrt {x d +c}}}{d^{5}}\) | \(219\) |
| default | \(\frac {\frac {2 b^{4} \left (x d +c \right )^{\frac {7}{2}}}{7}+\frac {8 a \,b^{3} d \left (x d +c \right )^{\frac {5}{2}}}{5}-\frac {8 b^{4} c \left (x d +c \right )^{\frac {5}{2}}}{5}+4 a^{2} b^{2} d^{2} \left (x d +c \right )^{\frac {3}{2}}-8 a \,b^{3} c d \left (x d +c \right )^{\frac {3}{2}}+4 b^{4} c^{2} \left (x d +c \right )^{\frac {3}{2}}+8 a^{3} b \,d^{3} \sqrt {x d +c}-24 a^{2} b^{2} c \,d^{2} \sqrt {x d +c}+24 a \,b^{3} c^{2} d \sqrt {x d +c}-8 b^{4} c^{3} \sqrt {x d +c}-\frac {2 \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}{\sqrt {x d +c}}}{d^{5}}\) | \(219\) |
Input:
int((b*x+a)^4/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/35*((5*b^4*x^4+28*a*b^3*x^3+70*a^2*b^2*x^2+140*a^3*b*x-35*a^4)*d^4+280*c *(-1/35*b^3*x^3-1/5*a*b^2*x^2-a^2*b*x+a^3)*b*d^3-560*c^2*(-1/35*b^2*x^2-2/ 5*a*b*x+a^2)*b^2*d^2+448*c^3*(-1/7*b*x+a)*b^3*d-128*c^4*b^4)/(d*x+c)^(1/2) /d^5
Time = 0.11 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.56 \[ \int \frac {(a+b x)^4}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (5 \, b^{4} d^{4} x^{4} - 128 \, b^{4} c^{4} + 448 \, a b^{3} c^{3} d - 560 \, a^{2} b^{2} c^{2} d^{2} + 280 \, a^{3} b c d^{3} - 35 \, a^{4} d^{4} - 4 \, {\left (2 \, b^{4} c d^{3} - 7 \, a b^{3} d^{4}\right )} x^{3} + 2 \, {\left (8 \, b^{4} c^{2} d^{2} - 28 \, a b^{3} c d^{3} + 35 \, a^{2} b^{2} d^{4}\right )} x^{2} - 4 \, {\left (16 \, b^{4} c^{3} d - 56 \, a b^{3} c^{2} d^{2} + 70 \, a^{2} b^{2} c d^{3} - 35 \, a^{3} b d^{4}\right )} x\right )} \sqrt {d x + c}}{35 \, {\left (d^{6} x + c d^{5}\right )}} \] Input:
integrate((b*x+a)^4/(d*x+c)^(3/2),x, algorithm="fricas")
Output:
2/35*(5*b^4*d^4*x^4 - 128*b^4*c^4 + 448*a*b^3*c^3*d - 560*a^2*b^2*c^2*d^2 + 280*a^3*b*c*d^3 - 35*a^4*d^4 - 4*(2*b^4*c*d^3 - 7*a*b^3*d^4)*x^3 + 2*(8* b^4*c^2*d^2 - 28*a*b^3*c*d^3 + 35*a^2*b^2*d^4)*x^2 - 4*(16*b^4*c^3*d - 56* a*b^3*c^2*d^2 + 70*a^2*b^2*c*d^3 - 35*a^3*b*d^4)*x)*sqrt(d*x + c)/(d^6*x + c*d^5)
Time = 3.07 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.59 \[ \int \frac {(a+b x)^4}{(c+d x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {b^{4} \left (c + d x\right )^{\frac {7}{2}}}{7 d^{4}} + \frac {\left (c + d x\right )^{\frac {5}{2}} \cdot \left (4 a b^{3} d - 4 b^{4} c\right )}{5 d^{4}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \cdot \left (6 a^{2} b^{2} d^{2} - 12 a b^{3} c d + 6 b^{4} c^{2}\right )}{3 d^{4}} + \frac {\sqrt {c + d x} \left (4 a^{3} b d^{3} - 12 a^{2} b^{2} c d^{2} + 12 a b^{3} c^{2} d - 4 b^{4} c^{3}\right )}{d^{4}} - \frac {\left (a d - b c\right )^{4}}{d^{4} \sqrt {c + d x}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\begin {cases} a^{4} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{5}}{5 b} & \text {otherwise} \end {cases}}{c^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)**4/(d*x+c)**(3/2),x)
Output:
Piecewise((2*(b**4*(c + d*x)**(7/2)/(7*d**4) + (c + d*x)**(5/2)*(4*a*b**3* d - 4*b**4*c)/(5*d**4) + (c + d*x)**(3/2)*(6*a**2*b**2*d**2 - 12*a*b**3*c* d + 6*b**4*c**2)/(3*d**4) + sqrt(c + d*x)*(4*a**3*b*d**3 - 12*a**2*b**2*c* d**2 + 12*a*b**3*c**2*d - 4*b**4*c**3)/d**4 - (a*d - b*c)**4/(d**4*sqrt(c + d*x)))/d, Ne(d, 0)), (Piecewise((a**4*x, Eq(b, 0)), ((a + b*x)**5/(5*b), True))/c**(3/2), True))
Time = 0.04 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.54 \[ \int \frac {(a+b x)^4}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {5 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{4} - 28 \, {\left (b^{4} c - a b^{3} d\right )} {\left (d x + c\right )}^{\frac {5}{2}} + 70 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} - 140 \, {\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} \sqrt {d x + c}}{d^{4}} - \frac {35 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )}}{\sqrt {d x + c} d^{4}}\right )}}{35 \, d} \] Input:
integrate((b*x+a)^4/(d*x+c)^(3/2),x, algorithm="maxima")
Output:
2/35*((5*(d*x + c)^(7/2)*b^4 - 28*(b^4*c - a*b^3*d)*(d*x + c)^(5/2) + 70*( b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*(d*x + c)^(3/2) - 140*(b^4*c^3 - 3*a* b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3)*sqrt(d*x + c))/d^4 - 35*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(sqrt(d*x + c)*d^4))/d
Leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (109) = 218\).
Time = 0.12 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.95 \[ \int \frac {(a+b x)^4}{(c+d x)^{3/2}} \, dx=-\frac {2 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )}}{\sqrt {d x + c} d^{5}} + \frac {2 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{4} d^{30} - 28 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4} c d^{30} + 70 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{4} c^{2} d^{30} - 140 \, \sqrt {d x + c} b^{4} c^{3} d^{30} + 28 \, {\left (d x + c\right )}^{\frac {5}{2}} a b^{3} d^{31} - 140 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{3} c d^{31} + 420 \, \sqrt {d x + c} a b^{3} c^{2} d^{31} + 70 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} b^{2} d^{32} - 420 \, \sqrt {d x + c} a^{2} b^{2} c d^{32} + 140 \, \sqrt {d x + c} a^{3} b d^{33}\right )}}{35 \, d^{35}} \] Input:
integrate((b*x+a)^4/(d*x+c)^(3/2),x, algorithm="giac")
Output:
-2*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4) /(sqrt(d*x + c)*d^5) + 2/35*(5*(d*x + c)^(7/2)*b^4*d^30 - 28*(d*x + c)^(5/ 2)*b^4*c*d^30 + 70*(d*x + c)^(3/2)*b^4*c^2*d^30 - 140*sqrt(d*x + c)*b^4*c^ 3*d^30 + 28*(d*x + c)^(5/2)*a*b^3*d^31 - 140*(d*x + c)^(3/2)*a*b^3*c*d^31 + 420*sqrt(d*x + c)*a*b^3*c^2*d^31 + 70*(d*x + c)^(3/2)*a^2*b^2*d^32 - 420 *sqrt(d*x + c)*a^2*b^2*c*d^32 + 140*sqrt(d*x + c)*a^3*b*d^33)/d^35
Time = 0.03 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.24 \[ \int \frac {(a+b x)^4}{(c+d x)^{3/2}} \, dx=\frac {2\,b^4\,{\left (c+d\,x\right )}^{7/2}}{7\,d^5}-\frac {\left (8\,b^4\,c-8\,a\,b^3\,d\right )\,{\left (c+d\,x\right )}^{5/2}}{5\,d^5}-\frac {2\,a^4\,d^4-8\,a^3\,b\,c\,d^3+12\,a^2\,b^2\,c^2\,d^2-8\,a\,b^3\,c^3\,d+2\,b^4\,c^4}{d^5\,\sqrt {c+d\,x}}+\frac {4\,b^2\,{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^{3/2}}{d^5}+\frac {8\,b\,{\left (a\,d-b\,c\right )}^3\,\sqrt {c+d\,x}}{d^5} \] Input:
int((a + b*x)^4/(c + d*x)^(3/2),x)
Output:
(2*b^4*(c + d*x)^(7/2))/(7*d^5) - ((8*b^4*c - 8*a*b^3*d)*(c + d*x)^(5/2))/ (5*d^5) - (2*a^4*d^4 + 2*b^4*c^4 + 12*a^2*b^2*c^2*d^2 - 8*a*b^3*c^3*d - 8* a^3*b*c*d^3)/(d^5*(c + d*x)^(1/2)) + (4*b^2*(a*d - b*c)^2*(c + d*x)^(3/2)) /d^5 + (8*b*(a*d - b*c)^3*(c + d*x)^(1/2))/d^5
Time = 0.16 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.51 \[ \int \frac {(a+b x)^4}{(c+d x)^{3/2}} \, dx=\frac {\frac {2}{7} b^{4} d^{4} x^{4}+\frac {8}{5} a \,b^{3} d^{4} x^{3}-\frac {16}{35} b^{4} c \,d^{3} x^{3}+4 a^{2} b^{2} d^{4} x^{2}-\frac {16}{5} a \,b^{3} c \,d^{3} x^{2}+\frac {32}{35} b^{4} c^{2} d^{2} x^{2}+8 a^{3} b \,d^{4} x -16 a^{2} b^{2} c \,d^{3} x +\frac {64}{5} a \,b^{3} c^{2} d^{2} x -\frac {128}{35} b^{4} c^{3} d x -2 a^{4} d^{4}+16 a^{3} b c \,d^{3}-32 a^{2} b^{2} c^{2} d^{2}+\frac {128}{5} a \,b^{3} c^{3} d -\frac {256}{35} b^{4} c^{4}}{\sqrt {d x +c}\, d^{5}} \] Input:
int((b*x+a)^4/(d*x+c)^(3/2),x)
Output:
(2*( - 35*a**4*d**4 + 280*a**3*b*c*d**3 + 140*a**3*b*d**4*x - 560*a**2*b** 2*c**2*d**2 - 280*a**2*b**2*c*d**3*x + 70*a**2*b**2*d**4*x**2 + 448*a*b**3 *c**3*d + 224*a*b**3*c**2*d**2*x - 56*a*b**3*c*d**3*x**2 + 28*a*b**3*d**4* x**3 - 128*b**4*c**4 - 64*b**4*c**3*d*x + 16*b**4*c**2*d**2*x**2 - 8*b**4* c*d**3*x**3 + 5*b**4*d**4*x**4))/(35*sqrt(c + d*x)*d**5)