Integrand size = 19, antiderivative size = 136 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^{11/2}} \, dx=-\frac {2 (c+d x)^{3/2}}{9 (b c-a d) (a+b x)^{9/2}}+\frac {4 d (c+d x)^{3/2}}{21 (b c-a d)^2 (a+b x)^{7/2}}-\frac {16 d^2 (c+d x)^{3/2}}{105 (b c-a d)^3 (a+b x)^{5/2}}+\frac {32 d^3 (c+d x)^{3/2}}{315 (b c-a d)^4 (a+b x)^{3/2}} \] Output:
-2/9*(d*x+c)^(3/2)/(-a*d+b*c)/(b*x+a)^(9/2)+4/21*d*(d*x+c)^(3/2)/(-a*d+b*c )^2/(b*x+a)^(7/2)-16/105*d^2*(d*x+c)^(3/2)/(-a*d+b*c)^3/(b*x+a)^(5/2)+32/3 15*d^3*(d*x+c)^(3/2)/(-a*d+b*c)^4/(b*x+a)^(3/2)
Time = 0.22 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.87 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^{11/2}} \, dx=\frac {2 (c+d x)^{3/2} \left (105 a^3 d^3+63 a^2 b d^2 (-3 c+2 d x)+9 a b^2 d \left (15 c^2-12 c d x+8 d^2 x^2\right )+b^3 \left (-35 c^3+30 c^2 d x-24 c d^2 x^2+16 d^3 x^3\right )\right )}{315 (b c-a d)^4 (a+b x)^{9/2}} \] Input:
Integrate[Sqrt[c + d*x]/(a + b*x)^(11/2),x]
Output:
(2*(c + d*x)^(3/2)*(105*a^3*d^3 + 63*a^2*b*d^2*(-3*c + 2*d*x) + 9*a*b^2*d* (15*c^2 - 12*c*d*x + 8*d^2*x^2) + b^3*(-35*c^3 + 30*c^2*d*x - 24*c*d^2*x^2 + 16*d^3*x^3)))/(315*(b*c - a*d)^4*(a + b*x)^(9/2))
Time = 0.20 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.19, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {55, 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.
| \(\displaystyle \int \frac {\sqrt {c+d x}}{(a+b x)^{11/2}} \, dx\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle -\frac {2 d \int \frac {\sqrt {c+d x}}{(a+b x)^{9/2}}dx}{3 (b c-a d)}-\frac {2 (c+d x)^{3/2}}{9 (a+b x)^{9/2} (b c-a d)}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle -\frac {2 d \left (-\frac {4 d \int \frac {\sqrt {c+d x}}{(a+b x)^{7/2}}dx}{7 (b c-a d)}-\frac {2 (c+d x)^{3/2}}{7 (a+b x)^{7/2} (b c-a d)}\right )}{3 (b c-a d)}-\frac {2 (c+d x)^{3/2}}{9 (a+b x)^{9/2} (b c-a d)}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle -\frac {2 d \left (-\frac {4 d \left (-\frac {2 d \int \frac {\sqrt {c+d x}}{(a+b x)^{5/2}}dx}{5 (b c-a d)}-\frac {2 (c+d x)^{3/2}}{5 (a+b x)^{5/2} (b c-a d)}\right )}{7 (b c-a d)}-\frac {2 (c+d x)^{3/2}}{7 (a+b x)^{7/2} (b c-a d)}\right )}{3 (b c-a d)}-\frac {2 (c+d x)^{3/2}}{9 (a+b x)^{9/2} (b c-a d)}\) |
| \(\Big \downarrow \) 48 |
| \(\displaystyle -\frac {2 (c+d x)^{3/2}}{9 (a+b x)^{9/2} (b c-a d)}-\frac {2 d \left (-\frac {2 (c+d x)^{3/2}}{7 (a+b x)^{7/2} (b c-a d)}-\frac {4 d \left (\frac {4 d (c+d x)^{3/2}}{15 (a+b x)^{3/2} (b c-a d)^2}-\frac {2 (c+d x)^{3/2}}{5 (a+b x)^{5/2} (b c-a d)}\right )}{7 (b c-a d)}\right )}{3 (b c-a d)}\) |
Input:
Int[Sqrt[c + d*x]/(a + b*x)^(11/2),x]
Output:
(-2*(c + d*x)^(3/2))/(9*(b*c - a*d)*(a + b*x)^(9/2)) - (2*d*((-2*(c + d*x) ^(3/2))/(7*(b*c - a*d)*(a + b*x)^(7/2)) - (4*d*((-2*(c + d*x)^(3/2))/(5*(b *c - a*d)*(a + b*x)^(5/2)) + (4*d*(c + d*x)^(3/2))/(15*(b*c - a*d)^2*(a + b*x)^(3/2))))/(7*(b*c - a*d))))/(3*(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 = 0.15 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.26
| method | result | size |
| gosper | \(\frac {2 \left (x d +c \right )^{\frac {3}{2}} \left (16 d^{3} x^{3} b^{3}+72 x^{2} a \,b^{2} d^{3}-24 x^{2} b^{3} c \,d^{2}+126 x \,a^{2} b \,d^{3}-108 x a \,b^{2} c \,d^{2}+30 x \,b^{3} c^{2} d +105 a^{3} d^{3}-189 a^{2} b c \,d^{2}+135 a \,b^{2} c^{2} d -35 b^{3} c^{3}\right )}{315 \left (b x +a \right )^{\frac {9}{2}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\) | \(171\) |
| orering | \(\frac {2 \left (x d +c \right )^{\frac {3}{2}} \left (16 d^{3} x^{3} b^{3}+72 x^{2} a \,b^{2} d^{3}-24 x^{2} b^{3} c \,d^{2}+126 x \,a^{2} b \,d^{3}-108 x a \,b^{2} c \,d^{2}+30 x \,b^{3} c^{2} d +105 a^{3} d^{3}-189 a^{2} b c \,d^{2}+135 a \,b^{2} c^{2} d -35 b^{3} c^{3}\right )}{315 \left (b x +a \right )^{\frac {9}{2}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\) | \(171\) |
| default | \(-\frac {\sqrt {x d +c}}{4 b \left (b x +a \right )^{\frac {9}{2}}}+\frac {\left (a d -b c \right ) \left (-\frac {2 \sqrt {x d +c}}{9 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {9}{2}}}-\frac {8 d \left (-\frac {2 \sqrt {x d +c}}{7 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {7}{2}}}-\frac {6 d \left (-\frac {2 \sqrt {x d +c}}{5 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {5}{2}}}-\frac {4 d \left (-\frac {2 \sqrt {x d +c}}{3 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {3}{2}}}+\frac {4 d \sqrt {x d +c}}{3 \left (-a d +b c \right )^{2} \sqrt {b x +a}}\right )}{5 \left (-a d +b c \right )}\right )}{7 \left (-a d +b c \right )}\right )}{9 \left (-a d +b c \right )}\right )}{8 b}\) | \(208\) |
Input:
int((d*x+c)^(1/2)/(b*x+a)^(11/2),x,method=_RETURNVERBOSE)
Output:
2/315*(d*x+c)^(3/2)*(16*b^3*d^3*x^3+72*a*b^2*d^3*x^2-24*b^3*c*d^2*x^2+126* a^2*b*d^3*x-108*a*b^2*c*d^2*x+30*b^3*c^2*d*x+105*a^3*d^3-189*a^2*b*c*d^2+1 35*a*b^2*c^2*d-35*b^3*c^3)/(b*x+a)^(9/2)/(a^4*d^4-4*a^3*b*c*d^3+6*a^2*b^2* c^2*d^2-4*a*b^3*c^3*d+b^4*c^4)
Leaf count of result is larger than twice the leaf count of optimal. 532 vs. \(2 (112) = 224\).
Time = 2.16 (sec) , antiderivative size = 532, normalized size of antiderivative = 3.91 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^{11/2}} \, dx=\frac {2 \, {\left (16 \, b^{3} d^{4} x^{4} - 35 \, b^{3} c^{4} + 135 \, a b^{2} c^{3} d - 189 \, a^{2} b c^{2} d^{2} + 105 \, a^{3} c d^{3} - 8 \, {\left (b^{3} c d^{3} - 9 \, a b^{2} d^{4}\right )} x^{3} + 6 \, {\left (b^{3} c^{2} d^{2} - 6 \, a b^{2} c d^{3} + 21 \, a^{2} b d^{4}\right )} x^{2} - {\left (5 \, b^{3} c^{3} d - 27 \, a b^{2} c^{2} d^{2} + 63 \, a^{2} b c d^{3} - 105 \, a^{3} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{315 \, {\left (a^{5} b^{4} c^{4} - 4 \, a^{6} b^{3} c^{3} d + 6 \, a^{7} b^{2} c^{2} d^{2} - 4 \, a^{8} b c d^{3} + a^{9} d^{4} + {\left (b^{9} c^{4} - 4 \, a b^{8} c^{3} d + 6 \, a^{2} b^{7} c^{2} d^{2} - 4 \, a^{3} b^{6} c d^{3} + a^{4} b^{5} d^{4}\right )} x^{5} + 5 \, {\left (a b^{8} c^{4} - 4 \, a^{2} b^{7} c^{3} d + 6 \, a^{3} b^{6} c^{2} d^{2} - 4 \, a^{4} b^{5} c d^{3} + a^{5} b^{4} d^{4}\right )} x^{4} + 10 \, {\left (a^{2} b^{7} c^{4} - 4 \, a^{3} b^{6} c^{3} d + 6 \, a^{4} b^{5} c^{2} d^{2} - 4 \, a^{5} b^{4} c d^{3} + a^{6} b^{3} d^{4}\right )} x^{3} + 10 \, {\left (a^{3} b^{6} c^{4} - 4 \, a^{4} b^{5} c^{3} d + 6 \, a^{5} b^{4} c^{2} d^{2} - 4 \, a^{6} b^{3} c d^{3} + a^{7} b^{2} d^{4}\right )} x^{2} + 5 \, {\left (a^{4} b^{5} c^{4} - 4 \, a^{5} b^{4} c^{3} d + 6 \, a^{6} b^{3} c^{2} d^{2} - 4 \, a^{7} b^{2} c d^{3} + a^{8} b d^{4}\right )} x\right )}} \] Input:
integrate((d*x+c)^(1/2)/(b*x+a)^(11/2),x, algorithm="fricas")
Output:
2/315*(16*b^3*d^4*x^4 - 35*b^3*c^4 + 135*a*b^2*c^3*d - 189*a^2*b*c^2*d^2 + 105*a^3*c*d^3 - 8*(b^3*c*d^3 - 9*a*b^2*d^4)*x^3 + 6*(b^3*c^2*d^2 - 6*a*b^ 2*c*d^3 + 21*a^2*b*d^4)*x^2 - (5*b^3*c^3*d - 27*a*b^2*c^2*d^2 + 63*a^2*b*c *d^3 - 105*a^3*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a^5*b^4*c^4 - 4*a^6*b^ 3*c^3*d + 6*a^7*b^2*c^2*d^2 - 4*a^8*b*c*d^3 + a^9*d^4 + (b^9*c^4 - 4*a*b^8 *c^3*d + 6*a^2*b^7*c^2*d^2 - 4*a^3*b^6*c*d^3 + a^4*b^5*d^4)*x^5 + 5*(a*b^8 *c^4 - 4*a^2*b^7*c^3*d + 6*a^3*b^6*c^2*d^2 - 4*a^4*b^5*c*d^3 + a^5*b^4*d^4 )*x^4 + 10*(a^2*b^7*c^4 - 4*a^3*b^6*c^3*d + 6*a^4*b^5*c^2*d^2 - 4*a^5*b^4* c*d^3 + a^6*b^3*d^4)*x^3 + 10*(a^3*b^6*c^4 - 4*a^4*b^5*c^3*d + 6*a^5*b^4*c ^2*d^2 - 4*a^6*b^3*c*d^3 + a^7*b^2*d^4)*x^2 + 5*(a^4*b^5*c^4 - 4*a^5*b^4*c ^3*d + 6*a^6*b^3*c^2*d^2 - 4*a^7*b^2*c*d^3 + a^8*b*d^4)*x)
\[ \int \frac {\sqrt {c+d x}}{(a+b x)^{11/2}} \, dx=\int \frac {\sqrt {c + d x}}{\left (a + b x\right )^{\frac {11}{2}}}\, dx \] Input:
integrate((d*x+c)**(1/2)/(b*x+a)**(11/2),x)
Output:
Integral(sqrt(c + d*x)/(a + b*x)**(11/2), x)
Exception generated. \[ \int \frac {\sqrt {c+d x}}{(a+b x)^{11/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((d*x+c)^(1/2)/(b*x+a)^(11/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 989 vs. \(2 (112) = 224\).
Time = 0.24 (sec) , antiderivative size = 989, normalized size of antiderivative = 7.27 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^{11/2}} \, dx =\text {Too large to display} \] Input:
integrate((d*x+c)^(1/2)/(b*x+a)^(11/2),x, algorithm="giac")
Output:
64/315*(sqrt(b*d)*b^13*c^5*d^4 - 5*sqrt(b*d)*a*b^12*c^4*d^5 + 10*sqrt(b*d) *a^2*b^11*c^3*d^6 - 10*sqrt(b*d)*a^3*b^10*c^2*d^7 + 5*sqrt(b*d)*a^4*b^9*c* d^8 - sqrt(b*d)*a^5*b^8*d^9 - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt( b^2*c + (b*x + a)*b*d - a*b*d))^2*b^11*c^4*d^4 + 36*sqrt(b*d)*(sqrt(b*d)*s qrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^10*c^3*d^5 - 54* sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^ 2*a^2*b^9*c^2*d^6 + 36*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + ( b*x + a)*b*d - a*b*d))^2*a^3*b^8*c*d^7 - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^7*d^8 + 36*sqrt(b*d)*(s qrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^9*c^3*d^ 4 - 108*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b^8*c^2*d^5 + 108*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^ 2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b^7*c*d^6 - 36*sqrt(b*d)*(sqrt(b*d)*sq rt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^3*b^6*d^7 - 84*sqrt (b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^ 7*c^2*d^4 + 168*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a )*b*d - a*b*d))^6*a*b^6*c*d^5 - 84*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sq rt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^2*b^5*d^6 - 189*sqrt(b*d)*(sqrt(b*d )*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*b^5*c*d^4 + 189*s qrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)...
Time = 0.60 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.15 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^{11/2}} \, dx=\frac {\sqrt {c+d\,x}\,\left (\frac {32\,d^4\,x^4}{315\,b\,{\left (a\,d-b\,c\right )}^4}-\frac {-210\,a^3\,c\,d^3+378\,a^2\,b\,c^2\,d^2-270\,a\,b^2\,c^3\,d+70\,b^3\,c^4}{315\,b^4\,{\left (a\,d-b\,c\right )}^4}+\frac {x\,\left (210\,a^3\,d^4-126\,a^2\,b\,c\,d^3+54\,a\,b^2\,c^2\,d^2-10\,b^3\,c^3\,d\right )}{315\,b^4\,{\left (a\,d-b\,c\right )}^4}+\frac {16\,d^3\,x^3\,\left (9\,a\,d-b\,c\right )}{315\,b^2\,{\left (a\,d-b\,c\right )}^4}+\frac {4\,d^2\,x^2\,\left (21\,a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2\right )}{105\,b^3\,{\left (a\,d-b\,c\right )}^4}\right )}{x^4\,\sqrt {a+b\,x}+\frac {a^4\,\sqrt {a+b\,x}}{b^4}+\frac {6\,a^2\,x^2\,\sqrt {a+b\,x}}{b^2}+\frac {4\,a\,x^3\,\sqrt {a+b\,x}}{b}+\frac {4\,a^3\,x\,\sqrt {a+b\,x}}{b^3}} \] Input:
int((c + d*x)^(1/2)/(a + b*x)^(11/2),x)
Output:
((c + d*x)^(1/2)*((32*d^4*x^4)/(315*b*(a*d - b*c)^4) - (70*b^3*c^4 - 210*a ^3*c*d^3 + 378*a^2*b*c^2*d^2 - 270*a*b^2*c^3*d)/(315*b^4*(a*d - b*c)^4) + (x*(210*a^3*d^4 - 10*b^3*c^3*d + 54*a*b^2*c^2*d^2 - 126*a^2*b*c*d^3))/(315 *b^4*(a*d - b*c)^4) + (16*d^3*x^3*(9*a*d - b*c))/(315*b^2*(a*d - b*c)^4) + (4*d^2*x^2*(21*a^2*d^2 + b^2*c^2 - 6*a*b*c*d))/(105*b^3*(a*d - b*c)^4)))/ (x^4*(a + b*x)^(1/2) + (a^4*(a + b*x)^(1/2))/b^4 + (6*a^2*x^2*(a + b*x)^(1 /2))/b^2 + (4*a*x^3*(a + b*x)^(1/2))/b + (4*a^3*x*(a + b*x)^(1/2))/b^3)
Time = 0.19 (sec) , antiderivative size = 706, normalized size of antiderivative = 5.19 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^{11/2}} \, dx=\frac {-\frac {32 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a^{4} d^{4}}{315}-\frac {128 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a^{3} b \,d^{4} x}{315}-\frac {64 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a^{2} b^{2} d^{4} x^{2}}{105}-\frac {128 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a \,b^{3} d^{4} x^{3}}{315}-\frac {32 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, b^{4} d^{4} x^{4}}{315}+\frac {2 \sqrt {d x +c}\, a^{3} b^{2} c \,d^{3}}{3}+\frac {2 \sqrt {d x +c}\, a^{3} b^{2} d^{4} x}{3}-\frac {6 \sqrt {d x +c}\, a^{2} b^{3} c^{2} d^{2}}{5}-\frac {2 \sqrt {d x +c}\, a^{2} b^{3} c \,d^{3} x}{5}+\frac {4 \sqrt {d x +c}\, a^{2} b^{3} d^{4} x^{2}}{5}+\frac {6 \sqrt {d x +c}\, a \,b^{4} c^{3} d}{7}+\frac {6 \sqrt {d x +c}\, a \,b^{4} c^{2} d^{2} x}{35}-\frac {8 \sqrt {d x +c}\, a \,b^{4} c \,d^{3} x^{2}}{35}+\frac {16 \sqrt {d x +c}\, a \,b^{4} d^{4} x^{3}}{35}-\frac {2 \sqrt {d x +c}\, b^{5} c^{4}}{9}-\frac {2 \sqrt {d x +c}\, b^{5} c^{3} d x}{63}+\frac {4 \sqrt {d x +c}\, b^{5} c^{2} d^{2} x^{2}}{105}-\frac {16 \sqrt {d x +c}\, b^{5} c \,d^{3} x^{3}}{315}+\frac {32 \sqrt {d x +c}\, b^{5} d^{4} x^{4}}{315}}{\sqrt {b x +a}\, b^{2} \left (a^{4} b^{4} d^{4} x^{4}-4 a^{3} b^{5} c \,d^{3} x^{4}+6 a^{2} b^{6} c^{2} d^{2} x^{4}-4 a \,b^{7} c^{3} d \,x^{4}+b^{8} c^{4} x^{4}+4 a^{5} b^{3} d^{4} x^{3}-16 a^{4} b^{4} c \,d^{3} x^{3}+24 a^{3} b^{5} c^{2} d^{2} x^{3}-16 a^{2} b^{6} c^{3} d \,x^{3}+4 a \,b^{7} c^{4} x^{3}+6 a^{6} b^{2} d^{4} x^{2}-24 a^{5} b^{3} c \,d^{3} x^{2}+36 a^{4} b^{4} c^{2} d^{2} x^{2}-24 a^{3} b^{5} c^{3} d \,x^{2}+6 a^{2} b^{6} c^{4} x^{2}+4 a^{7} b \,d^{4} x -16 a^{6} b^{2} c \,d^{3} x +24 a^{5} b^{3} c^{2} d^{2} x -16 a^{4} b^{4} c^{3} d x +4 a^{3} b^{5} c^{4} x +a^{8} d^{4}-4 a^{7} b c \,d^{3}+6 a^{6} b^{2} c^{2} d^{2}-4 a^{5} b^{3} c^{3} d +a^{4} b^{4} c^{4}\right )} \] Input:
int((d*x+c)^(1/2)/(b*x+a)^(11/2),x)
Output:
(2*( - 16*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**4*d**4 - 64*sqrt(d)*sqrt(b)*sqr t(a + b*x)*a**3*b*d**4*x - 96*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**2*b**2*d**4 *x**2 - 64*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b**3*d**4*x**3 - 16*sqrt(d)*sqr t(b)*sqrt(a + b*x)*b**4*d**4*x**4 + 105*sqrt(c + d*x)*a**3*b**2*c*d**3 + 1 05*sqrt(c + d*x)*a**3*b**2*d**4*x - 189*sqrt(c + d*x)*a**2*b**3*c**2*d**2 - 63*sqrt(c + d*x)*a**2*b**3*c*d**3*x + 126*sqrt(c + d*x)*a**2*b**3*d**4*x **2 + 135*sqrt(c + d*x)*a*b**4*c**3*d + 27*sqrt(c + d*x)*a*b**4*c**2*d**2* x - 36*sqrt(c + d*x)*a*b**4*c*d**3*x**2 + 72*sqrt(c + d*x)*a*b**4*d**4*x** 3 - 35*sqrt(c + d*x)*b**5*c**4 - 5*sqrt(c + d*x)*b**5*c**3*d*x + 6*sqrt(c + d*x)*b**5*c**2*d**2*x**2 - 8*sqrt(c + d*x)*b**5*c*d**3*x**3 + 16*sqrt(c + d*x)*b**5*d**4*x**4))/(315*sqrt(a + b*x)*b**2*(a**8*d**4 - 4*a**7*b*c*d* *3 + 4*a**7*b*d**4*x + 6*a**6*b**2*c**2*d**2 - 16*a**6*b**2*c*d**3*x + 6*a **6*b**2*d**4*x**2 - 4*a**5*b**3*c**3*d + 24*a**5*b**3*c**2*d**2*x - 24*a* *5*b**3*c*d**3*x**2 + 4*a**5*b**3*d**4*x**3 + a**4*b**4*c**4 - 16*a**4*b** 4*c**3*d*x + 36*a**4*b**4*c**2*d**2*x**2 - 16*a**4*b**4*c*d**3*x**3 + a**4 *b**4*d**4*x**4 + 4*a**3*b**5*c**4*x - 24*a**3*b**5*c**3*d*x**2 + 24*a**3* b**5*c**2*d**2*x**3 - 4*a**3*b**5*c*d**3*x**4 + 6*a**2*b**6*c**4*x**2 - 16 *a**2*b**6*c**3*d*x**3 + 6*a**2*b**6*c**2*d**2*x**4 + 4*a*b**7*c**4*x**3 - 4*a*b**7*c**3*d*x**4 + b**8*c**4*x**4))