Integrand size = 19, antiderivative size = 136 \[ \int \frac {1}{(a+b x)^{9/2} \sqrt {c+d x}} \, dx=-\frac {2 \sqrt {c+d x}}{7 (b c-a d) (a+b x)^{7/2}}+\frac {12 d \sqrt {c+d x}}{35 (b c-a d)^2 (a+b x)^{5/2}}-\frac {16 d^2 \sqrt {c+d x}}{35 (b c-a d)^3 (a+b x)^{3/2}}+\frac {32 d^3 \sqrt {c+d x}}{35 (b c-a d)^4 \sqrt {a+b x}} \] Output:
-2/7*(d*x+c)^(1/2)/(-a*d+b*c)/(b*x+a)^(7/2)+12/35*d*(d*x+c)^(1/2)/(-a*d+b* c)^2/(b*x+a)^(5/2)-16/35*d^2*(d*x+c)^(1/2)/(-a*d+b*c)^3/(b*x+a)^(3/2)+32/3 5*d^3*(d*x+c)^(1/2)/(-a*d+b*c)^4/(b*x+a)^(1/2)
Time = 0.17 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(a+b x)^{9/2} \sqrt {c+d x}} \, dx=\frac {2 \sqrt {c+d x} \left (35 a^3 d^3-35 a^2 b d^2 (c-2 d x)+7 a b^2 d \left (3 c^2-4 c d x+8 d^2 x^2\right )+b^3 \left (-5 c^3+6 c^2 d x-8 c d^2 x^2+16 d^3 x^3\right )\right )}{35 (b c-a d)^4 (a+b x)^{7/2}} \] Input:
Integrate[1/((a + b*x)^(9/2)*Sqrt[c + d*x]),x]
Output:
(2*Sqrt[c + d*x]*(35*a^3*d^3 - 35*a^2*b*d^2*(c - 2*d*x) + 7*a*b^2*d*(3*c^2 - 4*c*d*x + 8*d^2*x^2) + b^3*(-5*c^3 + 6*c^2*d*x - 8*c*d^2*x^2 + 16*d^3*x ^3)))/(35*(b*c - a*d)^4*(a + b*x)^(7/2))
Time = 0.21 (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.
| \(\displaystyle \int \frac {1}{(a+b x)^{9/2} \sqrt {c+d x}} \, dx\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle -\frac {6 d \int \frac {1}{(a+b x)^{7/2} \sqrt {c+d x}}dx}{7 (b c-a d)}-\frac {2 \sqrt {c+d x}}{7 (a+b x)^{7/2} (b c-a d)}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle -\frac {6 d \left (-\frac {4 d \int \frac {1}{(a+b x)^{5/2} \sqrt {c+d x}}dx}{5 (b c-a d)}-\frac {2 \sqrt {c+d x}}{5 (a+b x)^{5/2} (b c-a d)}\right )}{7 (b c-a d)}-\frac {2 \sqrt {c+d x}}{7 (a+b x)^{7/2} (b c-a d)}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle -\frac {6 d \left (-\frac {4 d \left (-\frac {2 d \int \frac {1}{(a+b x)^{3/2} \sqrt {c+d x}}dx}{3 (b c-a d)}-\frac {2 \sqrt {c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\right )}{5 (b c-a d)}-\frac {2 \sqrt {c+d x}}{5 (a+b x)^{5/2} (b c-a d)}\right )}{7 (b c-a d)}-\frac {2 \sqrt {c+d x}}{7 (a+b x)^{7/2} (b c-a d)}\) |
| \(\Big \downarrow \) 48 |
| \(\displaystyle -\frac {6 d \left (-\frac {4 d \left (\frac {4 d \sqrt {c+d x}}{3 \sqrt {a+b x} (b c-a d)^2}-\frac {2 \sqrt {c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\right )}{5 (b c-a d)}-\frac {2 \sqrt {c+d x}}{5 (a+b x)^{5/2} (b c-a d)}\right )}{7 (b c-a d)}-\frac {2 \sqrt {c+d x}}{7 (a+b x)^{7/2} (b c-a d)}\) |
Input:
Int[1/((a + b*x)^(9/2)*Sqrt[c + d*x]),x]
Output:
(-2*Sqrt[c + d*x])/(7*(b*c - a*d)*(a + b*x)^(7/2)) - (6*d*((-2*Sqrt[c + d* x])/(5*(b*c - a*d)*(a + b*x)^(5/2)) - (4*d*((-2*Sqrt[c + d*x])/(3*(b*c - a *d)*(a + b*x)^(3/2)) + (4*d*Sqrt[c + d*x])/(3*(b*c - a*d)^2*Sqrt[a + b*x]) ))/(5*(b*c - a*d))))/(7*(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 = 135, normalized size of antiderivative = 0.99
| method | result | size |
| default | \(-\frac {2 \sqrt {x d +c}}{7 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {7}{2}}}-\frac {6 d \left (-\frac {2 \sqrt {x d +c}}{5 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {5}{2}}}-\frac {4 d \left (-\frac {2 \sqrt {x d +c}}{3 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {3}{2}}}+\frac {4 d \sqrt {x d +c}}{3 \left (-a d +b c \right )^{2} \sqrt {b x +a}}\right )}{5 \left (-a d +b c \right )}\right )}{7 \left (-a d +b c \right )}\) | \(135\) |
| gosper | \(\frac {2 \sqrt {x d +c}\, \left (16 d^{3} x^{3} b^{3}+56 x^{2} a \,b^{2} d^{3}-8 x^{2} b^{3} c \,d^{2}+70 x \,a^{2} b \,d^{3}-28 x a \,b^{2} c \,d^{2}+6 x \,b^{3} c^{2} d +35 a^{3} d^{3}-35 a^{2} b c \,d^{2}+21 a \,b^{2} c^{2} d -5 b^{3} c^{3}\right )}{35 \left (b x +a \right )^{\frac {7}{2}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\) | \(171\) |
| orering | \(\frac {2 \sqrt {x d +c}\, \left (16 d^{3} x^{3} b^{3}+56 x^{2} a \,b^{2} d^{3}-8 x^{2} b^{3} c \,d^{2}+70 x \,a^{2} b \,d^{3}-28 x a \,b^{2} c \,d^{2}+6 x \,b^{3} c^{2} d +35 a^{3} d^{3}-35 a^{2} b c \,d^{2}+21 a \,b^{2} c^{2} d -5 b^{3} c^{3}\right )}{35 \left (b x +a \right )^{\frac {7}{2}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\) | \(171\) |
Input:
int(1/(b*x+a)^(9/2)/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/7*(d*x+c)^(1/2)/(-a*d+b*c)/(b*x+a)^(7/2)-6/7*d/(-a*d+b*c)*(-2/5*(d*x+c) ^(1/2)/(-a*d+b*c)/(b*x+a)^(5/2)-4/5*d/(-a*d+b*c)*(-2/3*(d*x+c)^(1/2)/(-a*d +b*c)/(b*x+a)^(3/2)+4/3*d*(d*x+c)^(1/2)/(-a*d+b*c)^2/(b*x+a)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 419 vs. \(2 (112) = 224\).
Time = 0.51 (sec) , antiderivative size = 419, normalized size of antiderivative = 3.08 \[ \int \frac {1}{(a+b x)^{9/2} \sqrt {c+d x}} \, dx=\frac {2 \, {\left (16 \, b^{3} d^{3} x^{3} - 5 \, b^{3} c^{3} + 21 \, a b^{2} c^{2} d - 35 \, a^{2} b c d^{2} + 35 \, a^{3} d^{3} - 8 \, {\left (b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{2} + 2 \, {\left (3 \, b^{3} c^{2} d - 14 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{35 \, {\left (a^{4} b^{4} c^{4} - 4 \, a^{5} b^{3} c^{3} d + 6 \, a^{6} b^{2} c^{2} d^{2} - 4 \, a^{7} b c d^{3} + a^{8} d^{4} + {\left (b^{8} c^{4} - 4 \, a b^{7} c^{3} d + 6 \, a^{2} b^{6} c^{2} d^{2} - 4 \, a^{3} b^{5} c d^{3} + a^{4} b^{4} d^{4}\right )} x^{4} + 4 \, {\left (a b^{7} c^{4} - 4 \, a^{2} b^{6} c^{3} d + 6 \, a^{3} b^{5} c^{2} d^{2} - 4 \, a^{4} b^{4} c d^{3} + a^{5} b^{3} d^{4}\right )} x^{3} + 6 \, {\left (a^{2} b^{6} c^{4} - 4 \, a^{3} b^{5} c^{3} d + 6 \, a^{4} b^{4} c^{2} d^{2} - 4 \, a^{5} b^{3} c d^{3} + a^{6} b^{2} d^{4}\right )} x^{2} + 4 \, {\left (a^{3} b^{5} c^{4} - 4 \, a^{4} b^{4} c^{3} d + 6 \, a^{5} b^{3} c^{2} d^{2} - 4 \, a^{6} b^{2} c d^{3} + a^{7} b d^{4}\right )} x\right )}} \] Input:
integrate(1/(b*x+a)^(9/2)/(d*x+c)^(1/2),x, algorithm="fricas")
Output:
2/35*(16*b^3*d^3*x^3 - 5*b^3*c^3 + 21*a*b^2*c^2*d - 35*a^2*b*c*d^2 + 35*a^ 3*d^3 - 8*(b^3*c*d^2 - 7*a*b^2*d^3)*x^2 + 2*(3*b^3*c^2*d - 14*a*b^2*c*d^2 + 35*a^2*b*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a^4*b^4*c^4 - 4*a^5*b^3*c^ 3*d + 6*a^6*b^2*c^2*d^2 - 4*a^7*b*c*d^3 + a^8*d^4 + (b^8*c^4 - 4*a*b^7*c^3 *d + 6*a^2*b^6*c^2*d^2 - 4*a^3*b^5*c*d^3 + a^4*b^4*d^4)*x^4 + 4*(a*b^7*c^4 - 4*a^2*b^6*c^3*d + 6*a^3*b^5*c^2*d^2 - 4*a^4*b^4*c*d^3 + a^5*b^3*d^4)*x^ 3 + 6*(a^2*b^6*c^4 - 4*a^3*b^5*c^3*d + 6*a^4*b^4*c^2*d^2 - 4*a^5*b^3*c*d^3 + a^6*b^2*d^4)*x^2 + 4*(a^3*b^5*c^4 - 4*a^4*b^4*c^3*d + 6*a^5*b^3*c^2*d^2 - 4*a^6*b^2*c*d^3 + a^7*b*d^4)*x)
\[ \int \frac {1}{(a+b x)^{9/2} \sqrt {c+d x}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {9}{2}} \sqrt {c + d x}}\, dx \] Input:
integrate(1/(b*x+a)**(9/2)/(d*x+c)**(1/2),x)
Output:
Integral(1/((a + b*x)**(9/2)*sqrt(c + d*x)), x)
Exception generated. \[ \int \frac {1}{(a+b x)^{9/2} \sqrt {c+d x}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(b*x+a)^(9/2)/(d*x+c)^(1/2),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. 386 vs. \(2 (112) = 224\).
Time = 0.17 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.84 \[ \int \frac {1}{(a+b x)^{9/2} \sqrt {c+d x}} \, dx=\frac {64 \, {\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3} - 7 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{4} c^{2} + 14 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{3} c d - 7 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b^{2} d^{2} + 21 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} b^{2} c - 21 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a b d - 35 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{6}\right )} \sqrt {b d} b^{4} d^{3}}{35 \, {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}^{7} {\left | b \right |}} \] Input:
integrate(1/(b*x+a)^(9/2)/(d*x+c)^(1/2),x, algorithm="giac")
Output:
64/35*(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 - 7*(sqrt(b *d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^4*c^2 + 14*(s qrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^3*c*d - 7*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2* b^2*d^2 + 21*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d ))^4*b^2*c - 21*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a* b*d))^4*a*b*d - 35*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6)*sqrt(b*d)*b^4*d^3/((b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)^7*abs(b))
Time = 0.68 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.54 \[ \int \frac {1}{(a+b x)^{9/2} \sqrt {c+d x}} \, dx=\frac {\sqrt {c+d\,x}\,\left (\frac {32\,d^3\,x^3}{35\,{\left (a\,d-b\,c\right )}^4}+\frac {70\,a^3\,d^3-70\,a^2\,b\,c\,d^2+42\,a\,b^2\,c^2\,d-10\,b^3\,c^3}{35\,b^3\,{\left (a\,d-b\,c\right )}^4}+\frac {4\,d\,x\,\left (35\,a^2\,d^2-14\,a\,b\,c\,d+3\,b^2\,c^2\right )}{35\,b^2\,{\left (a\,d-b\,c\right )}^4}+\frac {16\,d^2\,x^2\,\left (7\,a\,d-b\,c\right )}{35\,b\,{\left (a\,d-b\,c\right )}^4}\right )}{x^3\,\sqrt {a+b\,x}+\frac {a^3\,\sqrt {a+b\,x}}{b^3}+\frac {3\,a\,x^2\,\sqrt {a+b\,x}}{b}+\frac {3\,a^2\,x\,\sqrt {a+b\,x}}{b^2}} \] Input:
int(1/((a + b*x)^(9/2)*(c + d*x)^(1/2)),x)
Output:
((c + d*x)^(1/2)*((32*d^3*x^3)/(35*(a*d - b*c)^4) + (70*a^3*d^3 - 10*b^3*c ^3 + 42*a*b^2*c^2*d - 70*a^2*b*c*d^2)/(35*b^3*(a*d - b*c)^4) + (4*d*x*(35* a^2*d^2 + 3*b^2*c^2 - 14*a*b*c*d))/(35*b^2*(a*d - b*c)^4) + (16*d^2*x^2*(7 *a*d - b*c))/(35*b*(a*d - b*c)^4)))/(x^3*(a + b*x)^(1/2) + (a^3*(a + b*x)^ (1/2))/b^3 + (3*a*x^2*(a + b*x)^(1/2))/b + (3*a^2*x*(a + b*x)^(1/2))/b^2)
Time = 0.17 (sec) , antiderivative size = 522, normalized size of antiderivative = 3.84 \[ \int \frac {1}{(a+b x)^{9/2} \sqrt {c+d x}} \, dx=\frac {-\frac {32 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a^{3} d^{3}}{35}-\frac {96 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a^{2} b \,d^{3} x}{35}-\frac {96 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a \,b^{2} d^{3} x^{2}}{35}-\frac {32 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, b^{3} d^{3} x^{3}}{35}+2 \sqrt {d x +c}\, a^{3} b \,d^{3}-2 \sqrt {d x +c}\, a^{2} b^{2} c \,d^{2}+4 \sqrt {d x +c}\, a^{2} b^{2} d^{3} x +\frac {6 \sqrt {d x +c}\, a \,b^{3} c^{2} d}{5}-\frac {8 \sqrt {d x +c}\, a \,b^{3} c \,d^{2} x}{5}+\frac {16 \sqrt {d x +c}\, a \,b^{3} d^{3} x^{2}}{5}-\frac {2 \sqrt {d x +c}\, b^{4} c^{3}}{7}+\frac {12 \sqrt {d x +c}\, b^{4} c^{2} d x}{35}-\frac {16 \sqrt {d x +c}\, b^{4} c \,d^{2} x^{2}}{35}+\frac {32 \sqrt {d x +c}\, b^{4} d^{3} x^{3}}{35}}{\sqrt {b x +a}\, b \left (a^{4} b^{3} d^{4} x^{3}-4 a^{3} b^{4} c \,d^{3} x^{3}+6 a^{2} b^{5} c^{2} d^{2} x^{3}-4 a \,b^{6} c^{3} d \,x^{3}+b^{7} c^{4} x^{3}+3 a^{5} b^{2} d^{4} x^{2}-12 a^{4} b^{3} c \,d^{3} x^{2}+18 a^{3} b^{4} c^{2} d^{2} x^{2}-12 a^{2} b^{5} c^{3} d \,x^{2}+3 a \,b^{6} c^{4} x^{2}+3 a^{6} b \,d^{4} x -12 a^{5} b^{2} c \,d^{3} x +18 a^{4} b^{3} c^{2} d^{2} x -12 a^{3} b^{4} c^{3} d x +3 a^{2} b^{5} c^{4} x +a^{7} d^{4}-4 a^{6} b c \,d^{3}+6 a^{5} b^{2} c^{2} d^{2}-4 a^{4} b^{3} c^{3} d +a^{3} b^{4} c^{4}\right )} \] Input:
int(1/(b*x+a)^(9/2)/(d*x+c)^(1/2),x)
Output:
(2*( - 16*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**3*d**3 - 48*sqrt(d)*sqrt(b)*sqr t(a + b*x)*a**2*b*d**3*x - 48*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b**2*d**3*x* *2 - 16*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b**3*d**3*x**3 + 35*sqrt(c + d*x)*a* *3*b*d**3 - 35*sqrt(c + d*x)*a**2*b**2*c*d**2 + 70*sqrt(c + d*x)*a**2*b**2 *d**3*x + 21*sqrt(c + d*x)*a*b**3*c**2*d - 28*sqrt(c + d*x)*a*b**3*c*d**2* x + 56*sqrt(c + d*x)*a*b**3*d**3*x**2 - 5*sqrt(c + d*x)*b**4*c**3 + 6*sqrt (c + d*x)*b**4*c**2*d*x - 8*sqrt(c + d*x)*b**4*c*d**2*x**2 + 16*sqrt(c + d *x)*b**4*d**3*x**3))/(35*sqrt(a + b*x)*b*(a**7*d**4 - 4*a**6*b*c*d**3 + 3* a**6*b*d**4*x + 6*a**5*b**2*c**2*d**2 - 12*a**5*b**2*c*d**3*x + 3*a**5*b** 2*d**4*x**2 - 4*a**4*b**3*c**3*d + 18*a**4*b**3*c**2*d**2*x - 12*a**4*b**3 *c*d**3*x**2 + a**4*b**3*d**4*x**3 + a**3*b**4*c**4 - 12*a**3*b**4*c**3*d* x + 18*a**3*b**4*c**2*d**2*x**2 - 4*a**3*b**4*c*d**3*x**3 + 3*a**2*b**5*c* *4*x - 12*a**2*b**5*c**3*d*x**2 + 6*a**2*b**5*c**2*d**2*x**3 + 3*a*b**6*c* *4*x**2 - 4*a*b**6*c**3*d*x**3 + b**7*c**4*x**3))