Integrand size = 19, antiderivative size = 101 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^{9/2}} \, dx=-\frac {2 (c+d x)^{3/2}}{7 (b c-a d) (a+b x)^{7/2}}+\frac {8 d (c+d x)^{3/2}}{35 (b c-a d)^2 (a+b x)^{5/2}}-\frac {16 d^2 (c+d x)^{3/2}}{105 (b c-a d)^3 (a+b x)^{3/2}} \]
-2/7*(d*x+c)^(3/2)/(-a*d+b*c)/(b*x+a)^(7/2)+8/35*d*(d*x+c)^(3/2)/(-a*d+b*c )^2/(b*x+a)^(5/2)-16/105*d^2*(d*x+c)^(3/2)/(-a*d+b*c)^3/(b*x+a)^(3/2)
Time = 0.13 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^{9/2}} \, dx=-\frac {2 (c+d x)^{3/2} \left (35 a^2 d^2+14 a b d (-3 c+2 d x)+b^2 \left (15 c^2-12 c d x+8 d^2 x^2\right )\right )}{105 (b c-a d)^3 (a+b x)^{7/2}} \]
(-2*(c + d*x)^(3/2)*(35*a^2*d^2 + 14*a*b*d*(-3*c + 2*d*x) + b^2*(15*c^2 - 12*c*d*x + 8*d^2*x^2)))/(105*(b*c - a*d)^3*(a + b*x)^(7/2))
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.
\(\displaystyle \int \frac {\sqrt {c+d x}}{(a+b x)^{9/2}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {4 d \int \frac {\sqrt {c+d x}}{(a+b x)^{7/2}}dx}{7 (b c-a d)}-\frac {2 (c+d x)^{3/2}}{7 (a+b x)^{7/2} (b c-a d)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {4 d \left (-\frac {2 d \int \frac {\sqrt {c+d x}}{(a+b x)^{5/2}}dx}{5 (b c-a d)}-\frac {2 (c+d x)^{3/2}}{5 (a+b x)^{5/2} (b c-a d)}\right )}{7 (b c-a d)}-\frac {2 (c+d x)^{3/2}}{7 (a+b x)^{7/2} (b c-a d)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {2 (c+d x)^{3/2}}{7 (a+b x)^{7/2} (b c-a d)}-\frac {4 d \left (\frac {4 d (c+d x)^{3/2}}{15 (a+b x)^{3/2} (b c-a d)^2}-\frac {2 (c+d x)^{3/2}}{5 (a+b x)^{5/2} (b c-a d)}\right )}{7 (b c-a d)}\) |
(-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))
3.15.68.3.1 Defintions of rubi rules used
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 = 0.75 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(\frac {2 \left (d x +c \right )^{\frac {3}{2}} \left (8 d^{2} x^{2} b^{2}+28 x a b \,d^{2}-12 x \,b^{2} c d +35 a^{2} d^{2}-42 a b c d +15 b^{2} c^{2}\right )}{105 \left (b x +a \right )^{\frac {7}{2}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
default | \(-\frac {\sqrt {d x +c}}{3 b \left (b x +a \right )^{\frac {7}{2}}}+\frac {\left (a d -b c \right ) \left (-\frac {2 \sqrt {d x +c}}{7 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {7}{2}}}-\frac {6 d \left (-\frac {2 \sqrt {d x +c}}{5 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {5}{2}}}-\frac {4 d \left (-\frac {2 \sqrt {d x +c}}{3 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {3}{2}}}+\frac {4 d \sqrt {d x +c}}{3 \left (-a d +b c \right )^{2} \sqrt {b x +a}}\right )}{5 \left (-a d +b c \right )}\right )}{7 \left (-a d +b c \right )}\right )}{6 b}\) | \(168\) |
2/105*(d*x+c)^(3/2)*(8*b^2*d^2*x^2+28*a*b*d^2*x-12*b^2*c*d*x+35*a^2*d^2-42 *a*b*c*d+15*b^2*c^2)/(b*x+a)^(7/2)/(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. 337 vs. \(2 (83) = 166\).
Time = 0.66 (sec) , antiderivative size = 337, normalized size of antiderivative = 3.34 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^{9/2}} \, dx=-\frac {2 \, {\left (8 \, b^{2} d^{3} x^{3} + 15 \, b^{2} c^{3} - 42 \, a b c^{2} d + 35 \, a^{2} c d^{2} - 4 \, {\left (b^{2} c d^{2} - 7 \, a b d^{3}\right )} x^{2} + {\left (3 \, b^{2} c^{2} d - 14 \, a b c d^{2} + 35 \, a^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{105 \, {\left (a^{4} b^{3} c^{3} - 3 \, a^{5} b^{2} c^{2} d + 3 \, a^{6} b c d^{2} - a^{7} d^{3} + {\left (b^{7} c^{3} - 3 \, a b^{6} c^{2} d + 3 \, a^{2} b^{5} c d^{2} - a^{3} b^{4} d^{3}\right )} x^{4} + 4 \, {\left (a b^{6} c^{3} - 3 \, a^{2} b^{5} c^{2} d + 3 \, a^{3} b^{4} c d^{2} - a^{4} b^{3} d^{3}\right )} x^{3} + 6 \, {\left (a^{2} b^{5} c^{3} - 3 \, a^{3} b^{4} c^{2} d + 3 \, a^{4} b^{3} c d^{2} - a^{5} b^{2} d^{3}\right )} x^{2} + 4 \, {\left (a^{3} b^{4} c^{3} - 3 \, a^{4} b^{3} c^{2} d + 3 \, a^{5} b^{2} c d^{2} - a^{6} b d^{3}\right )} x\right )}} \]
-2/105*(8*b^2*d^3*x^3 + 15*b^2*c^3 - 42*a*b*c^2*d + 35*a^2*c*d^2 - 4*(b^2* c*d^2 - 7*a*b*d^3)*x^2 + (3*b^2*c^2*d - 14*a*b*c*d^2 + 35*a^2*d^3)*x)*sqrt (b*x + a)*sqrt(d*x + c)/(a^4*b^3*c^3 - 3*a^5*b^2*c^2*d + 3*a^6*b*c*d^2 - a ^7*d^3 + (b^7*c^3 - 3*a*b^6*c^2*d + 3*a^2*b^5*c*d^2 - a^3*b^4*d^3)*x^4 + 4 *(a*b^6*c^3 - 3*a^2*b^5*c^2*d + 3*a^3*b^4*c*d^2 - a^4*b^3*d^3)*x^3 + 6*(a^ 2*b^5*c^3 - 3*a^3*b^4*c^2*d + 3*a^4*b^3*c*d^2 - a^5*b^2*d^3)*x^2 + 4*(a^3* b^4*c^3 - 3*a^4*b^3*c^2*d + 3*a^5*b^2*c*d^2 - a^6*b*d^3)*x)
\[ \int \frac {\sqrt {c+d x}}{(a+b x)^{9/2}} \, dx=\int \frac {\sqrt {c + d x}}{\left (a + b x\right )^{\frac {9}{2}}}\, dx \]
Exception generated. \[ \int \frac {\sqrt {c+d x}}{(a+b x)^{9/2}} \, dx=\text {Exception raised: ValueError} \]
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. 689 vs. \(2 (83) = 166\).
Time = 0.41 (sec) , antiderivative size = 689, normalized size of antiderivative = 6.82 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^{9/2}} \, dx=-\frac {32 \, {\left (\sqrt {b d} b^{10} c^{4} d^{3} - 4 \, \sqrt {b d} a b^{9} c^{3} d^{4} + 6 \, \sqrt {b d} a^{2} b^{8} c^{2} d^{5} - 4 \, \sqrt {b d} a^{3} b^{7} c d^{6} + \sqrt {b d} a^{4} b^{6} d^{7} - 7 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{8} c^{3} d^{3} + 21 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{7} c^{2} d^{4} - 21 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b^{6} c d^{5} + 7 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{3} b^{5} d^{6} + 21 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} b^{6} c^{2} d^{3} - 42 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a b^{5} c d^{4} + 21 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a^{2} b^{4} d^{5} + 35 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{6} b^{4} c d^{3} - 35 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{6} a b^{3} d^{4} + 70 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{8} b^{2} d^{3}\right )} {\left | b \right |}}{105 \, {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}^{7} b^{2}} \]
-32/105*(sqrt(b*d)*b^10*c^4*d^3 - 4*sqrt(b*d)*a*b^9*c^3*d^4 + 6*sqrt(b*d)* a^2*b^8*c^2*d^5 - 4*sqrt(b*d)*a^3*b^7*c*d^6 + sqrt(b*d)*a^4*b^6*d^7 - 7*sq rt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2* b^8*c^3*d^3 + 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^7*c^2*d^4 - 21*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^6*c*d^5 + 7*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^5*d^6 + 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d ))^4*b^6*c^2*d^3 - 42*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^5*c*d^4 + 21*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^4*d^5 + 35*sqrt(b*d)*(sqr t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^4*c*d^3 - 35*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^3*d^4 + 70*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*b^2*d^3)*abs(b)/((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*b^2)
Time = 0.96 (sec) , antiderivative size = 203, normalized size of antiderivative = 2.01 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^{9/2}} \, dx=\frac {\sqrt {c+d\,x}\,\left (\frac {70\,a^2\,c\,d^2-84\,a\,b\,c^2\,d+30\,b^2\,c^3}{105\,b^3\,{\left (a\,d-b\,c\right )}^3}+\frac {x\,\left (70\,a^2\,d^3-28\,a\,b\,c\,d^2+6\,b^2\,c^2\,d\right )}{105\,b^3\,{\left (a\,d-b\,c\right )}^3}+\frac {16\,d^3\,x^3}{105\,b\,{\left (a\,d-b\,c\right )}^3}+\frac {8\,d^2\,x^2\,\left (7\,a\,d-b\,c\right )}{105\,b^2\,{\left (a\,d-b\,c\right )}^3}\right )}{x^3\,\sqrt {a+b\,x}+\frac {a^3\,\sqrt {a+b\,x}}{b^3}+\frac {3\,a\,x^2\,\sqrt {a+b\,x}}{b}+\frac {3\,a^2\,x\,\sqrt {a+b\,x}}{b^2}} \]
((c + d*x)^(1/2)*((30*b^2*c^3 + 70*a^2*c*d^2 - 84*a*b*c^2*d)/(105*b^3*(a*d - b*c)^3) + (x*(70*a^2*d^3 + 6*b^2*c^2*d - 28*a*b*c*d^2))/(105*b^3*(a*d - b*c)^3) + (16*d^3*x^3)/(105*b*(a*d - b*c)^3) + (8*d^2*x^2*(7*a*d - b*c))/ (105*b^2*(a*d - b*c)^3)))/(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)