Integrand size = 19, antiderivative size = 171 \[ \int \frac {1}{(a+b x)^{9/2} (c+d x)^{3/2}} \, dx=-\frac {2}{7 (b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}+\frac {16 d}{35 (b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}-\frac {32 d^2}{35 (b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}+\frac {128 d^3}{35 (b c-a d)^4 \sqrt {a+b x} \sqrt {c+d x}}+\frac {256 d^4 \sqrt {a+b x}}{35 (b c-a d)^5 \sqrt {c+d x}} \]
-2/7/(-a*d+b*c)/(b*x+a)^(7/2)/(d*x+c)^(1/2)+16/35*d/(-a*d+b*c)^2/(b*x+a)^( 5/2)/(d*x+c)^(1/2)-32/35*d^2/(-a*d+b*c)^3/(b*x+a)^(3/2)/(d*x+c)^(1/2)+128/ 35*d^3/(-a*d+b*c)^4/(b*x+a)^(1/2)/(d*x+c)^(1/2)+256/35*d^4*(b*x+a)^(1/2)/( -a*d+b*c)^5/(d*x+c)^(1/2)
Time = 0.24 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(a+b x)^{9/2} (c+d x)^{3/2}} \, dx=\frac {2 \left (35 a^4 d^4+140 a^3 b d^3 (c+2 d x)+70 a^2 b^2 d^2 \left (-c^2+4 c d x+8 d^2 x^2\right )+28 a b^3 d \left (c^3-2 c^2 d x+8 c d^2 x^2+16 d^3 x^3\right )+b^4 \left (-5 c^4+8 c^3 d x-16 c^2 d^2 x^2+64 c d^3 x^3+128 d^4 x^4\right )\right )}{35 (b c-a d)^5 (a+b x)^{7/2} \sqrt {c+d x}} \]
(2*(35*a^4*d^4 + 140*a^3*b*d^3*(c + 2*d*x) + 70*a^2*b^2*d^2*(-c^2 + 4*c*d* x + 8*d^2*x^2) + 28*a*b^3*d*(c^3 - 2*c^2*d*x + 8*c*d^2*x^2 + 16*d^3*x^3) + b^4*(-5*c^4 + 8*c^3*d*x - 16*c^2*d^2*x^2 + 64*c*d^3*x^3 + 128*d^4*x^4)))/ (35*(b*c - a*d)^5*(a + b*x)^(7/2)*Sqrt[c + d*x])
Time = 0.22 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.20, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {55, 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.
\(\displaystyle \int \frac {1}{(a+b x)^{9/2} (c+d x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {8 d \int \frac {1}{(a+b x)^{7/2} (c+d x)^{3/2}}dx}{7 (b c-a d)}-\frac {2}{7 (a+b x)^{7/2} \sqrt {c+d x} (b c-a d)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {8 d \left (-\frac {6 d \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/2}}dx}{5 (b c-a d)}-\frac {2}{5 (a+b x)^{5/2} \sqrt {c+d x} (b c-a d)}\right )}{7 (b c-a d)}-\frac {2}{7 (a+b x)^{7/2} \sqrt {c+d x} (b c-a d)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {8 d \left (-\frac {6 d \left (-\frac {4 d \int \frac {1}{(a+b x)^{3/2} (c+d x)^{3/2}}dx}{3 (b c-a d)}-\frac {2}{3 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}\right )}{5 (b c-a d)}-\frac {2}{5 (a+b x)^{5/2} \sqrt {c+d x} (b c-a d)}\right )}{7 (b c-a d)}-\frac {2}{7 (a+b x)^{7/2} \sqrt {c+d x} (b c-a d)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {8 d \left (-\frac {6 d \left (-\frac {4 d \left (-\frac {2 d \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}}dx}{b c-a d}-\frac {2}{\sqrt {a+b x} \sqrt {c+d x} (b c-a d)}\right )}{3 (b c-a d)}-\frac {2}{3 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}\right )}{5 (b c-a d)}-\frac {2}{5 (a+b x)^{5/2} \sqrt {c+d x} (b c-a d)}\right )}{7 (b c-a d)}-\frac {2}{7 (a+b x)^{7/2} \sqrt {c+d x} (b c-a d)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {8 d \left (-\frac {6 d \left (-\frac {4 d \left (-\frac {4 d \sqrt {a+b x}}{\sqrt {c+d x} (b c-a d)^2}-\frac {2}{\sqrt {a+b x} \sqrt {c+d x} (b c-a d)}\right )}{3 (b c-a d)}-\frac {2}{3 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}\right )}{5 (b c-a d)}-\frac {2}{5 (a+b x)^{5/2} \sqrt {c+d x} (b c-a d)}\right )}{7 (b c-a d)}-\frac {2}{7 (a+b x)^{7/2} \sqrt {c+d x} (b c-a d)}\) |
-2/(7*(b*c - a*d)*(a + b*x)^(7/2)*Sqrt[c + d*x]) - (8*d*(-2/(5*(b*c - a*d) *(a + b*x)^(5/2)*Sqrt[c + d*x]) - (6*d*(-2/(3*(b*c - a*d)*(a + b*x)^(3/2)* Sqrt[c + d*x]) - (4*d*(-2/((b*c - a*d)*Sqrt[a + b*x]*Sqrt[c + d*x]) - (4*d *Sqrt[a + b*x])/((b*c - a*d)^2*Sqrt[c + d*x])))/(3*(b*c - a*d))))/(5*(b*c - a*d))))/(7*(b*c - a*d))
3.16.10.3.1 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 = 0.53 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.08
method | result | size |
default | \(-\frac {2}{7 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {7}{2}} \sqrt {d x +c}}-\frac {8 d \left (-\frac {2}{5 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {5}{2}} \sqrt {d x +c}}-\frac {6 d \left (-\frac {2}{3 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {3}{2}} \sqrt {d x +c}}-\frac {4 d \left (-\frac {2}{\left (-a d +b c \right ) \sqrt {b x +a}\, \sqrt {d x +c}}+\frac {4 d \sqrt {b x +a}}{\left (-a d +b c \right ) \sqrt {d x +c}\, \left (a d -b c \right )}\right )}{3 \left (-a d +b c \right )}\right )}{5 \left (-a d +b c \right )}\right )}{7 \left (-a d +b c \right )}\) | \(185\) |
gosper | \(-\frac {2 \left (128 d^{4} x^{4} b^{4}+448 a \,b^{3} d^{4} x^{3}+64 b^{4} c \,d^{3} x^{3}+560 a^{2} b^{2} d^{4} x^{2}+224 a \,b^{3} c \,d^{3} x^{2}-16 b^{4} c^{2} d^{2} x^{2}+280 a^{3} b \,d^{4} x +280 a^{2} b^{2} c \,d^{3} x -56 a \,b^{3} c^{2} d^{2} x +8 b^{4} c^{3} d x +35 a^{4} d^{4}+140 a^{3} b c \,d^{3}-70 a^{2} b^{2} c^{2} d^{2}+28 a \,b^{3} c^{3} d -5 b^{4} c^{4}\right )}{35 \left (b x +a \right )^{\frac {7}{2}} \sqrt {d x +c}\, \left (a^{5} d^{5}-5 a^{4} b c \,d^{4}+10 a^{3} b^{2} c^{2} d^{3}-10 a^{2} b^{3} c^{3} d^{2}+5 a \,b^{4} c^{4} d -b^{5} c^{5}\right )}\) | \(256\) |
-2/7/(-a*d+b*c)/(b*x+a)^(7/2)/(d*x+c)^(1/2)-8/7*d/(-a*d+b*c)*(-2/5/(-a*d+b *c)/(b*x+a)^(5/2)/(d*x+c)^(1/2)-6/5*d/(-a*d+b*c)*(-2/3/(-a*d+b*c)/(b*x+a)^ (3/2)/(d*x+c)^(1/2)-4/3*d/(-a*d+b*c)*(-2/(-a*d+b*c)/(b*x+a)^(1/2)/(d*x+c)^ (1/2)+4*d/(-a*d+b*c)*(b*x+a)^(1/2)/(d*x+c)^(1/2)/(a*d-b*c))))
Leaf count of result is larger than twice the leaf count of optimal. 689 vs. \(2 (141) = 282\).
Time = 1.40 (sec) , antiderivative size = 689, normalized size of antiderivative = 4.03 \[ \int \frac {1}{(a+b x)^{9/2} (c+d x)^{3/2}} \, dx=\frac {2 \, {\left (128 \, b^{4} d^{4} x^{4} - 5 \, b^{4} c^{4} + 28 \, a b^{3} c^{3} d - 70 \, a^{2} b^{2} c^{2} d^{2} + 140 \, a^{3} b c d^{3} + 35 \, a^{4} d^{4} + 64 \, {\left (b^{4} c d^{3} + 7 \, a b^{3} d^{4}\right )} x^{3} - 16 \, {\left (b^{4} c^{2} d^{2} - 14 \, a b^{3} c d^{3} - 35 \, a^{2} b^{2} d^{4}\right )} x^{2} + 8 \, {\left (b^{4} c^{3} d - 7 \, a b^{3} c^{2} d^{2} + 35 \, a^{2} b^{2} c d^{3} + 35 \, a^{3} b d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{35 \, {\left (a^{4} b^{5} c^{6} - 5 \, a^{5} b^{4} c^{5} d + 10 \, a^{6} b^{3} c^{4} d^{2} - 10 \, a^{7} b^{2} c^{3} d^{3} + 5 \, a^{8} b c^{2} d^{4} - a^{9} c d^{5} + {\left (b^{9} c^{5} d - 5 \, a b^{8} c^{4} d^{2} + 10 \, a^{2} b^{7} c^{3} d^{3} - 10 \, a^{3} b^{6} c^{2} d^{4} + 5 \, a^{4} b^{5} c d^{5} - a^{5} b^{4} d^{6}\right )} x^{5} + {\left (b^{9} c^{6} - a b^{8} c^{5} d - 10 \, a^{2} b^{7} c^{4} d^{2} + 30 \, a^{3} b^{6} c^{3} d^{3} - 35 \, a^{4} b^{5} c^{2} d^{4} + 19 \, a^{5} b^{4} c d^{5} - 4 \, a^{6} b^{3} d^{6}\right )} x^{4} + 2 \, {\left (2 \, a b^{8} c^{6} - 7 \, a^{2} b^{7} c^{5} d + 5 \, a^{3} b^{6} c^{4} d^{2} + 10 \, a^{4} b^{5} c^{3} d^{3} - 20 \, a^{5} b^{4} c^{2} d^{4} + 13 \, a^{6} b^{3} c d^{5} - 3 \, a^{7} b^{2} d^{6}\right )} x^{3} + 2 \, {\left (3 \, a^{2} b^{7} c^{6} - 13 \, a^{3} b^{6} c^{5} d + 20 \, a^{4} b^{5} c^{4} d^{2} - 10 \, a^{5} b^{4} c^{3} d^{3} - 5 \, a^{6} b^{3} c^{2} d^{4} + 7 \, a^{7} b^{2} c d^{5} - 2 \, a^{8} b d^{6}\right )} x^{2} + {\left (4 \, a^{3} b^{6} c^{6} - 19 \, a^{4} b^{5} c^{5} d + 35 \, a^{5} b^{4} c^{4} d^{2} - 30 \, a^{6} b^{3} c^{3} d^{3} + 10 \, a^{7} b^{2} c^{2} d^{4} + a^{8} b c d^{5} - a^{9} d^{6}\right )} x\right )}} \]
2/35*(128*b^4*d^4*x^4 - 5*b^4*c^4 + 28*a*b^3*c^3*d - 70*a^2*b^2*c^2*d^2 + 140*a^3*b*c*d^3 + 35*a^4*d^4 + 64*(b^4*c*d^3 + 7*a*b^3*d^4)*x^3 - 16*(b^4* c^2*d^2 - 14*a*b^3*c*d^3 - 35*a^2*b^2*d^4)*x^2 + 8*(b^4*c^3*d - 7*a*b^3*c^ 2*d^2 + 35*a^2*b^2*c*d^3 + 35*a^3*b*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a ^4*b^5*c^6 - 5*a^5*b^4*c^5*d + 10*a^6*b^3*c^4*d^2 - 10*a^7*b^2*c^3*d^3 + 5 *a^8*b*c^2*d^4 - a^9*c*d^5 + (b^9*c^5*d - 5*a*b^8*c^4*d^2 + 10*a^2*b^7*c^3 *d^3 - 10*a^3*b^6*c^2*d^4 + 5*a^4*b^5*c*d^5 - a^5*b^4*d^6)*x^5 + (b^9*c^6 - a*b^8*c^5*d - 10*a^2*b^7*c^4*d^2 + 30*a^3*b^6*c^3*d^3 - 35*a^4*b^5*c^2*d ^4 + 19*a^5*b^4*c*d^5 - 4*a^6*b^3*d^6)*x^4 + 2*(2*a*b^8*c^6 - 7*a^2*b^7*c^ 5*d + 5*a^3*b^6*c^4*d^2 + 10*a^4*b^5*c^3*d^3 - 20*a^5*b^4*c^2*d^4 + 13*a^6 *b^3*c*d^5 - 3*a^7*b^2*d^6)*x^3 + 2*(3*a^2*b^7*c^6 - 13*a^3*b^6*c^5*d + 20 *a^4*b^5*c^4*d^2 - 10*a^5*b^4*c^3*d^3 - 5*a^6*b^3*c^2*d^4 + 7*a^7*b^2*c*d^ 5 - 2*a^8*b*d^6)*x^2 + (4*a^3*b^6*c^6 - 19*a^4*b^5*c^5*d + 35*a^5*b^4*c^4* d^2 - 30*a^6*b^3*c^3*d^3 + 10*a^7*b^2*c^2*d^4 + a^8*b*c*d^5 - a^9*d^6)*x)
\[ \int \frac {1}{(a+b x)^{9/2} (c+d x)^{3/2}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {9}{2}} \left (c + d x\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {1}{(a+b x)^{9/2} (c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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. 1518 vs. \(2 (141) = 282\).
Time = 0.73 (sec) , antiderivative size = 1518, normalized size of antiderivative = 8.88 \[ \int \frac {1}{(a+b x)^{9/2} (c+d x)^{3/2}} \, dx=\text {Too large to display} \]
2*sqrt(b*x + a)*b^2*d^4/((b^5*c^5*abs(b) - 5*a*b^4*c^4*d*abs(b) + 10*a^2*b ^3*c^3*d^2*abs(b) - 10*a^3*b^2*c^2*d^3*abs(b) + 5*a^4*b*c*d^4*abs(b) - a^5 *d^5*abs(b))*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)) + 4/35*(93*sqrt(b*d)*b^1 4*c^6*d^3 - 558*sqrt(b*d)*a*b^13*c^5*d^4 + 1395*sqrt(b*d)*a^2*b^12*c^4*d^5 - 1860*sqrt(b*d)*a^3*b^11*c^3*d^6 + 1395*sqrt(b*d)*a^4*b^10*c^2*d^7 - 558 *sqrt(b*d)*a^5*b^9*c*d^8 + 93*sqrt(b*d)*a^6*b^8*d^9 - 616*sqrt(b*d)*(sqrt( b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^12*c^5*d^3 + 3080*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^11*c^4*d^4 - 6160*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^10*c^3*d^5 + 6160*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^9*c^2*d^6 - 3080*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^8*c*d^7 + 616*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2 *c + (b*x + a)*b*d - a*b*d))^2*a^5*b^7*d^8 + 1673*sqrt(b*d)*(sqrt(b*d)*sqr t(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^10*c^4*d^3 - 6692*sq rt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4* a*b^9*c^3*d^4 + 10038*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^8*c^2*d^5 - 6692*sqrt(b*d)*(sqrt(b*d)*sqrt(b *x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^3*b^7*c*d^6 + 1673*sqrt (b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4...
Time = 1.52 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.97 \[ \int \frac {1}{(a+b x)^{9/2} (c+d x)^{3/2}} \, dx=-\frac {\sqrt {c+d\,x}\,\left (\frac {256\,b\,d^3\,x^4}{35\,{\left (a\,d-b\,c\right )}^5}+\frac {128\,d^2\,x^3\,\left (7\,a\,d+b\,c\right )}{35\,{\left (a\,d-b\,c\right )}^5}+\frac {70\,a^4\,d^4+280\,a^3\,b\,c\,d^3-140\,a^2\,b^2\,c^2\,d^2+56\,a\,b^3\,c^3\,d-10\,b^4\,c^4}{35\,b^3\,d\,{\left (a\,d-b\,c\right )}^5}+\frac {x\,\left (560\,a^3\,b\,d^4+560\,a^2\,b^2\,c\,d^3-112\,a\,b^3\,c^2\,d^2+16\,b^4\,c^3\,d\right )}{35\,b^3\,d\,{\left (a\,d-b\,c\right )}^5}+\frac {32\,d\,x^2\,\left (35\,a^2\,d^2+14\,a\,b\,c\,d-b^2\,c^2\right )}{35\,b\,{\left (a\,d-b\,c\right )}^5}\right )}{x^4\,\sqrt {a+b\,x}+\frac {a^3\,c\,\sqrt {a+b\,x}}{b^3\,d}+\frac {x^3\,\left (3\,a\,d+b\,c\right )\,\sqrt {a+b\,x}}{b\,d}+\frac {3\,a\,x^2\,\left (a\,d+b\,c\right )\,\sqrt {a+b\,x}}{b^2\,d}+\frac {a^2\,x\,\left (a\,d+3\,b\,c\right )\,\sqrt {a+b\,x}}{b^3\,d}} \]
-((c + d*x)^(1/2)*((256*b*d^3*x^4)/(35*(a*d - b*c)^5) + (128*d^2*x^3*(7*a* d + b*c))/(35*(a*d - b*c)^5) + (70*a^4*d^4 - 10*b^4*c^4 - 140*a^2*b^2*c^2* d^2 + 56*a*b^3*c^3*d + 280*a^3*b*c*d^3)/(35*b^3*d*(a*d - b*c)^5) + (x*(560 *a^3*b*d^4 + 16*b^4*c^3*d - 112*a*b^3*c^2*d^2 + 560*a^2*b^2*c*d^3))/(35*b^ 3*d*(a*d - b*c)^5) + (32*d*x^2*(35*a^2*d^2 - b^2*c^2 + 14*a*b*c*d))/(35*b* (a*d - b*c)^5)))/(x^4*(a + b*x)^(1/2) + (a^3*c*(a + b*x)^(1/2))/(b^3*d) + (x^3*(3*a*d + b*c)*(a + b*x)^(1/2))/(b*d) + (3*a*x^2*(a*d + b*c)*(a + b*x) ^(1/2))/(b^2*d) + (a^2*x*(a*d + 3*b*c)*(a + b*x)^(1/2))/(b^3*d))