Integrand size = 19, antiderivative size = 136 \[ \int \frac {1}{(a+b x)^{7/2} (c+d x)^{3/2}} \, dx=-\frac {2}{5 (b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}+\frac {4 d}{5 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}-\frac {16 d^2}{5 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}-\frac {32 d^3 \sqrt {a+b x}}{5 (b c-a d)^4 \sqrt {c+d x}} \]
-2/5/(-a*d+b*c)/(b*x+a)^(5/2)/(d*x+c)^(1/2)+4/5*d/(-a*d+b*c)^2/(b*x+a)^(3/ 2)/(d*x+c)^(1/2)-16/5*d^2/(-a*d+b*c)^3/(b*x+a)^(1/2)/(d*x+c)^(1/2)-32/5*d^ 3*(b*x+a)^(1/2)/(-a*d+b*c)^4/(d*x+c)^(1/2)
Time = 0.15 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(a+b x)^{7/2} (c+d x)^{3/2}} \, dx=-\frac {2 \left (5 a^3 d^3+15 a^2 b d^2 (c+2 d x)+5 a b^2 d \left (-c^2+4 c d x+8 d^2 x^2\right )+b^3 \left (c^3-2 c^2 d x+8 c d^2 x^2+16 d^3 x^3\right )\right )}{5 (b c-a d)^4 (a+b x)^{5/2} \sqrt {c+d x}} \]
(-2*(5*a^3*d^3 + 15*a^2*b*d^2*(c + 2*d*x) + 5*a*b^2*d*(-c^2 + 4*c*d*x + 8* d^2*x^2) + b^3*(c^3 - 2*c^2*d*x + 8*c*d^2*x^2 + 16*d^3*x^3)))/(5*(b*c - a* d)^4*(a + b*x)^(5/2)*Sqrt[c + d*x])
Time = 0.20 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.16, 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.
\(\displaystyle \int \frac {1}{(a+b x)^{7/2} (c+d x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {6 d \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/2}}dx}{5 (b c-a d)}-\frac {2}{5 (a+b x)^{5/2} \sqrt {c+d x} (b c-a d)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {6 d \left (-\frac {4 d \int \frac {1}{(a+b x)^{3/2} (c+d x)^{3/2}}dx}{3 (b c-a d)}-\frac {2}{3 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}\right )}{5 (b c-a d)}-\frac {2}{5 (a+b x)^{5/2} \sqrt {c+d x} (b c-a d)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {6 d \left (-\frac {4 d \left (-\frac {2 d \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}}dx}{b c-a d}-\frac {2}{\sqrt {a+b x} \sqrt {c+d x} (b c-a d)}\right )}{3 (b c-a d)}-\frac {2}{3 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}\right )}{5 (b c-a d)}-\frac {2}{5 (a+b x)^{5/2} \sqrt {c+d x} (b c-a d)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {6 d \left (-\frac {4 d \left (-\frac {4 d \sqrt {a+b x}}{\sqrt {c+d x} (b c-a d)^2}-\frac {2}{\sqrt {a+b x} \sqrt {c+d x} (b c-a d)}\right )}{3 (b c-a d)}-\frac {2}{3 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}\right )}{5 (b c-a d)}-\frac {2}{5 (a+b x)^{5/2} \sqrt {c+d x} (b c-a d)}\) |
-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))
3.16.9.3.1 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 = 0.52 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.07
method | result | size |
default | \(-\frac {2}{5 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {5}{2}} \sqrt {d x +c}}-\frac {6 d \left (-\frac {2}{3 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {3}{2}} \sqrt {d x +c}}-\frac {4 d \left (-\frac {2}{\left (-a d +b c \right ) \sqrt {b x +a}\, \sqrt {d x +c}}+\frac {4 d \sqrt {b x +a}}{\left (-a d +b c \right ) \sqrt {d x +c}\, \left (a d -b c \right )}\right )}{3 \left (-a d +b c \right )}\right )}{5 \left (-a d +b c \right )}\) | \(145\) |
gosper | \(-\frac {2 \left (16 d^{3} x^{3} b^{3}+40 x^{2} a \,b^{2} d^{3}+8 x^{2} b^{3} c \,d^{2}+30 x \,a^{2} b \,d^{3}+20 x a \,b^{2} c \,d^{2}-2 x \,b^{3} c^{2} d +5 a^{3} d^{3}+15 a^{2} b c \,d^{2}-5 a \,b^{2} c^{2} d +b^{3} c^{3}\right )}{5 \left (b x +a \right )^{\frac {5}{2}} \sqrt {d x +c}\, \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right )}\) | \(170\) |
-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. 455 vs. \(2 (112) = 224\).
Time = 0.53 (sec) , antiderivative size = 455, normalized size of antiderivative = 3.35 \[ \int \frac {1}{(a+b x)^{7/2} (c+d x)^{3/2}} \, dx=-\frac {2 \, {\left (16 \, b^{3} d^{3} x^{3} + b^{3} c^{3} - 5 \, a b^{2} c^{2} d + 15 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3} + 8 \, {\left (b^{3} c d^{2} + 5 \, a b^{2} d^{3}\right )} x^{2} - 2 \, {\left (b^{3} c^{2} d - 10 \, a b^{2} c d^{2} - 15 \, a^{2} b d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{5 \, {\left (a^{3} b^{4} c^{5} - 4 \, a^{4} b^{3} c^{4} d + 6 \, a^{5} b^{2} c^{3} d^{2} - 4 \, a^{6} b c^{2} d^{3} + a^{7} c d^{4} + {\left (b^{7} c^{4} d - 4 \, a b^{6} c^{3} d^{2} + 6 \, a^{2} b^{5} c^{2} d^{3} - 4 \, a^{3} b^{4} c d^{4} + a^{4} b^{3} d^{5}\right )} x^{4} + {\left (b^{7} c^{5} - a b^{6} c^{4} d - 6 \, a^{2} b^{5} c^{3} d^{2} + 14 \, a^{3} b^{4} c^{2} d^{3} - 11 \, a^{4} b^{3} c d^{4} + 3 \, a^{5} b^{2} d^{5}\right )} x^{3} + 3 \, {\left (a b^{6} c^{5} - 3 \, a^{2} b^{5} c^{4} d + 2 \, a^{3} b^{4} c^{3} d^{2} + 2 \, a^{4} b^{3} c^{2} d^{3} - 3 \, a^{5} b^{2} c d^{4} + a^{6} b d^{5}\right )} x^{2} + {\left (3 \, a^{2} b^{5} c^{5} - 11 \, a^{3} b^{4} c^{4} d + 14 \, a^{4} b^{3} c^{3} d^{2} - 6 \, a^{5} b^{2} c^{2} d^{3} - a^{6} b c d^{4} + a^{7} d^{5}\right )} x\right )}} \]
-2/5*(16*b^3*d^3*x^3 + b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*d^ 3 + 8*(b^3*c*d^2 + 5*a*b^2*d^3)*x^2 - 2*(b^3*c^2*d - 10*a*b^2*c*d^2 - 15*a ^2*b*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a^3*b^4*c^5 - 4*a^4*b^3*c^4*d + 6*a^5*b^2*c^3*d^2 - 4*a^6*b*c^2*d^3 + a^7*c*d^4 + (b^7*c^4*d - 4*a*b^6*c^3 *d^2 + 6*a^2*b^5*c^2*d^3 - 4*a^3*b^4*c*d^4 + a^4*b^3*d^5)*x^4 + (b^7*c^5 - a*b^6*c^4*d - 6*a^2*b^5*c^3*d^2 + 14*a^3*b^4*c^2*d^3 - 11*a^4*b^3*c*d^4 + 3*a^5*b^2*d^5)*x^3 + 3*(a*b^6*c^5 - 3*a^2*b^5*c^4*d + 2*a^3*b^4*c^3*d^2 + 2*a^4*b^3*c^2*d^3 - 3*a^5*b^2*c*d^4 + a^6*b*d^5)*x^2 + (3*a^2*b^5*c^5 - 1 1*a^3*b^4*c^4*d + 14*a^4*b^3*c^3*d^2 - 6*a^5*b^2*c^2*d^3 - a^6*b*c*d^4 + a ^7*d^5)*x)
\[ \int \frac {1}{(a+b x)^{7/2} (c+d x)^{3/2}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {7}{2}} \left (c + d x\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {1}{(a+b x)^{7/2} (c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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. 830 vs. \(2 (112) = 224\).
Time = 0.51 (sec) , antiderivative size = 830, normalized size of antiderivative = 6.10 \[ \int \frac {1}{(a+b x)^{7/2} (c+d x)^{3/2}} \, dx =\text {Too large to display} \]
-2*sqrt(b*x + a)*b^2*d^3/((b^4*c^4*abs(b) - 4*a*b^3*c^3*d*abs(b) + 6*a^2*b ^2*c^2*d^2*abs(b) - 4*a^3*b*c*d^3*abs(b) + a^4*d^4*abs(b))*sqrt(b^2*c + (b *x + a)*b*d - a*b*d)) - 4/5*(11*sqrt(b*d)*b^10*c^4*d^2 - 44*sqrt(b*d)*a*b^ 9*c^3*d^3 + 66*sqrt(b*d)*a^2*b^8*c^2*d^4 - 44*sqrt(b*d)*a^3*b^7*c*d^5 + 11 *sqrt(b*d)*a^4*b^6*d^6 - 50*sqrt(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^2 + 150*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^3 - 150*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^4 + 50*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^5 + 80*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - s qrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^6*c^2*d^2 - 160*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^3 + 80 *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^4 - 30*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^4*c*d^2 + 30*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^3 + 5*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^2)/((b^3*c^3* abs(b) - 3*a*b^2*c^2*d*abs(b) + 3*a^2*b*c*d^2*abs(b) - a^3*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)^5)
Time = 1.29 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.67 \[ \int \frac {1}{(a+b x)^{7/2} (c+d x)^{3/2}} \, dx=-\frac {\sqrt {c+d\,x}\,\left (\frac {16\,d\,x^2\,\left (5\,a\,d+b\,c\right )}{5\,{\left (a\,d-b\,c\right )}^4}+\frac {2\,a^3\,d^3+6\,a^2\,b\,c\,d^2-2\,a\,b^2\,c^2\,d+\frac {2\,b^3\,c^3}{5}}{b^2\,d\,{\left (a\,d-b\,c\right )}^4}+\frac {32\,b\,d^2\,x^3}{5\,{\left (a\,d-b\,c\right )}^4}+\frac {4\,x\,\left (15\,a^2\,d^2+10\,a\,b\,c\,d-b^2\,c^2\right )}{5\,b\,{\left (a\,d-b\,c\right )}^4}\right )}{x^3\,\sqrt {a+b\,x}+\frac {a^2\,c\,\sqrt {a+b\,x}}{b^2\,d}+\frac {x^2\,\left (2\,a\,d+b\,c\right )\,\sqrt {a+b\,x}}{b\,d}+\frac {a\,x\,\left (a\,d+2\,b\,c\right )\,\sqrt {a+b\,x}}{b^2\,d}} \]
-((c + d*x)^(1/2)*((16*d*x^2*(5*a*d + b*c))/(5*(a*d - b*c)^4) + (2*a^3*d^3 + (2*b^3*c^3)/5 - 2*a*b^2*c^2*d + 6*a^2*b*c*d^2)/(b^2*d*(a*d - b*c)^4) + (32*b*d^2*x^3)/(5*(a*d - b*c)^4) + (4*x*(15*a^2*d^2 - b^2*c^2 + 10*a*b*c*d ))/(5*b*(a*d - b*c)^4)))/(x^3*(a + b*x)^(1/2) + (a^2*c*(a + b*x)^(1/2))/(b ^2*d) + (x^2*(2*a*d + b*c)*(a + b*x)^(1/2))/(b*d) + (a*x*(a*d + 2*b*c)*(a + b*x)^(1/2))/(b^2*d))
Time = 0.02 (sec) , antiderivative size = 579, normalized size of antiderivative = 4.26 \[ \int \frac {1}{(a+b x)^{7/2} (c+d x)^{3/2}} \, dx=\frac {\frac {32 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a^{2} c \,d^{2}}{5}+\frac {32 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a^{2} d^{3} x}{5}+\frac {64 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a b c \,d^{2} x}{5}+\frac {64 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a b \,d^{3} x^{2}}{5}+\frac {32 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, b^{2} c \,d^{2} x^{2}}{5}+\frac {32 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, b^{2} d^{3} x^{3}}{5}-2 \sqrt {d x +c}\, a^{3} d^{3}-6 \sqrt {d x +c}\, a^{2} b c \,d^{2}-12 \sqrt {d x +c}\, a^{2} b \,d^{3} x +2 \sqrt {d x +c}\, a \,b^{2} c^{2} d -8 \sqrt {d x +c}\, a \,b^{2} c \,d^{2} x -16 \sqrt {d x +c}\, a \,b^{2} d^{3} x^{2}-\frac {2 \sqrt {d x +c}\, b^{3} c^{3}}{5}+\frac {4 \sqrt {d x +c}\, b^{3} c^{2} d x}{5}-\frac {16 \sqrt {d x +c}\, b^{3} c \,d^{2} x^{2}}{5}-\frac {32 \sqrt {d x +c}\, b^{3} d^{3} x^{3}}{5}}{\sqrt {b x +a}\, \left (a^{4} b^{2} d^{5} x^{3}-4 a^{3} b^{3} c \,d^{4} x^{3}+6 a^{2} b^{4} c^{2} d^{3} x^{3}-4 a \,b^{5} c^{3} d^{2} x^{3}+b^{6} c^{4} d \,x^{3}+2 a^{5} b \,d^{5} x^{2}-7 a^{4} b^{2} c \,d^{4} x^{2}+8 a^{3} b^{3} c^{2} d^{3} x^{2}-2 a^{2} b^{4} c^{3} d^{2} x^{2}-2 a \,b^{5} c^{4} d \,x^{2}+b^{6} c^{5} x^{2}+a^{6} d^{5} x -2 a^{5} b c \,d^{4} x -2 a^{4} b^{2} c^{2} d^{3} x +8 a^{3} b^{3} c^{3} d^{2} x -7 a^{2} b^{4} c^{4} d x +2 a \,b^{5} c^{5} x +a^{6} c \,d^{4}-4 a^{5} b \,c^{2} d^{3}+6 a^{4} b^{2} c^{3} d^{2}-4 a^{3} b^{3} c^{4} d +a^{2} b^{4} c^{5}\right )} \]
int(1/(sqrt(c + d*x)*sqrt(a + b*x)*(a**3*c + a**3*d*x + 3*a**2*b*c*x + 3*a **2*b*d*x**2 + 3*a*b**2*c*x**2 + 3*a*b**2*d*x**3 + b**3*c*x**3 + b**3*d*x* *4)),x)
(2*(16*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**2*c*d**2 + 16*sqrt(d)*sqrt(b)*sqrt (a + b*x)*a**2*d**3*x + 32*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b*c*d**2*x + 32 *sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b*d**3*x**2 + 16*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b**2*c*d**2*x**2 + 16*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b**2*d**3*x**3 - 5*sqrt(c + d*x)*a**3*d**3 - 15*sqrt(c + d*x)*a**2*b*c*d**2 - 30*sqrt(c + d*x)*a**2*b*d**3*x + 5*sqrt(c + d*x)*a*b**2*c**2*d - 20*sqrt(c + d*x)*a*b* *2*c*d**2*x - 40*sqrt(c + d*x)*a*b**2*d**3*x**2 - sqrt(c + d*x)*b**3*c**3 + 2*sqrt(c + d*x)*b**3*c**2*d*x - 8*sqrt(c + d*x)*b**3*c*d**2*x**2 - 16*sq rt(c + d*x)*b**3*d**3*x**3))/(5*sqrt(a + b*x)*(a**6*c*d**4 + a**6*d**5*x - 4*a**5*b*c**2*d**3 - 2*a**5*b*c*d**4*x + 2*a**5*b*d**5*x**2 + 6*a**4*b**2 *c**3*d**2 - 2*a**4*b**2*c**2*d**3*x - 7*a**4*b**2*c*d**4*x**2 + a**4*b**2 *d**5*x**3 - 4*a**3*b**3*c**4*d + 8*a**3*b**3*c**3*d**2*x + 8*a**3*b**3*c* *2*d**3*x**2 - 4*a**3*b**3*c*d**4*x**3 + a**2*b**4*c**5 - 7*a**2*b**4*c**4 *d*x - 2*a**2*b**4*c**3*d**2*x**2 + 6*a**2*b**4*c**2*d**3*x**3 + 2*a*b**5* c**5*x - 2*a*b**5*c**4*d*x**2 - 4*a*b**5*c**3*d**2*x**3 + b**6*c**5*x**2 + b**6*c**4*d*x**3))