Integrand size = 19, antiderivative size = 98 \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/2}} \, dx=\frac {2}{(b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}-\frac {8 b \sqrt {c+d x}}{3 (b c-a d)^2 (a+b x)^{3/2}}+\frac {16 b d \sqrt {c+d x}}{3 (b c-a d)^3 \sqrt {a+b x}} \] Output:
2/(-a*d+b*c)/(b*x+a)^(3/2)/(d*x+c)^(1/2)-8/3*b*(d*x+c)^(1/2)/(-a*d+b*c)^2/ (b*x+a)^(3/2)+16/3*b*d*(d*x+c)^(1/2)/(-a*d+b*c)^3/(b*x+a)^(1/2)
Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.77 \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/2}} \, dx=\frac {2 \left (3 a^2 d^2+6 a b d (c+2 d x)+b^2 \left (-c^2+4 c d x+8 d^2 x^2\right )\right )}{3 (b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}} \] Input:
Integrate[1/((a + b*x)^(5/2)*(c + d*x)^(3/2)),x]
Output:
(2*(3*a^2*d^2 + 6*a*b*d*(c + 2*d*x) + b^2*(-c^2 + 4*c*d*x + 8*d^2*x^2)))/( 3*(b*c - a*d)^3*(a + b*x)^(3/2)*Sqrt[c + d*x])
Time = 0.16 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.12, 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.
\(\displaystyle \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {4 d \int \frac {1}{(a+b x)^{3/2} (c+d x)^{3/2}}dx}{3 (b c-a d)}-\frac {2}{3 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {4 d \left (-\frac {2 d \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}}dx}{b c-a d}-\frac {2}{\sqrt {a+b x} \sqrt {c+d x} (b c-a d)}\right )}{3 (b c-a d)}-\frac {2}{3 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {4 d \left (-\frac {4 d \sqrt {a+b x}}{\sqrt {c+d x} (b c-a d)^2}-\frac {2}{\sqrt {a+b x} \sqrt {c+d x} (b c-a d)}\right )}{3 (b c-a d)}-\frac {2}{3 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}\) |
Input:
Int[1/((a + b*x)^(5/2)*(c + d*x)^(3/2)),x]
Output:
-2/(3*(b*c - a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x]) - (4*d*(-2/((b*c - a*d)*S qrt[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))
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 = 0.16 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.07
method | result | size |
gosper | \(-\frac {2 \left (8 d^{2} x^{2} b^{2}+12 x a b \,d^{2}+4 x \,b^{2} c d +3 a^{2} d^{2}+6 a b c d -b^{2} c^{2}\right )}{3 \left (b x +a \right )^{\frac {3}{2}} \sqrt {x d +c}\, \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
default | \(-\frac {2}{3 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {3}{2}} \sqrt {x d +c}}-\frac {4 d \left (-\frac {2}{\left (-a d +b c \right ) \sqrt {b x +a}\, \sqrt {x d +c}}+\frac {4 d \sqrt {b x +a}}{\left (-a d +b c \right ) \sqrt {x d +c}\, \left (a d -b c \right )}\right )}{3 \left (-a d +b c \right )}\) | \(105\) |
orering | \(-\frac {2 \left (8 d^{2} x^{2} b^{2}+12 x a b \,d^{2}+4 x \,b^{2} c d +3 a^{2} d^{2}+6 a b c d -b^{2} c^{2}\right )}{3 \left (b x +a \right )^{\frac {3}{2}} \sqrt {x d +c}\, \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
Input:
int(1/(b*x+a)^(5/2)/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/3*(8*b^2*d^2*x^2+12*a*b*d^2*x+4*b^2*c*d*x+3*a^2*d^2+6*a*b*c*d-b^2*c^2)/ (b*x+a)^(3/2)/(d*x+c)^(1/2)/(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. 273 vs. \(2 (82) = 164\).
Time = 0.21 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.79 \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/2}} \, dx=\frac {2 \, {\left (8 \, b^{2} d^{2} x^{2} - b^{2} c^{2} + 6 \, a b c d + 3 \, a^{2} d^{2} + 4 \, {\left (b^{2} c d + 3 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} + {\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} + {\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}} \] Input:
integrate(1/(b*x+a)^(5/2)/(d*x+c)^(3/2),x, algorithm="fricas")
Output:
2/3*(8*b^2*d^2*x^2 - b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2 + 4*(b^2*c*d + 3*a*b* d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a^2*b^3*c^4 - 3*a^3*b^2*c^3*d + 3*a^4 *b*c^2*d^2 - a^5*c*d^3 + (b^5*c^3*d - 3*a*b^4*c^2*d^2 + 3*a^2*b^3*c*d^3 - a^3*b^2*d^4)*x^3 + (b^5*c^4 - a*b^4*c^3*d - 3*a^2*b^3*c^2*d^2 + 5*a^3*b^2* c*d^3 - 2*a^4*b*d^4)*x^2 + (2*a*b^4*c^4 - 5*a^2*b^3*c^3*d + 3*a^3*b^2*c^2* d^2 + a^4*b*c*d^3 - a^5*d^4)*x)
\[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/2}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(b*x+a)**(5/2)/(d*x+c)**(3/2),x)
Output:
Integral(1/((a + b*x)**(5/2)*(c + d*x)**(3/2)), x)
Exception generated. \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(b*x+a)^(5/2)/(d*x+c)^(3/2),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. 368 vs. \(2 (82) = 164\).
Time = 0.20 (sec) , antiderivative size = 368, normalized size of antiderivative = 3.76 \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/2}} \, dx=\frac {2 \, \sqrt {b x + a} b^{2} d^{2}}{{\left (b^{3} c^{3} {\left | b \right |} - 3 \, a b^{2} c^{2} d {\left | b \right |} + 3 \, a^{2} b c d^{2} {\left | b \right |} - a^{3} d^{3} {\left | b \right |}\right )} \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} + \frac {4 \, {\left (5 \, \sqrt {b d} b^{6} c^{2} d - 10 \, \sqrt {b d} a b^{5} c d^{2} + 5 \, \sqrt {b d} a^{2} b^{4} d^{3} - 12 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{4} c d + 12 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{3} d^{2} + 3 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} b^{2} d\right )}}{3 \, {\left (b^{2} c^{2} {\left | b \right |} - 2 \, a b c d {\left | b \right |} + a^{2} d^{2} {\left | b \right |}\right )} {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}^{3}} \] Input:
integrate(1/(b*x+a)^(5/2)/(d*x+c)^(3/2),x, algorithm="giac")
Output:
2*sqrt(b*x + a)*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))*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)) + 4/3* (5*sqrt(b*d)*b^6*c^2*d - 10*sqrt(b*d)*a*b^5*c*d^2 + 5*sqrt(b*d)*a^2*b^4*d^ 3 - 12*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a *b*d))^2*b^4*c*d + 12*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^3*d^2 + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^2*d)/((b^2*c^2*abs(b) - 2*a*b*c* d*abs(b) + a^2*d^2*abs(b))*(b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqr t(b^2*c + (b*x + a)*b*d - a*b*d))^2)^3)
Time = 0.61 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.44 \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/2}} \, dx=-\frac {\sqrt {c+d\,x}\,\left (\frac {8\,x\,\left (3\,a\,d+b\,c\right )}{3\,{\left (a\,d-b\,c\right )}^3}+\frac {16\,b\,d\,x^2}{3\,{\left (a\,d-b\,c\right )}^3}+\frac {6\,a^2\,d^2+12\,a\,b\,c\,d-2\,b^2\,c^2}{3\,b\,d\,{\left (a\,d-b\,c\right )}^3}\right )}{x^2\,\sqrt {a+b\,x}+\frac {a\,c\,\sqrt {a+b\,x}}{b\,d}+\frac {x\,\left (a\,d+b\,c\right )\,\sqrt {a+b\,x}}{b\,d}} \] Input:
int(1/((a + b*x)^(5/2)*(c + d*x)^(3/2)),x)
Output:
-((c + d*x)^(1/2)*((8*x*(3*a*d + b*c))/(3*(a*d - b*c)^3) + (16*b*d*x^2)/(3 *(a*d - b*c)^3) + (6*a^2*d^2 - 2*b^2*c^2 + 12*a*b*c*d)/(3*b*d*(a*d - b*c)^ 3)))/(x^2*(a + b*x)^(1/2) + (a*c*(a + b*x)^(1/2))/(b*d) + (x*(a*d + b*c)*( a + b*x)^(1/2))/(b*d))
Time = 0.17 (sec) , antiderivative size = 298, normalized size of antiderivative = 3.04 \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/2}} \, dx=\frac {\frac {16 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a c d}{3}+\frac {16 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a \,d^{2} x}{3}+\frac {16 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, b c d x}{3}+\frac {16 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, b \,d^{2} x^{2}}{3}-2 \sqrt {d x +c}\, a^{2} d^{2}-4 \sqrt {d x +c}\, a b c d -8 \sqrt {d x +c}\, a b \,d^{2} x +\frac {2 \sqrt {d x +c}\, b^{2} c^{2}}{3}-\frac {8 \sqrt {d x +c}\, b^{2} c d x}{3}-\frac {16 \sqrt {d x +c}\, b^{2} d^{2} x^{2}}{3}}{\sqrt {b x +a}\, \left (a^{3} b \,d^{4} x^{2}-3 a^{2} b^{2} c \,d^{3} x^{2}+3 a \,b^{3} c^{2} d^{2} x^{2}-b^{4} c^{3} d \,x^{2}+a^{4} d^{4} x -2 a^{3} b c \,d^{3} x +2 a \,b^{3} c^{3} d x -b^{4} c^{4} x +a^{4} c \,d^{3}-3 a^{3} b \,c^{2} d^{2}+3 a^{2} b^{2} c^{3} d -a \,b^{3} c^{4}\right )} \] Input:
int(1/(b*x+a)^(5/2)/(d*x+c)^(3/2),x)
Output:
(2*(8*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*c*d + 8*sqrt(d)*sqrt(b)*sqrt(a + b*x )*a*d**2*x + 8*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b*c*d*x + 8*sqrt(d)*sqrt(b)*s qrt(a + b*x)*b*d**2*x**2 - 3*sqrt(c + d*x)*a**2*d**2 - 6*sqrt(c + d*x)*a*b *c*d - 12*sqrt(c + d*x)*a*b*d**2*x + sqrt(c + d*x)*b**2*c**2 - 4*sqrt(c + d*x)*b**2*c*d*x - 8*sqrt(c + d*x)*b**2*d**2*x**2))/(3*sqrt(a + b*x)*(a**4* c*d**3 + a**4*d**4*x - 3*a**3*b*c**2*d**2 - 2*a**3*b*c*d**3*x + a**3*b*d** 4*x**2 + 3*a**2*b**2*c**3*d - 3*a**2*b**2*c*d**3*x**2 - a*b**3*c**4 + 2*a* b**3*c**3*d*x + 3*a*b**3*c**2*d**2*x**2 - b**4*c**4*x - b**4*c**3*d*x**2))