Integrand size = 19, antiderivative size = 98 \[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=-\frac {2}{(b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}-\frac {8 d \sqrt {a+b x}}{3 (b c-a d)^2 (c+d x)^{3/2}}-\frac {16 b d \sqrt {a+b x}}{3 (b c-a d)^3 \sqrt {c+d x}} \] Output:
-2/(-a*d+b*c)/(b*x+a)^(1/2)/(d*x+c)^(3/2)-8/3*d*(b*x+a)^(1/2)/(-a*d+b*c)^2 /(d*x+c)^(3/2)-16/3*b*d*(b*x+a)^(1/2)/(-a*d+b*c)^3/(d*x+c)^(1/2)
Time = 0.01 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.80 \[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\frac {2 a^2 d^2-4 a b d (3 c+2 d x)-2 b^2 \left (3 c^2+12 c d x+8 d^2 x^2\right )}{3 (b c-a d)^3 \sqrt {a+b x} (c+d x)^{3/2}} \] Input:
Integrate[1/((a + b*x)^(3/2)*(c + d*x)^(5/2)),x]
Output:
(2*a^2*d^2 - 4*a*b*d*(3*c + 2*d*x) - 2*b^2*(3*c^2 + 12*c*d*x + 8*d^2*x^2)) /(3*(b*c - a*d)^3*Sqrt[a + b*x]*(c + d*x)^(3/2))
Time = 0.18 (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.
\(\displaystyle \int \frac {1}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {4 d \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/2}}dx}{b c-a d}-\frac {2}{\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {4 d \left (\frac {2 b \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}}dx}{3 (b c-a d)}+\frac {2 \sqrt {a+b x}}{3 (c+d x)^{3/2} (b c-a d)}\right )}{b c-a d}-\frac {2}{\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {4 d \left (\frac {4 b \sqrt {a+b x}}{3 \sqrt {c+d x} (b c-a d)^2}+\frac {2 \sqrt {a+b x}}{3 (c+d x)^{3/2} (b c-a d)}\right )}{b c-a d}-\frac {2}{\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}\) |
Input:
Int[1/((a + b*x)^(3/2)*(c + d*x)^(5/2)),x]
Output:
-2/((b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2)) - (4*d*((2*Sqrt[a + b*x])/( 3*(b*c - a*d)*(c + d*x)^(3/2)) + (4*b*Sqrt[a + b*x])/(3*(b*c - a*d)^2*Sqrt [c + d*x])))/(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 = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.97
method | result | size |
default | \(-\frac {2}{\left (-a d +b c \right ) \sqrt {b x +a}\, \left (x d +c \right )^{\frac {3}{2}}}-\frac {4 d \left (-\frac {2 \sqrt {b x +a}}{3 \left (a d -b c \right ) \left (x d +c \right )^{\frac {3}{2}}}+\frac {4 b \sqrt {b x +a}}{3 \left (a d -b c \right )^{2} \sqrt {x d +c}}\right )}{-a d +b c}\) | \(95\) |
gosper | \(-\frac {2 \left (-8 d^{2} x^{2} b^{2}-4 x a b \,d^{2}-12 x \,b^{2} c d +a^{2} d^{2}-6 a b c d -3 b^{2} c^{2}\right )}{3 \sqrt {b x +a}\, \left (x d +c \right )^{\frac {3}{2}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(104\) |
orering | \(-\frac {2 \left (-8 d^{2} x^{2} b^{2}-4 x a b \,d^{2}-12 x \,b^{2} c d +a^{2} d^{2}-6 a b c d -3 b^{2} c^{2}\right )}{3 \sqrt {b x +a}\, \left (x d +c \right )^{\frac {3}{2}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(104\) |
Input:
int(1/(b*x+a)^(3/2)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/(-a*d+b*c)/(b*x+a)^(1/2)/(d*x+c)^(3/2)-4*d/(-a*d+b*c)*(-2/3/(a*d-b*c)/( d*x+c)^(3/2)*(b*x+a)^(1/2)+4/3*b/(a*d-b*c)^2*(b*x+a)^(1/2)/(d*x+c)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (82) = 164\).
Time = 0.31 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.79 \[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=-\frac {2 \, {\left (8 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} + 6 \, a b c d - a^{2} d^{2} + 4 \, {\left (3 \, b^{2} c d + a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (a b^{3} c^{5} - 3 \, a^{2} b^{2} c^{4} d + 3 \, a^{3} b c^{3} d^{2} - a^{4} c^{2} d^{3} + {\left (b^{4} c^{3} d^{2} - 3 \, a b^{3} c^{2} d^{3} + 3 \, a^{2} b^{2} c d^{4} - a^{3} b d^{5}\right )} x^{3} + {\left (2 \, b^{4} c^{4} d - 5 \, a b^{3} c^{3} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{3} + a^{3} b c d^{4} - a^{4} d^{5}\right )} x^{2} + {\left (b^{4} c^{5} - a b^{3} c^{4} d - 3 \, a^{2} b^{2} c^{3} d^{2} + 5 \, a^{3} b c^{2} d^{3} - 2 \, a^{4} c d^{4}\right )} x\right )}} \] Input:
integrate(1/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="fricas")
Output:
-2/3*(8*b^2*d^2*x^2 + 3*b^2*c^2 + 6*a*b*c*d - a^2*d^2 + 4*(3*b^2*c*d + a*b *d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3* b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^3 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a ^3*b*c*d^4 - a^4*d^5)*x^2 + (b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5 *a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x)
\[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(b*x+a)**(3/2)/(d*x+c)**(5/2),x)
Output:
Integral(1/((a + b*x)**(3/2)*(c + d*x)**(5/2)), x)
Exception generated. \[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(b*x+a)^(3/2)/(d*x+c)^(5/2),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. 373 vs. \(2 (82) = 164\).
Time = 0.16 (sec) , antiderivative size = 373, normalized size of antiderivative = 3.81 \[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=-\frac {4 \, \sqrt {b d} b^{3}}{{\left (b^{2} c^{2} {\left | b \right |} - 2 \, a b c d {\left | b \right |} + a^{2} d^{2} {\left | b \right |}\right )} {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}} - \frac {2 \, \sqrt {b x + a} {\left (\frac {5 \, {\left (b^{6} c^{2} d^{3} {\left | b \right |} - 2 \, a b^{5} c d^{4} {\left | b \right |} + a^{2} b^{4} d^{5} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{7} c^{5} d - 5 \, a b^{6} c^{4} d^{2} + 10 \, a^{2} b^{5} c^{3} d^{3} - 10 \, a^{3} b^{4} c^{2} d^{4} + 5 \, a^{4} b^{3} c d^{5} - a^{5} b^{2} d^{6}} + \frac {6 \, {\left (b^{7} c^{3} d^{2} {\left | b \right |} - 3 \, a b^{6} c^{2} d^{3} {\left | b \right |} + 3 \, a^{2} b^{5} c d^{4} {\left | b \right |} - a^{3} b^{4} d^{5} {\left | b \right |}\right )}}{b^{7} c^{5} d - 5 \, a b^{6} c^{4} d^{2} + 10 \, a^{2} b^{5} c^{3} d^{3} - 10 \, a^{3} b^{4} c^{2} d^{4} + 5 \, a^{4} b^{3} c d^{5} - a^{5} b^{2} d^{6}}\right )}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} \] Input:
integrate(1/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="giac")
Output:
-4*sqrt(b*d)*b^3/((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) - sqrt(b^2*c + (b*x + a)*b*d - a*b* d))^2)) - 2/3*sqrt(b*x + a)*(5*(b^6*c^2*d^3*abs(b) - 2*a*b^5*c*d^4*abs(b) + a^2*b^4*d^5*abs(b))*(b*x + a)/(b^7*c^5*d - 5*a*b^6*c^4*d^2 + 10*a^2*b^5* c^3*d^3 - 10*a^3*b^4*c^2*d^4 + 5*a^4*b^3*c*d^5 - a^5*b^2*d^6) + 6*(b^7*c^3 *d^2*abs(b) - 3*a*b^6*c^2*d^3*abs(b) + 3*a^2*b^5*c*d^4*abs(b) - a^3*b^4*d^ 5*abs(b))/(b^7*c^5*d - 5*a*b^6*c^4*d^2 + 10*a^2*b^5*c^3*d^3 - 10*a^3*b^4*c ^2*d^4 + 5*a^4*b^3*c*d^5 - a^5*b^2*d^6))/(b^2*c + (b*x + a)*b*d - a*b*d)^( 3/2)
Time = 1.40 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.35 \[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\frac {\sqrt {c+d\,x}\,\left (\frac {16\,b^2\,x^2}{3\,{\left (a\,d-b\,c\right )}^3}+\frac {-2\,a^2\,d^2+12\,a\,b\,c\,d+6\,b^2\,c^2}{3\,d^2\,{\left (a\,d-b\,c\right )}^3}+\frac {8\,b\,x\,\left (a\,d+3\,b\,c\right )}{3\,d\,{\left (a\,d-b\,c\right )}^3}\right )}{x^2\,\sqrt {a+b\,x}+\frac {c^2\,\sqrt {a+b\,x}}{d^2}+\frac {2\,c\,x\,\sqrt {a+b\,x}}{d}} \] Input:
int(1/((a + b*x)^(3/2)*(c + d*x)^(5/2)),x)
Output:
((c + d*x)^(1/2)*((16*b^2*x^2)/(3*(a*d - b*c)^3) + (6*b^2*c^2 - 2*a^2*d^2 + 12*a*b*c*d)/(3*d^2*(a*d - b*c)^3) + (8*b*x*(a*d + 3*b*c))/(3*d*(a*d - b* c)^3)))/(x^2*(a + b*x)^(1/2) + (c^2*(a + b*x)^(1/2))/d^2 + (2*c*x*(a + b*x )^(1/2))/d)
Time = 0.20 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.94 \[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\frac {-\frac {16 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, b \,c^{2}}{3}-\frac {32 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, b c d x}{3}-\frac {16 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, b \,d^{2} x^{2}}{3}-\frac {2 \sqrt {d x +c}\, a^{2} d^{2}}{3}+4 \sqrt {d x +c}\, a b c d +\frac {8 \sqrt {d x +c}\, a b \,d^{2} x}{3}+2 \sqrt {d x +c}\, b^{2} c^{2}+8 \sqrt {d x +c}\, b^{2} c d x +\frac {16 \sqrt {d x +c}\, b^{2} d^{2} x^{2}}{3}}{\sqrt {b x +a}\, \left (a^{3} d^{5} x^{2}-3 a^{2} b c \,d^{4} x^{2}+3 a \,b^{2} c^{2} d^{3} x^{2}-b^{3} c^{3} d^{2} x^{2}+2 a^{3} c \,d^{4} x -6 a^{2} b \,c^{2} d^{3} x +6 a \,b^{2} c^{3} d^{2} x -2 b^{3} c^{4} d x +a^{3} c^{2} d^{3}-3 a^{2} b \,c^{3} d^{2}+3 a \,b^{2} c^{4} d -b^{3} c^{5}\right )} \] Input:
int(1/(b*x+a)^(3/2)/(d*x+c)^(5/2),x)
Output:
(2*( - 8*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b*c**2 - 16*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b*c*d*x - 8*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b*d**2*x**2 - sqrt(c + d* x)*a**2*d**2 + 6*sqrt(c + d*x)*a*b*c*d + 4*sqrt(c + d*x)*a*b*d**2*x + 3*sq rt(c + d*x)*b**2*c**2 + 12*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**3*c**2*d**3 + 2*a**3*c*d**4*x + a**3*d** 5*x**2 - 3*a**2*b*c**3*d**2 - 6*a**2*b*c**2*d**3*x - 3*a**2*b*c*d**4*x**2 + 3*a*b**2*c**4*d + 6*a*b**2*c**3*d**2*x + 3*a*b**2*c**2*d**3*x**2 - b**3* c**5 - 2*b**3*c**4*d*x - b**3*c**3*d**2*x**2))