Integrand size = 22, antiderivative size = 131 \[ \int \frac {\sqrt {c+d x}}{x^3 \sqrt {a+b x}} \, dx=\frac {(3 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{4 a^2 c x}-\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 a c x^2}-\frac {(b c-a d) (3 b c+a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{3/2}} \] Output:
1/4*(a*d+3*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a^2/c/x-1/2*(b*x+a)^(1/2)*(d*x +c)^(3/2)/a/c/x^2-1/4*(-a*d+b*c)*(a*d+3*b*c)*arctanh(c^(1/2)*(b*x+a)^(1/2) /a^(1/2)/(d*x+c)^(1/2))/a^(5/2)/c^(3/2)
Time = 10.05 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {c+d x}}{x^3 \sqrt {a+b x}} \, dx=-\frac {\sqrt {a+b x} \sqrt {c+d x} (2 a c-3 b c x+a d x)}{4 a^2 c x^2}+\frac {\left (-3 b^2 c^2+2 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{3/2}} \] Input:
Integrate[Sqrt[c + d*x]/(x^3*Sqrt[a + b*x]),x]
Output:
-1/4*(Sqrt[a + b*x]*Sqrt[c + d*x]*(2*a*c - 3*b*c*x + a*d*x))/(a^2*c*x^2) + ((-3*b^2*c^2 + 2*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt [a]*Sqrt[c + d*x])])/(4*a^(5/2)*c^(3/2))
Time = 0.23 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {107, 105, 104, 221}
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.
\(\displaystyle \int \frac {\sqrt {c+d x}}{x^3 \sqrt {a+b x}} \, dx\) |
\(\Big \downarrow \) 107 |
\(\displaystyle -\frac {(a d+3 b c) \int \frac {\sqrt {c+d x}}{x^2 \sqrt {a+b x}}dx}{4 a c}-\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 a c x^2}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle -\frac {(a d+3 b c) \left (-\frac {(b c-a d) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 a}-\frac {\sqrt {a+b x} \sqrt {c+d x}}{a x}\right )}{4 a c}-\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 a c x^2}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle -\frac {(a d+3 b c) \left (-\frac {(b c-a d) \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{a}-\frac {\sqrt {a+b x} \sqrt {c+d x}}{a x}\right )}{4 a c}-\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 a c x^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {(a d+3 b c) \left (\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2} \sqrt {c}}-\frac {\sqrt {a+b x} \sqrt {c+d x}}{a x}\right )}{4 a c}-\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 a c x^2}\) |
Input:
Int[Sqrt[c + d*x]/(x^3*Sqrt[a + b*x]),x]
Output:
-1/2*(Sqrt[a + b*x]*(c + d*x)^(3/2))/(a*c*x^2) - ((3*b*c + a*d)*(-((Sqrt[a + b*x]*Sqrt[c + d*x])/(a*x)) + ((b*c - a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x ])/(Sqrt[a]*Sqrt[c + d*x])])/(a^(3/2)*Sqrt[c])))/(4*a*c)
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] || !SumSimplerQ[p, 1]) && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x ] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Leaf count of result is larger than twice the leaf count of optimal. \(256\) vs. \(2(105)=210\).
Time = 0.22 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.96
method | result | size |
default | \(\frac {\sqrt {x d +c}\, \sqrt {b x +a}\, \left (\ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a^{2} d^{2} x^{2}+2 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a b c d \,x^{2}-3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) b^{2} c^{2} x^{2}-2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {a c}\, a d x +6 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {a c}\, b c x -4 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, a c \sqrt {a c}\right )}{8 a^{2} c \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, x^{2} \sqrt {a c}}\) | \(257\) |
Input:
int((d*x+c)^(1/2)/x^3/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/8*(d*x+c)^(1/2)*(b*x+a)^(1/2)/a^2/c*(ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x +a)*(d*x+c))^(1/2)+2*a*c)/x)*a^2*d^2*x^2+2*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*( (b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*a*b*c*d*x^2-3*ln((a*d*x+b*c*x+2*(a*c)^(1/ 2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*b^2*c^2*x^2-2*((b*x+a)*(d*x+c))^(1/2) *(a*c)^(1/2)*a*d*x+6*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*b*c*x-4*((b*x+a)* (d*x+c))^(1/2)*a*c*(a*c)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/x^2/(a*c)^(1/2)
Time = 0.20 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.56 \[ \int \frac {\sqrt {c+d x}}{x^3 \sqrt {a+b x}} \, dx=\left [-\frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \sqrt {a c} x^{2} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} + 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (2 \, a^{2} c^{2} - {\left (3 \, a b c^{2} - a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, a^{3} c^{2} x^{2}}, \frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \sqrt {-a c} x^{2} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) - 2 \, {\left (2 \, a^{2} c^{2} - {\left (3 \, a b c^{2} - a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, a^{3} c^{2} x^{2}}\right ] \] Input:
integrate((d*x+c)^(1/2)/x^3/(b*x+a)^(1/2),x, algorithm="fricas")
Output:
[-1/16*((3*b^2*c^2 - 2*a*b*c*d - a^2*d^2)*sqrt(a*c)*x^2*log((8*a^2*c^2 + ( b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 + 4*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*s qrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(2*a^2*c^2 - (3*a*b*c^2 - a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*c^2*x^2), 1/8 *((3*b^2*c^2 - 2*a*b*c*d - a^2*d^2)*sqrt(-a*c)*x^2*arctan(1/2*(2*a*c + (b* c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)) - 2*(2*a^2*c^2 - (3*a*b*c^2 - a^2*c*d)*x)*sqrt(b *x + a)*sqrt(d*x + c))/(a^3*c^2*x^2)]
\[ \int \frac {\sqrt {c+d x}}{x^3 \sqrt {a+b x}} \, dx=\int \frac {\sqrt {c + d x}}{x^{3} \sqrt {a + b x}}\, dx \] Input:
integrate((d*x+c)**(1/2)/x**3/(b*x+a)**(1/2),x)
Output:
Integral(sqrt(c + d*x)/(x**3*sqrt(a + b*x)), x)
Exception generated. \[ \int \frac {\sqrt {c+d x}}{x^3 \sqrt {a+b x}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((d*x+c)^(1/2)/x^3/(b*x+a)^(1/2),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. 1048 vs. \(2 (105) = 210\).
Time = 0.55 (sec) , antiderivative size = 1048, normalized size of antiderivative = 8.00 \[ \int \frac {\sqrt {c+d x}}{x^3 \sqrt {a+b x}} \, dx =\text {Too large to display} \] Input:
integrate((d*x+c)^(1/2)/x^3/(b*x+a)^(1/2),x, algorithm="giac")
Output:
-1/4*((3*sqrt(b*d)*b^5*c^2 - 2*sqrt(b*d)*a*b^4*c*d - sqrt(b*d)*a^2*b^3*d^2 )*arctan(-1/2*(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)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*a^2*b*c) - 2*( 3*sqrt(b*d)*b^11*c^5 - 13*sqrt(b*d)*a*b^10*c^4*d + 22*sqrt(b*d)*a^2*b^9*c^ 3*d^2 - 18*sqrt(b*d)*a^3*b^8*c^2*d^3 + 7*sqrt(b*d)*a^4*b^7*c*d^4 - sqrt(b* d)*a^5*b^6*d^5 - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^9*c^4 + 8*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^8*c^3*d + 14*sqrt(b*d)*(sqrt(b*d)*s qrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^7*c^2*d^2 - 16 *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^6*c*d^3 + 3*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^5*d^4 + 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^7*c^3 + 3*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^6*c^2*d + 7*sqr t(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^5*c*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*a^3*b^4*d^3 - 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sq rt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^5*c^2 + 2*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^4*c*d + sqrt(b*d)*( sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^2*b^...
Time = 12.71 (sec) , antiderivative size = 901, normalized size of antiderivative = 6.88 \[ \int \frac {\sqrt {c+d x}}{x^3 \sqrt {a+b x}} \, dx =\text {Too large to display} \] Input:
int((c + d*x)^(1/2)/(x^3*(a + b*x)^(1/2)),x)
Output:
(log(((a + b*x)^(1/2) - a^(1/2))/((c + d*x)^(1/2) - c^(1/2)))*(a^(5/2)*c^( 1/2)*d^2 - 3*a^(1/2)*b^2*c^(5/2) + 2*a^(3/2)*b*c^(3/2)*d))/(8*a^3*c^2) - ( ((a + b*x)^(1/2) - a^(1/2))*(d^2/(4*a*c) - (3*d*(a*d + b*c))/(16*a^2*c)))/ ((c + d*x)^(1/2) - c^(1/2)) - (log(((c^(1/2)*(a + b*x)^(1/2) - a^(1/2)*(c + d*x)^(1/2))*(b*c^(1/2) - (a^(1/2)*d*((a + b*x)^(1/2) - a^(1/2)))/((c + d *x)^(1/2) - c^(1/2))))/((c + d*x)^(1/2) - c^(1/2)))*(a^(5/2)*c^(1/2)*d^2 - 3*a^(1/2)*b^2*c^(5/2) + 2*a^(3/2)*b*c^(3/2)*d))/(8*a^3*c^2) - ((((a + b*x )^(1/2) - a^(1/2))^2*((11*b^4*c^2)/32 - (5*a^2*b^2*d^2)/32 + (a*b^3*c*d)/8 ))/(a^(5/2)*c^(3/2)*d^2*((c + d*x)^(1/2) - c^(1/2))^2) - b^4/(32*a^(3/2)*c ^(1/2)*d^2) + (((a + b*x)^(1/2) - a^(1/2))^3*((b^4*c^3)/16 + (a^3*b*d^3)/1 6 + (3*a^2*b^2*c*d^2)/8 - (9*a*b^3*c^2*d)/8))/(a^3*c^2*d^2*((c + d*x)^(1/2 ) - c^(1/2))^3) - (((b^4*c)/8 - (a*b^3*d)/8)*((a + b*x)^(1/2) - a^(1/2)))/ (a^2*c*d^2*((c + d*x)^(1/2) - c^(1/2))) + (((a + b*x)^(1/2) - a^(1/2))^5*( (b^3*c^2)/4 + (5*a^2*b*d^2)/16 - (11*a*b^2*c*d)/16))/(a^3*c*d*((c + d*x)^( 1/2) - c^(1/2))^5) + (((a + b*x)^(1/2) - a^(1/2))^4*((a^4*d^4)/32 - (7*b^4 *c^4)/32 + (21*a^2*b^2*c^2*d^2)/32 + (a*b^3*c^3*d)/2 - (a^3*b*c*d^3)/2))/( a^(7/2)*c^(5/2)*d^2*((c + d*x)^(1/2) - c^(1/2))^4))/(((a + b*x)^(1/2) - a^ (1/2))^6/((c + d*x)^(1/2) - c^(1/2))^6 + (b^2*((a + b*x)^(1/2) - a^(1/2))^ 2)/(d^2*((c + d*x)^(1/2) - c^(1/2))^2) + (((a + b*x)^(1/2) - a^(1/2))^4*(a ^2*d^2 + b^2*c^2 + 4*a*b*c*d))/(a*c*d^2*((c + d*x)^(1/2) - c^(1/2))^4) ...
Time = 0.31 (sec) , antiderivative size = 819, normalized size of antiderivative = 6.25 \[ \int \frac {\sqrt {c+d x}}{x^3 \sqrt {a+b x}} \, dx =\text {Too large to display} \] Input:
int((d*x+c)^(1/2)/x^3/(b*x+a)^(1/2),x)
Output:
( - 4*sqrt(c + d*x)*sqrt(a + b*x)*a**3*c**2*d - 2*sqrt(c + d*x)*sqrt(a + b *x)*a**3*c*d**2*x - 4*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b*c**3 + 4*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b*c**2*d*x + 6*sqrt(c + d*x)*sqrt(a + b*x)*a*b**2 *c**3*x - sqrt(c)*sqrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a**3*d**3*x**2 - 3*sqrt(c)*sqrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a**2*b*c*d**2*x**2 + sqrt(c)*sqrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b* c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a*b**2*c**2*d*x**2 + 3 *sqrt(c)*sqrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c ) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*b**3*c**3*x**2 - sqrt(c )*sqrt(a)*log(sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d )*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a**3*d**3*x**2 - 3*sqrt(c)*sqrt(a )*log(sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a**2*b*c*d**2*x**2 + sqrt(c)*sqrt(a)*log( sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x ) + sqrt(b)*sqrt(c + d*x))*a*b**2*c**2*d*x**2 + 3*sqrt(c)*sqrt(a)*log(sqrt (2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*b**3*c**3*x**2 + sqrt(c)*sqrt(a)*log(2*sqrt(d)*sqrt (b)*sqrt(c + d*x)*sqrt(a + b*x) + 2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + 2...