\(\int \frac {x \sqrt {c+d x}}{\sqrt {a+b x}} \, dx\) [318]

Optimal result

Integrand size = 20, antiderivative size = 125 \[ \int \frac {x \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=-\frac {(b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^2 d}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b d}-\frac {(b c-a d) (b c+3 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} d^{3/2}} \] Output:

-1/4*(3*a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^2/d+1/2*(b*x+a)^(1/2)*(d*x+ 
c)^(3/2)/b/d-1/4*(-a*d+b*c)*(3*a*d+b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1 
/2)/(d*x+c)^(1/2))/b^(5/2)/d^(3/2)
 

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.01 \[ \int \frac {x \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} (-3 a d+b (c+2 d x))}{4 b^2 d}+\frac {\left (b^2 c^2+2 a b c d-3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \left (\sqrt {a-\frac {b c}{d}}-\sqrt {a+b x}\right )}\right )}{2 b^{5/2} d^{3/2}} \] Input:

Integrate[(x*Sqrt[c + d*x])/Sqrt[a + b*x],x]
 

Output:

(Sqrt[a + b*x]*Sqrt[c + d*x]*(-3*a*d + b*(c + 2*d*x)))/(4*b^2*d) + ((b^2*c 
^2 + 2*a*b*c*d - 3*a^2*d^2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*(Sqrt 
[a - (b*c)/d] - Sqrt[a + b*x]))])/(2*b^(5/2)*d^(3/2))
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {90, 60, 66, 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.

\(\displaystyle \int \frac {x \sqrt {c+d x}}{\sqrt {a+b x}} \, dx\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b d}-\frac {(3 a d+b c) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}}dx}{4 b d}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b d}-\frac {(3 a d+b c) \left (\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b d}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b d}-\frac {(3 a d+b c) \left (\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b d}-\frac {(3 a d+b c) \left (\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} \sqrt {d}}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b d}\)

Input:

Int[(x*Sqrt[c + d*x])/Sqrt[a + b*x],x]
 

Output:

(Sqrt[a + b*x]*(c + d*x)^(3/2))/(2*b*d) - ((b*c + 3*a*d)*((Sqrt[a + b*x]*S 
qrt[c + d*x])/b + ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sq 
rt[c + d*x])])/(b^(3/2)*Sqrt[d])))/(4*b*d)
 

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 221

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(250\) vs. \(2(99)=198\).

Time = 0.22 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.01

method	result	size
default	\(\frac {\sqrt {x d +c}\, \sqrt {b x +a}\, \left (4 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, b d x +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{2} d^{2}-2 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a b c d -\ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) b^{2} c^{2}-6 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, a d +2 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, b c \right )}{8 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, b^{2} d \sqrt {d b}}\)	\(251\)

Input:

int(x*(d*x+c)^(1/2)/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/8*(d*x+c)^(1/2)*(b*x+a)^(1/2)*(4*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*b*d 
*x+3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^ 
(1/2))*a^2*d^2-2*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d 
+b*c)/(d*b)^(1/2))*a*b*c*d-ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b) 
^(1/2)+a*d+b*c)/(d*b)^(1/2))*b^2*c^2-6*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2) 
*a*d+2*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b*c)/((b*x+a)*(d*x+c))^(1/2)/b^ 
2/d/(d*b)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.43 \[ \int \frac {x \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\left [-\frac {{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (2 \, b^{2} d^{2} x + b^{2} c d - 3 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, b^{3} d^{2}}, \frac {{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (2 \, b^{2} d^{2} x + b^{2} c d - 3 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, b^{3} d^{2}}\right ] \] Input:

integrate(x*(d*x+c)^(1/2)/(b*x+a)^(1/2),x, algorithm="fricas")
 

Output:

[-1/16*((b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*sqrt(b*d)*log(8*b^2*d^2*x^2 + b^ 
2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + 
 a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(2*b^2*d^2*x + b^2*c*d - 
3*a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d^2), 1/8*((b^2*c^2 + 2*a*b*c 
*d - 3*a^2*d^2)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqr 
t(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) 
+ 2*(2*b^2*d^2*x + b^2*c*d - 3*a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3* 
d^2)]
 

Sympy [F]

\[ \int \frac {x \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\int \frac {x \sqrt {c + d x}}{\sqrt {a + b x}}\, dx \] Input:

integrate(x*(d*x+c)**(1/2)/(b*x+a)**(1/2),x)
 

Output:

Integral(x*sqrt(c + d*x)/sqrt(a + b*x), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x*(d*x+c)^(1/2)/(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
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.06 \[ \int \frac {x \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, b x + 2 \, a + \frac {b c d - 5 \, a d^{2}}{d^{2}}\right )} \sqrt {b x + a} + \frac {{\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d}\right )} {\left | b \right |}}{4 \, b^{4}} \] Input:

integrate(x*(d*x+c)^(1/2)/(b*x+a)^(1/2),x, algorithm="giac")
 

Output:

1/4*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*b*x + 2*a + (b*c*d - 5*a*d^2)/ 
d^2)*sqrt(b*x + a) + (b^3*c^2 + 2*a*b^2*c*d - 3*a^2*b*d^2)*log(abs(-sqrt(b 
*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d))*a 
bs(b)/b^4
 

Mupad [B] (verification not implemented)

Time = 11.95 (sec) , antiderivative size = 584, normalized size of antiderivative = 4.67 \[ \int \frac {x \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\frac {\frac {\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (-\frac {3\,a^2\,b\,d^2}{2}+a\,b^2\,c\,d+\frac {b^3\,c^2}{2}\right )}{d^5\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (\frac {11\,a^2\,d^2}{2}+23\,a\,b\,c\,d+\frac {7\,b^2\,c^2}{2}\right )}{d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}-\frac {8\,\sqrt {a}\,c^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}{d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7\,\left (-\frac {3\,a^2\,d^2}{2}+a\,b\,c\,d+\frac {b^2\,c^2}{2}\right )}{b^2\,d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^7}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5\,\left (\frac {11\,a^2\,d^2}{2}+23\,a\,b\,c\,d+\frac {7\,b^2\,c^2}{2}\right )}{b\,d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}-\frac {\sqrt {a}\,\sqrt {c}\,\left (32\,a\,d+16\,b\,c\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}-\frac {8\,\sqrt {a}\,b^2\,c^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}}{\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}+\frac {b^4}{d^4}-\frac {4\,b^3\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}+\frac {6\,b^2\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}-\frac {4\,b\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}{d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}}+\frac {\mathrm {atanh}\left (\frac {\sqrt {d}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}\right )\,\left (a\,d-b\,c\right )\,\left (3\,a\,d+b\,c\right )}{2\,b^{5/2}\,d^{3/2}} \] Input:

int((x*(c + d*x)^(1/2))/(a + b*x)^(1/2),x)
 

Output:

((((a + b*x)^(1/2) - a^(1/2))*((b^3*c^2)/2 - (3*a^2*b*d^2)/2 + a*b^2*c*d)) 
/(d^5*((c + d*x)^(1/2) - c^(1/2))) + (((a + b*x)^(1/2) - a^(1/2))^3*((11*a 
^2*d^2)/2 + (7*b^2*c^2)/2 + 23*a*b*c*d))/(d^4*((c + d*x)^(1/2) - c^(1/2))^ 
3) - (8*a^(1/2)*c^(3/2)*((a + b*x)^(1/2) - a^(1/2))^6)/(d^2*((c + d*x)^(1/ 
2) - c^(1/2))^6) + (((a + b*x)^(1/2) - a^(1/2))^7*((b^2*c^2)/2 - (3*a^2*d^ 
2)/2 + a*b*c*d))/(b^2*d^2*((c + d*x)^(1/2) - c^(1/2))^7) + (((a + b*x)^(1/ 
2) - a^(1/2))^5*((11*a^2*d^2)/2 + (7*b^2*c^2)/2 + 23*a*b*c*d))/(b*d^3*((c 
+ d*x)^(1/2) - c^(1/2))^5) - (a^(1/2)*c^(1/2)*(32*a*d + 16*b*c)*((a + b*x) 
^(1/2) - a^(1/2))^4)/(d^3*((c + d*x)^(1/2) - c^(1/2))^4) - (8*a^(1/2)*b^2* 
c^(3/2)*((a + b*x)^(1/2) - a^(1/2))^2)/(d^4*((c + d*x)^(1/2) - c^(1/2))^2) 
)/(((a + b*x)^(1/2) - a^(1/2))^8/((c + d*x)^(1/2) - c^(1/2))^8 + b^4/d^4 - 
 (4*b^3*((a + b*x)^(1/2) - a^(1/2))^2)/(d^3*((c + d*x)^(1/2) - c^(1/2))^2) 
 + (6*b^2*((a + b*x)^(1/2) - a^(1/2))^4)/(d^2*((c + d*x)^(1/2) - c^(1/2))^ 
4) - (4*b*((a + b*x)^(1/2) - a^(1/2))^6)/(d*((c + d*x)^(1/2) - c^(1/2))^6) 
) + (atanh((d^(1/2)*((a + b*x)^(1/2) - a^(1/2)))/(b^(1/2)*((c + d*x)^(1/2) 
 - c^(1/2))))*(a*d - b*c)*(3*a*d + b*c))/(2*b^(5/2)*d^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.58 \[ \int \frac {x \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\frac {-3 \sqrt {d x +c}\, \sqrt {b x +a}\, a b \,d^{2}+\sqrt {d x +c}\, \sqrt {b x +a}\, b^{2} c d +2 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{2} d^{2} x +3 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} d^{2}-2 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a b c d -\sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{2} c^{2}}{4 b^{3} d^{2}} \] Input:

int(x*(d*x+c)^(1/2)/(b*x+a)^(1/2),x)
 

Output:

( - 3*sqrt(c + d*x)*sqrt(a + b*x)*a*b*d**2 + sqrt(c + d*x)*sqrt(a + b*x)*b 
**2*c*d + 2*sqrt(c + d*x)*sqrt(a + b*x)*b**2*d**2*x + 3*sqrt(d)*sqrt(b)*lo 
g((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*d* 
*2 - 2*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x)) 
/sqrt(a*d - b*c))*a*b*c*d - sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + s 
qrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b**2*c**2)/(4*b**3*d**2)