\(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx\) [121]

Optimal result

Integrand size = 35, antiderivative size = 160 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx=\frac {(2 a+b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{2 (a+b x)}+\frac {b c \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} (a+b x)}-\frac {a \sqrt {c} \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a+b x} \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx=\frac {\sqrt {(a+b x)^2} \left (\sqrt {d} (2 a+b x) \sqrt {c+d x^2}+4 a \sqrt {c} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} x-\sqrt {c+d x^2}}{\sqrt {c}}\right )-b c \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )\right )}{2 \sqrt {d} (a+b x)} \] Input:

Integrate[(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + d*x^2])/x,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.68, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {1334, 27, 535, 538, 224, 219, 243, 73, 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.

\(\displaystyle \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx\)

\(\Big \downarrow \) 1334

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 535

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

\(\Big \downarrow \) 538

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{2} c \left (2 a \int \frac {1}{x \sqrt {d x^2+c}}dx+\frac {b \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{\sqrt {d}}\right )+\frac {1}{2} (2 a+b x) \sqrt {c+d x^2}\right )}{a+b x}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{2} c \left (a \int \frac {1}{x^2 \sqrt {d x^2+c}}dx^2+\frac {b \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{\sqrt {d}}\right )+\frac {1}{2} (2 a+b x) \sqrt {c+d x^2}\right )}{a+b x}\)

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{2} c \left (\frac {b \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{\sqrt {d}}-\frac {2 a \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}}\right )+\frac {1}{2} (2 a+b x) \sqrt {c+d x^2}\right )}{a+b x}\)

Input:

Int[(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + d*x^2])/x,x]
 

Output:

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(((2*a + b*x)*Sqrt[c + d*x^2])/2 + (c*((b*A 
rcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/Sqrt[d] - (2*a*ArcTanh[Sqrt[c + d*x^2 
]/Sqrt[c]])/Sqrt[c]))/2))/(a + b*x)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 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]
 

rule 224

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 535

Int[(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_))/(x_), x_Symbol] :> Sim 
p[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^p/(2*p*(2*p + 1))), x] + Simp[a/(2*p 
 + 1)   Int[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^(p - 1)/x), x], x] /; Free 
Q[{a, b, c, d}, x] && GtQ[p, 0] && IntegerQ[2*p]
 

rule 538

Int[((c_) + (d_.)*(x_))/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp 
[c   Int[1/(x*Sqrt[a + b*x^2]), x], x] + Simp[d   Int[1/Sqrt[a + b*x^2], x] 
, x] /; FreeQ[{a, b, c, d}, x]
 

rule 1334

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_ 
) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/((4 
*c)^IntPart[p]*(b + 2*c*x)^(2*FracPart[p]))   Int[(g + h*x)^m*(b + 2*c*x)^( 
2*p)*(d + f*x^2)^q, x], x] /; FreeQ[{a, b, c, d, f, g, h, m, p, q}, x] && E 
qQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.89 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.58

method	result	size
default	\(\frac {\operatorname {csgn}\left (b x +a \right ) \left (\sqrt {d \,x^{2}+c}\, \sqrt {d}\, b x -2 \sqrt {c}\, \sqrt {d}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) a +2 \sqrt {d \,x^{2}+c}\, \sqrt {d}\, a +\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right ) b c \right )}{2 \sqrt {d}}\)	\(92\)

Input:

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

Output:

1/2*csgn(b*x+a)*((d*x^2+c)^(1/2)*d^(1/2)*b*x-2*c^(1/2)*d^(1/2)*ln(2*(c^(1/ 
2)*(d*x^2+c)^(1/2)+c)/x)*a+2*(d*x^2+c)^(1/2)*d^(1/2)*a+ln(x*d^(1/2)+(d*x^2 
+c)^(1/2))*b*c)/d^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 347, normalized size of antiderivative = 2.17 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx=\left [\frac {b c \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, a \sqrt {c} d \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (b d x + 2 \, a d\right )} \sqrt {d x^{2} + c}}{4 \, d}, -\frac {b c \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - a \sqrt {c} d \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - {\left (b d x + 2 \, a d\right )} \sqrt {d x^{2} + c}}{2 \, d}, \frac {4 \, a \sqrt {-c} d \arctan \left (\frac {\sqrt {d x^{2} + c} \sqrt {-c}}{c}\right ) + b c \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (b d x + 2 \, a d\right )} \sqrt {d x^{2} + c}}{4 \, d}, -\frac {b c \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - 2 \, a \sqrt {-c} d \arctan \left (\frac {\sqrt {d x^{2} + c} \sqrt {-c}}{c}\right ) - {\left (b d x + 2 \, a d\right )} \sqrt {d x^{2} + c}}{2 \, d}\right ] \] Input:

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

Output:

[1/4*(b*c*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*a*sq 
rt(c)*d*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 2*(b*d*x + 2 
*a*d)*sqrt(d*x^2 + c))/d, -1/2*(b*c*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 
+ c)) - a*sqrt(c)*d*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) - 
(b*d*x + 2*a*d)*sqrt(d*x^2 + c))/d, 1/4*(4*a*sqrt(-c)*d*arctan(sqrt(d*x^2 
+ c)*sqrt(-c)/c) + b*c*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x 
- c) + 2*(b*d*x + 2*a*d)*sqrt(d*x^2 + c))/d, -1/2*(b*c*sqrt(-d)*arctan(sqr 
t(-d)*x/sqrt(d*x^2 + c)) - 2*a*sqrt(-c)*d*arctan(sqrt(d*x^2 + c)*sqrt(-c)/ 
c) - (b*d*x + 2*a*d)*sqrt(d*x^2 + c))/d]
 

Sympy [F]

\[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx=\int \frac {\sqrt {c + d x^{2}} \sqrt {\left (a + b x\right )^{2}}}{x}\, dx \] Input:

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

Output:

Integral(sqrt(c + d*x**2)*sqrt((a + b*x)**2)/x, x)
 

Maxima [F]

\[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx=\int { \frac {\sqrt {d x^{2} + c} \sqrt {{\left (b x + a\right )}^{2}}}{x} \,d x } \] Input:

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx=\int \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+c}}{x} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx=\frac {2 \sqrt {d \,x^{2}+c}\, a d +\sqrt {d \,x^{2}+c}\, b d x +2 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}-\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a d -2 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a d +\sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {d}\, x}{\sqrt {c}}\right ) b c}{2 d} \] Input:

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

Output:

(2*sqrt(c + d*x**2)*a*d + sqrt(c + d*x**2)*b*d*x + 2*sqrt(c)*log((sqrt(c + 
 d*x**2) - sqrt(c) + sqrt(d)*x)/sqrt(c))*a*d - 2*sqrt(c)*log((sqrt(c + d*x 
**2) + sqrt(c) + sqrt(d)*x)/sqrt(c))*a*d + sqrt(d)*log((sqrt(c + d*x**2) + 
 sqrt(d)*x)/sqrt(c))*b*c)/(2*d)