\(\int \frac {(a+b x^2)^2 \sqrt {c+d x^2}}{x^3} \, dx\) [833]

Optimal result

Integrand size = 24, antiderivative size = 94 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^3} \, dx=2 a b \sqrt {c+d x^2}-\frac {a^2 \sqrt {c+d x^2}}{2 x^2}+\frac {b^2 \left (c+d x^2\right )^{3/2}}{3 d}-\frac {a (4 b c+a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 \sqrt {c}} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^3} \, dx=\frac {1}{6} \left (\frac {\sqrt {c+d x^2} \left (-3 a^2 d+12 a b d x^2+2 b^2 x^2 \left (c+d x^2\right )\right )}{d x^2}-\frac {3 a (4 b c+a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}}\right ) \] Input:

Integrate[((a + b*x^2)^2*Sqrt[c + d*x^2])/x^3,x]
 

Output:

((Sqrt[c + d*x^2]*(-3*a^2*d + 12*a*b*d*x^2 + 2*b^2*x^2*(c + d*x^2)))/(d*x^ 
2) - (3*a*(4*b*c + a*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/Sqrt[c])/6
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {354, 100, 27, 90, 60, 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.

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[((a + b*x^2)^2*Sqrt[c + d*x^2])/x^3,x]
 

Output:

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

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 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 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 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 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

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 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.93

method	result	size
pseudoelliptic	\(-\frac {a d \,x^{2} \left (a d +4 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}}{\sqrt {c}}\right )+\sqrt {x^{2} d +c}\, \left (-\frac {2 c^{\frac {3}{2}} b^{2} x^{2}}{3}+\sqrt {c}\, d \left (-\frac {2}{3} b^{2} x^{4}-4 a b \,x^{2}+a^{2}\right )\right )}{2 \sqrt {c}\, d \,x^{2}}\)	\(87\)
risch	\(-\frac {a^{2} \sqrt {x^{2} d +c}}{2 x^{2}}-\frac {a \left (a d +4 b c \right ) \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {x^{2} d +c}}{x}\right )}{2 \sqrt {c}}+\frac {b^{2} c \sqrt {x^{2} d +c}}{d}+d \,b^{2} \left (\frac {x^{2} \sqrt {x^{2} d +c}}{3 d}-\frac {2 c \sqrt {x^{2} d +c}}{3 d^{2}}\right )+2 a b \sqrt {x^{2} d +c}\)	\(124\)
default	\(a^{2} \left (-\frac {\left (x^{2} d +c \right )^{\frac {3}{2}}}{2 c \,x^{2}}+\frac {d \left (\sqrt {x^{2} d +c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {x^{2} d +c}}{x}\right )\right )}{2 c}\right )+\frac {b^{2} \left (x^{2} d +c \right )^{\frac {3}{2}}}{3 d}+2 a b \left (\sqrt {x^{2} d +c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {x^{2} d +c}}{x}\right )\right )\)	\(127\)

Input:

int((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.28 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^3} \, dx=\left [\frac {3 \, {\left (4 \, a b c d + a^{2} d^{2}\right )} \sqrt {c} x^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (2 \, b^{2} c d x^{4} - 3 \, a^{2} c d + 2 \, {\left (b^{2} c^{2} + 6 \, a b c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, c d x^{2}}, \frac {3 \, {\left (4 \, a b c d + a^{2} d^{2}\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {d x^{2} + c} \sqrt {-c}}{c}\right ) + {\left (2 \, b^{2} c d x^{4} - 3 \, a^{2} c d + 2 \, {\left (b^{2} c^{2} + 6 \, a b c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, c d x^{2}}\right ] \] Input:

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^3,x, algorithm="fricas")
 

Output:

[1/12*(3*(4*a*b*c*d + a^2*d^2)*sqrt(c)*x^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c) 
*sqrt(c) + 2*c)/x^2) + 2*(2*b^2*c*d*x^4 - 3*a^2*c*d + 2*(b^2*c^2 + 6*a*b*c 
*d)*x^2)*sqrt(d*x^2 + c))/(c*d*x^2), 1/6*(3*(4*a*b*c*d + a^2*d^2)*sqrt(-c) 
*x^2*arctan(sqrt(d*x^2 + c)*sqrt(-c)/c) + (2*b^2*c*d*x^4 - 3*a^2*c*d + 2*( 
b^2*c^2 + 6*a*b*c*d)*x^2)*sqrt(d*x^2 + c))/(c*d*x^2)]
 

Sympy [A] (verification not implemented)

Time = 18.87 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.76 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^3} \, dx=- \frac {a^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{2 x} - \frac {a^{2} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{2 \sqrt {c}} - 2 a b \sqrt {c} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )} + \frac {2 a b c}{\sqrt {d} x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {2 a b \sqrt {d} x}{\sqrt {\frac {c}{d x^{2}} + 1}} + b^{2} \left (\begin {cases} \frac {c \sqrt {c + d x^{2}}}{3 d} + \frac {x^{2} \sqrt {c + d x^{2}}}{3} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((b*x**2+a)**2*(d*x**2+c)**(1/2)/x**3,x)
 

Output:

-a**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(2*x) - a**2*d*asinh(sqrt(c)/(sqrt(d)*x 
))/(2*sqrt(c)) - 2*a*b*sqrt(c)*asinh(sqrt(c)/(sqrt(d)*x)) + 2*a*b*c/(sqrt( 
d)*x*sqrt(c/(d*x**2) + 1)) + 2*a*b*sqrt(d)*x/sqrt(c/(d*x**2) + 1) + b**2*P 
iecewise((c*sqrt(c + d*x**2)/(3*d) + x**2*sqrt(c + d*x**2)/3, Ne(d, 0)), ( 
sqrt(c)*x**2/2, True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.16 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^3} \, dx=-2 \, a b \sqrt {c} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - \frac {a^{2} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{2 \, \sqrt {c}} + 2 \, \sqrt {d x^{2} + c} a b + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2}}{3 \, d} + \frac {\sqrt {d x^{2} + c} a^{2} d}{2 \, c} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2}}{2 \, c x^{2}} \] Input:

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^3,x, algorithm="maxima")
 

Output:

-2*a*b*sqrt(c)*arcsinh(c/(sqrt(c*d)*abs(x))) - 1/2*a^2*d*arcsinh(c/(sqrt(c 
*d)*abs(x)))/sqrt(c) + 2*sqrt(d*x^2 + c)*a*b + 1/3*(d*x^2 + c)^(3/2)*b^2/d 
 + 1/2*sqrt(d*x^2 + c)*a^2*d/c - 1/2*(d*x^2 + c)^(3/2)*a^2/(c*x^2)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^3} \, dx=\frac {1}{6} \, d {\left (\frac {3 \, {\left (4 \, a b c + a^{2} d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c} d} - \frac {3 \, \sqrt {d x^{2} + c} a^{2}}{d x^{2}} + \frac {2 \, {\left ({\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d^{4} + 6 \, \sqrt {d x^{2} + c} a b d^{5}\right )}}{d^{6}}\right )} \] Input:

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^3,x, algorithm="giac")
 

Output:

1/6*d*(3*(4*a*b*c + a^2*d)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(sqrt(-c)*d) - 
 3*sqrt(d*x^2 + c)*a^2/(d*x^2) + 2*((d*x^2 + c)^(3/2)*b^2*d^4 + 6*sqrt(d*x 
^2 + c)*a*b*d^5)/d^6)
 

Mupad [B] (verification not implemented)

Time = 1.16 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^3} \, dx=\frac {b^2\,{\left (d\,x^2+c\right )}^{3/2}}{3\,d}-\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d}-\frac {2\,b^2\,c}{d}\right )\,\sqrt {d\,x^2+c}-\frac {a^2\,\sqrt {d\,x^2+c}}{2\,x^2}+\frac {a\,\mathrm {atan}\left (\frac {\sqrt {d\,x^2+c}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,\left (a\,d+4\,b\,c\right )\,1{}\mathrm {i}}{2\,\sqrt {c}} \] Input:

int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/x^3,x)
                                                                                    
                                                                                    
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.31 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^3} \, dx=\frac {-3 \sqrt {d \,x^{2}+c}\, a^{2} c d +12 \sqrt {d \,x^{2}+c}\, a b c d \,x^{2}+2 \sqrt {d \,x^{2}+c}\, b^{2} c^{2} x^{2}+2 \sqrt {d \,x^{2}+c}\, b^{2} c d \,x^{4}+3 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}-\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a^{2} d^{2} x^{2}+12 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}-\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a b c d \,x^{2}-3 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a^{2} d^{2} x^{2}-12 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a b c d \,x^{2}}{6 c d \,x^{2}} \] Input:

int((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^3,x)
 

Output:

( - 3*sqrt(c + d*x**2)*a**2*c*d + 12*sqrt(c + d*x**2)*a*b*c*d*x**2 + 2*sqr 
t(c + d*x**2)*b**2*c**2*x**2 + 2*sqrt(c + d*x**2)*b**2*c*d*x**4 + 3*sqrt(c 
)*log((sqrt(c + d*x**2) - sqrt(c) + sqrt(d)*x)/sqrt(c))*a**2*d**2*x**2 + 1 
2*sqrt(c)*log((sqrt(c + d*x**2) - sqrt(c) + sqrt(d)*x)/sqrt(c))*a*b*c*d*x* 
*2 - 3*sqrt(c)*log((sqrt(c + d*x**2) + sqrt(c) + sqrt(d)*x)/sqrt(c))*a**2* 
d**2*x**2 - 12*sqrt(c)*log((sqrt(c + d*x**2) + sqrt(c) + sqrt(d)*x)/sqrt(c 
))*a*b*c*d*x**2)/(6*c*d*x**2)