\(\int \frac {(a+b x^2)^2 (c+d x^2)^{3/2}}{x^4} \, dx\) [854]

Optimal result

Integrand size = 24, antiderivative size = 181 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx=\frac {\left (3 b^2 c^2+8 a d (3 b c+a d)\right ) x \sqrt {c+d x^2}}{8 c}+\frac {1}{12} \left (3 b^2+\frac {8 a d (3 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}-\frac {2 a (3 b c+a d) \left (c+d x^2\right )^{5/2}}{3 c^2 x}+\frac {\left (3 b^2 c^2+8 a d (3 b c+a d)\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 \sqrt {d}} \] Output:

1/8*(3*b^2*c^2+8*a*d*(a*d+3*b*c))*x*(d*x^2+c)^(1/2)/c+1/12*(3*b^2+8*a*d*(a 
*d+3*b*c)/c^2)*x*(d*x^2+c)^(3/2)-1/3*a^2*(d*x^2+c)^(5/2)/c/x^3-2/3*a*(a*d+ 
3*b*c)*(d*x^2+c)^(5/2)/c^2/x+1/8*(3*b^2*c^2+8*a*d*(a*d+3*b*c))*arctanh(d^( 
1/2)*x/(d*x^2+c)^(1/2))/d^(1/2)
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.70 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx=\frac {1}{24} \left (\frac {\sqrt {c+d x^2} \left (24 a b x^2 \left (-2 c+d x^2\right )+3 b^2 x^4 \left (5 c+2 d x^2\right )-8 a^2 \left (c+4 d x^2\right )\right )}{x^3}+\frac {6 \left (3 b^2 c^2+24 a b c d+8 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{-\sqrt {c}+\sqrt {c+d x^2}}\right )}{\sqrt {d}}\right ) \] Input:

Integrate[((a + b*x^2)^2*(c + d*x^2)^(3/2))/x^4,x]
 

Output:

((Sqrt[c + d*x^2]*(24*a*b*x^2*(-2*c + d*x^2) + 3*b^2*x^4*(5*c + 2*d*x^2) - 
 8*a^2*(c + 4*d*x^2)))/x^3 + (6*(3*b^2*c^2 + 24*a*b*c*d + 8*a^2*d^2)*ArcTa 
nh[(Sqrt[d]*x)/(-Sqrt[c] + Sqrt[c + d*x^2])])/Sqrt[d])/24
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.85, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {365, 359, 211, 211, 224, 219}

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.

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

\(\Big \downarrow \) 365

\(\displaystyle \frac {\int \frac {\left (3 b^2 c x^2+2 a (3 b c+a d)\right ) \left (d x^2+c\right )^{3/2}}{x^2}dx}{3 c}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}\)

\(\Big \downarrow \) 359

\(\displaystyle \frac {\frac {\left (8 a d (a d+3 b c)+3 b^2 c^2\right ) \int \left (d x^2+c\right )^{3/2}dx}{c}-\frac {2 a \left (c+d x^2\right )^{5/2} (a d+3 b c)}{c x}}{3 c}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {\left (8 a d (a d+3 b c)+3 b^2 c^2\right ) \left (\frac {3}{4} c \int \sqrt {d x^2+c}dx+\frac {1}{4} x \left (c+d x^2\right )^{3/2}\right )}{c}-\frac {2 a \left (c+d x^2\right )^{5/2} (a d+3 b c)}{c x}}{3 c}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {\left (8 a d (a d+3 b c)+3 b^2 c^2\right ) \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{\sqrt {d x^2+c}}dx+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {1}{4} x \left (c+d x^2\right )^{3/2}\right )}{c}-\frac {2 a \left (c+d x^2\right )^{5/2} (a d+3 b c)}{c x}}{3 c}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {\left (8 a d (a d+3 b c)+3 b^2 c^2\right ) \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {1}{4} x \left (c+d x^2\right )^{3/2}\right )}{c}-\frac {2 a \left (c+d x^2\right )^{5/2} (a d+3 b c)}{c x}}{3 c}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\left (8 a d (a d+3 b c)+3 b^2 c^2\right ) \left (\frac {3}{4} c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}}+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {1}{4} x \left (c+d x^2\right )^{3/2}\right )}{c}-\frac {2 a \left (c+d x^2\right )^{5/2} (a d+3 b c)}{c x}}{3 c}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}\)

Input:

Int[((a + b*x^2)^2*(c + d*x^2)^(3/2))/x^4,x]
 

Output:

-1/3*(a^2*(c + d*x^2)^(5/2))/(c*x^3) + ((-2*a*(3*b*c + a*d)*(c + d*x^2)^(5 
/2))/(c*x) + ((3*b^2*c^2 + 8*a*d*(3*b*c + a*d))*((x*(c + d*x^2)^(3/2))/4 + 
 (3*c*((x*Sqrt[c + d*x^2])/2 + (c*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(2 
*Sqrt[d])))/4))/c)/(3*c)
 

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 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 359

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + 
Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] 
&& LtQ[m, -1] &&  !ILtQ[p, -1]
 

rule 365

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x 
_Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] 
- Simp[1/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p 
+ 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, 
 b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
 

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.60

method	result	size
risch	\(-\frac {\sqrt {x^{2} d +c}\, \left (-6 x^{6} b^{2} d -24 a b d \,x^{4}-15 b^{2} c \,x^{4}+32 a^{2} d \,x^{2}+48 a b c \,x^{2}+8 a^{2} c \right )}{24 x^{3}}+\frac {\left (a^{2} d^{2}+3 a b c d +\frac {3}{8} b^{2} c^{2}\right ) \ln \left (\sqrt {d}\, x +\sqrt {x^{2} d +c}\right )}{\sqrt {d}}\)	\(108\)
pseudoelliptic	\(\frac {x^{3} \left (a^{2} d^{2}+3 a b c d +\frac {3}{8} b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}}{x \sqrt {d}}\right )-\frac {\sqrt {x^{2} d +c}\, \left (\left (-\frac {3}{4} b^{2} x^{6}-3 a b \,x^{4}+4 a^{2} x^{2}\right ) d^{\frac {3}{2}}+c \sqrt {d}\, \left (-\frac {15}{8} b^{2} x^{4}+6 a b \,x^{2}+a^{2}\right )\right )}{3}}{\sqrt {d}\, x^{3}}\)	\(116\)
default	\(b^{2} \left (\frac {x \left (x^{2} d +c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {x^{2} d +c}}{2}+\frac {c \ln \left (\sqrt {d}\, x +\sqrt {x^{2} d +c}\right )}{2 \sqrt {d}}\right )}{4}\right )+a^{2} \left (-\frac {\left (x^{2} d +c \right )^{\frac {5}{2}}}{3 c \,x^{3}}+\frac {2 d \left (-\frac {\left (x^{2} d +c \right )^{\frac {5}{2}}}{c x}+\frac {4 d \left (\frac {x \left (x^{2} d +c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {x^{2} d +c}}{2}+\frac {c \ln \left (\sqrt {d}\, x +\sqrt {x^{2} d +c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{c}\right )}{3 c}\right )+2 a b \left (-\frac {\left (x^{2} d +c \right )^{\frac {5}{2}}}{c x}+\frac {4 d \left (\frac {x \left (x^{2} d +c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {x^{2} d +c}}{2}+\frac {c \ln \left (\sqrt {d}\, x +\sqrt {x^{2} d +c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{c}\right )\)	\(239\)

Input:

int((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/24*(d*x^2+c)^(1/2)*(-6*b^2*d*x^6-24*a*b*d*x^4-15*b^2*c*x^4+32*a^2*d*x^2 
+48*a*b*c*x^2+8*a^2*c)/x^3+(a^2*d^2+3*a*b*c*d+3/8*b^2*c^2)*ln(d^(1/2)*x+(d 
*x^2+c)^(1/2))/d^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx=\left [\frac {3 \, {\left (3 \, b^{2} c^{2} + 24 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {d} x^{3} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (6 \, b^{2} d^{2} x^{6} + 3 \, {\left (5 \, b^{2} c d + 8 \, a b d^{2}\right )} x^{4} - 8 \, a^{2} c d - 16 \, {\left (3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{48 \, d x^{3}}, -\frac {3 \, {\left (3 \, b^{2} c^{2} + 24 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {-d} x^{3} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (6 \, b^{2} d^{2} x^{6} + 3 \, {\left (5 \, b^{2} c d + 8 \, a b d^{2}\right )} x^{4} - 8 \, a^{2} c d - 16 \, {\left (3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{24 \, d x^{3}}\right ] \] Input:

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^4,x, algorithm="fricas")
 

Output:

[1/48*(3*(3*b^2*c^2 + 24*a*b*c*d + 8*a^2*d^2)*sqrt(d)*x^3*log(-2*d*x^2 - 2 
*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(6*b^2*d^2*x^6 + 3*(5*b^2*c*d + 8*a*b* 
d^2)*x^4 - 8*a^2*c*d - 16*(3*a*b*c*d + 2*a^2*d^2)*x^2)*sqrt(d*x^2 + c))/(d 
*x^3), -1/24*(3*(3*b^2*c^2 + 24*a*b*c*d + 8*a^2*d^2)*sqrt(-d)*x^3*arctan(s 
qrt(-d)*x/sqrt(d*x^2 + c)) - (6*b^2*d^2*x^6 + 3*(5*b^2*c*d + 8*a*b*d^2)*x^ 
4 - 8*a^2*c*d - 16*(3*a*b*c*d + 2*a^2*d^2)*x^2)*sqrt(d*x^2 + c))/(d*x^3)]
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 2.35 (sec) , antiderivative size = 435, normalized size of antiderivative = 2.40 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx=- \frac {a^{2} \sqrt {c} d}{x \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {a^{2} c \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{3 x^{2}} - \frac {a^{2} d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{3} + a^{2} d^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} - \frac {a^{2} d^{2} x}{\sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {2 a b c^{\frac {3}{2}}}{x \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {2 a b \sqrt {c} d x}{\sqrt {1 + \frac {d x^{2}}{c}}} + 2 a b c \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} + 2 a b d \left (\begin {cases} \frac {c \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {c + d x^{2}}}{2} & \text {for}\: d \neq 0 \\\sqrt {c} x & \text {otherwise} \end {cases}\right ) + b^{2} c \left (\begin {cases} \frac {c \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {c + d x^{2}}}{2} & \text {for}\: d \neq 0 \\\sqrt {c} x & \text {otherwise} \end {cases}\right ) + b^{2} d \left (\begin {cases} - \frac {c^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{8 d} + \frac {c x \sqrt {c + d x^{2}}}{8 d} + \frac {x^{3} \sqrt {c + d x^{2}}}{4} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{3}}{3} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((b*x**2+a)**2*(d*x**2+c)**(3/2)/x**4,x)
 

Output:

-a**2*sqrt(c)*d/(x*sqrt(1 + d*x**2/c)) - a**2*c*sqrt(d)*sqrt(c/(d*x**2) + 
1)/(3*x**2) - a**2*d**(3/2)*sqrt(c/(d*x**2) + 1)/3 + a**2*d**(3/2)*asinh(s 
qrt(d)*x/sqrt(c)) - a**2*d**2*x/(sqrt(c)*sqrt(1 + d*x**2/c)) - 2*a*b*c**(3 
/2)/(x*sqrt(1 + d*x**2/c)) - 2*a*b*sqrt(c)*d*x/sqrt(1 + d*x**2/c) + 2*a*b* 
c*sqrt(d)*asinh(sqrt(d)*x/sqrt(c)) + 2*a*b*d*Piecewise((c*Piecewise((log(2 
*sqrt(d)*sqrt(c + d*x**2) + 2*d*x)/sqrt(d), Ne(c, 0)), (x*log(x)/sqrt(d*x* 
*2), True))/2 + x*sqrt(c + d*x**2)/2, Ne(d, 0)), (sqrt(c)*x, True)) + b**2 
*c*Piecewise((c*Piecewise((log(2*sqrt(d)*sqrt(c + d*x**2) + 2*d*x)/sqrt(d) 
, Ne(c, 0)), (x*log(x)/sqrt(d*x**2), True))/2 + x*sqrt(c + d*x**2)/2, Ne(d 
, 0)), (sqrt(c)*x, True)) + b**2*d*Piecewise((-c**2*Piecewise((log(2*sqrt( 
d)*sqrt(c + d*x**2) + 2*d*x)/sqrt(d), Ne(c, 0)), (x*log(x)/sqrt(d*x**2), T 
rue))/(8*d) + c*x*sqrt(c + d*x**2)/(8*d) + x**3*sqrt(c + d*x**2)/4, Ne(d, 
0)), (sqrt(c)*x**3/3, True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx=\frac {1}{4} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} x + \frac {3}{8} \, \sqrt {d x^{2} + c} b^{2} c x + 3 \, \sqrt {d x^{2} + c} a b d x + \frac {\sqrt {d x^{2} + c} a^{2} d^{2} x}{c} + \frac {3 \, b^{2} c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {d}} + 3 \, a b c \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) + a^{2} d^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b}{x} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d}{3 \, c x} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2}}{3 \, c x^{3}} \] Input:

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^4,x, algorithm="maxima")
 

Output:

1/4*(d*x^2 + c)^(3/2)*b^2*x + 3/8*sqrt(d*x^2 + c)*b^2*c*x + 3*sqrt(d*x^2 + 
 c)*a*b*d*x + sqrt(d*x^2 + c)*a^2*d^2*x/c + 3/8*b^2*c^2*arcsinh(d*x/sqrt(c 
*d))/sqrt(d) + 3*a*b*c*sqrt(d)*arcsinh(d*x/sqrt(c*d)) + a^2*d^(3/2)*arcsin 
h(d*x/sqrt(c*d)) - 2*(d*x^2 + c)^(3/2)*a*b/x - 2/3*(d*x^2 + c)^(3/2)*a^2*d 
/(c*x) - 1/3*(d*x^2 + c)^(5/2)*a^2/(c*x^3)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.42 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx=\frac {1}{8} \, {\left (2 \, b^{2} d x^{2} + \frac {5 \, b^{2} c d^{2} + 8 \, a b d^{3}}{d^{2}}\right )} \sqrt {d x^{2} + c} x - \frac {{\left (3 \, b^{2} c^{2} + 24 \, a b c d + 8 \, a^{2} d^{2}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{16 \, \sqrt {d}} + \frac {4 \, {\left (3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c^{2} \sqrt {d} + 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} c d^{\frac {3}{2}} - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{3} \sqrt {d} - 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} c^{2} d^{\frac {3}{2}} + 3 \, a b c^{4} \sqrt {d} + 2 \, a^{2} c^{3} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3}} \] Input:

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^4,x, algorithm="giac")
 

Output:

1/8*(2*b^2*d*x^2 + (5*b^2*c*d^2 + 8*a*b*d^3)/d^2)*sqrt(d*x^2 + c)*x - 1/16 
*(3*b^2*c^2 + 24*a*b*c*d + 8*a^2*d^2)*log((sqrt(d)*x - sqrt(d*x^2 + c))^2) 
/sqrt(d) + 4/3*(3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*b*c^2*sqrt(d) + 3*(sqr 
t(d)*x - sqrt(d*x^2 + c))^4*a^2*c*d^(3/2) - 6*(sqrt(d)*x - sqrt(d*x^2 + c) 
)^2*a*b*c^3*sqrt(d) - 3*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*c^2*d^(3/2) + 
3*a*b*c^4*sqrt(d) + 2*a^2*c^3*d^(3/2))/((sqrt(d)*x - sqrt(d*x^2 + c))^2 - 
c)^3
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{3/2}}{x^4} \,d x \] Input:

int(((a + b*x^2)^2*(c + d*x^2)^(3/2))/x^4,x)
 

Output:

int(((a + b*x^2)^2*(c + d*x^2)^(3/2))/x^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.29 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx=\frac {-64 \sqrt {d \,x^{2}+c}\, a^{2} c d -256 \sqrt {d \,x^{2}+c}\, a^{2} d^{2} x^{2}-384 \sqrt {d \,x^{2}+c}\, a b c d \,x^{2}+192 \sqrt {d \,x^{2}+c}\, a b \,d^{2} x^{4}+120 \sqrt {d \,x^{2}+c}\, b^{2} c d \,x^{4}+48 \sqrt {d \,x^{2}+c}\, b^{2} d^{2} x^{6}+192 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a^{2} d^{2} x^{3}+576 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a b c d \,x^{3}+72 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {d}\, x}{\sqrt {c}}\right ) b^{2} c^{2} x^{3}+160 \sqrt {d}\, a b c d \,x^{3}+15 \sqrt {d}\, b^{2} c^{2} x^{3}}{192 d \,x^{3}} \] Input:

int((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^4,x)
 

Output:

( - 64*sqrt(c + d*x**2)*a**2*c*d - 256*sqrt(c + d*x**2)*a**2*d**2*x**2 - 3 
84*sqrt(c + d*x**2)*a*b*c*d*x**2 + 192*sqrt(c + d*x**2)*a*b*d**2*x**4 + 12 
0*sqrt(c + d*x**2)*b**2*c*d*x**4 + 48*sqrt(c + d*x**2)*b**2*d**2*x**6 + 19 
2*sqrt(d)*log((sqrt(c + d*x**2) + sqrt(d)*x)/sqrt(c))*a**2*d**2*x**3 + 576 
*sqrt(d)*log((sqrt(c + d*x**2) + sqrt(d)*x)/sqrt(c))*a*b*c*d*x**3 + 72*sqr 
t(d)*log((sqrt(c + d*x**2) + sqrt(d)*x)/sqrt(c))*b**2*c**2*x**3 + 160*sqrt 
(d)*a*b*c*d*x**3 + 15*sqrt(d)*b**2*c**2*x**3)/(192*d*x**3)