\(\int \frac {(a+b x^2)^2 (c+d x^2)^{3/2}}{x^4} \, dx\) [621]

Optimal result

Integrand size = 24, antiderivative size = 184 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx=\frac {\left (3 b^2 c^2+8 a d (3 b c+a d)\right ) x \sqrt {c+d x^2}}{8 c}+\frac {\left (3 b^2 c^2+8 a d (3 b c+a d)\right ) x \left (c+d x^2\right )^{3/2}}{12 c^2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}-\frac {2 a (3 b c+a d) \left (c+d x^2\right )^{5/2}}{3 c^2 x}+\frac {\left (3 b^2 c^2+8 a d (3 b c+a d)\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 \sqrt {d}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {473, 464, 201, 223, 212} \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx=-\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}+\frac {\left (8 a d (a d+3 b c)+3 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 \sqrt {d}}+\frac {1}{12} x \left (c+d x^2\right )^{3/2} \left (\frac {8 a d (a d+3 b c)}{c^2}+3 b^2\right )+\frac {x \sqrt {c+d x^2} \left (8 a d (a d+3 b c)+3 b^2 c^2\right )}{8 c}-\frac {2 a \left (c+d x^2\right )^{5/2} (a d+3 b c)}{3 c^2 x} \]

[In]

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

[Out]

((3*b^2*c^2 + 8*a*d*(3*b*c + a*d))*x*Sqrt[c + d*x^2])/(8*c) + ((3*b^2 + (8*a*d*(3*b*c + a*d))/c^2)*x*(c + d*x^
2)^(3/2))/12 - (a^2*(c + d*x^2)^(5/2))/(3*c*x^3) - (2*a*(3*b*c + a*d)*(c + d*x^2)^(5/2))/(3*c^2*x) + ((3*b^2*c
^2 + 8*a*d*(3*b*c + a*d))*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(8*Sqrt[d])

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 473

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

Rubi steps \begin{align*} \text {integral}& = -\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}+\frac {\int \frac {\left (2 a (3 b c+a d)+3 b^2 c x^2\right ) \left (c+d x^2\right )^{3/2}}{x^2} \, dx}{3 c} \\ & = -\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 {1}{3} \left (-3 b^2-\frac {8 a d (3 b c+a d)}{c^2}\right ) \int \left (c+d x^2\right )^{3/2} \, dx \\ & = \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 {1}{4} \left (c \left (-3 b^2-\frac {8 a d (3 b c+a d)}{c^2}\right )\right ) \int \sqrt {c+d x^2} \, dx \\ & = \frac {1}{8} c \left (3 b^2+\frac {8 a d (3 b c+a d)}{c^2}\right ) x \sqrt {c+d x^2}+\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 {1}{8} \left (-3 b^2 c^2-24 a b c d-8 a^2 d^2\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx \\ & = \frac {1}{8} c \left (3 b^2+\frac {8 a d (3 b c+a d)}{c^2}\right ) x \sqrt {c+d x^2}+\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 {1}{8} \left (-3 b^2 c^2-24 a b c d-8 a^2 d^2\right ) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right ) \\ & = \frac {1}{8} c \left (3 b^2+\frac {8 a d (3 b c+a d)}{c^2}\right ) x \sqrt {c+d x^2}+\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+24 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 \sqrt {d}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.69 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx=\frac {1}{24} \left (\frac {\sqrt {c+d x^2} \left (24 a b x^2 \left (-2 c+d x^2\right )+3 b^2 x^4 \left (5 c+2 d x^2\right )-8 a^2 \left (c+4 d x^2\right )\right )}{x^3}+\frac {6 \left (3 b^2 c^2+24 a b c d+8 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{-\sqrt {c}+\sqrt {c+d x^2}}\right )}{\sqrt {d}}\right ) \]

[In]

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

[Out]

((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)*ArcTanh[(Sqrt[d]*x)/(-Sqrt[c] + Sqrt[c + d*x^2])])/Sqrt[d])/24

Maple [A] (verified)

Time = 2.91 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.59

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.45 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx=\left [\frac {3 \, {\left (3 \, b^{2} c^{2} + 24 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {d} x^{3} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (6 \, b^{2} d^{2} x^{6} + 3 \, {\left (5 \, b^{2} c d + 8 \, a b d^{2}\right )} x^{4} - 8 \, a^{2} c d - 16 \, {\left (3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{48 \, d x^{3}}, -\frac {3 \, {\left (3 \, b^{2} c^{2} + 24 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {-d} x^{3} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (6 \, b^{2} d^{2} x^{6} + 3 \, {\left (5 \, b^{2} c d + 8 \, a b d^{2}\right )} x^{4} - 8 \, a^{2} c d - 16 \, {\left (3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{24 \, d x^{3}}\right ] \]

[In]

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

[Out]

[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(sqrt(-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 = 2.55 (sec) , antiderivative size = 435, normalized size of antiderivative = 2.36 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx=- \frac {a^{2} \sqrt {c} d}{x \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {a^{2} c \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{3 x^{2}} - \frac {a^{2} d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{3} + a^{2} d^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} - \frac {a^{2} d^{2} x}{\sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {2 a b c^{\frac {3}{2}}}{x \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {2 a b \sqrt {c} d x}{\sqrt {1 + \frac {d x^{2}}{c}}} + 2 a b c \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} + 2 a b d \left (\begin {cases} \frac {c \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {c + d x^{2}}}{2} & \text {for}\: d \neq 0 \\\sqrt {c} x & \text {otherwise} \end {cases}\right ) + b^{2} c \left (\begin {cases} \frac {c \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {c + d x^{2}}}{2} & \text {for}\: d \neq 0 \\\sqrt {c} x & \text {otherwise} \end {cases}\right ) + b^{2} d \left (\begin {cases} - \frac {c^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{8 d} + \frac {c x \sqrt {c + d x^{2}}}{8 d} + \frac {x^{3} \sqrt {c + d x^{2}}}{4} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{3}}{3} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

-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(sqrt(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)/s
qrt(d), Ne(c, 0)), (x*log(x)/sqrt(d*x**2), True))/(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)

none

Time = 0.19 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx=\frac {1}{4} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} x + \frac {3}{8} \, \sqrt {d x^{2} + c} b^{2} c x + 3 \, \sqrt {d x^{2} + c} a b d x + \frac {\sqrt {d x^{2} + c} a^{2} d^{2} x}{c} + \frac {3 \, b^{2} c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {d}} + 3 \, a b c \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) + a^{2} d^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b}{x} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d}{3 \, c x} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2}}{3 \, c x^{3}} \]

[In]

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

[Out]

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)*arcs
inh(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)

none

Time = 0.32 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.40 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx=\frac {1}{8} \, {\left (2 \, b^{2} d x^{2} + \frac {5 \, b^{2} c d^{2} + 8 \, a b d^{3}}{d^{2}}\right )} \sqrt {d x^{2} + c} x - \frac {{\left (3 \, b^{2} c^{2} + 24 \, a b c d + 8 \, a^{2} d^{2}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{16 \, \sqrt {d}} + \frac {4 \, {\left (3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c^{2} \sqrt {d} + 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} c d^{\frac {3}{2}} - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{3} \sqrt {d} - 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} c^{2} d^{\frac {3}{2}} + 3 \, a b c^{4} \sqrt {d} + 2 \, a^{2} c^{3} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3}} \]

[In]

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

[Out]

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*(s
qrt(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. \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{3/2}}{x^4} \,d x \]

[In]

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

[Out]

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