\(\int \frac {(a+b x^2)^2 \sqrt {c+d x^2}}{x^4} \, dx\) [608]

Optimal result

Integrand size = 24, antiderivative size = 111 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^4} \, dx=\frac {b (b c+4 a d) x \sqrt {c+d x^2}}{2 c}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{3 c x^3}-\frac {2 a b \left (c+d x^2\right )^{3/2}}{c x}+\frac {b (b c+4 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {473, 464, 201, 223, 212} \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^4} \, dx=-\frac {a^2 \left (c+d x^2\right )^{3/2}}{3 c x^3}+\frac {b (4 a d+b c) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}}-\frac {2 a b \left (c+d x^2\right )^{3/2}}{c x}+\frac {b x \sqrt {c+d x^2} (4 a d+b c)}{2 c} \]

[In]

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

[Out]

(b*(b*c + 4*a*d)*x*Sqrt[c + d*x^2])/(2*c) - (a^2*(c + d*x^2)^(3/2))/(3*c*x^3) - (2*a*b*(c + d*x^2)^(3/2))/(c*x
) + (b*(b*c + 4*a*d)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(2*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 )^{3/2}}{3 c x^3}+\frac {\int \frac {\left (6 a b c+3 b^2 c x^2\right ) \sqrt {c+d x^2}}{x^2} \, dx}{3 c} \\ & = -\frac {a^2 \left (c+d x^2\right )^{3/2}}{3 c x^3}-\frac {2 a b \left (c+d x^2\right )^{3/2}}{c x}+\frac {(b (b c+4 a d)) \int \sqrt {c+d x^2} \, dx}{c} \\ & = \frac {b (b c+4 a d) x \sqrt {c+d x^2}}{2 c}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{3 c x^3}-\frac {2 a b \left (c+d x^2\right )^{3/2}}{c x}+\frac {1}{2} (b (b c+4 a d)) \int \frac {1}{\sqrt {c+d x^2}} \, dx \\ & = \frac {b (b c+4 a d) x \sqrt {c+d x^2}}{2 c}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{3 c x^3}-\frac {2 a b \left (c+d x^2\right )^{3/2}}{c x}+\frac {1}{2} (b (b c+4 a d)) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right ) \\ & = \frac {b (b c+4 a d) x \sqrt {c+d x^2}}{2 c}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{3 c x^3}-\frac {2 a b \left (c+d x^2\right )^{3/2}}{c x}+\frac {b (b c+4 a d) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^4} \, dx=\frac {1}{6} \left (\frac {\sqrt {c+d x^2} \left (-12 a b c x^2+3 b^2 c x^4-2 a^2 \left (c+d x^2\right )\right )}{c x^3}-\frac {3 b (b c+4 a d) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{\sqrt {d}}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.94 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.74

method	result	size
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (-3 b^{2} c \,x^{4}+2 a^{2} d \,x^{2}+12 a b c \,x^{2}+2 a^{2} c \right )}{6 x^{3} c}+\frac {\left (4 a d +b c \right ) b \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\)	\(82\)
pseudoelliptic	\(-\frac {-6 x^{3} b \left (a d +\frac {b c}{4}\right ) c \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )+\left (d^{\frac {3}{2}} a^{2} x^{2}+c \sqrt {d}\, \left (-\frac {3}{2} b^{2} x^{4}+6 a b \,x^{2}+a^{2}\right )\right ) \sqrt {d \,x^{2}+c}}{3 \sqrt {d}\, x^{3} c}\)	\(90\)
default	\(b^{2} \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )-\frac {a^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3 c \,x^{3}}+2 a b \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{c x}+\frac {2 d \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{c}\right )\)	\(124\)

[In]

int((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/6*(d*x^2+c)^(1/2)*(-3*b^2*c*x^4+2*a^2*d*x^2+12*a*b*c*x^2+2*a^2*c)/x^3/c+1/2*(4*a*d+b*c)*b*ln(x*d^(1/2)+(d*x
^2+c)^(1/2))/d^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.89 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^4} \, dx=\left [\frac {3 \, {\left (b^{2} c^{2} + 4 \, a b c d\right )} \sqrt {d} x^{3} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (3 \, b^{2} c d x^{4} - 2 \, a^{2} c d - 2 \, {\left (6 \, a b c d + a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, c d x^{3}}, -\frac {3 \, {\left (b^{2} c^{2} + 4 \, a b c d\right )} \sqrt {-d} x^{3} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (3 \, b^{2} c d x^{4} - 2 \, a^{2} c d - 2 \, {\left (6 \, a b c d + a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, c d x^{3}}\right ] \]

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^4,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.52 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.71 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^4} \, dx=- \frac {a^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{3 x^{2}} - \frac {a^{2} d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{3 c} - \frac {2 a b \sqrt {c}}{x \sqrt {1 + \frac {d x^{2}}{c}}} + 2 a b \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} - \frac {2 a b d x}{\sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} + b^{2} \left (\begin {cases} \frac {c \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {c + d x^{2}}}{2} & \text {for}\: d \neq 0 \\\sqrt {c} x & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(1/2)/x**4,x)

[Out]

-a**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(3*x**2) - a**2*d**(3/2)*sqrt(c/(d*x**2) + 1)/(3*c) - 2*a*b*sqrt(c)/(x*sqrt
(1 + d*x**2/c)) + 2*a*b*sqrt(d)*asinh(sqrt(d)*x/sqrt(c)) - 2*a*b*d*x/(sqrt(c)*sqrt(1 + d*x**2/c)) + b**2*Piece
wise((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))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^4} \, dx=\frac {1}{2} \, \sqrt {d x^{2} + c} b^{2} x + \frac {b^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, \sqrt {d}} + 2 \, a b \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {2 \, \sqrt {d x^{2} + c} a b}{x} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2}}{3 \, c x^{3}} \]

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.65 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^4} \, dx=\frac {1}{2} \, \sqrt {d x^{2} + c} b^{2} x - \frac {{\left (b^{2} c + 4 \, a b d\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{4 \, \sqrt {d}} + \frac {2 \, {\left (6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c \sqrt {d} + 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} d^{\frac {3}{2}} - 12 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{2} \sqrt {d} + 6 \, a b c^{3} \sqrt {d} + a^{2} c^{2} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3}} \]

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^4,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/x^4,x)

[Out]

int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/x^4, x)