\(\int \frac {(a+b x^2)^2 \sqrt {c+d x^2}}{x^6} \, dx\) [609]

Optimal result

Integrand size = 24, antiderivative size = 103 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^6} \, dx=-\frac {b^2 \sqrt {c+d x^2}}{x}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{5 c x^5}-\frac {2 a (5 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^3}+b^2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right ) \]

[Out]

-1/5*a^2*(d*x^2+c)^(3/2)/c/x^5-2/15*a*(-a*d+5*b*c)*(d*x^2+c)^(3/2)/c^2/x^3+b^2*arctanh(x*d^(1/2)/(d*x^2+c)^(1/
2))*d^(1/2)-b^2*(d*x^2+c)^(1/2)/x

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 103, 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, 462, 283, 223, 212} \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^6} \, dx=-\frac {a^2 \left (c+d x^2\right )^{3/2}}{5 c x^5}-\frac {2 a \left (c+d x^2\right )^{3/2} (5 b c-a d)}{15 c^2 x^3}+b^2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )-\frac {b^2 \sqrt {c+d x^2}}{x} \]

[In]

Int[((a + b*x^2)^2*Sqrt[c + d*x^2])/x^6,x]

[Out]

-((b^2*Sqrt[c + d*x^2])/x) - (a^2*(c + d*x^2)^(3/2))/(5*c*x^5) - (2*a*(5*b*c - a*d)*(c + d*x^2)^(3/2))/(15*c^2
*x^3) + b^2*Sqrt[d]*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]]

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 283

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

Rule 462

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[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -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}}{5 c x^5}+\frac {\int \frac {\left (2 a (5 b c-a d)+5 b^2 c x^2\right ) \sqrt {c+d x^2}}{x^4} \, dx}{5 c} \\ & = -\frac {a^2 \left (c+d x^2\right )^{3/2}}{5 c x^5}-\frac {2 a (5 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^3}+b^2 \int \frac {\sqrt {c+d x^2}}{x^2} \, dx \\ & = -\frac {b^2 \sqrt {c+d x^2}}{x}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{5 c x^5}-\frac {2 a (5 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^3}+\left (b^2 d\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx \\ & = -\frac {b^2 \sqrt {c+d x^2}}{x}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{5 c x^5}-\frac {2 a (5 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^3}+\left (b^2 d\right ) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right ) \\ & = -\frac {b^2 \sqrt {c+d x^2}}{x}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{5 c x^5}-\frac {2 a (5 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^3}+b^2 \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^6} \, dx=-\frac {\sqrt {c+d x^2} \left (15 b^2 c^2 x^4+10 a b c x^2 \left (c+d x^2\right )+a^2 \left (3 c^2+c d x^2-2 d^2 x^4\right )\right )}{15 c^2 x^5}-b^2 \sqrt {d} \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right ) \]

[In]

Integrate[((a + b*x^2)^2*Sqrt[c + d*x^2])/x^6,x]

[Out]

-1/15*(Sqrt[c + d*x^2]*(15*b^2*c^2*x^4 + 10*a*b*c*x^2*(c + d*x^2) + a^2*(3*c^2 + c*d*x^2 - 2*d^2*x^4)))/(c^2*x
^5) - b^2*Sqrt[d]*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]]

Maple [A] (verified)

Time = 2.94 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.98

method	result	size
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (-2 a^{2} d^{2} x^{4}+10 x^{4} a b c d +15 b^{2} c^{2} x^{4}+a^{2} c d \,x^{2}+10 a b \,c^{2} x^{2}+3 a^{2} c^{2}\right )}{15 x^{5} c^{2}}+b^{2} \sqrt {d}\, \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )\)	\(101\)
pseudoelliptic	\(\frac {5 b^{2} c^{2} \sqrt {d}\, \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right ) x^{5}-\left (\left (5 b^{2} x^{4}+\frac {10}{3} a b \,x^{2}+a^{2}\right ) c^{2}+\frac {a d \,x^{2} \left (10 b \,x^{2}+a \right ) c}{3}-\frac {2 a^{2} d^{2} x^{4}}{3}\right ) \sqrt {d \,x^{2}+c}}{5 c^{2} x^{5}}\)	\(103\)
default	\(b^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{c x}+\frac {2 d \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{c}\right )+a^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{5 c \,x^{5}}+\frac {2 d \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{15 c^{2} x^{3}}\right )-\frac {2 a b \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3 c \,x^{3}}\)	\(124\)

[In]

int((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/15*(d*x^2+c)^(1/2)*(-2*a^2*d^2*x^4+10*a*b*c*d*x^4+15*b^2*c^2*x^4+a^2*c*d*x^2+10*a*b*c^2*x^2+3*a^2*c^2)/x^5/
c^2+b^2*d^(1/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.15 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^6} \, dx=\left [\frac {15 \, b^{2} c^{2} \sqrt {d} x^{5} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 2 \, {\left ({\left (15 \, b^{2} c^{2} + 10 \, a b c d - 2 \, a^{2} d^{2}\right )} x^{4} + 3 \, a^{2} c^{2} + {\left (10 \, a b c^{2} + a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{30 \, c^{2} x^{5}}, -\frac {15 \, b^{2} c^{2} \sqrt {-d} x^{5} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left ({\left (15 \, b^{2} c^{2} + 10 \, a b c d - 2 \, a^{2} d^{2}\right )} x^{4} + 3 \, a^{2} c^{2} + {\left (10 \, a b c^{2} + a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{15 \, c^{2} x^{5}}\right ] \]

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^6,x, algorithm="fricas")

[Out]

[1/30*(15*b^2*c^2*sqrt(d)*x^5*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 2*((15*b^2*c^2 + 10*a*b*c*d -
2*a^2*d^2)*x^4 + 3*a^2*c^2 + (10*a*b*c^2 + a^2*c*d)*x^2)*sqrt(d*x^2 + c))/(c^2*x^5), -1/15*(15*b^2*c^2*sqrt(-d
)*x^5*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + ((15*b^2*c^2 + 10*a*b*c*d - 2*a^2*d^2)*x^4 + 3*a^2*c^2 + (10*a*b*c^
2 + a^2*c*d)*x^2)*sqrt(d*x^2 + c))/(c^2*x^5)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (92) = 184\).

Time = 1.83 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.93 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^6} \, dx=- \frac {a^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{5 x^{4}} - \frac {a^{2} d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{15 c x^{2}} + \frac {2 a^{2} d^{\frac {5}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{15 c^{2}} - \frac {2 a b \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{3 x^{2}} - \frac {2 a b d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{3 c} - \frac {b^{2} \sqrt {c}}{x \sqrt {1 + \frac {d x^{2}}{c}}} + b^{2} \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} - \frac {b^{2} d x}{\sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} \]

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(1/2)/x**6,x)

[Out]

-a**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(5*x**4) - a**2*d**(3/2)*sqrt(c/(d*x**2) + 1)/(15*c*x**2) + 2*a**2*d**(5/2)
*sqrt(c/(d*x**2) + 1)/(15*c**2) - 2*a*b*sqrt(d)*sqrt(c/(d*x**2) + 1)/(3*x**2) - 2*a*b*d**(3/2)*sqrt(c/(d*x**2)
+ 1)/(3*c) - b**2*sqrt(c)/(x*sqrt(1 + d*x**2/c)) + b**2*sqrt(d)*asinh(sqrt(d)*x/sqrt(c)) - b**2*d*x/(sqrt(c)*
sqrt(1 + d*x**2/c))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^6} \, dx=b^{2} \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {\sqrt {d x^{2} + c} b^{2}}{x} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b}{3 \, c x^{3}} + \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d}{15 \, c^{2} x^{3}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2}}{5 \, c x^{5}} \]

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^6,x, algorithm="maxima")

[Out]

b^2*sqrt(d)*arcsinh(d*x/sqrt(c*d)) - sqrt(d*x^2 + c)*b^2/x - 2/3*(d*x^2 + c)^(3/2)*a*b/(c*x^3) + 2/15*(d*x^2 +
c)^(3/2)*a^2*d/(c^2*x^3) - 1/5*(d*x^2 + c)^(3/2)*a^2/(c*x^5)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 403 vs. \(2 (87) = 174\).

Time = 0.30 (sec) , antiderivative size = 403, normalized size of antiderivative = 3.91 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^6} \, dx=-\frac {1}{2} \, b^{2} \sqrt {d} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right ) + \frac {2 \, {\left (15 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} b^{2} c \sqrt {d} + 30 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} a b d^{\frac {3}{2}} - 60 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} b^{2} c^{2} \sqrt {d} - 60 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} a b c d^{\frac {3}{2}} + 30 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} a^{2} d^{\frac {5}{2}} + 90 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b^{2} c^{3} \sqrt {d} + 40 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c^{2} d^{\frac {3}{2}} + 10 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} c d^{\frac {5}{2}} - 60 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{2} c^{4} \sqrt {d} - 20 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{3} d^{\frac {3}{2}} + 10 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} c^{2} d^{\frac {5}{2}} + 15 \, b^{2} c^{5} \sqrt {d} + 10 \, a b c^{4} d^{\frac {3}{2}} - 2 \, a^{2} c^{3} d^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{5}} \]

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^6,x, algorithm="giac")

[Out]

-1/2*b^2*sqrt(d)*log((sqrt(d)*x - sqrt(d*x^2 + c))^2) + 2/15*(15*(sqrt(d)*x - sqrt(d*x^2 + c))^8*b^2*c*sqrt(d)
+ 30*(sqrt(d)*x - sqrt(d*x^2 + c))^8*a*b*d^(3/2) - 60*(sqrt(d)*x - sqrt(d*x^2 + c))^6*b^2*c^2*sqrt(d) - 60*(s
qrt(d)*x - sqrt(d*x^2 + c))^6*a*b*c*d^(3/2) + 30*(sqrt(d)*x - sqrt(d*x^2 + c))^6*a^2*d^(5/2) + 90*(sqrt(d)*x -
sqrt(d*x^2 + c))^4*b^2*c^3*sqrt(d) + 40*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*b*c^2*d^(3/2) + 10*(sqrt(d)*x - sqr
t(d*x^2 + c))^4*a^2*c*d^(5/2) - 60*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b^2*c^4*sqrt(d) - 20*(sqrt(d)*x - sqrt(d*x^
2 + c))^2*a*b*c^3*d^(3/2) + 10*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*c^2*d^(5/2) + 15*b^2*c^5*sqrt(d) + 10*a*b*c
^4*d^(3/2) - 2*a^2*c^3*d^(5/2))/((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^5

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^6} \, dx=\int \frac {{\left (b\,x^2+a\right )}^2\,\sqrt {d\,x^2+c}}{x^6} \,d x \]

[In]

int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/x^6,x)

[Out]

int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/x^6, x)