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

Optimal. Leaf size=111 \[ -\frac {a^2 \left (c+d x^2\right )^{3/2}}{3 c x^3}-\frac {2 a b \left (c+d x^2\right )^{3/2}}{c x}+\frac {b x \sqrt {c+d x^2} (4 a d+b c)}{2 c}+\frac {b (4 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}} \]

[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

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Rubi [A]  time = 0.07, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {462, 453, 195, 217, 206} \[ -\frac {a^2 \left (c+d x^2\right )^{3/2}}{3 c x^3}-\frac {2 a b \left (c+d x^2\right )^{3/2}}{c x}+\frac {b x \sqrt {c+d x^2} (4 a d+b c)}{2 c}+\frac {b (4 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}} \]

Antiderivative was successfully verified.

[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 195

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

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 453

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 462

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*} \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 {\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)) \operatorname {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*}

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Mathematica [A]  time = 0.17, size = 91, normalized size = 0.82 \[ \sqrt {c+d x^2} \left (-\frac {a^2}{3 x^3}-\frac {a (a d+6 b c)}{3 c x}+\frac {b^2 x}{2}\right )+\frac {b (4 a d+b c) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{2 \sqrt {d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1/3*a^2/x^3 - (a*(6*b*c + a*d))/(3*c*x) + (b^2*x)/2)*Sqrt[c + d*x^2] + (b*(b*c + 4*a*d)*Log[d*x + Sqrt[d]*Sq
rt[c + d*x^2]])/(2*Sqrt[d])

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fricas [A]  time = 0.89, size = 210, normalized size = 1.89 \[ \left [\frac {3 \, {\left (b^{2} c^{2} + 4 \, a b c d\right )} \sqrt {d} x^{3} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (3 \, b^{2} c d x^{4} - 2 \, a^{2} c d - 2 \, {\left (6 \, a b c d + a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, c d x^{3}}, -\frac {3 \, {\left (b^{2} c^{2} + 4 \, a b c d\right )} \sqrt {-d} x^{3} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (3 \, b^{2} c d x^{4} - 2 \, a^{2} c d - 2 \, {\left (6 \, a b c d + a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, c d x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [B]  time = 0.46, size = 188, normalized size = 1.69 \[ \frac {1}{2} \, \sqrt {d x^{2} + c} b^{2} x - \frac {{\left (b^{2} c \sqrt {d} + 4 \, a b d^{\frac {3}{2}}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{4 \, d} + \frac {2 \, {\left (6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c \sqrt {d} + 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} d^{\frac {3}{2}} - 12 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{2} \sqrt {d} + 6 \, a b c^{3} \sqrt {d} + a^{2} c^{2} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

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maple [A]  time = 0.01, size = 122, normalized size = 1.10 \[ 2 a b \sqrt {d}\, \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )+\frac {b^{2} c \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}+\frac {2 \sqrt {d \,x^{2}+c}\, a b d x}{c}+\frac {\sqrt {d \,x^{2}+c}\, b^{2} x}{2}-\frac {2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a b}{c x}-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2}}{3 c \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.08, size = 86, normalized size = 0.77 \[ \frac {1}{2} \, \sqrt {d x^{2} + c} b^{2} x + \frac {b^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, \sqrt {d}} + 2 \, a b \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {2 \, \sqrt {d x^{2} + c} a b}{x} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2}}{3 \, c x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^2+a\right )}^2\,\sqrt {d\,x^2+c}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 5.46, size = 170, normalized size = 1.53 \[ - \frac {a^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{3 x^{2}} - \frac {a^{2} d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{3 c} - \frac {2 a b \sqrt {c}}{x \sqrt {1 + \frac {d x^{2}}{c}}} + 2 a b \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} - \frac {2 a b d x}{\sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {b^{2} \sqrt {c} x \sqrt {1 + \frac {d x^{2}}{c}}}{2} + \frac {b^{2} c \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{2 \sqrt {d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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*sqrt(
c)*x*sqrt(1 + d*x**2/c)/2 + b**2*c*asinh(sqrt(d)*x/sqrt(c))/(2*sqrt(d))

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