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

Optimal. Leaf size=177 \[ -\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4} \left (c+d x^2\right )^{3/2}}{c x \left (a+b x^2\right )}+\frac {x \sqrt {a^2+2 a b x^2+b^2 x^4} \sqrt {c+d x^2} (2 a d+b c)}{2 c \left (a+b x^2\right )}+\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} (2 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {1250, 453, 195, 217, 206} \[ -\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4} \left (c+d x^2\right )^{3/2}}{c x \left (a+b x^2\right )}+\frac {x \sqrt {a^2+2 a b x^2+b^2 x^4} \sqrt {c+d x^2} (2 a d+b c)}{2 c \left (a+b x^2\right )}+\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} (2 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c + 2*a*d)*x*Sqrt[c + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*c*(a + b*x^2)) - (a*(c + d*x^2)^(3/2)*Sqr
t[a^2 + 2*a*b*x^2 + b^2*x^4])/(c*x*(a + b*x^2)) + ((b*c + 2*a*d)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*ArcTanh[(Sqrt
[d]*x)/Sqrt[c + d*x^2]])/(2*Sqrt[d]*(a + b*x^2))

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 1250

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dis
t[(a + b*x^2 + c*x^4)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(f*x)^m*(d + e*x^2)^q*(b/2
+ c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 122, normalized size = 0.69 \[ \frac {\sqrt {\left (a+b x^2\right )^2} \sqrt {c+d x^2} \left (\sqrt {c} \sqrt {d} \left (b x^2-2 a\right ) \sqrt {\frac {d x^2}{c}+1}+x (2 a d+b c) \sinh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )\right )}{2 \sqrt {c} \sqrt {d} x \left (a+b x^2\right ) \sqrt {\frac {d x^2}{c}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^2)^2]*Sqrt[c + d*x^2]*(Sqrt[c]*Sqrt[d]*(-2*a + b*x^2)*Sqrt[1 + (d*x^2)/c] + (b*c + 2*a*d)*x*Arc
Sinh[(Sqrt[d]*x)/Sqrt[c]]))/(2*Sqrt[c]*Sqrt[d]*x*(a + b*x^2)*Sqrt[1 + (d*x^2)/c])

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fricas [A]  time = 0.80, size = 134, normalized size = 0.76 \[ \left [\frac {{\left (b c + 2 \, a d\right )} \sqrt {d} x \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (b d x^{2} - 2 \, a d\right )} \sqrt {d x^{2} + c}}{4 \, d x}, -\frac {{\left (b c + 2 \, a d\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (b d x^{2} - 2 \, a d\right )} \sqrt {d x^{2} + c}}{2 \, d x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.47, size = 116, normalized size = 0.66 \[ \frac {1}{2} \, \sqrt {d x^{2} + c} b x \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {2 \, a c \sqrt {d} \mathrm {sgn}\left (b x^{2} + a\right )}{{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c} - \frac {{\left (b c \sqrt {d} \mathrm {sgn}\left (b x^{2} + a\right ) + 2 \, a d^{\frac {3}{2}} \mathrm {sgn}\left (b x^{2} + a\right )\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 128, normalized size = 0.72 \[ \frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (2 a c d x \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )+b \,c^{2} x \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )+2 \sqrt {d \,x^{2}+c}\, a \,d^{\frac {3}{2}} x^{2}+\sqrt {d \,x^{2}+c}\, b c \sqrt {d}\, x^{2}-2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a \sqrt {d}\right )}{2 \left (b \,x^{2}+a \right ) c \sqrt {d}\, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.08, size = 59, normalized size = 0.33 \[ \frac {1}{2} \, \sqrt {d x^{2} + c} b x + \frac {b c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, \sqrt {d}} + a \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {\sqrt {d x^{2} + c} a}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c + d x^{2}} \sqrt {\left (a + b x^{2}\right )^{2}}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(c + d*x**2)*sqrt((a + b*x**2)**2)/x**2, x)

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