3.1.45 \(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^3} \, dx\) [45]

Optimal. Leaf size=161 \[ -\frac {(a+2 b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{2 x^2 (a+b x)}+\frac {b \sqrt {d} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{a+b x}-\frac {a d \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 \sqrt {c} (a+b x)} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {1015, 825, 858, 223, 212, 272, 65, 214} \begin {gather*} -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+2 b x) \sqrt {c+d x^2}}{2 x^2 (a+b x)}+\frac {b \sqrt {d} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{a+b x}-\frac {a d \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 \sqrt {c} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((a + 2*b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + d*x^2])/(x^2*(a + b*x)) + (b*Sqrt[d]*Sqrt[a^2 + 2*a*b
*x + b^2*x^2]*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(a + b*x) - (a*d*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[Sqr
t[c + d*x^2]/Sqrt[c]])/(2*Sqrt[c]*(a + b*x))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 214

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

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 825

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(d + e*x)^
(m + 1))*((a + c*x^2)^p/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)))*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*
p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e^2) + 2*c*d*p*(e*f - d*g))*x), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^
2 + a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p +
1) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e
^2, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1015

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

Rubi steps

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

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Mathematica [A]
time = 0.25, size = 124, normalized size = 0.77 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (2 a d x^2 \tanh ^{-1}\left (\frac {\sqrt {d} x-\sqrt {c+d x^2}}{\sqrt {c}}\right )-\sqrt {c} \left ((a+2 b x) \sqrt {c+d x^2}+2 b \sqrt {d} x^2 \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )\right )\right )}{2 \sqrt {c} x^2 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.11, size = 141, normalized size = 0.88

method	result	size
risch	\(-\frac {\left (2 b x +a \right ) \sqrt {\left (b x +a \right )^{2}}\, \sqrt {d \,x^{2}+c}}{2 x^{2} \left (b x +a \right )}+\frac {\left (\sqrt {d}\, b \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )-\frac {d a \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{2 \sqrt {c}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b x +a}\)	\(107\)
default	\(-\frac {\mathrm {csgn}\left (b x +a \right ) \left (-2 \sqrt {d \,x^{2}+c}\, d^{\frac {3}{2}} b \,x^{3}+\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) d^{\frac {3}{2}} a \,x^{2}+2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} \sqrt {d}\, b x -\sqrt {d \,x^{2}+c}\, d^{\frac {3}{2}} a \,x^{2}-2 \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) b c d \,x^{2}+a \left (d \,x^{2}+c \right )^{\frac {3}{2}} \sqrt {d}\right )}{2 c \,x^{2} \sqrt {d}}\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x+a)^2)^(1/2)*(d*x^2+c)^(1/2)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x+a)^2)^(1/2)*(d*x^2+c)^(1/2)/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.41, size = 377, normalized size = 2.34 \begin {gather*} \left [\frac {2 \, b c \sqrt {d} x^{2} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + a \sqrt {c} d x^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (2 \, b c x + a c\right )} \sqrt {d x^{2} + c}}{4 \, c x^{2}}, -\frac {4 \, b c \sqrt {-d} x^{2} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - a \sqrt {c} d x^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (2 \, b c x + a c\right )} \sqrt {d x^{2} + c}}{4 \, c x^{2}}, \frac {a \sqrt {-c} d x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + b c \sqrt {d} x^{2} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - {\left (2 \, b c x + a c\right )} \sqrt {d x^{2} + c}}{2 \, c x^{2}}, -\frac {2 \, b c \sqrt {-d} x^{2} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - a \sqrt {-c} d x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (2 \, b c x + a c\right )} \sqrt {d x^{2} + c}}{2 \, c x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x+a)^2)^(1/2)*(d*x^2+c)^(1/2)/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {c + d x^{2}} \sqrt {\left (a + b x\right )^{2}}}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x+a)**2)**(1/2)*(d*x**2+c)**(1/2)/x**3,x)

[Out]

Integral(sqrt(c + d*x**2)*sqrt((a + b*x)**2)/x**3, x)

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Giac [A]
time = 4.63, size = 199, normalized size = 1.24 \begin {gather*} \frac {a d \arctan \left (-\frac {\sqrt {d} x - \sqrt {d x^{2} + c}}{\sqrt {-c}}\right ) \mathrm {sgn}\left (b x + a\right )}{\sqrt {-c}} - b \sqrt {d} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right ) \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{3} a d \mathrm {sgn}\left (b x + a\right ) + 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c \sqrt {d} \mathrm {sgn}\left (b x + a\right ) + {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )} a c d \mathrm {sgn}\left (b x + a\right ) - 2 \, b c^{2} \sqrt {d} \mathrm {sgn}\left (b x + a\right )}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x+a)^2)^(1/2)*(d*x^2+c)^(1/2)/x^3,x, algorithm="giac")

[Out]

a*d*arctan(-(sqrt(d)*x - sqrt(d*x^2 + c))/sqrt(-c))*sgn(b*x + a)/sqrt(-c) - b*sqrt(d)*log(abs(-sqrt(d)*x + sqr
t(d*x^2 + c)))*sgn(b*x + a) + ((sqrt(d)*x - sqrt(d*x^2 + c))^3*a*d*sgn(b*x + a) + 2*(sqrt(d)*x - sqrt(d*x^2 +
c))^2*b*c*sqrt(d)*sgn(b*x + a) + (sqrt(d)*x - sqrt(d*x^2 + c))*a*c*d*sgn(b*x + a) - 2*b*c^2*sqrt(d)*sgn(b*x +
a))/((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^2

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+c}}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((a + b*x)^2)^(1/2)*(c + d*x^2)^(1/2))/x^3,x)

[Out]

int((((a + b*x)^2)^(1/2)*(c + d*x^2)^(1/2))/x^3, x)

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