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

Optimal. Leaf size=160 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (2 a+b x) \sqrt {c+d x^2}}{2 (a+b x)}+\frac {b c \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} (a+b x)}-\frac {a \sqrt {c} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a+b x} \]

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Rubi [A]  time = 0.12, antiderivative size = 160, 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 = {1001, 815, 844, 217, 206, 266, 63, 208} \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2} (2 a+b x) \sqrt {c+d x^2}}{2 (a+b x)}+\frac {b c \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} (a+b x)}-\frac {a \sqrt {c} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{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,x]

[Out]

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

Rule 63

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 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 208

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

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 266

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 815

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

Rule 844

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 1001

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

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Mathematica [A]  time = 0.14, size = 139, normalized size = 0.87 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (\sqrt {d} \sqrt {\frac {d x^2}{c}+1} \left ((2 a+b x) \sqrt {c+d x^2}-2 a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )\right )+b \sqrt {c} \sqrt {c+d x^2} \sinh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )\right )}{2 \sqrt {d} (a+b x) \sqrt {\frac {d x^2}{c}+1}} \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,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.31, size = 114, normalized size = 0.71 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (\frac {1}{2} (2 a+b x) \sqrt {c+d x^2}+2 a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}-\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )-\frac {b c \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{2 \sqrt {d}}\right )}{a+b x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 341, normalized size = 2.13 \begin {gather*} \left [\frac {b c \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, a \sqrt {c} d \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (b d x + 2 \, a d\right )} \sqrt {d x^{2} + c}}{4 \, d}, -\frac {b c \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - a \sqrt {c} d \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - {\left (b d x + 2 \, a d\right )} \sqrt {d x^{2} + c}}{2 \, d}, \frac {4 \, a \sqrt {-c} d \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + b c \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (b d x + 2 \, a d\right )} \sqrt {d x^{2} + c}}{4 \, d}, -\frac {b c \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - 2 \, a \sqrt {-c} d \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - {\left (b d x + 2 \, a d\right )} \sqrt {d x^{2} + c}}{2 \, d}\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,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.25, size = 102, normalized size = 0.64 \begin {gather*} \frac {2 \, a c \arctan \left (-\frac {\sqrt {d} x - \sqrt {d x^{2} + c}}{\sqrt {-c}}\right ) \mathrm {sgn}\left (b x + a\right )}{\sqrt {-c}} - \frac {b c \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right ) \mathrm {sgn}\left (b x + a\right )}{2 \, \sqrt {d}} + \frac {1}{2} \, \sqrt {d x^{2} + c} {\left (b x \mathrm {sgn}\left (b x + a\right ) + 2 \, a \mathrm {sgn}\left (b x + a\right )\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,x, algorithm="giac")

[Out]

2*a*c*arctan(-(sqrt(d)*x - sqrt(d*x^2 + c))/sqrt(-c))*sgn(b*x + a)/sqrt(-c) - 1/2*b*c*log(abs(-sqrt(d)*x + sqr
t(d*x^2 + c)))*sgn(b*x + a)/sqrt(d) + 1/2*sqrt(d*x^2 + c)*(b*x*sgn(b*x + a) + 2*a*sgn(b*x + a))

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maple [C]  time = 0.01, size = 94, normalized size = 0.59 \begin {gather*} -\frac {\left (2 a \sqrt {c}\, \sqrt {d}\, \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )-b c \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )-\sqrt {d \,x^{2}+c}\, b \sqrt {d}\, x -2 \sqrt {d \,x^{2}+c}\, a \sqrt {d}\right ) \mathrm {csgn}\left (b x +a \right )}{2 \sqrt {d}} \end {gather*}

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,x)

[Out]

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

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

[Out]

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

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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} \,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,x)

[Out]

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

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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}\, 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,x)

[Out]

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

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