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

Optimal. Leaf size=202 \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a-b x) \sqrt {c+d x^2+e x}}{x (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (2 a d+b e) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{2 \sqrt {d} (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a e+2 b c) \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+d x^2+e x}}\right )}{2 \sqrt {c} (a+b x)} \]

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Rubi [A]  time = 0.19, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1000, 812, 843, 621, 206, 724} \begin {gather*} -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a-b x) \sqrt {c+d x^2+e x}}{x (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (2 a d+b e) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{2 \sqrt {d} (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a e+2 b c) \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+d x^2+e x}}\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 + e*x + d*x^2])/x^2,x]

[Out]

-(((a - b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + e*x + d*x^2])/(x*(a + b*x))) + ((2*a*d + b*e)*Sqrt[a^2 + 2
*a*b*x + b^2*x^2]*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + e*x + d*x^2])])/(2*Sqrt[d]*(a + b*x)) - ((2*b*c + a*
e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(2*c + e*x)/(2*Sqrt[c]*Sqrt[c + e*x + d*x^2])])/(2*Sqrt[c]*(a + b*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 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 812

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + b*x + c*x^2)^p)/(e^2*(m + 1)*(m
+ 2*p + 2)), x] + Dist[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(
b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p
+ 2))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 1000

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (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 + e*x + f*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, 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+e x+d x^2}}{x^2} \, 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+e x+d x^2}}{x^2} \, dx}{2 a b+2 b^2 x}\\ &=-\frac {(a-b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {-2 b (2 b c+a e)-2 b (2 a d+b e) x}{x \sqrt {c+e x+d x^2}} \, dx}{2 \left (2 a b+2 b^2 x\right )}\\ &=-\frac {(a-b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x (a+b x)}+\frac {\left (b (2 b c+a e) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {1}{x \sqrt {c+e x+d x^2}} \, dx}{2 a b+2 b^2 x}+\frac {\left (b (2 a d+b e) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {1}{\sqrt {c+e x+d x^2}} \, dx}{2 a b+2 b^2 x}\\ &=-\frac {(a-b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x (a+b x)}-\frac {\left (2 b (2 b c+a e) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {2 c+e x}{\sqrt {c+e x+d x^2}}\right )}{2 a b+2 b^2 x}+\frac {\left (2 b (2 a d+b e) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{4 d-x^2} \, dx,x,\frac {e+2 d x}{\sqrt {c+e x+d x^2}}\right )}{2 a b+2 b^2 x}\\ &=-\frac {(a-b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x (a+b x)}+\frac {(2 a d+b e) \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+e x+d x^2}}\right )}{2 \sqrt {d} (a+b x)}-\frac {(2 b c+a e) \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+e x+d x^2}}\right )}{2 \sqrt {c} (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 155, normalized size = 0.77 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (\sqrt {c} x (2 a d+b e) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+x (d x+e)}}\right )+\sqrt {d} \left (2 \sqrt {c} (b x-a) \sqrt {c+x (d x+e)}-x (a e+2 b c) \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+x (d x+e)}}\right )\right )\right )}{2 \sqrt {c} \sqrt {d} x (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.62, size = 136, normalized size = 0.67 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (\frac {(b x-a) \sqrt {c+d x^2+e x}}{x}+\frac {(-2 a d-b e) \log \left (-2 \sqrt {d} \sqrt {c+d x^2+e x}+2 d x+e\right )}{2 \sqrt {d}}+\frac {(-a e-2 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2+e x}-\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c}}\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 + e*x + d*x^2])/x^2,x]

[Out]

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

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fricas [A]  time = 0.69, size = 647, normalized size = 3.20 \begin {gather*} \left [\frac {{\left (2 \, a c d + b c e\right )} \sqrt {d} x \log \left (8 \, d^{2} x^{2} + 8 \, d e x + 4 \, \sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) + {\left (2 \, b c d + a d e\right )} \sqrt {c} x \log \left (\frac {8 \, c e x + {\left (4 \, c d + e^{2}\right )} x^{2} - 4 \, \sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {c} + 8 \, c^{2}}{x^{2}}\right ) + 4 \, {\left (b c d x - a c d\right )} \sqrt {d x^{2} + e x + c}}{4 \, c d x}, -\frac {2 \, {\left (2 \, a c d + b c e\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d e x + c d\right )}}\right ) - {\left (2 \, b c d + a d e\right )} \sqrt {c} x \log \left (\frac {8 \, c e x + {\left (4 \, c d + e^{2}\right )} x^{2} - 4 \, \sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {c} + 8 \, c^{2}}{x^{2}}\right ) - 4 \, {\left (b c d x - a c d\right )} \sqrt {d x^{2} + e x + c}}{4 \, c d x}, \frac {2 \, {\left (2 \, b c d + a d e\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {-c}}{2 \, {\left (c d x^{2} + c e x + c^{2}\right )}}\right ) + {\left (2 \, a c d + b c e\right )} \sqrt {d} x \log \left (8 \, d^{2} x^{2} + 8 \, d e x + 4 \, \sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) + 4 \, {\left (b c d x - a c d\right )} \sqrt {d x^{2} + e x + c}}{4 \, c d x}, \frac {{\left (2 \, b c d + a d e\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {-c}}{2 \, {\left (c d x^{2} + c e x + c^{2}\right )}}\right ) - {\left (2 \, a c d + b c e\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d e x + c d\right )}}\right ) + 2 \, {\left (b c d x - a c d\right )} \sqrt {d x^{2} + e x + c}}{2 \, c d x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((2*a*c*d + b*c*e)*sqrt(d)*x*log(8*d^2*x^2 + 8*d*e*x + 4*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(d) + 4*c*
d + e^2) + (2*b*c*d + a*d*e)*sqrt(c)*x*log((8*c*e*x + (4*c*d + e^2)*x^2 - 4*sqrt(d*x^2 + e*x + c)*(e*x + 2*c)*
sqrt(c) + 8*c^2)/x^2) + 4*(b*c*d*x - a*c*d)*sqrt(d*x^2 + e*x + c))/(c*d*x), -1/4*(2*(2*a*c*d + b*c*e)*sqrt(-d)
*x*arctan(1/2*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(-d)/(d^2*x^2 + d*e*x + c*d)) - (2*b*c*d + a*d*e)*sqrt(c)*
x*log((8*c*e*x + (4*c*d + e^2)*x^2 - 4*sqrt(d*x^2 + e*x + c)*(e*x + 2*c)*sqrt(c) + 8*c^2)/x^2) - 4*(b*c*d*x -
a*c*d)*sqrt(d*x^2 + e*x + c))/(c*d*x), 1/4*(2*(2*b*c*d + a*d*e)*sqrt(-c)*x*arctan(1/2*sqrt(d*x^2 + e*x + c)*(e
*x + 2*c)*sqrt(-c)/(c*d*x^2 + c*e*x + c^2)) + (2*a*c*d + b*c*e)*sqrt(d)*x*log(8*d^2*x^2 + 8*d*e*x + 4*sqrt(d*x
^2 + e*x + c)*(2*d*x + e)*sqrt(d) + 4*c*d + e^2) + 4*(b*c*d*x - a*c*d)*sqrt(d*x^2 + e*x + c))/(c*d*x), 1/2*((2
*b*c*d + a*d*e)*sqrt(-c)*x*arctan(1/2*sqrt(d*x^2 + e*x + c)*(e*x + 2*c)*sqrt(-c)/(c*d*x^2 + c*e*x + c^2)) - (2
*a*c*d + b*c*e)*sqrt(-d)*x*arctan(1/2*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(-d)/(d^2*x^2 + d*e*x + c*d)) + 2*
(b*c*d*x - a*c*d)*sqrt(d*x^2 + e*x + c))/(c*d*x)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(b*x+a)]Unable to divide, perhaps due to rounding error%%%{2,[4,4,0]%%%}+%%%{%%%{-16,[1]%%%},[4,2,1]%%%}+%%%{
%%%{32,[2]%%%},[4,0,2]%%%}+%%%{-4,[2,4,1]%%%}+%%%{%%%{32,[1]%%%},[2,2,2]%%%}+%%%{%%%{-64,[2]%%%},[2,0,3]%%%}+%
%%{2,[0,4,2]%%%}+%%%{%%%{-16,[1]%%%},[0,2,3]%%%}+%%%{%%%{32,[2]%%%},[0,0,4]%%%} / %%%{4,[4,0,0]%%%}+%%%{-8,[2,
0,1]%%%}+%%%{4,[0,0,2]%%%} Error: Bad Argument Value

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maple [C]  time = 0.01, size = 249, normalized size = 1.23 \begin {gather*} \frac {\left (2 a c \,d^{2} x \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )-a \sqrt {c}\, d^{\frac {3}{2}} e x \ln \left (\frac {e x +2 c +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {c}}{x}\right )-2 b \,c^{\frac {3}{2}} d^{\frac {3}{2}} x \ln \left (\frac {e x +2 c +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {c}}{x}\right )+b c d e x \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )+2 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {5}{2}} x^{2}+2 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {3}{2}} e x +2 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{\frac {3}{2}} x -2 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a \,d^{\frac {3}{2}}\right ) \mathrm {csgn}\left (b x +a \right )}{2 c \,d^{\frac {3}{2}} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*csgn(b*x+a)*(2*d^(5/2)*(d*x^2+e*x+c)^(1/2)*x^2*a-2*d^(3/2)*c^(3/2)*ln((e*x+2*c+2*(d*x^2+e*x+c)^(1/2)*c^(1/
2))/x)*x*b-d^(3/2)*c^(1/2)*ln((e*x+2*c+2*(d*x^2+e*x+c)^(1/2)*c^(1/2))/x)*x*a*e-2*d^(3/2)*(d*x^2+e*x+c)^(3/2)*a
+2*d^(3/2)*(d*x^2+e*x+c)^(1/2)*x*a*e+2*d^(3/2)*(d*x^2+e*x+c)^(1/2)*x*b*c+2*ln(1/2*(2*d*x+e+2*(d*x^2+e*x+c)^(1/
2)*d^(1/2))/d^(1/2))*x*a*c*d^2+ln(1/2*(2*d*x+e+2*(d*x^2+e*x+c)^(1/2)*d^(1/2))/d^(1/2))*d*x*b*c*e)/c/x/d^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {d x^{2} + e x + c} \sqrt {{\left (b x + a\right )}^{2}}}{x^{2}}\,{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+e*x+c)^(1/2)/x^2,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((((a + b*x)^2)^(1/2)*(c + e*x + d*x^2)^(1/2))/x^2, 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} + e x} \sqrt {\left (a + b x\right )^{2}}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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