3.1.50 \(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x^2} \, dx\) [50]

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

[Out]

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

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Rubi [A]
time = 0.10, 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 = {1014, 826, 857, 635, 212, 738} \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 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 635

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 738

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 826

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 857

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 1014

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 ) \text {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 ) \text {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 = 0.49, size = 151, normalized size = 0.75 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (2 \sqrt {d} (2 b c+a e) x \tanh ^{-1}\left (\frac {-\sqrt {d} x+\sqrt {c+x (e+d x)}}{\sqrt {c}}\right )+\sqrt {c} \left (2 \sqrt {d} (a-b x) \sqrt {c+x (e+d x)}+(2 a d+b e) x \log \left (e+2 d x-2 \sqrt {d} \sqrt {c+x (e+d x)}\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]

-1/2*(Sqrt[(a + b*x)^2]*(2*Sqrt[d]*(2*b*c + a*e)*x*ArcTanh[(-(Sqrt[d]*x) + Sqrt[c + x*(e + d*x)])/Sqrt[c]] + S
qrt[c]*(2*Sqrt[d]*(a - b*x)*Sqrt[c + x*(e + d*x)] + (2*a*d + b*e)*x*Log[e + 2*d*x - 2*Sqrt[d]*Sqrt[c + x*(e +
d*x)]])))/(Sqrt[c]*Sqrt[d]*x*(a + b*x))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.13, size = 249, normalized size = 1.23

method	result	size
risch	\(-\frac {a \sqrt {d \,x^{2}+e x +c}\, \sqrt {\left (b x +a \right )^{2}}}{x \left (b x +a \right )}+\frac {\left (b \sqrt {d \,x^{2}+e x +c}+\frac {e b \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right )}{2 \sqrt {d}}+a \sqrt {d}\, \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right )-\frac {\ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) a e}{2 \sqrt {c}}-\sqrt {c}\, \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) b \right ) \sqrt {\left (b x +a \right )^{2}}}{b x +a}\)	\(201\)
default	\(-\frac {\mathrm {csgn}\left (b x +a \right ) \left (-2 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} a \,x^{2}+2 d^{\frac {3}{2}} c^{\frac {3}{2}} \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) b x +d^{\frac {3}{2}} \sqrt {c}\, \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) a e x +2 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {3}{2}} a -2 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {3}{2}} a e x -2 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {3}{2}} b c x -2 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a c \,d^{2} x -\ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) d b c e x \right )}{2 c x \,d^{\frac {3}{2}}}\)	\(249\)

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,method=_RETURNVERBOSE)

[Out]

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

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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+e*x+c)^(1/2)/x^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.58, size = 691, normalized size = 3.42 \begin {gather*} \left [\frac {{\left (2 \, a c d x + b c x e\right )} \sqrt {d} \log \left (8 \, d^{2} x^{2} + 8 \, d x e + 4 \, \sqrt {d x^{2} + x e + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) + {\left (2 \, b c d x + a d x e\right )} \sqrt {c} \log \left (\frac {4 \, c d x^{2} + x^{2} e^{2} + 8 \, c x e - 4 \, \sqrt {d x^{2} + x e + c} {\left (x e + 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} + x e + c}}{4 \, c d x}, -\frac {2 \, {\left (2 \, a c d x + b c x e\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d x^{2} + x e + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d x e + c d\right )}}\right ) - {\left (2 \, b c d x + a d x e\right )} \sqrt {c} \log \left (\frac {4 \, c d x^{2} + x^{2} e^{2} + 8 \, c x e - 4 \, \sqrt {d x^{2} + x e + c} {\left (x e + 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} + x e + c}}{4 \, c d x}, \frac {2 \, {\left (2 \, b c d x + a d x e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{2} + x e + c} {\left (x e + 2 \, c\right )} \sqrt {-c}}{2 \, {\left (c d x^{2} + c x e + c^{2}\right )}}\right ) + {\left (2 \, a c d x + b c x e\right )} \sqrt {d} \log \left (8 \, d^{2} x^{2} + 8 \, d x e + 4 \, \sqrt {d x^{2} + x e + 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} + x e + c}}{4 \, c d x}, \frac {{\left (2 \, b c d x + a d x e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{2} + x e + c} {\left (x e + 2 \, c\right )} \sqrt {-c}}{2 \, {\left (c d x^{2} + c x e + c^{2}\right )}}\right ) - {\left (2 \, a c d x + b c x e\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d x^{2} + x e + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d x e + c d\right )}}\right ) + 2 \, {\left (b c d x - a c d\right )} \sqrt {d x^{2} + x e + 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*x + b*c*x*e)*sqrt(d)*log(8*d^2*x^2 + 8*d*x*e + 4*sqrt(d*x^2 + x*e + c)*(2*d*x + e)*sqrt(d) + 4*
c*d + e^2) + (2*b*c*d*x + a*d*x*e)*sqrt(c)*log((4*c*d*x^2 + x^2*e^2 + 8*c*x*e - 4*sqrt(d*x^2 + x*e + c)*(x*e +
2*c)*sqrt(c) + 8*c^2)/x^2) + 4*(b*c*d*x - a*c*d)*sqrt(d*x^2 + x*e + c))/(c*d*x), -1/4*(2*(2*a*c*d*x + b*c*x*e
)*sqrt(-d)*arctan(1/2*sqrt(d*x^2 + x*e + c)*(2*d*x + e)*sqrt(-d)/(d^2*x^2 + d*x*e + c*d)) - (2*b*c*d*x + a*d*x
*e)*sqrt(c)*log((4*c*d*x^2 + x^2*e^2 + 8*c*x*e - 4*sqrt(d*x^2 + x*e + c)*(x*e + 2*c)*sqrt(c) + 8*c^2)/x^2) - 4
*(b*c*d*x - a*c*d)*sqrt(d*x^2 + x*e + c))/(c*d*x), 1/4*(2*(2*b*c*d*x + a*d*x*e)*sqrt(-c)*arctan(1/2*sqrt(d*x^2
+ x*e + c)*(x*e + 2*c)*sqrt(-c)/(c*d*x^2 + c*x*e + c^2)) + (2*a*c*d*x + b*c*x*e)*sqrt(d)*log(8*d^2*x^2 + 8*d*
x*e + 4*sqrt(d*x^2 + x*e + c)*(2*d*x + e)*sqrt(d) + 4*c*d + e^2) + 4*(b*c*d*x - a*c*d)*sqrt(d*x^2 + x*e + c))/
(c*d*x), 1/2*((2*b*c*d*x + a*d*x*e)*sqrt(-c)*arctan(1/2*sqrt(d*x^2 + x*e + c)*(x*e + 2*c)*sqrt(-c)/(c*d*x^2 +
c*x*e + c^2)) - (2*a*c*d*x + b*c*x*e)*sqrt(-d)*arctan(1/2*sqrt(d*x^2 + x*e + c)*(2*d*x + e)*sqrt(-d)/(d^2*x^2
+ d*x*e + c*d)) + 2*(b*c*d*x - a*c*d)*sqrt(d*x^2 + x*e + c))/(c*d*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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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,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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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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