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

Optimal. Leaf size=211 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (4 a d e+4 b c d-b e^2\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{8 d^{3/2} (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2+e x} (4 a d+2 b d x+b e)}{4 d (a+b x)}-\frac {a \sqrt {c} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+d x^2+e x}}\right )}{a+b x} \]

[Out]

1/8*(4*a*d*e+4*b*c*d-b*e^2)*arctanh(1/2*(2*d*x+e)/d^(1/2)/(d*x^2+e*x+c)^(1/2))*((b*x+a)^2)^(1/2)/d^(3/2)/(b*x+
a)-a*arctanh(1/2*(e*x+2*c)/c^(1/2)/(d*x^2+e*x+c)^(1/2))*c^(1/2)*((b*x+a)^2)^(1/2)/(b*x+a)+1/4*(2*b*d*x+4*a*d+b
*e)*((b*x+a)^2)^(1/2)*(d*x^2+e*x+c)^(1/2)/d/(b*x+a)

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Rubi [A]  time = 0.22, antiderivative size = 211, 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, 814, 843, 621, 206, 724} \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (4 a d e+4 b c d-b e^2\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{8 d^{3/2} (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2+e x} (4 a d+2 b d x+b e)}{4 d (a+b x)}-\frac {a \sqrt {c} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+d x^2+e x}}\right )}{a+b x} \]

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,x]

[Out]

((4*a*d + b*e + 2*b*d*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + e*x + d*x^2])/(4*d*(a + b*x)) + ((4*b*c*d + 4*
a*d*e - b*e^2)*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])])/(8*d^(3/2
)*(a + b*x)) - (a*Sqrt[c]*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])]
)/(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 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*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] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 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} \, 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} \, dx}{2 a b+2 b^2 x}\\ &=\frac {(4 a d+b e+2 b d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{4 d (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {-8 a b c d-b \left (4 a d e+b \left (4 c d-e^2\right )\right ) x}{x \sqrt {c+e x+d x^2}} \, dx}{4 d \left (2 a b+2 b^2 x\right )}\\ &=\frac {(4 a d+b e+2 b d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{4 d (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+e x+d x^2}} \, dx}{2 a b+2 b^2 x}+\frac {\left (b \left (4 b c d+4 a d e-b e^2\right ) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {1}{\sqrt {c+e x+d x^2}} \, dx}{4 d \left (2 a b+2 b^2 x\right )}\\ &=\frac {(4 a d+b e+2 b d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{4 d (a+b x)}-\frac {\left (4 a b c \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 (b \left (4 b c d+4 a d e-b e^2\right ) \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 d \left (2 a b+2 b^2 x\right )}\\ &=\frac {(4 a d+b e+2 b d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{4 d (a+b x)}+\frac {\left (4 b c d+4 a d e-b e^2\right ) \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 )}{8 d^{3/2} (a+b x)}-\frac {a \sqrt {c} \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 )}{a+b x}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 149, normalized size = 0.71 \[ \frac {\sqrt {(a+b x)^2} \left (\left (4 a d e+4 b c d-b e^2\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+x (d x+e)}}\right )+2 \sqrt {d} \left (\sqrt {c+x (d x+e)} (4 a d+b (2 d x+e))-4 a \sqrt {c} d \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+x (d x+e)}}\right )\right )\right )}{8 d^{3/2} (a+b x)} \]

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,x]

[Out]

(Sqrt[(a + b*x)^2]*((4*b*c*d + 4*a*d*e - b*e^2)*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + x*(e + d*x)])] + 2*Sqr
t[d]*(Sqrt[c + x*(e + d*x)]*(4*a*d + b*(e + 2*d*x)) - 4*a*Sqrt[c]*d*ArcTanh[(2*c + e*x)/(2*Sqrt[c]*Sqrt[c + x*
(e + d*x)])])))/(8*d^(3/2)*(a + b*x))

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fricas [A]  time = 2.54, size = 651, normalized size = 3.09 \[ \left [\frac {8 \, a \sqrt {c} d^{2} \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 ) - {\left (4 \, b c d + 4 \, a d e - b e^{2}\right )} \sqrt {d} \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 (2 \, b d^{2} x + 4 \, a d^{2} + b d e\right )} \sqrt {d x^{2} + e x + c}}{16 \, d^{2}}, \frac {4 \, a \sqrt {c} d^{2} \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 ) - {\left (4 \, b c d + 4 \, a d e - b e^{2}\right )} \sqrt {-d} \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 (2 \, b d^{2} x + 4 \, a d^{2} + b d e\right )} \sqrt {d x^{2} + e x + c}}{8 \, d^{2}}, \frac {16 \, a \sqrt {-c} d^{2} \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 (4 \, b c d + 4 \, a d e - b e^{2}\right )} \sqrt {d} \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 (2 \, b d^{2} x + 4 \, a d^{2} + b d e\right )} \sqrt {d x^{2} + e x + c}}{16 \, d^{2}}, \frac {8 \, a \sqrt {-c} d^{2} \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 (4 \, b c d + 4 \, a d e - b e^{2}\right )} \sqrt {-d} \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 (2 \, b d^{2} x + 4 \, a d^{2} + b d e\right )} \sqrt {d x^{2} + e x + c}}{8 \, d^{2}}\right ] \]

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,x, algorithm="fricas")

[Out]

[1/16*(8*a*sqrt(c)*d^2*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 + 4*a*d*e - b*e^2)*sqrt(d)*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*(2*b*d^2*x + 4*a*d^2 + b*d*e)*sqrt(d*x^2 + e*x + c))/d^2, 1/8*(4*a*sqrt(c)*d^2*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 + 4*a*d*e - b*
e^2)*sqrt(-d)*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*(2*b*d^2*x +
4*a*d^2 + b*d*e)*sqrt(d*x^2 + e*x + c))/d^2, 1/16*(16*a*sqrt(-c)*d^2*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)) - (4*b*c*d + 4*a*d*e - b*e^2)*sqrt(d)*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*(2*b*d^2*x + 4*a*d^2 + b*d*e)*sqrt(d*x^2 + e*x + c))/d^
2, 1/8*(8*a*sqrt(-c)*d^2*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)) - (4*b
*c*d + 4*a*d*e - b*e^2)*sqrt(-d)*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*(2*b*d^2*x + 4*a*d^2 + b*d*e)*sqrt(d*x^2 + e*x + c))/d^2]

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

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,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)]Warning, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 b
y ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution variable
should perhaps be purged.index.cc index_m operator + Error: Bad Argument Value

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maple [C]  time = 0.01, size = 214, normalized size = 1.01 \[ -\frac {\left (8 a \sqrt {c}\, d^{\frac {5}{2}} \ln \left (\frac {e x +2 c +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {c}}{x}\right )-4 a \,d^{2} e \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )-4 b c \,d^{2} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )+b d \,e^{2} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )-4 \sqrt {d \,x^{2}+e x +c}\, b \,d^{\frac {5}{2}} x -8 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {5}{2}}-2 \sqrt {d \,x^{2}+e x +c}\, b \,d^{\frac {3}{2}} e \right ) \mathrm {csgn}\left (b x +a \right )}{8 d^{\frac {5}{2}}} \]

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,x)

[Out]

-1/8*csgn(b*x+a)*(8*c^(1/2)*d^(5/2)*ln((2*c+e*x+2*c^(1/2)*(d*x^2+e*x+c)^(1/2))/x)*a-4*d^(5/2)*(d*x^2+e*x+c)^(1
/2)*x*b-8*d^(5/2)*(d*x^2+e*x+c)^(1/2)*a-2*d^(3/2)*(d*x^2+e*x+c)^(1/2)*b*e-4*d^2*ln(1/2*(2*d*x+e+2*(d*x^2+e*x+c
)^(1/2)*d^(1/2))/d^(1/2))*a*e-4*ln(1/2*(2*d*x+e+2*(d*x^2+e*x+c)^(1/2)*d^(1/2))/d^(1/2))*b*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))*b*d*e^2)/d^(5/2)

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

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,x, algorithm="maxima")

[Out]

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

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

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,x)

[Out]

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

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

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,x)

[Out]

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

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