3.51 \(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x^3} \, dx\)

Optimal. Leaf size=215 \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (4 a c d-a e^2+4 b c e\right ) \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+d x^2+e x}}\right )}{8 c^{3/2} (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2+e x} (x (a e+4 b c)+2 a c)}{4 c x^2 (a+b x)}+\frac {b \sqrt {d} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{a+b x} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 215, 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, 810, 843, 621, 206, 724} \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (4 a c d-a e^2+4 b c e\right ) \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+d x^2+e x}}\right )}{8 c^{3/2} (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2+e x} (x (a e+4 b c)+2 a c)}{4 c x^2 (a+b x)}+\frac {b \sqrt {d} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{a+b x} \]

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^3,x]

[Out]

-((2*a*c + (4*b*c + a*e)*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + e*x + d*x^2])/(4*c*x^2*(a + b*x)) + (b*Sqrt
[d]*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])])/(a + b*x) - ((4*a*c*
d + 4*b*c*e - a*e^2)*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])])/(8*
c^(3/2)*(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 810

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
- b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
- c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(m + 2*p + 2))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*
c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

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

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Mathematica [A]  time = 0.22, size = 161, normalized size = 0.75 \[ -\frac {\sqrt {(a+b x)^2} \left (x^2 \left (4 a c d-a e^2+4 b c e\right ) \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+x (d x+e)}}\right )+2 \sqrt {c} \sqrt {c+x (d x+e)} (2 a c+a e x+4 b c x)-8 b c^{3/2} \sqrt {d} x^2 \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+x (d x+e)}}\right )\right )}{8 c^{3/2} x^2 (a+b x)} \]

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^3,x]

[Out]

-1/8*(Sqrt[(a + b*x)^2]*(2*Sqrt[c]*(2*a*c + 4*b*c*x + a*e*x)*Sqrt[c + x*(e + d*x)] - 8*b*c^(3/2)*Sqrt[d]*x^2*A
rcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + x*(e + d*x)])] + (4*a*c*d + 4*b*c*e - a*e^2)*x^2*ArcTanh[(2*c + e*x)/(2
*Sqrt[c]*Sqrt[c + x*(e + d*x)])]))/(c^(3/2)*x^2*(a + b*x))

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fricas [A]  time = 1.81, size = 693, normalized size = 3.22 \[ \left [\frac {8 \, b c^{2} \sqrt {d} x^{2} \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 (4 \, a c d + 4 \, b c e - a e^{2}\right )} \sqrt {c} x^{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 ) - 4 \, {\left (2 \, a c^{2} + {\left (4 \, b c^{2} + a c e\right )} x\right )} \sqrt {d x^{2} + e x + c}}{16 \, c^{2} x^{2}}, -\frac {16 \, b c^{2} \sqrt {-d} x^{2} \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 (4 \, a c d + 4 \, b c e - a e^{2}\right )} \sqrt {c} x^{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 ) + 4 \, {\left (2 \, a c^{2} + {\left (4 \, b c^{2} + a c e\right )} x\right )} \sqrt {d x^{2} + e x + c}}{16 \, c^{2} x^{2}}, \frac {4 \, b c^{2} \sqrt {d} x^{2} \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 (4 \, a c d + 4 \, b c e - a e^{2}\right )} \sqrt {-c} x^{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 ) - 2 \, {\left (2 \, a c^{2} + {\left (4 \, b c^{2} + a c e\right )} x\right )} \sqrt {d x^{2} + e x + c}}{8 \, c^{2} x^{2}}, -\frac {8 \, b c^{2} \sqrt {-d} x^{2} \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 (4 \, a c d + 4 \, b c e - a e^{2}\right )} \sqrt {-c} x^{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 ) + 2 \, {\left (2 \, a c^{2} + {\left (4 \, b c^{2} + a c e\right )} x\right )} \sqrt {d x^{2} + e x + c}}{8 \, c^{2} x^{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^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.41, size = 450, normalized size = 2.09 \[ -b \sqrt {d} \log \left ({\left | -2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )} d - \sqrt {d} e \right |}\right ) \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (4 \, a c d \mathrm {sgn}\left (b x + a\right ) + 4 \, b c e \mathrm {sgn}\left (b x + a\right ) - a e^{2} \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (-\frac {\sqrt {d} x - \sqrt {d x^{2} + x e + c}}{\sqrt {-c}}\right )}{4 \, \sqrt {-c} c} + \frac {4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )}^{3} a c d \mathrm {sgn}\left (b x + a\right ) + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )}^{3} b c e \mathrm {sgn}\left (b x + a\right ) + 8 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )}^{2} b c^{2} \sqrt {d} \mathrm {sgn}\left (b x + a\right ) + 8 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )}^{2} a c \sqrt {d} e \mathrm {sgn}\left (b x + a\right ) + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )} a c^{2} d \mathrm {sgn}\left (b x + a\right ) + {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )}^{3} a e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )} b c^{2} e \mathrm {sgn}\left (b x + a\right ) - 8 \, b c^{3} \sqrt {d} \mathrm {sgn}\left (b x + a\right ) + {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )} a c e^{2} \mathrm {sgn}\left (b x + a\right )}{4 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )}^{2} - c\right )}^{2} c} \]

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^3,x, algorithm="giac")

[Out]

-b*sqrt(d)*log(abs(-2*(sqrt(d)*x - sqrt(d*x^2 + x*e + c))*d - sqrt(d)*e))*sgn(b*x + a) + 1/4*(4*a*c*d*sgn(b*x
+ a) + 4*b*c*e*sgn(b*x + a) - a*e^2*sgn(b*x + a))*arctan(-(sqrt(d)*x - sqrt(d*x^2 + x*e + c))/sqrt(-c))/(sqrt(
-c)*c) + 1/4*(4*(sqrt(d)*x - sqrt(d*x^2 + x*e + c))^3*a*c*d*sgn(b*x + a) + 4*(sqrt(d)*x - sqrt(d*x^2 + x*e + c
))^3*b*c*e*sgn(b*x + a) + 8*(sqrt(d)*x - sqrt(d*x^2 + x*e + c))^2*b*c^2*sqrt(d)*sgn(b*x + a) + 8*(sqrt(d)*x -
sqrt(d*x^2 + x*e + c))^2*a*c*sqrt(d)*e*sgn(b*x + a) + 4*(sqrt(d)*x - sqrt(d*x^2 + x*e + c))*a*c^2*d*sgn(b*x +
a) + (sqrt(d)*x - sqrt(d*x^2 + x*e + c))^3*a*e^2*sgn(b*x + a) - 4*(sqrt(d)*x - sqrt(d*x^2 + x*e + c))*b*c^2*e*
sgn(b*x + a) - 8*b*c^3*sqrt(d)*sgn(b*x + a) + (sqrt(d)*x - sqrt(d*x^2 + x*e + c))*a*c*e^2*sgn(b*x + a))/(((sqr
t(d)*x - sqrt(d*x^2 + x*e + c))^2 - c)^2*c)

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maple [C]  time = 0.02, size = 358, normalized size = 1.67 \[ \frac {\left (-4 a \,c^{\frac {3}{2}} d^{\frac {5}{2}} x^{2} \ln \left (\frac {e x +2 c +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {c}}{x}\right )+a \sqrt {c}\, d^{\frac {3}{2}} e^{2} x^{2} \ln \left (\frac {e x +2 c +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {c}}{x}\right )+8 b \,c^{2} d^{2} x^{2} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )-4 b \,c^{\frac {3}{2}} d^{\frac {3}{2}} e \,x^{2} \ln \left (\frac {e x +2 c +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {c}}{x}\right )-2 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {5}{2}} e \,x^{3}+8 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{\frac {5}{2}} x^{3}+4 \sqrt {d \,x^{2}+e x +c}\, a c \,d^{\frac {5}{2}} x^{2}-2 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {3}{2}} e^{2} x^{2}+8 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{\frac {3}{2}} e \,x^{2}+2 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a \,d^{\frac {3}{2}} e x -8 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b c \,d^{\frac {3}{2}} x -4 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a c \,d^{\frac {3}{2}}\right ) \mathrm {csgn}\left (b x +a \right )}{8 c^{2} d^{\frac {3}{2}} x^{2}} \]

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^3,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d x^{2} + e x + c} \sqrt {{\left (b x + a\right )}^{2}}}{x^{3}}\,{d x} \]

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^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+e\,x+c}}{x^3} \,d x \]

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^3,x)

[Out]

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

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

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**3,x)

[Out]

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

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