3.1.51 \(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x^3} \, dx\) [51]

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

[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 = 0.11, 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 = {1014, 824, 857, 635, 212, 738} \begin {gather*} -\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} \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^3,x]

[Out]

-1/4*((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])/(c*x^2*(a + b*x)) + (b*Sq
rt[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 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 824

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(-(d + e*x)^(m + 1))*((a + b*x + c*x^2)^p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)))*((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), 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 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^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 ) \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 {\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 ) \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 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 = 0.60, size = 155, normalized size = 0.72 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (\left (4 a c d+4 b c e-a e^2\right ) x^2 \tanh ^{-1}\left (\frac {-\sqrt {d} x+\sqrt {c+x (e+d x)}}{\sqrt {c}}\right )+\sqrt {c} \left ((2 a c+4 b c x+a e x) \sqrt {c+x (e+d x)}+4 b c \sqrt {d} x^2 \log \left (e+2 d x-2 \sqrt {d} \sqrt {c+x (e+d x)}\right )\right )\right )}{4 c^{3/2} x^2 (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^3,x]

[Out]

-1/4*(Sqrt[(a + b*x)^2]*((4*a*c*d + 4*b*c*e - a*e^2)*x^2*ArcTanh[(-(Sqrt[d]*x) + Sqrt[c + x*(e + d*x)])/Sqrt[c
]] + Sqrt[c]*((2*a*c + 4*b*c*x + a*e*x)*Sqrt[c + x*(e + d*x)] + 4*b*c*Sqrt[d]*x^2*Log[e + 2*d*x - 2*Sqrt[d]*Sq
rt[c + x*(e + d*x)]])))/(c^(3/2)*x^2*(a + b*x))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.12, size = 359, normalized size = 1.67

method	result	size
risch	\(-\frac {\sqrt {d \,x^{2}+e x +c}\, \left (a e x +4 b c x +2 a c \right ) \sqrt {\left (b x +a \right )^{2}}}{4 x^{2} c \left (b x +a \right )}+\frac {\left (b \sqrt {d}\, \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right )-\frac {\ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) a d}{2 \sqrt {c}}+\frac {\ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) a \,e^{2}}{8 c^{\frac {3}{2}}}-\frac {\ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) b e}{2 \sqrt {c}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b x +a}\)	\(210\)
default	\(-\frac {\mathrm {csgn}\left (b x +a \right ) \left (4 d^{\frac {5}{2}} c^{\frac {3}{2}} \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) a \,x^{2}+2 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} a e \,x^{3}-8 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} b c \,x^{3}+4 d^{\frac {3}{2}} c^{\frac {3}{2}} \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) b e \,x^{2}-4 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} a c \,x^{2}-d^{\frac {3}{2}} \sqrt {c}\, \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) a \,e^{2} x^{2}-2 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {3}{2}} a e x +8 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {3}{2}} b c x +2 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {3}{2}} a \,e^{2} x^{2}-8 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {3}{2}} b c e \,x^{2}+4 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {3}{2}} a c -8 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b \,c^{2} d^{2} x^{2}\right )}{8 c^{2} x^{2} d^{\frac {3}{2}}}\)	\(359\)

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,method=_RETURNVERBOSE)

[Out]

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

[Out]

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

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Fricas [A]
time = 0.61, size = 745, normalized size = 3.47 \begin {gather*} \left [\frac {8 \, b c^{2} \sqrt {d} x^{2} \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 (4 \, a c d x^{2} + 4 \, b c x^{2} e - a x^{2} e^{2}\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 (4 \, b c^{2} x + a c x e + 2 \, a c^{2}\right )} \sqrt {d x^{2} + x e + c}}{16 \, c^{2} x^{2}}, -\frac {16 \, b c^{2} \sqrt {-d} x^{2} \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 (4 \, a c d x^{2} + 4 \, b c x^{2} e - a x^{2} e^{2}\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 (4 \, b c^{2} x + a c x e + 2 \, a c^{2}\right )} \sqrt {d x^{2} + x e + c}}{16 \, c^{2} x^{2}}, \frac {4 \, b c^{2} \sqrt {d} x^{2} \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 (4 \, a c d x^{2} + 4 \, b c x^{2} e - a x^{2} e^{2}\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 ) - 2 \, {\left (4 \, b c^{2} x + a c x e + 2 \, a c^{2}\right )} \sqrt {d x^{2} + x e + c}}{8 \, c^{2} x^{2}}, -\frac {8 \, b c^{2} \sqrt {-d} x^{2} \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 (4 \, a c d x^{2} + 4 \, b c x^{2} e - a x^{2} e^{2}\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 ) + 2 \, {\left (4 \, b c^{2} x + a c x e + 2 \, a c^{2}\right )} \sqrt {d x^{2} + x e + c}}{8 \, c^{2} x^{2}}\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^3,x, algorithm="fricas")

[Out]

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

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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^{3}}\, 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**3,x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 450 vs. \(2 (166) = 332\).
time = 5.10, size = 450, normalized size = 2.09 \begin {gather*} -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} \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^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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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^3} \,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^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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