3.39 \(\int \frac{\sqrt{c+e x+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{x} \, dx\)

Optimal. Leaf size=286 \[ \frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \sqrt{c+d x^2+e x} \left (8 a d^2-2 b d e x-b e^2\right )}{8 d^2 \left (a+b x^2\right )}+\frac{e \sqrt{a^2+2 a b x^2+b^2 x^4} \left (8 a d^2-b \left (4 c d-e^2\right )\right ) \tanh ^{-1}\left (\frac{2 d x+e}{2 \sqrt{d} \sqrt{c+d x^2+e x}}\right )}{16 d^{5/2} \left (a+b x^2\right )}+\frac{b \sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2}}{3 d \left (a+b x^2\right )}-\frac{a \sqrt{c} \sqrt{a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac{2 c+e x}{2 \sqrt{c} \sqrt{c+d x^2+e x}}\right )}{a+b x^2} \]

[Out]

((8*a*d^2 - b*e^2 - 2*b*d*e*x)*Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*d^2*(a + b*x^2)) + (b
*(c + e*x + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d*(a + b*x^2)) + (e*(8*a*d^2 - b*(4*c*d - e^2))*S
qrt[a^2 + 2*a*b*x^2 + b^2*x^4]*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + e*x + d*x^2])])/(16*d^(5/2)*(a + b*x^2)
) - (a*Sqrt[c]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*ArcTanh[(2*c + e*x)/(2*Sqrt[c]*Sqrt[c + e*x + d*x^2])])/(a + b*
x^2)

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Rubi [A]  time = 0.75443, antiderivative size = 286, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {6744, 1653, 814, 843, 621, 206, 724} \[ \frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \sqrt{c+d x^2+e x} \left (8 a d^2-2 b d e x-b e^2\right )}{8 d^2 \left (a+b x^2\right )}+\frac{e \sqrt{a^2+2 a b x^2+b^2 x^4} \left (8 a d^2-b \left (4 c d-e^2\right )\right ) \tanh ^{-1}\left (\frac{2 d x+e}{2 \sqrt{d} \sqrt{c+d x^2+e x}}\right )}{16 d^{5/2} \left (a+b x^2\right )}+\frac{b \sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2}}{3 d \left (a+b x^2\right )}-\frac{a \sqrt{c} \sqrt{a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac{2 c+e x}{2 \sqrt{c} \sqrt{c+d x^2+e x}}\right )}{a+b x^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/x,x]

[Out]

((8*a*d^2 - b*e^2 - 2*b*d*e*x)*Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*d^2*(a + b*x^2)) + (b
*(c + e*x + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d*(a + b*x^2)) + (e*(8*a*d^2 - b*(4*c*d - e^2))*S
qrt[a^2 + 2*a*b*x^2 + b^2*x^4]*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + e*x + d*x^2])])/(16*d^(5/2)*(a + b*x^2)
) - (a*Sqrt[c]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*ArcTanh[(2*c + e*x)/(2*Sqrt[c]*Sqrt[c + e*x + d*x^2])])/(a + b*
x^2)

Rule 6744

Int[(u_)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[Sqrt[a + b*x^n + c*x^(2*n)]/((4
*c)^(p - 1/2)*(b + 2*c*x^n)), Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] &
& EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\sqrt{c+e x+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{x} \, dx &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \int \frac{\left (2 a b+2 b^2 x^2\right ) \sqrt{c+e x+d x^2}}{x} \, dx}{2 a b+2 b^2 x^2}\\ &=\frac{b \left (c+e x+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )}+\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \int \frac{\left (6 a b d-3 b^2 e x\right ) \sqrt{c+e x+d x^2}}{x} \, dx}{3 d \left (2 a b+2 b^2 x^2\right )}\\ &=\frac{\left (8 a d^2-b e^2-2 b d e x\right ) \sqrt{c+e x+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 d^2 \left (a+b x^2\right )}+\frac{b \left (c+e x+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )}-\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \int \frac{-24 a b c d^2-\frac{3}{2} b e \left (8 a d^2-b \left (4 c d-e^2\right )\right ) x}{x \sqrt{c+e x+d x^2}} \, dx}{12 d^2 \left (2 a b+2 b^2 x^2\right )}\\ &=\frac{\left (8 a d^2-b e^2-2 b d e x\right ) \sqrt{c+e x+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 d^2 \left (a+b x^2\right )}+\frac{b \left (c+e x+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )}+\frac{\left (2 a b c \sqrt{a^2+2 a b x^2+b^2 x^4}\right ) \int \frac{1}{x \sqrt{c+e x+d x^2}} \, dx}{2 a b+2 b^2 x^2}+\frac{\left (b e \left (8 a d^2-b \left (4 c d-e^2\right )\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}\right ) \int \frac{1}{\sqrt{c+e x+d x^2}} \, dx}{8 d^2 \left (2 a b+2 b^2 x^2\right )}\\ &=\frac{\left (8 a d^2-b e^2-2 b d e x\right ) \sqrt{c+e x+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 d^2 \left (a+b x^2\right )}+\frac{b \left (c+e x+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )}-\frac{\left (4 a b c \sqrt{a^2+2 a b x^2+b^2 x^4}\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^2}+\frac{\left (b e \left (8 a d^2-b \left (4 c d-e^2\right )\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}\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 )}{4 d^2 \left (2 a b+2 b^2 x^2\right )}\\ &=\frac{\left (8 a d^2-b e^2-2 b d e x\right ) \sqrt{c+e x+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 d^2 \left (a+b x^2\right )}+\frac{b \left (c+e x+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )}+\frac{e \left (8 a d^2-b \left (4 c d-e^2\right )\right ) \sqrt{a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac{e+2 d x}{2 \sqrt{d} \sqrt{c+e x+d x^2}}\right )}{16 d^{5/2} \left (a+b x^2\right )}-\frac{a \sqrt{c} \sqrt{a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac{2 c+e x}{2 \sqrt{c} \sqrt{c+e x+d x^2}}\right )}{a+b x^2}\\ \end{align*}

Mathematica [A]  time = 0.322493, size = 176, normalized size = 0.62 \[ \frac{\sqrt{\left (a+b x^2\right )^2} \left (2 \sqrt{d} \left (\sqrt{c+x (d x+e)} \left (24 a d^2+b \left (8 c d+8 d^2 x^2+2 d e x-3 e^2\right )\right )-24 a \sqrt{c} d^2 \tanh ^{-1}\left (\frac{2 c+e x}{2 \sqrt{c} \sqrt{c+x (d x+e)}}\right )\right )+3 e \left (8 a d^2+b \left (e^2-4 c d\right )\right ) \tanh ^{-1}\left (\frac{2 d x+e}{2 \sqrt{d} \sqrt{c+x (d x+e)}}\right )\right )}{48 d^{5/2} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/x,x]

[Out]

(Sqrt[(a + b*x^2)^2]*(3*e*(8*a*d^2 + b*(-4*c*d + e^2))*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + x*(e + d*x)])]
+ 2*Sqrt[d]*(Sqrt[c + x*(e + d*x)]*(24*a*d^2 + b*(8*c*d - 3*e^2 + 2*d*e*x + 8*d^2*x^2)) - 24*a*Sqrt[c]*d^2*Arc
Tanh[(2*c + e*x)/(2*Sqrt[c]*Sqrt[c + x*(e + d*x)])])))/(48*d^(5/2)*(a + b*x^2))

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Maple [A]  time = 0.007, size = 251, normalized size = 0.9 \begin{align*} -{\frac{1}{48\,b{x}^{2}+48\,a}\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}} \left ( 48\,{d}^{7/2}\ln \left ({\frac{2\,c+ex+2\,\sqrt{c}\sqrt{{x}^{2}d+ex+c}}{x}} \right ) \sqrt{c}a-16\,{d}^{5/2} \left ({x}^{2}d+ex+c \right ) ^{3/2}b+12\,{d}^{5/2}\sqrt{{x}^{2}d+ex+c}xbe-48\,{d}^{7/2}\sqrt{{x}^{2}d+ex+c}a+6\,{d}^{3/2}\sqrt{{x}^{2}d+ex+c}b{e}^{2}-24\,{d}^{3}\ln \left ( 1/2\,{\frac{2\,\sqrt{{x}^{2}d+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) ae+12\,\ln \left ( 1/2\,{\frac{2\,\sqrt{{x}^{2}d+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) bc{d}^{2}e-3\,\ln \left ( 1/2\,{\frac{2\,\sqrt{{x}^{2}d+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) bd{e}^{3} \right ){d}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*((b*x^2+a)^2)^(1/2)*(48*d^(7/2)*ln((2*c+e*x+2*c^(1/2)*(d*x^2+e*x+c)^(1/2))/x)*c^(1/2)*a-16*d^(5/2)*(d*x^
2+e*x+c)^(3/2)*b+12*d^(5/2)*(d*x^2+e*x+c)^(1/2)*x*b*e-48*d^(7/2)*(d*x^2+e*x+c)^(1/2)*a+6*d^(3/2)*(d*x^2+e*x+c)
^(1/2)*b*e^2-24*d^3*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1/2))*a*e+12*ln(1/2*(2*(d*x^2+e*x+c)^(1/
2)*d^(1/2)+2*d*x+e)/d^(1/2))*b*c*d^2*e-3*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1/2))*b*d*e^3)/(b*x
^2+a)/d^(7/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d x^{2} + e x + c} \sqrt{{\left (b x^{2} + a\right )}^{2}}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 7.15379, size = 1798, normalized size = 6.29 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(48*a*sqrt(c)*d^3*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) + 3*(b*e^3 - 4*(b*c*d - 2*a*d^2)*e)*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*(8*b*d^3*x^2 + 2*b*d^2*e*x + 8*b*c*d^2 + 24*a*d^3 - 3*b*d*e^2)*sqrt(d*x^2 + e*x
+ c))/d^3, 1/48*(24*a*sqrt(c)*d^3*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) - 3*(b*e^3 - 4*(b*c*d - 2*a*d^2)*e)*sqrt(-d)*arctan(1/2*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqr
t(-d)/(d^2*x^2 + d*e*x + c*d)) + 2*(8*b*d^3*x^2 + 2*b*d^2*e*x + 8*b*c*d^2 + 24*a*d^3 - 3*b*d*e^2)*sqrt(d*x^2 +
e*x + c))/d^3, 1/96*(96*a*sqrt(-c)*d^3*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)) + 3*(b*e^3 - 4*(b*c*d - 2*a*d^2)*e)*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*(8*b*d^3*x^2 + 2*b*d^2*e*x + 8*b*c*d^2 + 24*a*d^3 - 3*b*d*e^2)*sqrt(d*x^2 + e*
x + c))/d^3, 1/48*(48*a*sqrt(-c)*d^3*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)) - 3*(b*e^3 - 4*(b*c*d - 2*a*d^2)*e)*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*(8*b*d^3*x^2 + 2*b*d^2*e*x + 8*b*c*d^2 + 24*a*d^3 - 3*b*d*e^2)*sqrt(d*x^2 + e*x + c))/
d^3]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError