3.2297 \(\int \frac{(d+e x)^{3/2}}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=363 \[ \frac{\left (-2 c e \left (-d \sqrt{b^2-4 a c}-2 a e+4 b d\right )+b e^2 \left (b-\sqrt{b^2-4 a c}\right )+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} \sqrt{c} \left (b^2-4 a c\right )^{3/2} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{\left (-2 c e \left (d \sqrt{b^2-4 a c}-2 a e+4 b d\right )+b e^2 \left (\sqrt{b^2-4 a c}+b\right )+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} \sqrt{c} \left (b^2-4 a c\right )^{3/2} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{\sqrt{d+e x} (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.29296, antiderivative size = 363, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {738, 826, 1166, 208} \[ \frac{\left (-2 c e \left (-d \sqrt{b^2-4 a c}-2 a e+4 b d\right )+b e^2 \left (b-\sqrt{b^2-4 a c}\right )+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} \sqrt{c} \left (b^2-4 a c\right )^{3/2} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{\left (-2 c e \left (d \sqrt{b^2-4 a c}-2 a e+4 b d\right )+b e^2 \left (\sqrt{b^2-4 a c}+b\right )+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} \sqrt{c} \left (b^2-4 a c\right )^{3/2} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{\sqrt{d+e x} (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(3/2)/(a + b*x + c*x^2)^2,x]

[Out]

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

Rule 738

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

Rule 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*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]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 3.18578, size = 418, normalized size = 1.15 \[ \frac{\frac{1}{2} \left (e (a e-b d)+c d^2\right ) \left (\frac{\sqrt{2} \left (\frac{\left (2 c e \left (d \sqrt{b^2-4 a c}+2 a e-4 b d\right )+b e^2 \left (b-\sqrt{b^2-4 a c}\right )+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )}{\sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}-\frac{\left (-2 c e \left (d \sqrt{b^2-4 a c}-2 a e+4 b d\right )+b e^2 \left (\sqrt{b^2-4 a c}+b\right )+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{c} \sqrt{b^2-4 a c}}+4 e \sqrt{d+e x}\right )+\frac{(d+e x)^{5/2} \left (-2 c (a e+c d x)+b^2 e+b c (e x-d)\right )}{a+x (b+c x)}+e (d+e x)^{3/2} (2 c d-b e)}{\left (b^2-4 a c\right ) \left (e (a e-b d)+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(3/2)/(a + b*x + c*x^2)^2,x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.293, size = 1503, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{\frac{3}{2}}}{{\left (c x^{2} + b x + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)/(c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 2.45584, size = 5069, normalized size = 13.96 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

1/2*(sqrt(1/2)*(a*b^2 - 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 + (b^3 - 4*a*b*c)*x)*sqrt((32*c^3*d^3 - 48*b*c^2*d^2*e
+ 6*(3*b^2*c + 4*a*c^2)*d*e^2 - (b^3 + 12*a*b*c)*e^3 + (b^6*c - 12*a*b^4*c^2 + 48*a^2*b^2*c^3 - 64*a^3*c^4)*s
qrt(e^6/(b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)))/(b^6*c - 12*a*b^4*c^2 + 48*a^2*b^2*c^3 - 64*a
^3*c^4))*log(sqrt(1/2)*((b^4 - 8*a*b^2*c + 16*a^2*c^2)*e^4 + 2*sqrt(e^6/(b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c
^4 - 64*a^3*c^5))*(2*(b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*d - (b^7*c - 12*a*b^5*c^2 + 48*a^2
*b^3*c^3 - 64*a^3*b*c^4)*e))*sqrt((32*c^3*d^3 - 48*b*c^2*d^2*e + 6*(3*b^2*c + 4*a*c^2)*d*e^2 - (b^3 + 12*a*b*c
)*e^3 + (b^6*c - 12*a*b^4*c^2 + 48*a^2*b^2*c^3 - 64*a^3*c^4)*sqrt(e^6/(b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4
- 64*a^3*c^5)))/(b^6*c - 12*a*b^4*c^2 + 48*a^2*b^2*c^3 - 64*a^3*c^4)) + (16*c^2*d^2*e^3 - 16*b*c*d*e^4 + (3*b
^2 + 4*a*c)*e^5)*sqrt(e*x + d)) - sqrt(1/2)*(a*b^2 - 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 + (b^3 - 4*a*b*c)*x)*sqrt
((32*c^3*d^3 - 48*b*c^2*d^2*e + 6*(3*b^2*c + 4*a*c^2)*d*e^2 - (b^3 + 12*a*b*c)*e^3 + (b^6*c - 12*a*b^4*c^2 + 4
8*a^2*b^2*c^3 - 64*a^3*c^4)*sqrt(e^6/(b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)))/(b^6*c - 12*a*b^
4*c^2 + 48*a^2*b^2*c^3 - 64*a^3*c^4))*log(-sqrt(1/2)*((b^4 - 8*a*b^2*c + 16*a^2*c^2)*e^4 + 2*sqrt(e^6/(b^6*c^2
- 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5))*(2*(b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*d -
(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*e))*sqrt((32*c^3*d^3 - 48*b*c^2*d^2*e + 6*(3*b^2*c + 4*
a*c^2)*d*e^2 - (b^3 + 12*a*b*c)*e^3 + (b^6*c - 12*a*b^4*c^2 + 48*a^2*b^2*c^3 - 64*a^3*c^4)*sqrt(e^6/(b^6*c^2 -
12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)))/(b^6*c - 12*a*b^4*c^2 + 48*a^2*b^2*c^3 - 64*a^3*c^4)) + (16*c^2
*d^2*e^3 - 16*b*c*d*e^4 + (3*b^2 + 4*a*c)*e^5)*sqrt(e*x + d)) + sqrt(1/2)*(a*b^2 - 4*a^2*c + (b^2*c - 4*a*c^2)
*x^2 + (b^3 - 4*a*b*c)*x)*sqrt((32*c^3*d^3 - 48*b*c^2*d^2*e + 6*(3*b^2*c + 4*a*c^2)*d*e^2 - (b^3 + 12*a*b*c)*e
^3 - (b^6*c - 12*a*b^4*c^2 + 48*a^2*b^2*c^3 - 64*a^3*c^4)*sqrt(e^6/(b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 -
64*a^3*c^5)))/(b^6*c - 12*a*b^4*c^2 + 48*a^2*b^2*c^3 - 64*a^3*c^4))*log(sqrt(1/2)*((b^4 - 8*a*b^2*c + 16*a^2*c
^2)*e^4 - 2*sqrt(e^6/(b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5))*(2*(b^6*c^2 - 12*a*b^4*c^3 + 48*a
^2*b^2*c^4 - 64*a^3*c^5)*d - (b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*e))*sqrt((32*c^3*d^3 - 48*
b*c^2*d^2*e + 6*(3*b^2*c + 4*a*c^2)*d*e^2 - (b^3 + 12*a*b*c)*e^3 - (b^6*c - 12*a*b^4*c^2 + 48*a^2*b^2*c^3 - 64
*a^3*c^4)*sqrt(e^6/(b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)))/(b^6*c - 12*a*b^4*c^2 + 48*a^2*b^2
*c^3 - 64*a^3*c^4)) + (16*c^2*d^2*e^3 - 16*b*c*d*e^4 + (3*b^2 + 4*a*c)*e^5)*sqrt(e*x + d)) - sqrt(1/2)*(a*b^2
- 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 + (b^3 - 4*a*b*c)*x)*sqrt((32*c^3*d^3 - 48*b*c^2*d^2*e + 6*(3*b^2*c + 4*a*c^
2)*d*e^2 - (b^3 + 12*a*b*c)*e^3 - (b^6*c - 12*a*b^4*c^2 + 48*a^2*b^2*c^3 - 64*a^3*c^4)*sqrt(e^6/(b^6*c^2 - 12*
a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)))/(b^6*c - 12*a*b^4*c^2 + 48*a^2*b^2*c^3 - 64*a^3*c^4))*log(-sqrt(1/2
)*((b^4 - 8*a*b^2*c + 16*a^2*c^2)*e^4 - 2*sqrt(e^6/(b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5))*(2*
(b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*d - (b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c
^4)*e))*sqrt((32*c^3*d^3 - 48*b*c^2*d^2*e + 6*(3*b^2*c + 4*a*c^2)*d*e^2 - (b^3 + 12*a*b*c)*e^3 - (b^6*c - 12*a
*b^4*c^2 + 48*a^2*b^2*c^3 - 64*a^3*c^4)*sqrt(e^6/(b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)))/(b^6
*c - 12*a*b^4*c^2 + 48*a^2*b^2*c^3 - 64*a^3*c^4)) + (16*c^2*d^2*e^3 - 16*b*c*d*e^4 + (3*b^2 + 4*a*c)*e^5)*sqrt
(e*x + d)) - 2*(b*d - 2*a*e + (2*c*d - b*e)*x)*sqrt(e*x + d))/(a*b^2 - 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 + (b^3
- 4*a*b*c)*x)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [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((e*x+d)^(3/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

Timed out