3.1618 \(\int \frac{(b+2 c x) (d+e x)^{5/2}}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=350 \[ -\frac{5 e \left (-2 c e \left (-d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (b-\sqrt{b^2-4 a c}\right )+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} c^{3/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{5 e \left (-2 c e \left (d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (\sqrt{b^2-4 a c}+b\right )+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} c^{3/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{(d+e x)^{5/2}}{a+b x+c x^2}+\frac{5 e^2 \sqrt{d+e x}}{c} \]

[Out]

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

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Rubi [A]  time = 1.28653, antiderivative size = 350, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {768, 703, 826, 1166, 208} \[ -\frac{5 e \left (-2 c e \left (-d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (b-\sqrt{b^2-4 a c}\right )+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} c^{3/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{5 e \left (-2 c e \left (d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (\sqrt{b^2-4 a c}+b\right )+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} c^{3/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{(d+e x)^{5/2}}{a+b x+c x^2}+\frac{5 e^2 \sqrt{d+e x}}{c} \]

Antiderivative was successfully verified.

[In]

Int[((b + 2*c*x)*(d + e*x)^(5/2))/(a + b*x + c*x^2)^2,x]

[Out]

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

Rule 768

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

Rule 703

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1))/(c*
(m - 1)), x] + Dist[1/c, Int[((d + e*x)^(m - 2)*Simp[c*d^2 - a*e^2 + e*(2*c*d - b*e)*x, x])/(a + b*x + c*x^2),
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] && GtQ[m, 1]

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{(b+2 c x) (d+e x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx &=-\frac{(d+e x)^{5/2}}{a+b x+c x^2}+\frac{1}{2} (5 e) \int \frac{(d+e x)^{3/2}}{a+b x+c x^2} \, dx\\ &=\frac{5 e^2 \sqrt{d+e x}}{c}-\frac{(d+e x)^{5/2}}{a+b x+c x^2}+\frac{(5 e) \int \frac{c d^2-a e^2+e (2 c d-b e) x}{\sqrt{d+e x} \left (a+b x+c x^2\right )} \, dx}{2 c}\\ &=\frac{5 e^2 \sqrt{d+e x}}{c}-\frac{(d+e x)^{5/2}}{a+b x+c x^2}+\frac{(5 e) \operatorname{Subst}\left (\int \frac{-d e (2 c d-b e)+e \left (c d^2-a e^2\right )+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 )}{c}\\ &=\frac{5 e^2 \sqrt{d+e x}}{c}-\frac{(d+e x)^{5/2}}{a+b x+c x^2}+\frac{\left (5 e \left (2 c^2 d^2+b \left (b-\sqrt{b^2-4 a c}\right ) e^2-2 c e \left (b d-\sqrt{b^2-4 a c} d+a e\right )\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 c \sqrt{b^2-4 a c}}-\frac{\left (5 e \left (2 c^2 d^2+b \left (b+\sqrt{b^2-4 a c}\right ) e^2-2 c e \left (b d+\sqrt{b^2-4 a c} d+a e\right )\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 c \sqrt{b^2-4 a c}}\\ &=\frac{5 e^2 \sqrt{d+e x}}{c}-\frac{(d+e x)^{5/2}}{a+b x+c x^2}-\frac{5 e \left (2 c^2 d^2+b \left (b-\sqrt{b^2-4 a c}\right ) e^2-2 c e \left (b d-\sqrt{b^2-4 a c} d+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} c^{3/2} \sqrt{b^2-4 a c} \sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}}+\frac{5 e \left (2 c^2 d^2+b \left (b+\sqrt{b^2-4 a c}\right ) e^2-2 c e \left (b d+\sqrt{b^2-4 a c} d+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} c^{3/2} \sqrt{b^2-4 a c} \sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}\\ \end{align*}

Mathematica [A]  time = 2.24728, size = 297, normalized size = 0.85 \[ -\frac{5 e \left (-4 \sqrt{c} e \sqrt{b^2-4 a c} \sqrt{d+e x}+\sqrt{2} \left (e \left (\sqrt{b^2-4 a c}-b\right )+2 c d\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )-\sqrt{2} \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )\right )}{4 c^{3/2} \sqrt{b^2-4 a c}}-\frac{e^2 (d+e x)^{5/2}}{e (a e-b d)+c d^2}+\frac{(d+e x)^{7/2} (b e-c d+c e x)}{(a+x (b+c x)) \left (e (a e-b d)+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[((b + 2*c*x)*(d + e*x)^(5/2))/(a + b*x + c*x^2)^2,x]

[Out]

-((e^2*(d + e*x)^(5/2))/(c*d^2 + e*(-(b*d) + a*e))) + ((d + e*x)^(7/2)*(-(c*d) + b*e + c*e*x))/((c*d^2 + e*(-(
b*d) + a*e))*(a + x*(b + c*x))) - (5*e*(-4*Sqrt[c]*Sqrt[b^2 - 4*a*c]*e*Sqrt[d + e*x] + Sqrt[2]*(2*c*d + (-b +
Sqrt[b^2 - 4*a*c])*e)^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e]] -
Sqrt[2]*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b + S
qrt[b^2 - 4*a*c])*e]]))/(4*c^(3/2)*Sqrt[b^2 - 4*a*c])

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Maple [B]  time = 0.042, size = 1333, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*x+b)*(e*x+d)^(5/2)/(c*x^2+b*x+a)^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, c x + b\right )}{\left (e x + d\right )}^{\frac{5}{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((2*c*x+b)*(e*x+d)^(5/2)/(c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

integrate((2*c*x + b)*(e*x + d)^(5/2)/(c*x^2 + b*x + a)^2, x)

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Fricas [B]  time = 2.13014, size = 5959, normalized size = 17.03 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^(5/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

-1/2*(5*sqrt(1/2)*(c^2*x^2 + b*c*x + a*c)*sqrt((2*c^3*d^3*e^2 - 3*b*c^2*d^2*e^3 + 3*(b^2*c - 2*a*c^2)*d*e^4 -
(b^3 - 3*a*b*c)*e^5 + (b^2*c^3 - 4*a*c^4)*sqrt((9*c^4*d^4*e^6 - 18*b*c^3*d^3*e^7 + 3*(5*b^2*c^2 - 2*a*c^3)*d^2
*e^8 - 6*(b^3*c - a*b*c^2)*d*e^9 + (b^4 - 2*a*b^2*c + a^2*c^2)*e^10)/(b^2*c^6 - 4*a*c^7)))/(b^2*c^3 - 4*a*c^4)
)*log(125*sqrt(1/2)*(3*(b^2*c^2 - 4*a*c^3)*d^2*e^4 - 3*(b^3*c - 4*a*b*c^2)*d*e^5 + (b^4 - 5*a*b^2*c + 4*a^2*c^
2)*e^6 - (2*(b^2*c^4 - 4*a*c^5)*d - (b^3*c^3 - 4*a*b*c^4)*e)*sqrt((9*c^4*d^4*e^6 - 18*b*c^3*d^3*e^7 + 3*(5*b^2
*c^2 - 2*a*c^3)*d^2*e^8 - 6*(b^3*c - a*b*c^2)*d*e^9 + (b^4 - 2*a*b^2*c + a^2*c^2)*e^10)/(b^2*c^6 - 4*a*c^7)))*
sqrt((2*c^3*d^3*e^2 - 3*b*c^2*d^2*e^3 + 3*(b^2*c - 2*a*c^2)*d*e^4 - (b^3 - 3*a*b*c)*e^5 + (b^2*c^3 - 4*a*c^4)*
sqrt((9*c^4*d^4*e^6 - 18*b*c^3*d^3*e^7 + 3*(5*b^2*c^2 - 2*a*c^3)*d^2*e^8 - 6*(b^3*c - a*b*c^2)*d*e^9 + (b^4 -
2*a*b^2*c + a^2*c^2)*e^10)/(b^2*c^6 - 4*a*c^7)))/(b^2*c^3 - 4*a*c^4)) - 250*(3*c^3*d^4*e^4 - 6*b*c^2*d^3*e^5 +
2*(2*b^2*c + a*c^2)*d^2*e^6 - (b^3 + 2*a*b*c)*d*e^7 + (a*b^2 - a^2*c)*e^8)*sqrt(e*x + d)) - 5*sqrt(1/2)*(c^2*
x^2 + b*c*x + a*c)*sqrt((2*c^3*d^3*e^2 - 3*b*c^2*d^2*e^3 + 3*(b^2*c - 2*a*c^2)*d*e^4 - (b^3 - 3*a*b*c)*e^5 + (
b^2*c^3 - 4*a*c^4)*sqrt((9*c^4*d^4*e^6 - 18*b*c^3*d^3*e^7 + 3*(5*b^2*c^2 - 2*a*c^3)*d^2*e^8 - 6*(b^3*c - a*b*c
^2)*d*e^9 + (b^4 - 2*a*b^2*c + a^2*c^2)*e^10)/(b^2*c^6 - 4*a*c^7)))/(b^2*c^3 - 4*a*c^4))*log(-125*sqrt(1/2)*(3
*(b^2*c^2 - 4*a*c^3)*d^2*e^4 - 3*(b^3*c - 4*a*b*c^2)*d*e^5 + (b^4 - 5*a*b^2*c + 4*a^2*c^2)*e^6 - (2*(b^2*c^4 -
4*a*c^5)*d - (b^3*c^3 - 4*a*b*c^4)*e)*sqrt((9*c^4*d^4*e^6 - 18*b*c^3*d^3*e^7 + 3*(5*b^2*c^2 - 2*a*c^3)*d^2*e^
8 - 6*(b^3*c - a*b*c^2)*d*e^9 + (b^4 - 2*a*b^2*c + a^2*c^2)*e^10)/(b^2*c^6 - 4*a*c^7)))*sqrt((2*c^3*d^3*e^2 -
3*b*c^2*d^2*e^3 + 3*(b^2*c - 2*a*c^2)*d*e^4 - (b^3 - 3*a*b*c)*e^5 + (b^2*c^3 - 4*a*c^4)*sqrt((9*c^4*d^4*e^6 -
18*b*c^3*d^3*e^7 + 3*(5*b^2*c^2 - 2*a*c^3)*d^2*e^8 - 6*(b^3*c - a*b*c^2)*d*e^9 + (b^4 - 2*a*b^2*c + a^2*c^2)*e
^10)/(b^2*c^6 - 4*a*c^7)))/(b^2*c^3 - 4*a*c^4)) - 250*(3*c^3*d^4*e^4 - 6*b*c^2*d^3*e^5 + 2*(2*b^2*c + a*c^2)*d
^2*e^6 - (b^3 + 2*a*b*c)*d*e^7 + (a*b^2 - a^2*c)*e^8)*sqrt(e*x + d)) + 5*sqrt(1/2)*(c^2*x^2 + b*c*x + a*c)*sqr
t((2*c^3*d^3*e^2 - 3*b*c^2*d^2*e^3 + 3*(b^2*c - 2*a*c^2)*d*e^4 - (b^3 - 3*a*b*c)*e^5 - (b^2*c^3 - 4*a*c^4)*sqr
t((9*c^4*d^4*e^6 - 18*b*c^3*d^3*e^7 + 3*(5*b^2*c^2 - 2*a*c^3)*d^2*e^8 - 6*(b^3*c - a*b*c^2)*d*e^9 + (b^4 - 2*a
*b^2*c + a^2*c^2)*e^10)/(b^2*c^6 - 4*a*c^7)))/(b^2*c^3 - 4*a*c^4))*log(125*sqrt(1/2)*(3*(b^2*c^2 - 4*a*c^3)*d^
2*e^4 - 3*(b^3*c - 4*a*b*c^2)*d*e^5 + (b^4 - 5*a*b^2*c + 4*a^2*c^2)*e^6 + (2*(b^2*c^4 - 4*a*c^5)*d - (b^3*c^3
- 4*a*b*c^4)*e)*sqrt((9*c^4*d^4*e^6 - 18*b*c^3*d^3*e^7 + 3*(5*b^2*c^2 - 2*a*c^3)*d^2*e^8 - 6*(b^3*c - a*b*c^2)
*d*e^9 + (b^4 - 2*a*b^2*c + a^2*c^2)*e^10)/(b^2*c^6 - 4*a*c^7)))*sqrt((2*c^3*d^3*e^2 - 3*b*c^2*d^2*e^3 + 3*(b^
2*c - 2*a*c^2)*d*e^4 - (b^3 - 3*a*b*c)*e^5 - (b^2*c^3 - 4*a*c^4)*sqrt((9*c^4*d^4*e^6 - 18*b*c^3*d^3*e^7 + 3*(5
*b^2*c^2 - 2*a*c^3)*d^2*e^8 - 6*(b^3*c - a*b*c^2)*d*e^9 + (b^4 - 2*a*b^2*c + a^2*c^2)*e^10)/(b^2*c^6 - 4*a*c^7
)))/(b^2*c^3 - 4*a*c^4)) - 250*(3*c^3*d^4*e^4 - 6*b*c^2*d^3*e^5 + 2*(2*b^2*c + a*c^2)*d^2*e^6 - (b^3 + 2*a*b*c
)*d*e^7 + (a*b^2 - a^2*c)*e^8)*sqrt(e*x + d)) - 5*sqrt(1/2)*(c^2*x^2 + b*c*x + a*c)*sqrt((2*c^3*d^3*e^2 - 3*b*
c^2*d^2*e^3 + 3*(b^2*c - 2*a*c^2)*d*e^4 - (b^3 - 3*a*b*c)*e^5 - (b^2*c^3 - 4*a*c^4)*sqrt((9*c^4*d^4*e^6 - 18*b
*c^3*d^3*e^7 + 3*(5*b^2*c^2 - 2*a*c^3)*d^2*e^8 - 6*(b^3*c - a*b*c^2)*d*e^9 + (b^4 - 2*a*b^2*c + a^2*c^2)*e^10)
/(b^2*c^6 - 4*a*c^7)))/(b^2*c^3 - 4*a*c^4))*log(-125*sqrt(1/2)*(3*(b^2*c^2 - 4*a*c^3)*d^2*e^4 - 3*(b^3*c - 4*a
*b*c^2)*d*e^5 + (b^4 - 5*a*b^2*c + 4*a^2*c^2)*e^6 + (2*(b^2*c^4 - 4*a*c^5)*d - (b^3*c^3 - 4*a*b*c^4)*e)*sqrt((
9*c^4*d^4*e^6 - 18*b*c^3*d^3*e^7 + 3*(5*b^2*c^2 - 2*a*c^3)*d^2*e^8 - 6*(b^3*c - a*b*c^2)*d*e^9 + (b^4 - 2*a*b^
2*c + a^2*c^2)*e^10)/(b^2*c^6 - 4*a*c^7)))*sqrt((2*c^3*d^3*e^2 - 3*b*c^2*d^2*e^3 + 3*(b^2*c - 2*a*c^2)*d*e^4 -
(b^3 - 3*a*b*c)*e^5 - (b^2*c^3 - 4*a*c^4)*sqrt((9*c^4*d^4*e^6 - 18*b*c^3*d^3*e^7 + 3*(5*b^2*c^2 - 2*a*c^3)*d^
2*e^8 - 6*(b^3*c - a*b*c^2)*d*e^9 + (b^4 - 2*a*b^2*c + a^2*c^2)*e^10)/(b^2*c^6 - 4*a*c^7)))/(b^2*c^3 - 4*a*c^4
)) - 250*(3*c^3*d^4*e^4 - 6*b*c^2*d^3*e^5 + 2*(2*b^2*c + a*c^2)*d^2*e^6 - (b^3 + 2*a*b*c)*d*e^7 + (a*b^2 - a^2
*c)*e^8)*sqrt(e*x + d)) - 2*(4*c*e^2*x^2 - c*d^2 + 5*a*e^2 - (2*c*d*e - 5*b*e^2)*x)*sqrt(e*x + d))/(c^2*x^2 +
b*c*x + a*c)

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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((2*c*x+b)*(e*x+d)**(5/2)/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

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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((2*c*x+b)*(e*x+d)^(5/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

Timed out