### 3.2294 $$\int \frac{1}{(d+e x)^{5/2} (a+b x+c x^2)} \, dx$$

Optimal. Leaf size=414 $-\frac{\sqrt{2} \sqrt{c} \left (-2 c e \left (d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (\sqrt{b^2-4 a c}+b\right )+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )^2}+\frac{\sqrt{2} \sqrt{c} \left (-2 c e \left (-d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (b-\sqrt{b^2-4 a c}\right )+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )^2}-\frac{2 e (2 c d-b e)}{\sqrt{d+e x} \left (a e^2-b d e+c d^2\right )^2}-\frac{2 e}{3 (d+e x)^{3/2} \left (a e^2-b d e+c d^2\right )}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.54601, antiderivative size = 414, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.227, Rules used = {709, 828, 826, 1166, 208} $-\frac{\sqrt{2} \sqrt{c} \left (-2 c e \left (d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (\sqrt{b^2-4 a c}+b\right )+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )^2}+\frac{\sqrt{2} \sqrt{c} \left (-2 c e \left (-d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (b-\sqrt{b^2-4 a c}\right )+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )^2}-\frac{2 e (2 c d-b e)}{\sqrt{d+e x} \left (a e^2-b d e+c d^2\right )^2}-\frac{2 e}{3 (d+e x)^{3/2} \left (a e^2-b d e+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((d + e*x)^(5/2)*(a + b*x + c*x^2)),x]

[Out]

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

Rule 709

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

Rule 828

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[((
e*f - d*g)*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[((d
+ e*x)^(m + 1)*Simp[c*d*f - f*b*e + a*e*g - c*(e*f - d*g)*x, x])/(a + b*x + c*x^2), 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] && FractionQ[m] && LtQ[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{1}{(d+e x)^{5/2} \left (a+b x+c x^2\right )} \, dx &=-\frac{2 e}{3 \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}+\frac{\int \frac{c d-b e-c e x}{(d+e x)^{3/2} \left (a+b x+c x^2\right )} \, dx}{c d^2-b d e+a e^2}\\ &=-\frac{2 e}{3 \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}-\frac{2 e (2 c d-b e)}{\left (c d^2-b d e+a e^2\right )^2 \sqrt{d+e x}}+\frac{\int \frac{c^2 d^2+b^2 e^2-c e (2 b d+a e)-c e (2 c d-b e) x}{\sqrt{d+e x} \left (a+b x+c x^2\right )} \, dx}{\left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac{2 e}{3 \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}-\frac{2 e (2 c d-b e)}{\left (c d^2-b d e+a e^2\right )^2 \sqrt{d+e x}}+\frac{2 \operatorname{Subst}\left (\int \frac{c d e (2 c d-b e)+e \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)\right )-c 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 )}{\left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac{2 e}{3 \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}-\frac{2 e (2 c d-b e)}{\left (c d^2-b d e+a e^2\right )^2 \sqrt{d+e x}}-\frac{\left (c \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 )}{\sqrt{b^2-4 a c} \left (c d^2-b d e+a e^2\right )^2}+\frac{\left (c \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 )}{\sqrt{b^2-4 a c} \left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac{2 e}{3 \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}-\frac{2 e (2 c d-b e)}{\left (c d^2-b d e+a e^2\right )^2 \sqrt{d+e x}}-\frac{\sqrt{2} \sqrt{c} \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{b^2-4 a c} \sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e} \left (c d^2-b d e+a e^2\right )^2}+\frac{\sqrt{2} \sqrt{c} \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{b^2-4 a c} \sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e} \left (c d^2-b d e+a e^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 1.33603, size = 377, normalized size = 0.91 $\frac{2 \left (-\frac{3 \sqrt{c} \left (\frac{\left (-2 c e \left (-d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (b-\sqrt{b^2-4 a c}\right )+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{\left (-2 c e \left (d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (\sqrt{b^2-4 a c}+b\right )+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )}{\sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}\right )}{\sqrt{2} \sqrt{b^2-4 a c} \left (e (b d-a e)-c d^2\right )}+\frac{3 e (b e-2 c d)}{\sqrt{d+e x} \left (e (a e-b d)+c d^2\right )}-\frac{e}{(d+e x)^{3/2}}\right )}{3 \left (e (a e-b d)+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((d + e*x)^(5/2)*(a + b*x + c*x^2)),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.256, size = 1444, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(5/2)/(c*x^2+b*x+a),x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + b x + a\right )}{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(5/2)/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

integrate(1/((c*x^2 + b*x + a)*(e*x + d)^(5/2)), x)

________________________________________________________________________________________

Fricas [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(1/(e*x+d)^(5/2)/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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(1/(e*x+d)**(5/2)/(c*x**2+b*x+a),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(1/(e*x+d)^(5/2)/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

Timed out