### 3.369 $$\int \frac{(d+e x)^{9/2}}{(b x+c x^2)^2} \, dx$$

Optimal. Leaf size=251 $\frac{e (d+e x)^{3/2} \left (5 b^2 e^2-6 b c d e+6 c^2 d^2\right )}{3 b^2 c^2}+\frac{e \sqrt{d+e x} (2 c d-b e) \left (5 b^2 e^2-b c d e+c^2 d^2\right )}{b^2 c^3}-\frac{(c d-b e)^{7/2} (5 b e+4 c d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 c^{7/2}}+\frac{d^{7/2} (4 c d-9 b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3}-\frac{(d+e x)^{7/2} (x (2 c d-b e)+b d)}{b^2 \left (b x+c x^2\right )}+\frac{e (d+e x)^{5/2} (2 c d-b e)}{b^2 c}$

[Out]

(e*(2*c*d - b*e)*(c^2*d^2 - b*c*d*e + 5*b^2*e^2)*Sqrt[d + e*x])/(b^2*c^3) + (e*(6*c^2*d^2 - 6*b*c*d*e + 5*b^2*
e^2)*(d + e*x)^(3/2))/(3*b^2*c^2) + (e*(2*c*d - b*e)*(d + e*x)^(5/2))/(b^2*c) - ((d + e*x)^(7/2)*(b*d + (2*c*d
- b*e)*x))/(b^2*(b*x + c*x^2)) + (d^(7/2)*(4*c*d - 9*b*e)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/b^3 - ((c*d - b*e)^
(7/2)*(4*c*d + 5*b*e)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b^3*c^(7/2))

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Rubi [A]  time = 0.534316, antiderivative size = 251, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.238, Rules used = {738, 824, 826, 1166, 208} $\frac{e (d+e x)^{3/2} \left (5 b^2 e^2-6 b c d e+6 c^2 d^2\right )}{3 b^2 c^2}+\frac{e \sqrt{d+e x} (2 c d-b e) \left (5 b^2 e^2-b c d e+c^2 d^2\right )}{b^2 c^3}-\frac{(c d-b e)^{7/2} (5 b e+4 c d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 c^{7/2}}+\frac{d^{7/2} (4 c d-9 b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3}-\frac{(d+e x)^{7/2} (x (2 c d-b e)+b d)}{b^2 \left (b x+c x^2\right )}+\frac{e (d+e x)^{5/2} (2 c d-b e)}{b^2 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^(9/2)/(b*x + c*x^2)^2,x]

[Out]

(e*(2*c*d - b*e)*(c^2*d^2 - b*c*d*e + 5*b^2*e^2)*Sqrt[d + e*x])/(b^2*c^3) + (e*(6*c^2*d^2 - 6*b*c*d*e + 5*b^2*
e^2)*(d + e*x)^(3/2))/(3*b^2*c^2) + (e*(2*c*d - b*e)*(d + e*x)^(5/2))/(b^2*c) - ((d + e*x)^(7/2)*(b*d + (2*c*d
- b*e)*x))/(b^2*(b*x + c*x^2)) + (d^(7/2)*(4*c*d - 9*b*e)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/b^3 - ((c*d - b*e)^
(7/2)*(4*c*d + 5*b*e)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b^3*c^(7/2))

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 824

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(g
*(d + e*x)^m)/(c*m), x] + Dist[1/c, Int[((d + e*x)^(m - 1)*Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x])
/(a + b*x + c*x^2), 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] && FractionQ[m] && GtQ[m, 0]

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)^{9/2}}{\left (b x+c x^2\right )^2} \, dx &=-\frac{(d+e x)^{7/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}-\frac{\int \frac{(d+e x)^{5/2} \left (\frac{1}{2} d (4 c d-9 b e)-\frac{5}{2} e (2 c d-b e) x\right )}{b x+c x^2} \, dx}{b^2}\\ &=\frac{e (2 c d-b e) (d+e x)^{5/2}}{b^2 c}-\frac{(d+e x)^{7/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}-\frac{\int \frac{(d+e x)^{3/2} \left (\frac{1}{2} c d^2 (4 c d-9 b e)-\frac{1}{2} e \left (6 c^2 d^2-6 b c d e+5 b^2 e^2\right ) x\right )}{b x+c x^2} \, dx}{b^2 c}\\ &=\frac{e \left (6 c^2 d^2-6 b c d e+5 b^2 e^2\right ) (d+e x)^{3/2}}{3 b^2 c^2}+\frac{e (2 c d-b e) (d+e x)^{5/2}}{b^2 c}-\frac{(d+e x)^{7/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}-\frac{\int \frac{\sqrt{d+e x} \left (\frac{1}{2} c^2 d^3 (4 c d-9 b e)-\frac{1}{2} e (2 c d-b e) \left (c^2 d^2-b c d e+5 b^2 e^2\right ) x\right )}{b x+c x^2} \, dx}{b^2 c^2}\\ &=\frac{e (2 c d-b e) \left (c^2 d^2-b c d e+5 b^2 e^2\right ) \sqrt{d+e x}}{b^2 c^3}+\frac{e \left (6 c^2 d^2-6 b c d e+5 b^2 e^2\right ) (d+e x)^{3/2}}{3 b^2 c^2}+\frac{e (2 c d-b e) (d+e x)^{5/2}}{b^2 c}-\frac{(d+e x)^{7/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}-\frac{\int \frac{\frac{1}{2} c^3 d^4 (4 c d-9 b e)+\frac{1}{2} e \left (2 c^4 d^4-4 b c^3 d^3 e-14 b^2 c^2 d^2 e^2+16 b^3 c d e^3-5 b^4 e^4\right ) x}{\sqrt{d+e x} \left (b x+c x^2\right )} \, dx}{b^2 c^3}\\ &=\frac{e (2 c d-b e) \left (c^2 d^2-b c d e+5 b^2 e^2\right ) \sqrt{d+e x}}{b^2 c^3}+\frac{e \left (6 c^2 d^2-6 b c d e+5 b^2 e^2\right ) (d+e x)^{3/2}}{3 b^2 c^2}+\frac{e (2 c d-b e) (d+e x)^{5/2}}{b^2 c}-\frac{(d+e x)^{7/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}-\frac{2 \operatorname{Subst}\left (\int \frac{\frac{1}{2} c^3 d^4 e (4 c d-9 b e)-\frac{1}{2} d e \left (2 c^4 d^4-4 b c^3 d^3 e-14 b^2 c^2 d^2 e^2+16 b^3 c d e^3-5 b^4 e^4\right )+\frac{1}{2} e \left (2 c^4 d^4-4 b c^3 d^3 e-14 b^2 c^2 d^2 e^2+16 b^3 c d e^3-5 b^4 e^4\right ) x^2}{c d^2-b d e+(-2 c d+b e) x^2+c x^4} \, dx,x,\sqrt{d+e x}\right )}{b^2 c^3}\\ &=\frac{e (2 c d-b e) \left (c^2 d^2-b c d e+5 b^2 e^2\right ) \sqrt{d+e x}}{b^2 c^3}+\frac{e \left (6 c^2 d^2-6 b c d e+5 b^2 e^2\right ) (d+e x)^{3/2}}{3 b^2 c^2}+\frac{e (2 c d-b e) (d+e x)^{5/2}}{b^2 c}-\frac{(d+e x)^{7/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}-\frac{\left (c d^4 (4 c d-9 b e)\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{b e}{2}+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{b^3}+\frac{\left ((c d-b e)^4 (4 c d+5 b e)\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{b e}{2}+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{b^3 c^3}\\ &=\frac{e (2 c d-b e) \left (c^2 d^2-b c d e+5 b^2 e^2\right ) \sqrt{d+e x}}{b^2 c^3}+\frac{e \left (6 c^2 d^2-6 b c d e+5 b^2 e^2\right ) (d+e x)^{3/2}}{3 b^2 c^2}+\frac{e (2 c d-b e) (d+e x)^{5/2}}{b^2 c}-\frac{(d+e x)^{7/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}+\frac{d^{7/2} (4 c d-9 b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3}-\frac{(c d-b e)^{7/2} (4 c d+5 b e) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.395775, size = 202, normalized size = 0.8 $\frac{\frac{b \sqrt{d+e x} \left (2 b^2 c^2 e^2 x \left (-9 d^2+13 d e x+e^2 x^2\right )+2 b^3 c e^3 x (19 d-5 e x)-15 b^4 e^4 x-3 b c^3 d^3 (d-4 e x)-6 c^4 d^4 x\right )}{c^3 x (b+c x)}-\frac{3 (c d-b e)^{7/2} (5 b e+4 c d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{c^{7/2}}+3 d^{7/2} (4 c d-9 b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{3 b^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^(9/2)/(b*x + c*x^2)^2,x]

[Out]

((b*Sqrt[d + e*x]*(-6*c^4*d^4*x - 15*b^4*e^4*x + 2*b^3*c*e^3*x*(19*d - 5*e*x) - 3*b*c^3*d^3*(d - 4*e*x) + 2*b^
2*c^2*e^2*x*(-9*d^2 + 13*d*e*x + e^2*x^2)))/(c^3*x*(b + c*x)) + 3*d^(7/2)*(4*c*d - 9*b*e)*ArcTanh[Sqrt[d + e*x
]/Sqrt[d]] - (3*(c*d - b*e)^(7/2)*(4*c*d + 5*b*e)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/c^(7/2))/(
3*b^3)

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Maple [B]  time = 0.28, size = 515, normalized size = 2.1 \begin{align*}{\frac{2\,{e}^{3}}{3\,{c}^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-4\,{\frac{{e}^{4}\sqrt{ex+d}b}{{c}^{3}}}+8\,{\frac{{e}^{3}d\sqrt{ex+d}}{{c}^{2}}}-{\frac{{e}^{5}{b}^{2}}{{c}^{3} \left ( cex+be \right ) }\sqrt{ex+d}}+4\,{\frac{{e}^{4}\sqrt{ex+d}bd}{{c}^{2} \left ( cex+be \right ) }}-6\,{\frac{{e}^{3}\sqrt{ex+d}{d}^{2}}{c \left ( cex+be \right ) }}+4\,{\frac{{e}^{2}\sqrt{ex+d}{d}^{3}}{b \left ( cex+be \right ) }}-{\frac{ce{d}^{4}}{{b}^{2} \left ( cex+be \right ) }\sqrt{ex+d}}+5\,{\frac{{e}^{5}{b}^{2}}{{c}^{3}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-16\,{\frac{{e}^{4}bd}{{c}^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+14\,{\frac{{e}^{3}{d}^{2}}{c\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+4\,{\frac{{e}^{2}{d}^{3}}{b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-11\,{\frac{ce{d}^{4}}{{b}^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+4\,{\frac{{c}^{2}{d}^{5}}{{b}^{3}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-{\frac{{d}^{4}}{{b}^{2}x}\sqrt{ex+d}}-9\,{\frac{e{d}^{7/2}}{{b}^{2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }+4\,{\frac{{d}^{9/2}c}{{b}^{3}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(9/2)/(c*x^2+b*x)^2,x)

[Out]

2/3*e^3/c^2*(e*x+d)^(3/2)-4*e^4/c^3*(e*x+d)^(1/2)*b+8*e^3/c^2*d*(e*x+d)^(1/2)-e^5/c^3*b^2*(e*x+d)^(1/2)/(c*e*x
+b*e)+4*e^4/c^2*b*(e*x+d)^(1/2)/(c*e*x+b*e)*d-6*e^3/c*(e*x+d)^(1/2)/(c*e*x+b*e)*d^2+4*e^2/b*(e*x+d)^(1/2)/(c*e
*x+b*e)*d^3-e*c/b^2*(e*x+d)^(1/2)/(c*e*x+b*e)*d^4+5*e^5/c^3*b^2/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b
*e-c*d)*c)^(1/2))-16*e^4/c^2*b/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*d+14*e^3/c/((b*
e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*d^2+4*e^2/b/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/
2)*c/((b*e-c*d)*c)^(1/2))*d^3-11*e*c/b^2/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*d^4+4
*c^2/b^3/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*d^5-d^4/b^2*(e*x+d)^(1/2)/x-9*e*d^(7/
2)/b^2*arctanh((e*x+d)^(1/2)/d^(1/2))+4*d^(9/2)/b^3*arctanh((e*x+d)^(1/2)/d^(1/2))*c

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(9/2)/(c*x^2+b*x)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 22.8602, size = 3251, normalized size = 12.95 \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)^(9/2)/(c*x^2+b*x)^2,x, algorithm="fricas")

[Out]

[-1/6*(3*((4*c^5*d^4 - 7*b*c^4*d^3*e - 3*b^2*c^3*d^2*e^2 + 11*b^3*c^2*d*e^3 - 5*b^4*c*e^4)*x^2 + (4*b*c^4*d^4
- 7*b^2*c^3*d^3*e - 3*b^3*c^2*d^2*e^2 + 11*b^4*c*d*e^3 - 5*b^5*e^4)*x)*sqrt((c*d - b*e)/c)*log((c*e*x + 2*c*d
- b*e + 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)) + 3*((4*c^5*d^4 - 9*b*c^4*d^3*e)*x^2 + (4*b*c^4*d^4
- 9*b^2*c^3*d^3*e)*x)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 2*(2*b^3*c^2*e^4*x^3 - 3*b^2*c^3*
d^4 + 2*(13*b^3*c^2*d*e^3 - 5*b^4*c*e^4)*x^2 - (6*b*c^4*d^4 - 12*b^2*c^3*d^3*e + 18*b^3*c^2*d^2*e^2 - 38*b^4*c
*d*e^3 + 15*b^5*e^4)*x)*sqrt(e*x + d))/(b^3*c^4*x^2 + b^4*c^3*x), -1/6*(6*((4*c^5*d^4 - 7*b*c^4*d^3*e - 3*b^2*
c^3*d^2*e^2 + 11*b^3*c^2*d*e^3 - 5*b^4*c*e^4)*x^2 + (4*b*c^4*d^4 - 7*b^2*c^3*d^3*e - 3*b^3*c^2*d^2*e^2 + 11*b^
4*c*d*e^3 - 5*b^5*e^4)*x)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)) + 3*(
(4*c^5*d^4 - 9*b*c^4*d^3*e)*x^2 + (4*b*c^4*d^4 - 9*b^2*c^3*d^3*e)*x)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d
) + 2*d)/x) - 2*(2*b^3*c^2*e^4*x^3 - 3*b^2*c^3*d^4 + 2*(13*b^3*c^2*d*e^3 - 5*b^4*c*e^4)*x^2 - (6*b*c^4*d^4 - 1
2*b^2*c^3*d^3*e + 18*b^3*c^2*d^2*e^2 - 38*b^4*c*d*e^3 + 15*b^5*e^4)*x)*sqrt(e*x + d))/(b^3*c^4*x^2 + b^4*c^3*x
), -1/6*(6*((4*c^5*d^4 - 9*b*c^4*d^3*e)*x^2 + (4*b*c^4*d^4 - 9*b^2*c^3*d^3*e)*x)*sqrt(-d)*arctan(sqrt(e*x + d)
*sqrt(-d)/d) + 3*((4*c^5*d^4 - 7*b*c^4*d^3*e - 3*b^2*c^3*d^2*e^2 + 11*b^3*c^2*d*e^3 - 5*b^4*c*e^4)*x^2 + (4*b*
c^4*d^4 - 7*b^2*c^3*d^3*e - 3*b^3*c^2*d^2*e^2 + 11*b^4*c*d*e^3 - 5*b^5*e^4)*x)*sqrt((c*d - b*e)/c)*log((c*e*x
+ 2*c*d - b*e + 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)) - 2*(2*b^3*c^2*e^4*x^3 - 3*b^2*c^3*d^4 + 2*(
13*b^3*c^2*d*e^3 - 5*b^4*c*e^4)*x^2 - (6*b*c^4*d^4 - 12*b^2*c^3*d^3*e + 18*b^3*c^2*d^2*e^2 - 38*b^4*c*d*e^3 +
15*b^5*e^4)*x)*sqrt(e*x + d))/(b^3*c^4*x^2 + b^4*c^3*x), -1/3*(3*((4*c^5*d^4 - 7*b*c^4*d^3*e - 3*b^2*c^3*d^2*e
^2 + 11*b^3*c^2*d*e^3 - 5*b^4*c*e^4)*x^2 + (4*b*c^4*d^4 - 7*b^2*c^3*d^3*e - 3*b^3*c^2*d^2*e^2 + 11*b^4*c*d*e^3
- 5*b^5*e^4)*x)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)) + 3*((4*c^5*d^
4 - 9*b*c^4*d^3*e)*x^2 + (4*b*c^4*d^4 - 9*b^2*c^3*d^3*e)*x)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) - (2*b^3
*c^2*e^4*x^3 - 3*b^2*c^3*d^4 + 2*(13*b^3*c^2*d*e^3 - 5*b^4*c*e^4)*x^2 - (6*b*c^4*d^4 - 12*b^2*c^3*d^3*e + 18*b
^3*c^2*d^2*e^2 - 38*b^4*c*d*e^3 + 15*b^5*e^4)*x)*sqrt(e*x + d))/(b^3*c^4*x^2 + b^4*c^3*x)]

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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((e*x+d)**(9/2)/(c*x**2+b*x)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.43982, size = 589, normalized size = 2.35 \begin{align*} -\frac{{\left (4 \, c d^{5} - 9 \, b d^{4} e\right )} \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b^{3} \sqrt{-d}} + \frac{{\left (4 \, c^{5} d^{5} - 11 \, b c^{4} d^{4} e + 4 \, b^{2} c^{3} d^{3} e^{2} + 14 \, b^{3} c^{2} d^{2} e^{3} - 16 \, b^{4} c d e^{4} + 5 \, b^{5} e^{5}\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{\sqrt{-c^{2} d + b c e} b^{3} c^{3}} + \frac{2 \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} c^{4} e^{3} + 12 \, \sqrt{x e + d} c^{4} d e^{3} - 6 \, \sqrt{x e + d} b c^{3} e^{4}\right )}}{3 \, c^{6}} - \frac{2 \,{\left (x e + d\right )}^{\frac{3}{2}} c^{4} d^{4} e - 2 \, \sqrt{x e + d} c^{4} d^{5} e - 4 \,{\left (x e + d\right )}^{\frac{3}{2}} b c^{3} d^{3} e^{2} + 5 \, \sqrt{x e + d} b c^{3} d^{4} e^{2} + 6 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{2} c^{2} d^{2} e^{3} - 6 \, \sqrt{x e + d} b^{2} c^{2} d^{3} e^{3} - 4 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{3} c d e^{4} + 4 \, \sqrt{x e + d} b^{3} c d^{2} e^{4} +{\left (x e + d\right )}^{\frac{3}{2}} b^{4} e^{5} - \sqrt{x e + d} b^{4} d e^{5}}{{\left ({\left (x e + d\right )}^{2} c - 2 \,{\left (x e + d\right )} c d + c d^{2} +{\left (x e + d\right )} b e - b d e\right )} b^{2} c^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(9/2)/(c*x^2+b*x)^2,x, algorithm="giac")

[Out]

-(4*c*d^5 - 9*b*d^4*e)*arctan(sqrt(x*e + d)/sqrt(-d))/(b^3*sqrt(-d)) + (4*c^5*d^5 - 11*b*c^4*d^4*e + 4*b^2*c^3
*d^3*e^2 + 14*b^3*c^2*d^2*e^3 - 16*b^4*c*d*e^4 + 5*b^5*e^5)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/(sqrt
(-c^2*d + b*c*e)*b^3*c^3) + 2/3*((x*e + d)^(3/2)*c^4*e^3 + 12*sqrt(x*e + d)*c^4*d*e^3 - 6*sqrt(x*e + d)*b*c^3*
e^4)/c^6 - (2*(x*e + d)^(3/2)*c^4*d^4*e - 2*sqrt(x*e + d)*c^4*d^5*e - 4*(x*e + d)^(3/2)*b*c^3*d^3*e^2 + 5*sqrt
(x*e + d)*b*c^3*d^4*e^2 + 6*(x*e + d)^(3/2)*b^2*c^2*d^2*e^3 - 6*sqrt(x*e + d)*b^2*c^2*d^3*e^3 - 4*(x*e + d)^(3
/2)*b^3*c*d*e^4 + 4*sqrt(x*e + d)*b^3*c*d^2*e^4 + (x*e + d)^(3/2)*b^4*e^5 - sqrt(x*e + d)*b^4*d*e^5)/(((x*e +
d)^2*c - 2*(x*e + d)*c*d + c*d^2 + (x*e + d)*b*e - b*d*e)*b^2*c^3)