### 3.375 $$\int \frac{1}{(d+e x)^{3/2} (b x+c x^2)^2} \, dx$$

Optimal. Leaf size=206 $-\frac{e \left (3 b^2 e^2-2 b c d e+2 c^2 d^2\right )}{b^2 d^2 \sqrt{d+e x} (c d-b e)^2}-\frac{c^{5/2} (4 c d-7 b e) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 (c d-b e)^{5/2}}+\frac{(3 b e+4 c d) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3 d^{5/2}}-\frac{c x (2 c d-b e)+b (c d-b e)}{b^2 d \left (b x+c x^2\right ) \sqrt{d+e x} (c d-b e)}$

[Out]

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

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Rubi [A]  time = 0.352746, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.238, Rules used = {740, 828, 826, 1166, 208} $-\frac{e \left (3 b^2 e^2-2 b c d e+2 c^2 d^2\right )}{b^2 d^2 \sqrt{d+e x} (c d-b e)^2}-\frac{c^{5/2} (4 c d-7 b e) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 (c d-b e)^{5/2}}+\frac{(3 b e+4 c d) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3 d^{5/2}}-\frac{c x (2 c d-b e)+b (c d-b e)}{b^2 d \left (b x+c x^2\right ) \sqrt{d+e x} (c d-b e)}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((d + e*x)^(3/2)*(b*x + c*x^2)^2),x]

[Out]

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

Rule 740

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

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

Mathematica [C]  time = 0.107133, size = 166, normalized size = 0.81 $\frac{c^2 d^2 x (b+c x) (4 c d-7 b e) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c (d+e x)}{c d-b e}\right )-(b e-c d) \left (x (b+c x) \left (3 b^2 e^2+b c d e-4 c^2 d^2\right ) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{e x}{d}+1\right )+b d \left (b^2 e+b c (e x-d)-2 c^2 d x\right )\right )}{b^3 d^2 x (b+c x) \sqrt{d+e x} (c d-b e)^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((d + e*x)^(3/2)*(b*x + c*x^2)^2),x]

[Out]

(c^2*d^2*(4*c*d - 7*b*e)*x*(b + c*x)*Hypergeometric2F1[-1/2, 1, 1/2, (c*(d + e*x))/(c*d - b*e)] - (-(c*d) + b*
e)*(b*d*(b^2*e - 2*c^2*d*x + b*c*(-d + e*x)) + (-4*c^2*d^2 + b*c*d*e + 3*b^2*e^2)*x*(b + c*x)*Hypergeometric2F
1[-1/2, 1, 1/2, 1 + (e*x)/d]))/(b^3*d^2*(c*d - b*e)^2*x*(b + c*x)*Sqrt[d + e*x])

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Maple [A]  time = 0.243, size = 229, normalized size = 1.1 \begin{align*} -2\,{\frac{{e}^{3}}{{d}^{2} \left ( be-cd \right ) ^{2}\sqrt{ex+d}}}-{\frac{e{c}^{3}}{{b}^{2} \left ( be-cd \right ) ^{2} \left ( cex+be \right ) }\sqrt{ex+d}}-7\,{\frac{e{c}^{3}}{{b}^{2} \left ( be-cd \right ) ^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+4\,{\frac{{c}^{4}d}{{b}^{3} \left ( be-cd \right ) ^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-{\frac{1}{{b}^{2}{d}^{2}x}\sqrt{ex+d}}+3\,{\frac{e}{{b}^{2}{d}^{5/2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }+4\,{\frac{c}{{b}^{3}{d}^{3/2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(3/2)/(c*x^2+b*x)^2,x)

[Out]

-2*e^3/d^2/(b*e-c*d)^2/(e*x+d)^(1/2)-e*c^3/b^2/(b*e-c*d)^2*(e*x+d)^(1/2)/(c*e*x+b*e)-7*e*c^3/b^2/(b*e-c*d)^2/(
(b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))+4*c^4/b^3/(b*e-c*d)^2/((b*e-c*d)*c)^(1/2)*arcta
n((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*d-1/b^2/d^2*(e*x+d)^(1/2)/x+3*e/b^2/d^(5/2)*arctanh((e*x+d)^(1/2)/d^(1/
2))+4/b^3/d^(3/2)*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(1/(e*x+d)^(3/2)/(c*x^2+b*x)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 6.29635, size = 4633, normalized size = 22.49 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(3/2)/(c*x^2+b*x)^2,x, algorithm="fricas")

[Out]

[-1/2*(((4*c^4*d^4*e - 7*b*c^3*d^3*e^2)*x^3 + (4*c^4*d^5 - 3*b*c^3*d^4*e - 7*b^2*c^2*d^3*e^2)*x^2 + (4*b*c^3*d
^5 - 7*b^2*c^2*d^4*e)*x)*sqrt(c/(c*d - b*e))*log((c*e*x + 2*c*d - b*e + 2*(c*d - b*e)*sqrt(e*x + d)*sqrt(c/(c*
d - b*e)))/(c*x + b)) - ((4*c^4*d^3*e - 5*b*c^3*d^2*e^2 - 2*b^2*c^2*d*e^3 + 3*b^3*c*e^4)*x^3 + (4*c^4*d^4 - b*
c^3*d^3*e - 7*b^2*c^2*d^2*e^2 + b^3*c*d*e^3 + 3*b^4*e^4)*x^2 + (4*b*c^3*d^4 - 5*b^2*c^2*d^3*e - 2*b^3*c*d^2*e^
2 + 3*b^4*d*e^3)*x)*sqrt(d)*log((e*x + 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*(b^2*c^2*d^4 - 2*b^3*c*d^3*e + b^
4*d^2*e^2 + (2*b*c^3*d^3*e - 2*b^2*c^2*d^2*e^2 + 3*b^3*c*d*e^3)*x^2 + (2*b*c^3*d^4 - b^2*c^2*d^3*e - b^3*c*d^2
*e^2 + 3*b^4*d*e^3)*x)*sqrt(e*x + d))/((b^3*c^3*d^5*e - 2*b^4*c^2*d^4*e^2 + b^5*c*d^3*e^3)*x^3 + (b^3*c^3*d^6
- b^4*c^2*d^5*e - b^5*c*d^4*e^2 + b^6*d^3*e^3)*x^2 + (b^4*c^2*d^6 - 2*b^5*c*d^5*e + b^6*d^4*e^2)*x), -1/2*(2*(
(4*c^4*d^4*e - 7*b*c^3*d^3*e^2)*x^3 + (4*c^4*d^5 - 3*b*c^3*d^4*e - 7*b^2*c^2*d^3*e^2)*x^2 + (4*b*c^3*d^5 - 7*b
^2*c^2*d^4*e)*x)*sqrt(-c/(c*d - b*e))*arctan(-(c*d - b*e)*sqrt(e*x + d)*sqrt(-c/(c*d - b*e))/(c*e*x + c*d)) -
((4*c^4*d^3*e - 5*b*c^3*d^2*e^2 - 2*b^2*c^2*d*e^3 + 3*b^3*c*e^4)*x^3 + (4*c^4*d^4 - b*c^3*d^3*e - 7*b^2*c^2*d^
2*e^2 + b^3*c*d*e^3 + 3*b^4*e^4)*x^2 + (4*b*c^3*d^4 - 5*b^2*c^2*d^3*e - 2*b^3*c*d^2*e^2 + 3*b^4*d*e^3)*x)*sqrt
(d)*log((e*x + 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*(b^2*c^2*d^4 - 2*b^3*c*d^3*e + b^4*d^2*e^2 + (2*b*c^3*d^3
*e - 2*b^2*c^2*d^2*e^2 + 3*b^3*c*d*e^3)*x^2 + (2*b*c^3*d^4 - b^2*c^2*d^3*e - b^3*c*d^2*e^2 + 3*b^4*d*e^3)*x)*s
qrt(e*x + d))/((b^3*c^3*d^5*e - 2*b^4*c^2*d^4*e^2 + b^5*c*d^3*e^3)*x^3 + (b^3*c^3*d^6 - b^4*c^2*d^5*e - b^5*c*
d^4*e^2 + b^6*d^3*e^3)*x^2 + (b^4*c^2*d^6 - 2*b^5*c*d^5*e + b^6*d^4*e^2)*x), -1/2*(2*((4*c^4*d^3*e - 5*b*c^3*d
^2*e^2 - 2*b^2*c^2*d*e^3 + 3*b^3*c*e^4)*x^3 + (4*c^4*d^4 - b*c^3*d^3*e - 7*b^2*c^2*d^2*e^2 + b^3*c*d*e^3 + 3*b
^4*e^4)*x^2 + (4*b*c^3*d^4 - 5*b^2*c^2*d^3*e - 2*b^3*c*d^2*e^2 + 3*b^4*d*e^3)*x)*sqrt(-d)*arctan(sqrt(e*x + d)
*sqrt(-d)/d) + ((4*c^4*d^4*e - 7*b*c^3*d^3*e^2)*x^3 + (4*c^4*d^5 - 3*b*c^3*d^4*e - 7*b^2*c^2*d^3*e^2)*x^2 + (4
*b*c^3*d^5 - 7*b^2*c^2*d^4*e)*x)*sqrt(c/(c*d - b*e))*log((c*e*x + 2*c*d - b*e + 2*(c*d - b*e)*sqrt(e*x + d)*sq
rt(c/(c*d - b*e)))/(c*x + b)) + 2*(b^2*c^2*d^4 - 2*b^3*c*d^3*e + b^4*d^2*e^2 + (2*b*c^3*d^3*e - 2*b^2*c^2*d^2*
e^2 + 3*b^3*c*d*e^3)*x^2 + (2*b*c^3*d^4 - b^2*c^2*d^3*e - b^3*c*d^2*e^2 + 3*b^4*d*e^3)*x)*sqrt(e*x + d))/((b^3
*c^3*d^5*e - 2*b^4*c^2*d^4*e^2 + b^5*c*d^3*e^3)*x^3 + (b^3*c^3*d^6 - b^4*c^2*d^5*e - b^5*c*d^4*e^2 + b^6*d^3*e
^3)*x^2 + (b^4*c^2*d^6 - 2*b^5*c*d^5*e + b^6*d^4*e^2)*x), -(((4*c^4*d^4*e - 7*b*c^3*d^3*e^2)*x^3 + (4*c^4*d^5
- 3*b*c^3*d^4*e - 7*b^2*c^2*d^3*e^2)*x^2 + (4*b*c^3*d^5 - 7*b^2*c^2*d^4*e)*x)*sqrt(-c/(c*d - b*e))*arctan(-(c*
d - b*e)*sqrt(e*x + d)*sqrt(-c/(c*d - b*e))/(c*e*x + c*d)) + ((4*c^4*d^3*e - 5*b*c^3*d^2*e^2 - 2*b^2*c^2*d*e^3
+ 3*b^3*c*e^4)*x^3 + (4*c^4*d^4 - b*c^3*d^3*e - 7*b^2*c^2*d^2*e^2 + b^3*c*d*e^3 + 3*b^4*e^4)*x^2 + (4*b*c^3*d
^4 - 5*b^2*c^2*d^3*e - 2*b^3*c*d^2*e^2 + 3*b^4*d*e^3)*x)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) + (b^2*c^2*
d^4 - 2*b^3*c*d^3*e + b^4*d^2*e^2 + (2*b*c^3*d^3*e - 2*b^2*c^2*d^2*e^2 + 3*b^3*c*d*e^3)*x^2 + (2*b*c^3*d^4 - b
^2*c^2*d^3*e - b^3*c*d^2*e^2 + 3*b^4*d*e^3)*x)*sqrt(e*x + d))/((b^3*c^3*d^5*e - 2*b^4*c^2*d^4*e^2 + b^5*c*d^3*
e^3)*x^3 + (b^3*c^3*d^6 - b^4*c^2*d^5*e - b^5*c*d^4*e^2 + b^6*d^3*e^3)*x^2 + (b^4*c^2*d^6 - 2*b^5*c*d^5*e + b^
6*d^4*e^2)*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(1/(e*x+d)**(3/2)/(c*x**2+b*x)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.36135, size = 477, normalized size = 2.32 \begin{align*} \frac{{\left (4 \, c^{4} d - 7 \, b c^{3} e\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{{\left (b^{3} c^{2} d^{2} - 2 \, b^{4} c d e + b^{5} e^{2}\right )} \sqrt{-c^{2} d + b c e}} - \frac{2 \,{\left (x e + d\right )}^{2} c^{3} d^{2} e - 2 \,{\left (x e + d\right )} c^{3} d^{3} e - 2 \,{\left (x e + d\right )}^{2} b c^{2} d e^{2} + 3 \,{\left (x e + d\right )} b c^{2} d^{2} e^{2} + 3 \,{\left (x e + d\right )}^{2} b^{2} c e^{3} - 7 \,{\left (x e + d\right )} b^{2} c d e^{3} + 2 \, b^{2} c d^{2} e^{3} + 3 \,{\left (x e + d\right )} b^{3} e^{4} - 2 \, b^{3} d e^{4}}{{\left (b^{2} c^{2} d^{4} - 2 \, b^{3} c d^{3} e + b^{4} d^{2} e^{2}\right )}{\left ({\left (x e + d\right )}^{\frac{5}{2}} c - 2 \,{\left (x e + d\right )}^{\frac{3}{2}} c d + \sqrt{x e + d} c d^{2} +{\left (x e + d\right )}^{\frac{3}{2}} b e - \sqrt{x e + d} b d e\right )}} - \frac{{\left (4 \, c d + 3 \, b e\right )} \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b^{3} \sqrt{-d} d^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(3/2)/(c*x^2+b*x)^2,x, algorithm="giac")

[Out]

(4*c^4*d - 7*b*c^3*e)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/((b^3*c^2*d^2 - 2*b^4*c*d*e + b^5*e^2)*sqrt
(-c^2*d + b*c*e)) - (2*(x*e + d)^2*c^3*d^2*e - 2*(x*e + d)*c^3*d^3*e - 2*(x*e + d)^2*b*c^2*d*e^2 + 3*(x*e + d)
*b*c^2*d^2*e^2 + 3*(x*e + d)^2*b^2*c*e^3 - 7*(x*e + d)*b^2*c*d*e^3 + 2*b^2*c*d^2*e^3 + 3*(x*e + d)*b^3*e^4 - 2
*b^3*d*e^4)/((b^2*c^2*d^4 - 2*b^3*c*d^3*e + b^4*d^2*e^2)*((x*e + d)^(5/2)*c - 2*(x*e + d)^(3/2)*c*d + sqrt(x*e
+ d)*c*d^2 + (x*e + d)^(3/2)*b*e - sqrt(x*e + d)*b*d*e)) - (4*c*d + 3*b*e)*arctan(sqrt(x*e + d)/sqrt(-d))/(b^
3*sqrt(-d)*d^2)