### 3.382 $$\int \frac{\sqrt{d+e x}}{(b x+c x^2)^3} \, dx$$

Optimal. Leaf size=245 $\frac{\sqrt{d+e x} \left (c x \left (b^2 e^2-24 b c d e+24 c^2 d^2\right )+b (c d-b e) (12 c d-b e)\right )}{4 b^4 d \left (b x+c x^2\right ) (c d-b e)}+\frac{c^{3/2} \left (35 b^2 e^2-84 b c d e+48 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{4 b^5 (c d-b e)^{3/2}}-\frac{\left (-b^2 e^2-12 b c d e+48 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{4 b^5 d^{3/2}}-\frac{(b+2 c x) \sqrt{d+e x}}{2 b^2 \left (b x+c x^2\right )^2}$

[Out]

-((b + 2*c*x)*Sqrt[d + e*x])/(2*b^2*(b*x + c*x^2)^2) + (Sqrt[d + e*x]*(b*(c*d - b*e)*(12*c*d - b*e) + c*(24*c^
2*d^2 - 24*b*c*d*e + b^2*e^2)*x))/(4*b^4*d*(c*d - b*e)*(b*x + c*x^2)) - ((48*c^2*d^2 - 12*b*c*d*e - b^2*e^2)*A
rcTanh[Sqrt[d + e*x]/Sqrt[d]])/(4*b^5*d^(3/2)) + (c^(3/2)*(48*c^2*d^2 - 84*b*c*d*e + 35*b^2*e^2)*ArcTanh[(Sqrt
[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(4*b^5*(c*d - b*e)^(3/2))

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Rubi [A]  time = 0.377742, antiderivative size = 245, 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 = {736, 822, 826, 1166, 208} $\frac{\sqrt{d+e x} \left (c x \left (b^2 e^2-24 b c d e+24 c^2 d^2\right )+b (c d-b e) (12 c d-b e)\right )}{4 b^4 d \left (b x+c x^2\right ) (c d-b e)}+\frac{c^{3/2} \left (35 b^2 e^2-84 b c d e+48 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{4 b^5 (c d-b e)^{3/2}}-\frac{\left (-b^2 e^2-12 b c d e+48 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{4 b^5 d^{3/2}}-\frac{(b+2 c x) \sqrt{d+e x}}{2 b^2 \left (b x+c x^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[d + e*x]/(b*x + c*x^2)^3,x]

[Out]

-((b + 2*c*x)*Sqrt[d + e*x])/(2*b^2*(b*x + c*x^2)^2) + (Sqrt[d + e*x]*(b*(c*d - b*e)*(12*c*d - b*e) + c*(24*c^
2*d^2 - 24*b*c*d*e + b^2*e^2)*x))/(4*b^4*d*(c*d - b*e)*(b*x + c*x^2)) - ((48*c^2*d^2 - 12*b*c*d*e - b^2*e^2)*A
rcTanh[Sqrt[d + e*x]/Sqrt[d]])/(4*b^5*d^(3/2)) + (c^(3/2)*(48*c^2*d^2 - 84*b*c*d*e + 35*b^2*e^2)*ArcTanh[(Sqrt
[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(4*b^5*(c*d - b*e)^(3/2))

Rule 736

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*(b + 2*
c*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
- 1)*(b*e*m + 2*c*d*(2*p + 3) + 2*c*e*(m + 2*p + 3)*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
, 0] && (LtQ[m, 1] || (ILtQ[m + 2*p + 3, 0] && NeQ[m, 2])) && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*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*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], 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] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

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{\sqrt{d+e x}}{\left (b x+c x^2\right )^3} \, dx &=-\frac{(b+2 c x) \sqrt{d+e x}}{2 b^2 \left (b x+c x^2\right )^2}+\frac{\int \frac{-6 c d+\frac{b e}{2}-5 c e x}{\sqrt{d+e x} \left (b x+c x^2\right )^2} \, dx}{2 b^2}\\ &=-\frac{(b+2 c x) \sqrt{d+e x}}{2 b^2 \left (b x+c x^2\right )^2}+\frac{\sqrt{d+e x} \left (b (c d-b e) (12 c d-b e)+c \left (24 c^2 d^2-24 b c d e+b^2 e^2\right ) x\right )}{4 b^4 d (c d-b e) \left (b x+c x^2\right )}-\frac{\int \frac{-\frac{1}{4} (c d-b e) \left (48 c^2 d^2-12 b c d e-b^2 e^2\right )-\frac{1}{4} c e \left (24 c^2 d^2-24 b c d e+b^2 e^2\right ) x}{\sqrt{d+e x} \left (b x+c x^2\right )} \, dx}{2 b^4 d (c d-b e)}\\ &=-\frac{(b+2 c x) \sqrt{d+e x}}{2 b^2 \left (b x+c x^2\right )^2}+\frac{\sqrt{d+e x} \left (b (c d-b e) (12 c d-b e)+c \left (24 c^2 d^2-24 b c d e+b^2 e^2\right ) x\right )}{4 b^4 d (c d-b e) \left (b x+c x^2\right )}-\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{4} e (c d-b e) \left (48 c^2 d^2-12 b c d e-b^2 e^2\right )+\frac{1}{4} c d e \left (24 c^2 d^2-24 b c d e+b^2 e^2\right )-\frac{1}{4} c e \left (24 c^2 d^2-24 b c d e+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^4 d (c d-b e)}\\ &=-\frac{(b+2 c x) \sqrt{d+e x}}{2 b^2 \left (b x+c x^2\right )^2}+\frac{\sqrt{d+e x} \left (b (c d-b e) (12 c d-b e)+c \left (24 c^2 d^2-24 b c d e+b^2 e^2\right ) x\right )}{4 b^4 d (c d-b e) \left (b x+c x^2\right )}+\frac{\left (c \left (48 c^2 d^2-12 b c d e-b^2 e^2\right )\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 )}{4 b^5 d}-\frac{\left (c^2 \left (48 c^2 d^2-84 b c d e+35 b^2 e^2\right )\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 )}{4 b^5 (c d-b e)}\\ &=-\frac{(b+2 c x) \sqrt{d+e x}}{2 b^2 \left (b x+c x^2\right )^2}+\frac{\sqrt{d+e x} \left (b (c d-b e) (12 c d-b e)+c \left (24 c^2 d^2-24 b c d e+b^2 e^2\right ) x\right )}{4 b^4 d (c d-b e) \left (b x+c x^2\right )}-\frac{\left (48 c^2 d^2-12 b c d e-b^2 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{4 b^5 d^{3/2}}+\frac{c^{3/2} \left (48 c^2 d^2-84 b c d e+35 b^2 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{4 b^5 (c d-b e)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.975623, size = 361, normalized size = 1.47 $\frac{-\frac{2 c (d+e x)^{3/2} \left (-b^2 e^2-9 b c d e+12 c^2 d^2\right )}{b^2 d (b e-c d)}+\frac{(b+c x) \left (2 b c^{5/2} (d+e x)^{3/2} \left (10 b^2 c d e^2+b^3 e^3-36 b c^2 d^2 e+24 c^3 d^3\right )+(b+c x) \left (2 c^{3/2} (c d-b e)^2 \left (-b^2 e^2-12 b c d e+48 c^2 d^2\right ) \left (\sqrt{d+e x}-\sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )\right )-2 c^3 d^2 \left (35 b^2 e^2-84 b c d e+48 c^2 d^2\right ) \left (\sqrt{c} \sqrt{d+e x}-\sqrt{c d-b e} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )\right )\right )\right )}{b^4 c^{3/2} d (c d-b e)^2}+\frac{2 (d+e x)^{3/2} (b e+8 c d)}{b d x}-\frac{4 (d+e x)^{3/2}}{x^2}}{8 b d (b+c x)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((-2*c*(12*c^2*d^2 - 9*b*c*d*e - b^2*e^2)*(d + e*x)^(3/2))/(b^2*d*(-(c*d) + b*e)) - (4*(d + e*x)^(3/2))/x^2 +
(2*(8*c*d + b*e)*(d + e*x)^(3/2))/(b*d*x) + ((b + c*x)*(2*b*c^(5/2)*(24*c^3*d^3 - 36*b*c^2*d^2*e + 10*b^2*c*d*
e^2 + b^3*e^3)*(d + e*x)^(3/2) + (b + c*x)*(2*c^(3/2)*(c*d - b*e)^2*(48*c^2*d^2 - 12*b*c*d*e - b^2*e^2)*(Sqrt[
d + e*x] - Sqrt[d]*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]) - 2*c^3*d^2*(48*c^2*d^2 - 84*b*c*d*e + 35*b^2*e^2)*(Sqrt[c]
*Sqrt[d + e*x] - Sqrt[c*d - b*e]*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]]))))/(b^4*c^(3/2)*d*(c*d - b*
e)^2))/(8*b*d*(b + c*x)^2)

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Maple [B]  time = 0.234, size = 436, normalized size = 1.8 \begin{align*}{\frac{11\,{e}^{2}{c}^{3}}{4\,{b}^{3} \left ( cex+be \right ) ^{2} \left ( be-cd \right ) } \left ( ex+d \right ) ^{{\frac{3}{2}}}}-3\,{\frac{e{c}^{4} \left ( ex+d \right ) ^{3/2}d}{{b}^{4} \left ( cex+be \right ) ^{2} \left ( be-cd \right ) }}+{\frac{13\,{e}^{2}{c}^{2}}{4\,{b}^{3} \left ( cex+be \right ) ^{2}}\sqrt{ex+d}}-3\,{\frac{e{c}^{3}\sqrt{ex+d}d}{{b}^{4} \left ( cex+be \right ) ^{2}}}+{\frac{35\,{e}^{2}{c}^{2}}{4\,{b}^{3} \left ( be-cd \right ) }\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( be-cd \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( be-cd \right ) c}}}}-21\,{\frac{e{c}^{3}d}{{b}^{4} \left ( be-cd \right ) \sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+12\,{\frac{{c}^{4}{d}^{2}}{{b}^{5} \left ( be-cd \right ) \sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-{\frac{1}{4\,{b}^{3}{x}^{2}d} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+3\,{\frac{ \left ( ex+d \right ) ^{3/2}c}{e{b}^{4}{x}^{2}}}-3\,{\frac{\sqrt{ex+d}cd}{e{b}^{4}{x}^{2}}}-{\frac{1}{4\,{b}^{3}{x}^{2}}\sqrt{ex+d}}+{\frac{{e}^{2}}{4\,{b}^{3}}{\it Artanh} \left ({\sqrt{ex+d}{\frac{1}{\sqrt{d}}}} \right ){d}^{-{\frac{3}{2}}}}+3\,{\frac{ce}{{b}^{4}\sqrt{d}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }-12\,{\frac{\sqrt{d}{c}^{2}}{{b}^{5}}{\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)^(1/2)/(c*x^2+b*x)^3,x)

[Out]

11/4*e^2/b^3*c^3/(c*e*x+b*e)^2/(b*e-c*d)*(e*x+d)^(3/2)-3*e/b^4*c^4/(c*e*x+b*e)^2/(b*e-c*d)*(e*x+d)^(3/2)*d+13/
4*e^2/b^3*c^2/(c*e*x+b*e)^2*(e*x+d)^(1/2)-3*e/b^4*c^3/(c*e*x+b*e)^2*(e*x+d)^(1/2)*d+35/4*e^2/b^3*c^2/(b*e-c*d)
/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))-21*e/b^4*c^3/(b*e-c*d)/((b*e-c*d)*c)^(1/2)*ar
ctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*d+12/b^5*c^4/(b*e-c*d)/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((
b*e-c*d)*c)^(1/2))*d^2-1/4/b^3/x^2/d*(e*x+d)^(3/2)+3/e/b^4/x^2*(e*x+d)^(3/2)*c-3/e/b^4/x^2*(e*x+d)^(1/2)*c*d-1
/4/b^3/x^2*(e*x+d)^(1/2)+1/4*e^2/b^3/d^(3/2)*arctanh((e*x+d)^(1/2)/d^(1/2))+3*e/b^4/d^(1/2)*arctanh((e*x+d)^(1
/2)/d^(1/2))*c-12/b^5*d^(1/2)*arctanh((e*x+d)^(1/2)/d^(1/2))*c^2

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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)^(1/2)/(c*x^2+b*x)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.48442, size = 4645, normalized size = 18.96 \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)^(1/2)/(c*x^2+b*x)^3,x, algorithm="fricas")

[Out]

[-1/8*(((48*c^5*d^4 - 84*b*c^4*d^3*e + 35*b^2*c^3*d^2*e^2)*x^4 + 2*(48*b*c^4*d^4 - 84*b^2*c^3*d^3*e + 35*b^3*c
^2*d^2*e^2)*x^3 + (48*b^2*c^3*d^4 - 84*b^3*c^2*d^3*e + 35*b^4*c*d^2*e^2)*x^2)*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)) + ((48*c^5*d^3 - 60*b*c^4*d^2*e + 1
1*b^2*c^3*d*e^2 + b^3*c^2*e^3)*x^4 + 2*(48*b*c^4*d^3 - 60*b^2*c^3*d^2*e + 11*b^3*c^2*d*e^2 + b^4*c*e^3)*x^3 +
(48*b^2*c^3*d^3 - 60*b^3*c^2*d^2*e + 11*b^4*c*d*e^2 + b^5*e^3)*x^2)*sqrt(d)*log((e*x + 2*sqrt(e*x + d)*sqrt(d)
+ 2*d)/x) + 2*(2*b^4*c*d^3 - 2*b^5*d^2*e - (24*b*c^4*d^3 - 24*b^2*c^3*d^2*e + b^3*c^2*d*e^2)*x^3 - (36*b^2*c^
3*d^3 - 37*b^3*c^2*d^2*e + 2*b^4*c*d*e^2)*x^2 - (8*b^3*c^2*d^3 - 9*b^4*c*d^2*e + b^5*d*e^2)*x)*sqrt(e*x + d))/
((b^5*c^3*d^3 - b^6*c^2*d^2*e)*x^4 + 2*(b^6*c^2*d^3 - b^7*c*d^2*e)*x^3 + (b^7*c*d^3 - b^8*d^2*e)*x^2), 1/8*(2*
((48*c^5*d^4 - 84*b*c^4*d^3*e + 35*b^2*c^3*d^2*e^2)*x^4 + 2*(48*b*c^4*d^4 - 84*b^2*c^3*d^3*e + 35*b^3*c^2*d^2*
e^2)*x^3 + (48*b^2*c^3*d^4 - 84*b^3*c^2*d^3*e + 35*b^4*c*d^2*e^2)*x^2)*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)) - ((48*c^5*d^3 - 60*b*c^4*d^2*e + 11*b^2*c^3*d*e^2 + b^3*c
^2*e^3)*x^4 + 2*(48*b*c^4*d^3 - 60*b^2*c^3*d^2*e + 11*b^3*c^2*d*e^2 + b^4*c*e^3)*x^3 + (48*b^2*c^3*d^3 - 60*b^
3*c^2*d^2*e + 11*b^4*c*d*e^2 + b^5*e^3)*x^2)*sqrt(d)*log((e*x + 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 2*(2*b^4*c
*d^3 - 2*b^5*d^2*e - (24*b*c^4*d^3 - 24*b^2*c^3*d^2*e + b^3*c^2*d*e^2)*x^3 - (36*b^2*c^3*d^3 - 37*b^3*c^2*d^2*
e + 2*b^4*c*d*e^2)*x^2 - (8*b^3*c^2*d^3 - 9*b^4*c*d^2*e + b^5*d*e^2)*x)*sqrt(e*x + d))/((b^5*c^3*d^3 - b^6*c^2
*d^2*e)*x^4 + 2*(b^6*c^2*d^3 - b^7*c*d^2*e)*x^3 + (b^7*c*d^3 - b^8*d^2*e)*x^2), 1/8*(2*((48*c^5*d^3 - 60*b*c^4
*d^2*e + 11*b^2*c^3*d*e^2 + b^3*c^2*e^3)*x^4 + 2*(48*b*c^4*d^3 - 60*b^2*c^3*d^2*e + 11*b^3*c^2*d*e^2 + b^4*c*e
^3)*x^3 + (48*b^2*c^3*d^3 - 60*b^3*c^2*d^2*e + 11*b^4*c*d*e^2 + b^5*e^3)*x^2)*sqrt(-d)*arctan(sqrt(e*x + d)*sq
rt(-d)/d) - ((48*c^5*d^4 - 84*b*c^4*d^3*e + 35*b^2*c^3*d^2*e^2)*x^4 + 2*(48*b*c^4*d^4 - 84*b^2*c^3*d^3*e + 35*
b^3*c^2*d^2*e^2)*x^3 + (48*b^2*c^3*d^4 - 84*b^3*c^2*d^3*e + 35*b^4*c*d^2*e^2)*x^2)*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)) - 2*(2*b^4*c*d^3 - 2*b^5*d^2*e
- (24*b*c^4*d^3 - 24*b^2*c^3*d^2*e + b^3*c^2*d*e^2)*x^3 - (36*b^2*c^3*d^3 - 37*b^3*c^2*d^2*e + 2*b^4*c*d*e^2)
*x^2 - (8*b^3*c^2*d^3 - 9*b^4*c*d^2*e + b^5*d*e^2)*x)*sqrt(e*x + d))/((b^5*c^3*d^3 - b^6*c^2*d^2*e)*x^4 + 2*(b
^6*c^2*d^3 - b^7*c*d^2*e)*x^3 + (b^7*c*d^3 - b^8*d^2*e)*x^2), 1/4*(((48*c^5*d^4 - 84*b*c^4*d^3*e + 35*b^2*c^3*
d^2*e^2)*x^4 + 2*(48*b*c^4*d^4 - 84*b^2*c^3*d^3*e + 35*b^3*c^2*d^2*e^2)*x^3 + (48*b^2*c^3*d^4 - 84*b^3*c^2*d^3
*e + 35*b^4*c*d^2*e^2)*x^2)*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)) + ((48*c^5*d^3 - 60*b*c^4*d^2*e + 11*b^2*c^3*d*e^2 + b^3*c^2*e^3)*x^4 + 2*(48*b*c^4*d^3 - 60*b^2*c^3*
d^2*e + 11*b^3*c^2*d*e^2 + b^4*c*e^3)*x^3 + (48*b^2*c^3*d^3 - 60*b^3*c^2*d^2*e + 11*b^4*c*d*e^2 + b^5*e^3)*x^2
)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) - (2*b^4*c*d^3 - 2*b^5*d^2*e - (24*b*c^4*d^3 - 24*b^2*c^3*d^2*e +
b^3*c^2*d*e^2)*x^3 - (36*b^2*c^3*d^3 - 37*b^3*c^2*d^2*e + 2*b^4*c*d*e^2)*x^2 - (8*b^3*c^2*d^3 - 9*b^4*c*d^2*e
+ b^5*d*e^2)*x)*sqrt(e*x + d))/((b^5*c^3*d^3 - b^6*c^2*d^2*e)*x^4 + 2*(b^6*c^2*d^3 - b^7*c*d^2*e)*x^3 + (b^7*c
*d^3 - b^8*d^2*e)*x^2)]

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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)**(1/2)/(c*x**2+b*x)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.27475, size = 686, normalized size = 2.8 \begin{align*} -\frac{{\left (48 \, c^{4} d^{2} - 84 \, b c^{3} d e + 35 \, b^{2} c^{2} e^{2}\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{4 \,{\left (b^{5} c d - b^{6} e\right )} \sqrt{-c^{2} d + b c e}} + \frac{24 \,{\left (x e + d\right )}^{\frac{7}{2}} c^{4} d^{2} e - 72 \,{\left (x e + d\right )}^{\frac{5}{2}} c^{4} d^{3} e + 72 \,{\left (x e + d\right )}^{\frac{3}{2}} c^{4} d^{4} e - 24 \, \sqrt{x e + d} c^{4} d^{5} e - 24 \,{\left (x e + d\right )}^{\frac{7}{2}} b c^{3} d e^{2} + 108 \,{\left (x e + d\right )}^{\frac{5}{2}} b c^{3} d^{2} e^{2} - 144 \,{\left (x e + d\right )}^{\frac{3}{2}} b c^{3} d^{3} e^{2} + 60 \, \sqrt{x e + d} b c^{3} d^{4} e^{2} +{\left (x e + d\right )}^{\frac{7}{2}} b^{2} c^{2} e^{3} - 40 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{2} c^{2} d e^{3} + 85 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{2} c^{2} d^{2} e^{3} - 46 \, \sqrt{x e + d} b^{2} c^{2} d^{3} e^{3} + 2 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{3} c e^{4} - 13 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{3} c d e^{4} + 9 \, \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}}{4 \,{\left (b^{4} c d^{2} - b^{5} d e\right )}{\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 )}^{2}} + \frac{{\left (48 \, c^{2} d^{2} - 12 \, b c d e - b^{2} e^{2}\right )} \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{4 \, b^{5} \sqrt{-d} d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(48*c^4*d^2 - 84*b*c^3*d*e + 35*b^2*c^2*e^2)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/((b^5*c*d - b^6
*e)*sqrt(-c^2*d + b*c*e)) + 1/4*(24*(x*e + d)^(7/2)*c^4*d^2*e - 72*(x*e + d)^(5/2)*c^4*d^3*e + 72*(x*e + d)^(3
/2)*c^4*d^4*e - 24*sqrt(x*e + d)*c^4*d^5*e - 24*(x*e + d)^(7/2)*b*c^3*d*e^2 + 108*(x*e + d)^(5/2)*b*c^3*d^2*e^
2 - 144*(x*e + d)^(3/2)*b*c^3*d^3*e^2 + 60*sqrt(x*e + d)*b*c^3*d^4*e^2 + (x*e + d)^(7/2)*b^2*c^2*e^3 - 40*(x*e
+ d)^(5/2)*b^2*c^2*d*e^3 + 85*(x*e + d)^(3/2)*b^2*c^2*d^2*e^3 - 46*sqrt(x*e + d)*b^2*c^2*d^3*e^3 + 2*(x*e + d
)^(5/2)*b^3*c*e^4 - 13*(x*e + d)^(3/2)*b^3*c*d*e^4 + 9*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)/((b^4*c*d^2 - b^5*d*e)*((x*e + d)^2*c - 2*(x*e + d)*c*d + c*d^2 + (x*e + d)*b*e - b*
d*e)^2) + 1/4*(48*c^2*d^2 - 12*b*c*d*e - b^2*e^2)*arctan(sqrt(x*e + d)/sqrt(-d))/(b^5*sqrt(-d)*d)