### 3.381 $$\int \frac{(d+e x)^{3/2}}{(b x+c x^2)^3} \, dx$$

Optimal. Leaf size=246 $-\frac{3 \left (b^2 e^2-12 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{4 b^5 \sqrt{d}}+\frac{3 \sqrt{c} \left (5 b^2 e^2-20 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{4 b^5 \sqrt{c d-b e}}-\frac{\sqrt{d+e x} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}+\frac{\sqrt{d+e x} (12 c x (2 c d-b e) (c d-b e)+b (12 c d-7 b e) (c d-b e))}{4 b^4 \left (b x+c x^2\right ) (c d-b e)}$

[Out]

-(Sqrt[d + e*x]*(b*d + (2*c*d - b*e)*x))/(2*b^2*(b*x + c*x^2)^2) + (Sqrt[d + e*x]*(b*(12*c*d - 7*b*e)*(c*d - b
*e) + 12*c*(c*d - b*e)*(2*c*d - b*e)*x))/(4*b^4*(c*d - b*e)*(b*x + c*x^2)) - (3*(16*c^2*d^2 - 12*b*c*d*e + b^2
*e^2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(4*b^5*Sqrt[d]) + (3*Sqrt[c]*(16*c^2*d^2 - 20*b*c*d*e + 5*b^2*e^2)*ArcTa
nh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(4*b^5*Sqrt[c*d - b*e])

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Rubi [A]  time = 0.464637, antiderivative size = 246, 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 = {738, 822, 826, 1166, 208} $-\frac{3 \left (b^2 e^2-12 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{4 b^5 \sqrt{d}}+\frac{3 \sqrt{c} \left (5 b^2 e^2-20 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{4 b^5 \sqrt{c d-b e}}-\frac{\sqrt{d+e x} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}+\frac{\sqrt{d+e x} (12 c x (2 c d-b e) (c d-b e)+b (12 c d-7 b e) (c d-b e))}{4 b^4 \left (b x+c x^2\right ) (c d-b e)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(Sqrt[d + e*x]*(b*d + (2*c*d - b*e)*x))/(2*b^2*(b*x + c*x^2)^2) + (Sqrt[d + e*x]*(b*(12*c*d - 7*b*e)*(c*d - b
*e) + 12*c*(c*d - b*e)*(2*c*d - b*e)*x))/(4*b^4*(c*d - b*e)*(b*x + c*x^2)) - (3*(16*c^2*d^2 - 12*b*c*d*e + b^2
*e^2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(4*b^5*Sqrt[d]) + (3*Sqrt[c]*(16*c^2*d^2 - 20*b*c*d*e + 5*b^2*e^2)*ArcTa
nh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(4*b^5*Sqrt[c*d - b*e])

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 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{(d+e x)^{3/2}}{\left (b x+c x^2\right )^3} \, dx &=-\frac{\sqrt{d+e x} (b d+(2 c d-b e) x)}{2 b^2 \left (b x+c x^2\right )^2}-\frac{\int \frac{\frac{1}{2} d (12 c d-7 b e)+\frac{5}{2} e (2 c d-b e) x}{\sqrt{d+e x} \left (b x+c x^2\right )^2} \, dx}{2 b^2}\\ &=-\frac{\sqrt{d+e x} (b d+(2 c d-b e) x)}{2 b^2 \left (b x+c x^2\right )^2}+\frac{\sqrt{d+e x} (b (12 c d-7 b e) (c d-b e)+12 c (c d-b e) (2 c d-b e) x)}{4 b^4 (c d-b e) \left (b x+c x^2\right )}+\frac{\int \frac{\frac{3}{4} d (c d-b e) \left (16 c^2 d^2-12 b c d e+b^2 e^2\right )+3 c d e (c d-b e) (2 c d-b e) x}{\sqrt{d+e x} \left (b x+c x^2\right )} \, dx}{2 b^4 d (c d-b e)}\\ &=-\frac{\sqrt{d+e x} (b d+(2 c d-b e) x)}{2 b^2 \left (b x+c x^2\right )^2}+\frac{\sqrt{d+e x} (b (12 c d-7 b e) (c d-b e)+12 c (c d-b e) (2 c d-b e) x)}{4 b^4 (c d-b e) \left (b x+c x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{-3 c d^2 e (c d-b e) (2 c d-b e)+\frac{3}{4} d e (c d-b e) \left (16 c^2 d^2-12 b c d e+b^2 e^2\right )+3 c d e (c d-b e) (2 c d-b e) 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{\sqrt{d+e x} (b d+(2 c d-b e) x)}{2 b^2 \left (b x+c x^2\right )^2}+\frac{\sqrt{d+e x} (b (12 c d-7 b e) (c d-b e)+12 c (c d-b e) (2 c d-b e) x)}{4 b^4 (c d-b e) \left (b x+c x^2\right )}+\frac{\left (3 c \left (16 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}-\frac{\left (3 c \left (16 c^2 d^2-20 b c d e+5 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}\\ &=-\frac{\sqrt{d+e x} (b d+(2 c d-b e) x)}{2 b^2 \left (b x+c x^2\right )^2}+\frac{\sqrt{d+e x} (b (12 c d-7 b e) (c d-b e)+12 c (c d-b e) (2 c d-b e) x)}{4 b^4 (c d-b e) \left (b x+c x^2\right )}-\frac{3 \left (16 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 \sqrt{d}}+\frac{3 \sqrt{c} \left (16 c^2 d^2-20 b c d e+5 b^2 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{4 b^5 \sqrt{c d-b e}}\\ \end{align*}

Mathematica [A]  time = 0.53347, size = 289, normalized size = 1.17 $\frac{3 x^2 (b+c x)^2 \left (13 b^2 c d e^2-b^3 e^3-28 b c^2 d^2 e+16 c^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )-\sqrt{d} \left (b \sqrt{d+e x} \left (b^2 c^2 x \left (8 d^2-55 d e x+12 e^2 x^2\right )+b^3 c \left (-2 d^2-13 d e x+19 e^2 x^2\right )+b^4 e (2 d+5 e x)+36 b c^3 d x^2 (d-e x)+24 c^4 d^2 x^3\right )+3 \sqrt{c} x^2 (b+c x)^2 \sqrt{c d-b e} \left (5 b^2 e^2-20 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )\right )}{4 b^5 \sqrt{d} x^2 (b+c x)^2 (b e-c d)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(3*(16*c^3*d^3 - 28*b*c^2*d^2*e + 13*b^2*c*d*e^2 - b^3*e^3)*x^2*(b + c*x)^2*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] - S
qrt[d]*(b*Sqrt[d + e*x]*(24*c^4*d^2*x^3 + 36*b*c^3*d*x^2*(d - e*x) + b^4*e*(2*d + 5*e*x) + b^2*c^2*x*(8*d^2 -
55*d*e*x + 12*e^2*x^2) + b^3*c*(-2*d^2 - 13*d*e*x + 19*e^2*x^2)) + 3*Sqrt[c]*Sqrt[c*d - b*e]*(16*c^2*d^2 - 20*
b*c*d*e + 5*b^2*e^2)*x^2*(b + c*x)^2*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]]))/(4*b^5*Sqrt[d]*(-(c*d)
+ b*e)*x^2*(b + c*x)^2)

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Maple [A]  time = 0.232, size = 414, normalized size = 1.7 \begin{align*} -{\frac{7\,{e}^{2}{c}^{2}}{4\,{b}^{3} \left ( cex+be \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+3\,{\frac{e{c}^{3} \left ( ex+d \right ) ^{3/2}d}{{b}^{4} \left ( cex+be \right ) ^{2}}}-{\frac{9\,{e}^{3}c}{4\,{b}^{2} \left ( cex+be \right ) ^{2}}\sqrt{ex+d}}+{\frac{21\,{e}^{2}{c}^{2}d}{4\,{b}^{3} \left ( cex+be \right ) ^{2}}\sqrt{ex+d}}-3\,{\frac{e{c}^{3}\sqrt{ex+d}{d}^{2}}{{b}^{4} \left ( cex+be \right ) ^{2}}}-{\frac{15\,{e}^{2}c}{4\,{b}^{3}}\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( be-cd \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( be-cd \right ) c}}}}+15\,{\frac{{c}^{2}ed}{{b}^{4}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-12\,{\frac{{d}^{2}{c}^{3}}{{b}^{5}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-{\frac{5}{4\,{b}^{3}{x}^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+3\,{\frac{ \left ( ex+d \right ) ^{3/2}cd}{e{b}^{4}{x}^{2}}}-3\,{\frac{\sqrt{ex+d}c{d}^{2}}{e{b}^{4}{x}^{2}}}+{\frac{3\,d}{4\,{b}^{3}{x}^{2}}\sqrt{ex+d}}-{\frac{3\,{e}^{2}}{4\,{b}^{3}}{\it Artanh} \left ({\sqrt{ex+d}{\frac{1}{\sqrt{d}}}} \right ){\frac{1}{\sqrt{d}}}}+9\,{\frac{e\sqrt{d}c}{{b}^{4}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }-12\,{\frac{{d}^{3/2}{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)^(3/2)/(c*x^2+b*x)^3,x)

[Out]

-7/4*e^2*c^2/b^3/(c*e*x+b*e)^2*(e*x+d)^(3/2)+3*e*c^3/b^4/(c*e*x+b*e)^2*(e*x+d)^(3/2)*d-9/4*e^3*c/b^2/(c*e*x+b*
e)^2*(e*x+d)^(1/2)+21/4*e^2*c^2/b^3/(c*e*x+b*e)^2*(e*x+d)^(1/2)*d-3*e*c^3/b^4/(c*e*x+b*e)^2*(e*x+d)^(1/2)*d^2-
15/4*e^2*c/b^3/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))+15*e*c^2/b^4/((b*e-c*d)*c)^(1/2
)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*d-12*c^3/b^5/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*
d)*c)^(1/2))*d^2-5/4/b^3/x^2*(e*x+d)^(3/2)+3/e/b^4/x^2*(e*x+d)^(3/2)*c*d-3/e/b^4/x^2*(e*x+d)^(1/2)*c*d^2+3/4/b
^3/x^2*(e*x+d)^(1/2)*d-3/4*e^2/b^3/d^(1/2)*arctanh((e*x+d)^(1/2)/d^(1/2))+9*e/b^4*d^(1/2)*arctanh((e*x+d)^(1/2
)/d^(1/2))*c-12/b^5*d^(3/2)*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)^(3/2)/(c*x^2+b*x)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.01614, size = 3532, normalized size = 14.36 \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)^(3/2)/(c*x^2+b*x)^3,x, algorithm="fricas")

[Out]

[1/8*(3*((16*c^4*d^3 - 20*b*c^3*d^2*e + 5*b^2*c^2*d*e^2)*x^4 + 2*(16*b*c^3*d^3 - 20*b^2*c^2*d^2*e + 5*b^3*c*d*
e^2)*x^3 + (16*b^2*c^2*d^3 - 20*b^3*c*d^2*e + 5*b^4*d*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)) + 3*((16*c^4*d^2 - 12*b*c^3*d*e + b^2*c^2*e^2)*x^
4 + 2*(16*b*c^3*d^2 - 12*b^2*c^2*d*e + b^3*c*e^2)*x^3 + (16*b^2*c^2*d^2 - 12*b^3*c*d*e + b^4*e^2)*x^2)*sqrt(d)
*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 2*(2*b^4*d^2 - 12*(2*b*c^3*d^2 - b^2*c^2*d*e)*x^3 - (36*b^2*c^
2*d^2 - 19*b^3*c*d*e)*x^2 - (8*b^3*c*d^2 - 5*b^4*d*e)*x)*sqrt(e*x + d))/(b^5*c^2*d*x^4 + 2*b^6*c*d*x^3 + b^7*d
*x^2), 1/8*(6*((16*c^4*d^3 - 20*b*c^3*d^2*e + 5*b^2*c^2*d*e^2)*x^4 + 2*(16*b*c^3*d^3 - 20*b^2*c^2*d^2*e + 5*b^
3*c*d*e^2)*x^3 + (16*b^2*c^2*d^3 - 20*b^3*c*d^2*e + 5*b^4*d*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)) + 3*((16*c^4*d^2 - 12*b*c^3*d*e + b^2*c^2*e^2)*x^4 + 2*(16*
b*c^3*d^2 - 12*b^2*c^2*d*e + b^3*c*e^2)*x^3 + (16*b^2*c^2*d^2 - 12*b^3*c*d*e + b^4*e^2)*x^2)*sqrt(d)*log((e*x
- 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 2*(2*b^4*d^2 - 12*(2*b*c^3*d^2 - b^2*c^2*d*e)*x^3 - (36*b^2*c^2*d^2 - 19
*b^3*c*d*e)*x^2 - (8*b^3*c*d^2 - 5*b^4*d*e)*x)*sqrt(e*x + d))/(b^5*c^2*d*x^4 + 2*b^6*c*d*x^3 + b^7*d*x^2), 1/8
*(6*((16*c^4*d^2 - 12*b*c^3*d*e + b^2*c^2*e^2)*x^4 + 2*(16*b*c^3*d^2 - 12*b^2*c^2*d*e + b^3*c*e^2)*x^3 + (16*b
^2*c^2*d^2 - 12*b^3*c*d*e + b^4*e^2)*x^2)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) + 3*((16*c^4*d^3 - 20*b*c^
3*d^2*e + 5*b^2*c^2*d*e^2)*x^4 + 2*(16*b*c^3*d^3 - 20*b^2*c^2*d^2*e + 5*b^3*c*d*e^2)*x^3 + (16*b^2*c^2*d^3 - 2
0*b^3*c*d^2*e + 5*b^4*d*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)*s
qrt(c/(c*d - b*e)))/(c*x + b)) - 2*(2*b^4*d^2 - 12*(2*b*c^3*d^2 - b^2*c^2*d*e)*x^3 - (36*b^2*c^2*d^2 - 19*b^3*
c*d*e)*x^2 - (8*b^3*c*d^2 - 5*b^4*d*e)*x)*sqrt(e*x + d))/(b^5*c^2*d*x^4 + 2*b^6*c*d*x^3 + b^7*d*x^2), 1/4*(3*(
(16*c^4*d^3 - 20*b*c^3*d^2*e + 5*b^2*c^2*d*e^2)*x^4 + 2*(16*b*c^3*d^3 - 20*b^2*c^2*d^2*e + 5*b^3*c*d*e^2)*x^3
+ (16*b^2*c^2*d^3 - 20*b^3*c*d^2*e + 5*b^4*d*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)) + 3*((16*c^4*d^2 - 12*b*c^3*d*e + b^2*c^2*e^2)*x^4 + 2*(16*b*c^3*d^2 - 12*
b^2*c^2*d*e + b^3*c*e^2)*x^3 + (16*b^2*c^2*d^2 - 12*b^3*c*d*e + b^4*e^2)*x^2)*sqrt(-d)*arctan(sqrt(e*x + d)*sq
rt(-d)/d) - (2*b^4*d^2 - 12*(2*b*c^3*d^2 - b^2*c^2*d*e)*x^3 - (36*b^2*c^2*d^2 - 19*b^3*c*d*e)*x^2 - (8*b^3*c*d
^2 - 5*b^4*d*e)*x)*sqrt(e*x + d))/(b^5*c^2*d*x^4 + 2*b^6*c*d*x^3 + b^7*d*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)**(3/2)/(c*x**2+b*x)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.37057, size = 529, normalized size = 2.15 \begin{align*} -\frac{3 \,{\left (16 \, c^{3} d^{2} - 20 \, b c^{2} d e + 5 \, b^{2} c e^{2}\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{4 \, \sqrt{-c^{2} d + b c e} b^{5}} + \frac{3 \,{\left (16 \, 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}} + \frac{24 \,{\left (x e + d\right )}^{\frac{7}{2}} c^{3} d e - 72 \,{\left (x e + d\right )}^{\frac{5}{2}} c^{3} d^{2} e + 72 \,{\left (x e + d\right )}^{\frac{3}{2}} c^{3} d^{3} e - 24 \, \sqrt{x e + d} c^{3} d^{4} e - 12 \,{\left (x e + d\right )}^{\frac{7}{2}} b c^{2} e^{2} + 72 \,{\left (x e + d\right )}^{\frac{5}{2}} b c^{2} d e^{2} - 108 \,{\left (x e + d\right )}^{\frac{3}{2}} b c^{2} d^{2} e^{2} + 48 \, \sqrt{x e + d} b c^{2} d^{3} e^{2} - 19 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{2} c e^{3} + 46 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{2} c d e^{3} - 27 \, \sqrt{x e + d} b^{2} c d^{2} e^{3} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{3} e^{4} + 3 \, \sqrt{x e + d} b^{3} d e^{4}}{4 \,{\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} b^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*(16*c^3*d^2 - 20*b*c^2*d*e + 5*b^2*c*e^2)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/(sqrt(-c^2*d + b*c
*e)*b^5) + 3/4*(16*c^2*d^2 - 12*b*c*d*e + b^2*e^2)*arctan(sqrt(x*e + d)/sqrt(-d))/(b^5*sqrt(-d)) + 1/4*(24*(x*
e + d)^(7/2)*c^3*d*e - 72*(x*e + d)^(5/2)*c^3*d^2*e + 72*(x*e + d)^(3/2)*c^3*d^3*e - 24*sqrt(x*e + d)*c^3*d^4*
e - 12*(x*e + d)^(7/2)*b*c^2*e^2 + 72*(x*e + d)^(5/2)*b*c^2*d*e^2 - 108*(x*e + d)^(3/2)*b*c^2*d^2*e^2 + 48*sqr
t(x*e + d)*b*c^2*d^3*e^2 - 19*(x*e + d)^(5/2)*b^2*c*e^3 + 46*(x*e + d)^(3/2)*b^2*c*d*e^3 - 27*sqrt(x*e + d)*b^
2*c*d^2*e^3 - 5*(x*e + d)^(3/2)*b^3*e^4 + 3*sqrt(x*e + d)*b^3*d*e^4)/(((x*e + d)^2*c - 2*(x*e + d)*c*d + c*d^2
+ (x*e + d)*b*e - b*d*e)^2*b^4)