### 3.380 $$\int \frac{(d+e x)^{5/2}}{(b x+c x^2)^3} \, dx$$

Optimal. Leaf size=232 $\frac{3 \sqrt{d+e x} \left (x \left (b^2 e^2-8 b c d e+8 c^2 d^2\right )+b d (4 c d-3 b e)\right )}{4 b^4 \left (b x+c x^2\right )}-\frac{3 \sqrt{d} \left (5 b^2 e^2-20 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{4 b^5}+\frac{3 \sqrt{c d-b e} \left (b^2 e^2-12 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{4 b^5 \sqrt{c}}-\frac{(d+e x)^{3/2} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}$

[Out]

-((d + e*x)^(3/2)*(b*d + (2*c*d - b*e)*x))/(2*b^2*(b*x + c*x^2)^2) + (3*Sqrt[d + e*x]*(b*d*(4*c*d - 3*b*e) + (
8*c^2*d^2 - 8*b*c*d*e + b^2*e^2)*x))/(4*b^4*(b*x + c*x^2)) - (3*Sqrt[d]*(16*c^2*d^2 - 20*b*c*d*e + 5*b^2*e^2)*
ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(4*b^5) + (3*Sqrt[c*d - b*e]*(16*c^2*d^2 - 12*b*c*d*e + b^2*e^2)*ArcTanh[(Sqrt
[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(4*b^5*Sqrt[c])

________________________________________________________________________________________

Rubi [A]  time = 0.307329, antiderivative size = 232, 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, 820, 826, 1166, 208} $\frac{3 \sqrt{d+e x} \left (x \left (b^2 e^2-8 b c d e+8 c^2 d^2\right )+b d (4 c d-3 b e)\right )}{4 b^4 \left (b x+c x^2\right )}-\frac{3 \sqrt{d} \left (5 b^2 e^2-20 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{4 b^5}+\frac{3 \sqrt{c d-b e} \left (b^2 e^2-12 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{4 b^5 \sqrt{c}}-\frac{(d+e x)^{3/2} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^(5/2)/(b*x + c*x^2)^3,x]

[Out]

-((d + e*x)^(3/2)*(b*d + (2*c*d - b*e)*x))/(2*b^2*(b*x + c*x^2)^2) + (3*Sqrt[d + e*x]*(b*d*(4*c*d - 3*b*e) + (
8*c^2*d^2 - 8*b*c*d*e + b^2*e^2)*x))/(4*b^4*(b*x + c*x^2)) - (3*Sqrt[d]*(16*c^2*d^2 - 20*b*c*d*e + 5*b^2*e^2)*
ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(4*b^5) + (3*Sqrt[c*d - b*e]*(16*c^2*d^2 - 12*b*c*d*e + b^2*e^2)*ArcTanh[(Sqrt
[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(4*b^5*Sqrt[c])

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 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*(f*b - 2*a*g + (2*c*f - b*g)*x))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/
((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*Simp[g*(2*a*e*m + b*d*(2*p + 3)) - f*
(b*e*m + 2*c*d*(2*p + 3)) - e*(2*c*f - b*g)*(m + 2*p + 3)*x, x], 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] && LtQ[p, -1] && GtQ[m, 0] && (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)^{5/2}}{\left (b x+c x^2\right )^3} \, dx &=-\frac{(d+e x)^{3/2} (b d+(2 c d-b e) x)}{2 b^2 \left (b x+c x^2\right )^2}-\frac{\int \frac{\sqrt{d+e x} \left (\frac{3}{2} d (4 c d-3 b e)+\frac{3}{2} e (2 c d-b e) x\right )}{\left (b x+c x^2\right )^2} \, dx}{2 b^2}\\ &=-\frac{(d+e x)^{3/2} (b d+(2 c d-b e) x)}{2 b^2 \left (b x+c x^2\right )^2}+\frac{3 \sqrt{d+e x} \left (b d (4 c d-3 b e)+\left (8 c^2 d^2-8 b c d e+b^2 e^2\right ) x\right )}{4 b^4 \left (b x+c x^2\right )}+\frac{\int \frac{\frac{3}{4} d \left (16 c^2 d^2-20 b c d e+5 b^2 e^2\right )+\frac{3}{4} e \left (8 c^2 d^2-8 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}\\ &=-\frac{(d+e x)^{3/2} (b d+(2 c d-b e) x)}{2 b^2 \left (b x+c x^2\right )^2}+\frac{3 \sqrt{d+e x} \left (b d (4 c d-3 b e)+\left (8 c^2 d^2-8 b c d e+b^2 e^2\right ) x\right )}{4 b^4 \left (b x+c x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{3}{4} d e \left (8 c^2 d^2-8 b c d e+b^2 e^2\right )+\frac{3}{4} d e \left (16 c^2 d^2-20 b c d e+5 b^2 e^2\right )+\frac{3}{4} e \left (8 c^2 d^2-8 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}\\ &=-\frac{(d+e x)^{3/2} (b d+(2 c d-b e) x)}{2 b^2 \left (b x+c x^2\right )^2}+\frac{3 \sqrt{d+e x} \left (b d (4 c d-3 b e)+\left (8 c^2 d^2-8 b c d e+b^2 e^2\right ) x\right )}{4 b^4 \left (b x+c x^2\right )}-\frac{\left (3 (c d-b e) \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 d \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{(d+e x)^{3/2} (b d+(2 c d-b e) x)}{2 b^2 \left (b x+c x^2\right )^2}+\frac{3 \sqrt{d+e x} \left (b d (4 c d-3 b e)+\left (8 c^2 d^2-8 b c d e+b^2 e^2\right ) x\right )}{4 b^4 \left (b x+c x^2\right )}-\frac{3 \sqrt{d} \left (16 c^2 d^2-20 b c d e+5 b^2 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{4 b^5}+\frac{3 \sqrt{c d-b e} \left (16 c^2 d^2-12 b c d e+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}}\\ \end{align*}

Mathematica [A]  time = 0.426926, size = 222, normalized size = 0.96 $\frac{\frac{b \sqrt{d+e x} \left (b^2 c x \left (8 d^2-37 d e x+3 e^2 x^2\right )+b^3 \left (-2 d^2-9 d e x+5 e^2 x^2\right )+12 b c^2 d x^2 (3 d-2 e x)+24 c^3 d^2 x^3\right )}{x^2 (b+c x)^2}-3 \sqrt{d} \left (5 b^2 e^2-20 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )+\frac{3 \sqrt{c d-b e} \left (b^2 e^2-12 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{\sqrt{c}}}{4 b^5}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^(5/2)/(b*x + c*x^2)^3,x]

[Out]

((b*Sqrt[d + e*x]*(24*c^3*d^2*x^3 + 12*b*c^2*d*x^2*(3*d - 2*e*x) + b^2*c*x*(8*d^2 - 37*d*e*x + 3*e^2*x^2) + b^
3*(-2*d^2 - 9*d*e*x + 5*e^2*x^2)))/(x^2*(b + c*x)^2) - 3*Sqrt[d]*(16*c^2*d^2 - 20*b*c*d*e + 5*b^2*e^2)*ArcTanh
[Sqrt[d + e*x]/Sqrt[d]] + (3*Sqrt[c*d - b*e]*(16*c^2*d^2 - 12*b*c*d*e + b^2*e^2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x
])/Sqrt[c*d - b*e]])/Sqrt[c])/(4*b^5)

________________________________________________________________________________________

Maple [B]  time = 0.244, size = 521, normalized size = 2.3 \begin{align*}{\frac{3\,{e}^{3}c}{4\,{b}^{2} \left ( cex+be \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{15\,{e}^{2}{c}^{2}d}{4\,{b}^{3} \left ( cex+be \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+3\,{\frac{e \left ( ex+d \right ) ^{3/2}{d}^{2}{c}^{3}}{{b}^{4} \left ( cex+be \right ) ^{2}}}+{\frac{5\,{e}^{4}}{4\,b \left ( cex+be \right ) ^{2}}\sqrt{ex+d}}-{\frac{11\,{e}^{3}cd}{2\,{b}^{2} \left ( cex+be \right ) ^{2}}\sqrt{ex+d}}+{\frac{29\,{e}^{2}{c}^{2}{d}^{2}}{4\,{b}^{3} \left ( cex+be \right ) ^{2}}\sqrt{ex+d}}-3\,{\frac{e\sqrt{ex+d}{c}^{3}{d}^{3}}{{b}^{4} \left ( cex+be \right ) ^{2}}}+{\frac{3\,{e}^{3}}{4\,{b}^{2}}\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( be-cd \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( be-cd \right ) c}}}}-{\frac{39\,{e}^{2}cd}{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}}}}+21\,{\frac{{c}^{2}e{d}^{2}}{{b}^{4}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-12\,{\frac{{c}^{3}{d}^{3}}{{b}^{5}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-{\frac{9\,d}{4\,{b}^{3}{x}^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+3\,{\frac{{d}^{2} \left ( ex+d \right ) ^{3/2}c}{e{b}^{4}{x}^{2}}}+{\frac{7\,{d}^{2}}{4\,{b}^{3}{x}^{2}}\sqrt{ex+d}}-3\,{\frac{{d}^{3}\sqrt{ex+d}c}{e{b}^{4}{x}^{2}}}-{\frac{15\,{e}^{2}}{4\,{b}^{3}}\sqrt{d}{\it Artanh} \left ({\sqrt{ex+d}{\frac{1}{\sqrt{d}}}} \right ) }+15\,{\frac{e{d}^{3/2}c}{{b}^{4}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }-12\,{\frac{{d}^{5/2}{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)^(5/2)/(c*x^2+b*x)^3,x)

[Out]

3/4*e^3/b^2/(c*e*x+b*e)^2*(e*x+d)^(3/2)*c-15/4*e^2/b^3/(c*e*x+b*e)^2*(e*x+d)^(3/2)*c^2*d+3*e/b^4/(c*e*x+b*e)^2
*(e*x+d)^(3/2)*d^2*c^3+5/4*e^4/b/(c*e*x+b*e)^2*(e*x+d)^(1/2)-11/2*e^3/b^2/(c*e*x+b*e)^2*(e*x+d)^(1/2)*c*d+29/4
*e^2/b^3/(c*e*x+b*e)^2*(e*x+d)^(1/2)*c^2*d^2-3*e/b^4/(c*e*x+b*e)^2*(e*x+d)^(1/2)*c^3*d^3+3/4*e^3/b^2/((b*e-c*d
)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))-39/4*e^2/b^3/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c
/((b*e-c*d)*c)^(1/2))*c*d+21*e/b^4/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*c^2*d^2-12/
b^5/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*c^3*d^3-9/4*d/b^3/x^2*(e*x+d)^(3/2)+3/e*d^
2/b^4/x^2*(e*x+d)^(3/2)*c+7/4*d^2/b^3/x^2*(e*x+d)^(1/2)-3/e*d^3/b^4/x^2*(e*x+d)^(1/2)*c-15/4*e^2*d^(1/2)/b^3*a
rctanh((e*x+d)^(1/2)/d^(1/2))+15*e*d^(3/2)/b^4*arctanh((e*x+d)^(1/2)/d^(1/2))*c-12*d^(5/2)/b^5*arctanh((e*x+d)
^(1/2)/d^(1/2))*c^2

________________________________________________________________________________________

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)^(5/2)/(c*x^2+b*x)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.91168, size = 3510, normalized size = 15.13 \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)^(5/2)/(c*x^2+b*x)^3,x, algorithm="fricas")

[Out]

[1/8*(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((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*((16*c^4*d^2 - 20*b*c^3*d*e + 5*b^2*c^2*e^2)*x^4 + 2*(16*b*c^3*d^2 - 20*b
^2*c^2*d*e + 5*b^3*c*e^2)*x^3 + (16*b^2*c^2*d^2 - 20*b^3*c*d*e + 5*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 - 3*(8*b*c^3*d^2 - 8*b^2*c^2*d*e + b^3*c*e^2)*x^3 - (36*b^2*c^2*d^2 - 3
7*b^3*c*d*e + 5*b^4*e^2)*x^2 - (8*b^3*c*d^2 - 9*b^4*d*e)*x)*sqrt(e*x + d))/(b^5*c^2*x^4 + 2*b^6*c*x^3 + b^7*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(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e
)/c)/(c*d - b*e)) + 3*((16*c^4*d^2 - 20*b*c^3*d*e + 5*b^2*c^2*e^2)*x^4 + 2*(16*b*c^3*d^2 - 20*b^2*c^2*d*e + 5*
b^3*c*e^2)*x^3 + (16*b^2*c^2*d^2 - 20*b^3*c*d*e + 5*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 - 3*(8*b*c^3*d^2 - 8*b^2*c^2*d*e + b^3*c*e^2)*x^3 - (36*b^2*c^2*d^2 - 37*b^3*c*d*e + 5
*b^4*e^2)*x^2 - (8*b^3*c*d^2 - 9*b^4*d*e)*x)*sqrt(e*x + d))/(b^5*c^2*x^4 + 2*b^6*c*x^3 + b^7*x^2), 1/8*(6*((16
*c^4*d^2 - 20*b*c^3*d*e + 5*b^2*c^2*e^2)*x^4 + 2*(16*b*c^3*d^2 - 20*b^2*c^2*d*e + 5*b^3*c*e^2)*x^3 + (16*b^2*c
^2*d^2 - 20*b^3*c*d*e + 5*b^4*e^2)*x^2)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/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((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^4*d^2 - 3*(8*b*c^3*d^2 - 8*b^2*c^2*d*e + b^3*c*e^2)*x^3 - (36*b^2*c^2*d^2 - 37*b^3*c*d*e + 5*b^4*e^2
)*x^2 - (8*b^3*c*d^2 - 9*b^4*d*e)*x)*sqrt(e*x + d))/(b^5*c^2*x^4 + 2*b^6*c*x^3 + b^7*x^2), 1/4*(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(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)) + 3*(
(16*c^4*d^2 - 20*b*c^3*d*e + 5*b^2*c^2*e^2)*x^4 + 2*(16*b*c^3*d^2 - 20*b^2*c^2*d*e + 5*b^3*c*e^2)*x^3 + (16*b^
2*c^2*d^2 - 20*b^3*c*d*e + 5*b^4*e^2)*x^2)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) - (2*b^4*d^2 - 3*(8*b*c^3
*d^2 - 8*b^2*c^2*d*e + b^3*c*e^2)*x^3 - (36*b^2*c^2*d^2 - 37*b^3*c*d*e + 5*b^4*e^2)*x^2 - (8*b^3*c*d^2 - 9*b^4
*d*e)*x)*sqrt(e*x + d))/(b^5*c^2*x^4 + 2*b^6*c*x^3 + b^7*x^2)]

________________________________________________________________________________________

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)**(5/2)/(c*x**2+b*x)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.44165, size = 605, normalized size = 2.61 \begin{align*} -\frac{3 \,{\left (16 \, c^{3} d^{3} - 28 \, b c^{2} d^{2} e + 13 \, b^{2} c d e^{2} - b^{3} e^{3}\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^{3} - 20 \, b c d^{2} e + 5 \, b^{2} d 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^{2} e - 72 \,{\left (x e + d\right )}^{\frac{5}{2}} c^{3} d^{3} e + 72 \,{\left (x e + d\right )}^{\frac{3}{2}} c^{3} d^{4} e - 24 \, \sqrt{x e + d} c^{3} d^{5} e - 24 \,{\left (x e + d\right )}^{\frac{7}{2}} b c^{2} d e^{2} + 108 \,{\left (x e + d\right )}^{\frac{5}{2}} b c^{2} d^{2} e^{2} - 144 \,{\left (x e + d\right )}^{\frac{3}{2}} b c^{2} d^{3} e^{2} + 60 \, \sqrt{x e + d} b c^{2} d^{4} e^{2} + 3 \,{\left (x e + d\right )}^{\frac{7}{2}} b^{2} c e^{3} - 46 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{2} c d e^{3} + 91 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{2} c d^{2} e^{3} - 48 \, \sqrt{x e + d} b^{2} c d^{3} e^{3} + 5 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{3} e^{4} - 19 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{3} d e^{4} + 12 \, \sqrt{x e + d} b^{3} d^{2} 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)^(5/2)/(c*x^2+b*x)^3,x, algorithm="giac")

[Out]

-3/4*(16*c^3*d^3 - 28*b*c^2*d^2*e + 13*b^2*c*d*e^2 - b^3*e^3)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/(sq
rt(-c^2*d + b*c*e)*b^5) + 3/4*(16*c^2*d^3 - 20*b*c*d^2*e + 5*b^2*d*e^2)*arctan(sqrt(x*e + d)/sqrt(-d))/(b^5*sq
rt(-d)) + 1/4*(24*(x*e + d)^(7/2)*c^3*d^2*e - 72*(x*e + d)^(5/2)*c^3*d^3*e + 72*(x*e + d)^(3/2)*c^3*d^4*e - 24
*sqrt(x*e + d)*c^3*d^5*e - 24*(x*e + d)^(7/2)*b*c^2*d*e^2 + 108*(x*e + d)^(5/2)*b*c^2*d^2*e^2 - 144*(x*e + d)^
(3/2)*b*c^2*d^3*e^2 + 60*sqrt(x*e + d)*b*c^2*d^4*e^2 + 3*(x*e + d)^(7/2)*b^2*c*e^3 - 46*(x*e + d)^(5/2)*b^2*c*
d*e^3 + 91*(x*e + d)^(3/2)*b^2*c*d^2*e^3 - 48*sqrt(x*e + d)*b^2*c*d^3*e^3 + 5*(x*e + d)^(5/2)*b^3*e^4 - 19*(x*
e + d)^(3/2)*b^3*d*e^4 + 12*sqrt(x*e + d)*b^3*d^2*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)