3.423 \(\int \frac{(d+e x)^{5/2}}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=343 \[ \frac{16 d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) (2 c d-b e) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right ),\frac{b e}{c d}\right )}{3 (-b)^{7/2} \sqrt{c} \sqrt{b x+c x^2} \sqrt{d+e x}}+\frac{2 \sqrt{d+e x} \left (x \left (b^2 e^2-16 b c d e+16 c^2 d^2\right )+b d (8 c d-7 b e)\right )}{3 b^4 \sqrt{b x+c x^2}}-\frac{2 \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (b^2 e^2-16 b c d e+16 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{3 (-b)^{7/2} \sqrt{c} \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}-\frac{2 (d+e x)^{3/2} (x (2 c d-b e)+b d)}{3 b^2 \left (b x+c x^2\right )^{3/2}} \]

[Out]

(-2*(d + e*x)^(3/2)*(b*d + (2*c*d - b*e)*x))/(3*b^2*(b*x + c*x^2)^(3/2)) + (2*Sqrt[d + e*x]*(b*d*(8*c*d - 7*b*
e) + (16*c^2*d^2 - 16*b*c*d*e + b^2*e^2)*x))/(3*b^4*Sqrt[b*x + c*x^2]) - (2*(16*c^2*d^2 - 16*b*c*d*e + b^2*e^2
)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[d + e*x]*EllipticE[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(3*(-b)^
(7/2)*Sqrt[c]*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) + (16*d*(c*d - b*e)*(2*c*d - b*e)*Sqrt[x]*Sqrt[1 + (c*x)/b]
*Sqrt[1 + (e*x)/d]*EllipticF[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(3*(-b)^(7/2)*Sqrt[c]*Sqrt[d +
e*x]*Sqrt[b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.356276, antiderivative size = 343, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {738, 820, 843, 715, 112, 110, 117, 116} \[ \frac{2 \sqrt{d+e x} \left (x \left (b^2 e^2-16 b c d e+16 c^2 d^2\right )+b d (8 c d-7 b e)\right )}{3 b^4 \sqrt{b x+c x^2}}-\frac{2 \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (b^2 e^2-16 b c d e+16 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{3 (-b)^{7/2} \sqrt{c} \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}-\frac{2 (d+e x)^{3/2} (x (2 c d-b e)+b d)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{16 d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) (2 c d-b e) F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{3 (-b)^{7/2} \sqrt{c} \sqrt{b x+c x^2} \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)^(3/2)*(b*d + (2*c*d - b*e)*x))/(3*b^2*(b*x + c*x^2)^(3/2)) + (2*Sqrt[d + e*x]*(b*d*(8*c*d - 7*b*
e) + (16*c^2*d^2 - 16*b*c*d*e + b^2*e^2)*x))/(3*b^4*Sqrt[b*x + c*x^2]) - (2*(16*c^2*d^2 - 16*b*c*d*e + b^2*e^2
)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[d + e*x]*EllipticE[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(3*(-b)^
(7/2)*Sqrt[c]*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) + (16*d*(c*d - b*e)*(2*c*d - b*e)*Sqrt[x]*Sqrt[1 + (c*x)/b]
*Sqrt[1 + (e*x)/d]*EllipticF[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(3*(-b)^(7/2)*Sqrt[c]*Sqrt[d +
e*x]*Sqrt[b*x + c*x^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 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 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 715

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(Sqrt[x]*Sqrt[b + c*x])/Sqrt[
b*x + c*x^2], Int[(d + e*x)^m/(Sqrt[x]*Sqrt[b + c*x]), x], x] /; FreeQ[{b, c, d, e}, x] && NeQ[c*d - b*e, 0] &
& NeQ[2*c*d - b*e, 0] && EqQ[m^2, 1/4]

Rule 112

Int[Sqrt[(e_) + (f_.)*(x_)]/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Dist[(Sqrt[e + f*x]*Sqrt[
1 + (d*x)/c])/(Sqrt[c + d*x]*Sqrt[1 + (f*x)/e]), Int[Sqrt[1 + (f*x)/e]/(Sqrt[b*x]*Sqrt[1 + (d*x)/c]), x], x] /
; FreeQ[{b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] &&  !(GtQ[c, 0] && GtQ[e, 0])

Rule 110

Int[Sqrt[(e_) + (f_.)*(x_)]/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[(2*Sqrt[e]*Rt[-(b/d)
, 2]*EllipticE[ArcSin[Sqrt[b*x]/(Sqrt[c]*Rt[-(b/d), 2])], (c*f)/(d*e)])/b, x] /; FreeQ[{b, c, d, e, f}, x] &&
NeQ[d*e - c*f, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !LtQ[-(b/d), 0]

Rule 117

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Dist[(Sqrt[1 + (d*x)/c]
*Sqrt[1 + (f*x)/e])/(Sqrt[c + d*x]*Sqrt[e + f*x]), Int[1/(Sqrt[b*x]*Sqrt[1 + (d*x)/c]*Sqrt[1 + (f*x)/e]), x],
x] /; FreeQ[{b, c, d, e, f}, x] &&  !(GtQ[c, 0] && GtQ[e, 0])

Rule 116

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*Rt[-(b/d), 2]*E
llipticF[ArcSin[Sqrt[b*x]/(Sqrt[c]*Rt[-(b/d), 2])], (c*f)/(d*e)])/(b*Sqrt[e]), x] /; FreeQ[{b, c, d, e, f}, x]
&& GtQ[c, 0] && GtQ[e, 0] && (PosQ[-(b/d)] || NegQ[-(b/f)])

Rubi steps

\begin{align*} \int \frac{(d+e x)^{5/2}}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 (d+e x)^{3/2} (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}-\frac{2 \int \frac{\sqrt{d+e x} \left (\frac{1}{2} d (8 c d-7 b e)+\frac{1}{2} e (2 c d-b e) x\right )}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 b^2}\\ &=-\frac{2 (d+e x)^{3/2} (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{2 \sqrt{d+e x} \left (b d (8 c d-7 b e)+\left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) x\right )}{3 b^4 \sqrt{b x+c x^2}}+\frac{4 \int \frac{-\frac{1}{4} b d e (8 c d-7 b e)-\frac{1}{4} e \left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) x}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{3 b^4}\\ &=-\frac{2 (d+e x)^{3/2} (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{2 \sqrt{d+e x} \left (b d (8 c d-7 b e)+\left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) x\right )}{3 b^4 \sqrt{b x+c x^2}}+\frac{(8 d (c d-b e) (2 c d-b e)) \int \frac{1}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{3 b^4}-\frac{\left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) \int \frac{\sqrt{d+e x}}{\sqrt{b x+c x^2}} \, dx}{3 b^4}\\ &=-\frac{2 (d+e x)^{3/2} (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{2 \sqrt{d+e x} \left (b d (8 c d-7 b e)+\left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) x\right )}{3 b^4 \sqrt{b x+c x^2}}+\frac{\left (8 d (c d-b e) (2 c d-b e) \sqrt{x} \sqrt{b+c x}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x} \sqrt{d+e x}} \, dx}{3 b^4 \sqrt{b x+c x^2}}-\frac{\left (\left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) \sqrt{x} \sqrt{b+c x}\right ) \int \frac{\sqrt{d+e x}}{\sqrt{x} \sqrt{b+c x}} \, dx}{3 b^4 \sqrt{b x+c x^2}}\\ &=-\frac{2 (d+e x)^{3/2} (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{2 \sqrt{d+e x} \left (b d (8 c d-7 b e)+\left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) x\right )}{3 b^4 \sqrt{b x+c x^2}}-\frac{\left (\left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) \sqrt{x} \sqrt{1+\frac{c x}{b}} \sqrt{d+e x}\right ) \int \frac{\sqrt{1+\frac{e x}{d}}}{\sqrt{x} \sqrt{1+\frac{c x}{b}}} \, dx}{3 b^4 \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}+\frac{\left (8 d (c d-b e) (2 c d-b e) \sqrt{x} \sqrt{1+\frac{c x}{b}} \sqrt{1+\frac{e x}{d}}\right ) \int \frac{1}{\sqrt{x} \sqrt{1+\frac{c x}{b}} \sqrt{1+\frac{e x}{d}}} \, dx}{3 b^4 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ &=-\frac{2 (d+e x)^{3/2} (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{2 \sqrt{d+e x} \left (b d (8 c d-7 b e)+\left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) x\right )}{3 b^4 \sqrt{b x+c x^2}}-\frac{2 \left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) \sqrt{x} \sqrt{1+\frac{c x}{b}} \sqrt{d+e x} E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{3 (-b)^{7/2} \sqrt{c} \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}+\frac{16 d (c d-b e) (2 c d-b e) \sqrt{x} \sqrt{1+\frac{c x}{b}} \sqrt{1+\frac{e x}{d}} F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{3 (-b)^{7/2} \sqrt{c} \sqrt{d+e x} \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 1.10205, size = 353, normalized size = 1.03 \[ \frac{2 \left (b (d+e x) \left (b^2 c x \left (6 d^2-25 d e x+e^2 x^2\right )+b^3 \left (-\left (d^2+7 d e x-2 e^2 x^2\right )\right )+8 b c^2 d x^2 (3 d-2 e x)+16 c^3 d^2 x^3\right )-x \sqrt{\frac{b}{c}} (b+c x) \left (-i b e x^{3/2} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (b^2 e^2-9 b c d e+8 c^2 d^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right ),\frac{c d}{b e}\right )+i b e x^{3/2} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (b^2 e^2-16 b c d e+16 c^2 d^2\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right )|\frac{c d}{b e}\right )+\sqrt{\frac{b}{c}} (b+c x) (d+e x) \left (b^2 e^2-16 b c d e+16 c^2 d^2\right )\right )\right )}{3 b^5 (x (b+c x))^{3/2} \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*(d + e*x)*(16*c^3*d^2*x^3 + 8*b*c^2*d*x^2*(3*d - 2*e*x) - b^3*(d^2 + 7*d*e*x - 2*e^2*x^2) + b^2*c*x*(6*d
^2 - 25*d*e*x + e^2*x^2)) - Sqrt[b/c]*x*(b + c*x)*(Sqrt[b/c]*(16*c^2*d^2 - 16*b*c*d*e + b^2*e^2)*(b + c*x)*(d
+ e*x) + I*b*e*(16*c^2*d^2 - 16*b*c*d*e + b^2*e^2)*Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x^(3/2)*EllipticE[I*Arc
Sinh[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)] - I*b*e*(8*c^2*d^2 - 9*b*c*d*e + b^2*e^2)*Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(
e*x)]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)])))/(3*b^5*(x*(b + c*x))^(3/2)*Sqrt[d + e*x]
)

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Maple [B]  time = 0.336, size = 1318, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(5/2)/(c*x^2+b*x)^(5/2),x)

[Out]

2/3*(-17*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c
*d))^(1/2))*x*b^4*c*d*e^2+32*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/
b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^3*c^2*d^2*e+8*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2
)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^4*c*d*e^2-24*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))
^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^3*c^2*d^2*e+8*((c*x+b)/b)^(1/2)*(
-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b^3*c^2*d*e^
2-24*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))
^(1/2))*x^2*b^2*c^3*d^2*e-24*x^3*b^2*c^3*d*e^2-5*x^2*b^3*c^2*d*e^2-19*x^2*b^2*c^3*d^2*e-16*x^4*b*c^4*d*e^2+8*x
^3*b*c^4*d^2*e+x^4*b^2*c^3*e^3+2*x^3*b^3*c^2*e^3+24*x^2*b*c^4*d^3+16*x^4*c^5*d^2*e+6*x*b^2*c^3*d^3-17*((c*x+b)
/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b
^3*c^2*d*e^2+32*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e
/(b*e-c*d))^(1/2))*x^2*b^2*c^3*d^2*e-16*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*Elliptic
E(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b*c^4*d^3+16*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c
*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b*c^4*d^3-16*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/
(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^2*c^3*d^3+16*((c*x+b)/b
)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^2*c
^3*d^3-8*x*b^3*c^2*d^2*e+((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(
1/2),(b*e/(b*e-c*d))^(1/2))*x*b^5*e^3-b^3*c^2*d^3+16*x^3*c^5*d^3+((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2
)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b^4*c*e^3)/x^2*(x*(c*x+b))^(1/2)/b^4/c
^2/(c*x+b)^2/(e*x+d)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{\frac{5}{2}}}{{\left (c x^{2} + b x\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)^(5/2)/(c*x^2 + b*x)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} \sqrt{c x^{2} + b x} \sqrt{e x + d}}{c^{3} x^{6} + 3 \, b c^{2} x^{5} + 3 \, b^{2} c x^{4} + b^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

integral((e^2*x^2 + 2*d*e*x + d^2)*sqrt(c*x^2 + b*x)*sqrt(e*x + d)/(c^3*x^6 + 3*b*c^2*x^5 + 3*b^2*c*x^4 + b^3*
x^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)**(5/2)/(c*x**2+b*x)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{\frac{5}{2}}}{{\left (c x^{2} + b x\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

integrate((e*x + d)^(5/2)/(c*x^2 + b*x)^(5/2), x)