### 3.415 $$\int \frac{(d+e x)^{5/2}}{(b x+c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=310 $-\frac{2 d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) (2 c d-b e) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right ),\frac{b e}{c d}\right )}{(-b)^{3/2} c^{3/2} \sqrt{b x+c x^2} \sqrt{d+e x}}+\frac{4 \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (b^2 e^2-b c d e+c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{(-b)^{3/2} c^{3/2} \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}-\frac{2 (d+e x)^{3/2} (x (2 c d-b e)+b d)}{b^2 \sqrt{b x+c x^2}}+\frac{2 e \sqrt{b x+c x^2} \sqrt{d+e x} (2 c d-b e)}{b^2 c}$

[Out]

(-2*(d + e*x)^(3/2)*(b*d + (2*c*d - b*e)*x))/(b^2*Sqrt[b*x + c*x^2]) + (2*e*(2*c*d - b*e)*Sqrt[d + e*x]*Sqrt[b
*x + c*x^2])/(b^2*c) + (4*(c^2*d^2 - b*c*d*e + b^2*e^2)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[d + e*x]*EllipticE[ArcS
in[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/((-b)^(3/2)*c^(3/2)*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) - (2*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)])/((-b)^(3/2)*c^(3/2)*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])

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Rubi [A]  time = 0.33904, antiderivative size = 310, 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, 832, 843, 715, 112, 110, 117, 116} $\frac{4 \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (b^2 e^2-b c d e+c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{(-b)^{3/2} c^{3/2} \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}-\frac{2 (d+e x)^{3/2} (x (2 c d-b e)+b d)}{b^2 \sqrt{b x+c x^2}}+\frac{2 e \sqrt{b x+c x^2} \sqrt{d+e x} (2 c d-b e)}{b^2 c}-\frac{2 d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) (2 c d-b e) F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{(-b)^{3/2} c^{3/2} \sqrt{b x+c x^2} \sqrt{d+e x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*(d + e*x)^(3/2)*(b*d + (2*c*d - b*e)*x))/(b^2*Sqrt[b*x + c*x^2]) + (2*e*(2*c*d - b*e)*Sqrt[d + e*x]*Sqrt[b
*x + c*x^2])/(b^2*c) + (4*(c^2*d^2 - b*c*d*e + b^2*e^2)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[d + e*x]*EllipticE[ArcS
in[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/((-b)^(3/2)*c^(3/2)*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) - (2*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)])/((-b)^(3/2)*c^(3/2)*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 832

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

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

Mathematica [C]  time = 1.16548, size = 262, normalized size = 0.85 $\frac{-2 i c e x^{3/2} \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (2 b^2 e^2-3 b c d e+c^2 d^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right ),\frac{c d}{b e}\right )+4 i c e x^{3/2} \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (b^2 e^2-b c d e+c^2 d^2\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right )|\frac{c d}{b e}\right )+2 b (d+e x) \left (2 b^2 e^2+b c e (e x-2 d)+c^2 d^2\right )}{b^2 c^2 \sqrt{x (b+c x)} \sqrt{d+e x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*b*(d + e*x)*(c^2*d^2 + 2*b^2*e^2 + b*c*e*(-2*d + e*x)) + (4*I)*Sqrt[b/c]*c*e*(c^2*d^2 - b*c*d*e + b^2*e^2)*
Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)] - (2*I)*Sqrt[
b/c]*c*e*(c^2*d^2 - 3*b*c*d*e + 2*b^2*e^2)*Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x^(3/2)*EllipticF[I*ArcSinh[Sqr
t[b/c]/Sqrt[x]], (c*d)/(b*e)])/(b^2*c^2*Sqrt[x*(b + c*x)]*Sqrt[d + e*x])

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Maple [B]  time = 0.31, size = 652, normalized size = 2.1 \begin{align*} -2\,{\frac{\sqrt{x \left ( cx+b \right ) }}{x \left ( cx+b \right ){c}^{3}{b}^{2}\sqrt{ex+d}} \left ( \sqrt{-{\frac{cx}{b}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+b}{b}}},\sqrt{{\frac{be}{be-cd}}} \right ) \sqrt{{\frac{cx+b}{b}}}\sqrt{-{\frac{c \left ( ex+d \right ) }{be-cd}}}{b}^{3}cd{e}^{2}-3\,\sqrt{-{\frac{cx}{b}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+b}{b}}},\sqrt{{\frac{be}{be-cd}}} \right ) \sqrt{{\frac{cx+b}{b}}}\sqrt{-{\frac{c \left ( ex+d \right ) }{be-cd}}}{b}^{2}{c}^{2}{d}^{2}e+2\,\sqrt{-{\frac{cx}{b}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+b}{b}}},\sqrt{{\frac{be}{be-cd}}} \right ) \sqrt{{\frac{cx+b}{b}}}\sqrt{-{\frac{c \left ( ex+d \right ) }{be-cd}}}b{c}^{3}{d}^{3}+2\,\sqrt{-{\frac{cx}{b}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+b}{b}}},\sqrt{{\frac{be}{be-cd}}} \right ) \sqrt{{\frac{cx+b}{b}}}\sqrt{-{\frac{c \left ( ex+d \right ) }{be-cd}}}{b}^{4}{e}^{3}-4\,\sqrt{-{\frac{cx}{b}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+b}{b}}},\sqrt{{\frac{be}{be-cd}}} \right ) \sqrt{{\frac{cx+b}{b}}}\sqrt{-{\frac{c \left ( ex+d \right ) }{be-cd}}}{b}^{3}cd{e}^{2}+4\,\sqrt{-{\frac{cx}{b}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+b}{b}}},\sqrt{{\frac{be}{be-cd}}} \right ) \sqrt{{\frac{cx+b}{b}}}\sqrt{-{\frac{c \left ( ex+d \right ) }{be-cd}}}{b}^{2}{c}^{2}{d}^{2}e-2\,\sqrt{-{\frac{cx}{b}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+b}{b}}},\sqrt{{\frac{be}{be-cd}}} \right ) \sqrt{{\frac{cx+b}{b}}}\sqrt{-{\frac{c \left ( ex+d \right ) }{be-cd}}}b{c}^{3}{d}^{3}+{x}^{2}{b}^{2}{c}^{2}{e}^{3}-2\,{x}^{2}b{c}^{3}d{e}^{2}+2\,{x}^{2}{c}^{4}{d}^{2}e+x{b}^{2}{c}^{2}d{e}^{2}-xb{c}^{3}{d}^{2}e+2\,x{c}^{4}{d}^{3}+b{c}^{3}{d}^{3} \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)^(3/2),x)

[Out]

-2*((-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))
^(1/2)*b^3*c*d*e^2-3*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*((c*x+b)/b)^(1/2)*(-(e*
x+d)*c/(b*e-c*d))^(1/2)*b^2*c^2*d^2*e+2*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*((c*
x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*b*c^3*d^3+2*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*
d))^(1/2))*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*b^4*e^3-4*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2)
,(b*e/(b*e-c*d))^(1/2))*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*b^3*c*d*e^2+4*(-c*x/b)^(1/2)*EllipticE(
((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*b^2*c^2*d^2*e-2*(-c*x/
b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*b*c
^3*d^3+x^2*b^2*c^2*e^3-2*x^2*b*c^3*d*e^2+2*x^2*c^4*d^2*e+x*b^2*c^2*d*e^2-x*b*c^3*d^2*e+2*x*c^4*d^3+b*c^3*d^3)/
x*(x*(c*x+b))^(1/2)/(c*x+b)/c^3/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{3}{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)^(3/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^(5/2)/(c*x^2 + b*x)^(3/2), 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^{2} x^{4} + 2 \, b c x^{3} + b^{2} x^{2}}, 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)^(3/2),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^2*x^4 + 2*b*c*x^3 + b^2*x^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((e*x+d)**(5/2)/(c*x**2+b*x)**(3/2),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{3}{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)^(3/2),x, algorithm="giac")

[Out]

integrate((e*x + d)^(5/2)/(c*x^2 + b*x)^(3/2), x)