### 3.396 $$\int \frac{(b x+c x^2)^{3/2}}{(d+e x)^{5/2}} \, dx$$

Optimal. Leaf size=298 $\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (4 c d-3 b e) (4 c d-b e) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right ),\frac{b e}{c d}\right )}{3 \sqrt{c} e^4 \sqrt{b x+c x^2} \sqrt{d+e x}}+\frac{2 \sqrt{b x+c x^2} (-3 b e+8 c d+2 c e x)}{3 e^3 \sqrt{d+e x}}-\frac{16 \sqrt{-b} \sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} (2 c d-b e) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{3 e^4 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}-\frac{2 \left (b x+c x^2\right )^{3/2}}{3 e (d+e x)^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.299729, antiderivative size = 298, 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 = {732, 812, 843, 715, 112, 110, 117, 116} $\frac{2 \sqrt{b x+c x^2} (-3 b e+8 c d+2 c e x)}{3 e^3 \sqrt{d+e x}}+\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (4 c d-3 b e) (4 c d-b e) F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{3 \sqrt{c} e^4 \sqrt{b x+c x^2} \sqrt{d+e x}}-\frac{16 \sqrt{-b} \sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} (2 c d-b e) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{3 e^4 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}-\frac{2 \left (b x+c x^2\right )^{3/2}}{3 e (d+e x)^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 732

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 1)), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^
(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 812

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

Mathematica [C]  time = 1.20718, size = 279, normalized size = 0.94 $\frac{2 (x (b+c x))^{3/2} \left (-i c e x^{3/2} \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} (5 b e-8 c d) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right ),\frac{c d}{b e}\right )+\frac{e x (b+c x) \left (c \left (8 d^2+10 d e x+e^2 x^2\right )-b e (3 d+4 e x)\right )}{d+e x}+8 i c e x^{3/2} \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} (b e-2 c d) E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right )|\frac{c d}{b e}\right )+8 (b+c x) (d+e x) (b e-2 c d)\right )}{3 e^4 x^2 (b+c x)^2 \sqrt{d+e x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(x*(b + c*x))^(3/2)*(8*(-2*c*d + b*e)*(b + c*x)*(d + e*x) + (e*x*(b + c*x)*(-(b*e*(3*d + 4*e*x)) + c*(8*d^2
+ 10*d*e*x + e^2*x^2)))/(d + e*x) + (8*I)*Sqrt[b/c]*c*e*(-2*c*d + b*e)*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)] - I*Sqrt[b/c]*c*e*(-8*c*d + 5*b*e)*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*e^4*x^2*(b + c*x)^2*Sqrt
[d + e*x])

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Maple [B]  time = 0.285, size = 1051, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(x*(c*x+b))^(1/2)*(3*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^3*e^3*((c*x+b)/b)^(1/2)*(-(e*x
+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-16*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^2*c*d*e^2*((c*
x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+16*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2)
)*x*b*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)-8*EllipticE(((c*x+b)/b)^(1/2),(b
*e/(b*e-c*d))^(1/2))*x*b^3*e^3*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+24*EllipticE(((c*
x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^2*c*d*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/
2)-16*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b*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)+3*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*d*e^2*((c*x+b)/b)^(1/2)*(-(e*x
+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-16*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c*d^2*e*((c*x+
b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+16*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*
b*c^2*d^3*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-8*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*
e-c*d))^(1/2))*b^3*d*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+24*EllipticE(((c*x+b)/b
)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c*d^2*e*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-16*El
lipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^2*d^3*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*
x/b)^(1/2)+x^4*c^3*e^3-3*x^3*b*c^2*e^3+10*x^3*c^3*d*e^2-4*x^2*b^2*c*e^3+7*x^2*b*c^2*d*e^2+8*x^2*c^3*d^2*e-3*x*
b^2*c*d*e^2+8*x*b*c^2*d^2*e)/(c*x+b)/x/(e*x+d)^(3/2)/c/e^4

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x \left (b + c x\right )\right )^{\frac{3}{2}}}{\left (d + e x\right )^{\frac{5}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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