### 3.397 $$\int \frac{(b x+c x^2)^{3/2}}{(d+e x)^{7/2}} \, dx$$

Optimal. Leaf size=354 $-\frac{16 \sqrt{-b} \sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (2 c d-b e) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right ),\frac{b e}{c d}\right )}{5 e^4 \sqrt{b x+c x^2} \sqrt{d+e x}}-\frac{2 \sqrt{b x+c x^2} \left (e x \left (b^2 e^2-10 b c d e+10 c^2 d^2\right )+c d^2 (8 c d-7 b e)\right )}{5 d e^3 (d+e x)^{3/2} (c d-b e)}+\frac{2 \sqrt{-b} \sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (b^2 e^2-16 b c d e+16 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{5 d e^4 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1} (c d-b e)}-\frac{2 \left (b x+c x^2\right )^{3/2}}{5 e (d+e x)^{5/2}}$

[Out]

(-2*(c*d^2*(8*c*d - 7*b*e) + e*(10*c^2*d^2 - 10*b*c*d*e + b^2*e^2)*x)*Sqrt[b*x + c*x^2])/(5*d*e^3*(c*d - b*e)*
(d + e*x)^(3/2)) - (2*(b*x + c*x^2)^(3/2))/(5*e*(d + e*x)^(5/2)) + (2*Sqrt[-b]*Sqrt[c]*(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)
])/(5*d*e^4*(c*d - b*e)*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) - (16*Sqrt[-b]*Sqrt[c]*(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)])/(5*e^4*Sqrt[d + e*
x]*Sqrt[b*x + c*x^2])

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Rubi [A]  time = 0.373561, antiderivative size = 354, 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, 810, 843, 715, 112, 110, 117, 116} $-\frac{2 \sqrt{b x+c x^2} \left (e x \left (b^2 e^2-10 b c d e+10 c^2 d^2\right )+c d^2 (8 c d-7 b e)\right )}{5 d e^3 (d+e x)^{3/2} (c d-b e)}+\frac{2 \sqrt{-b} \sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (b^2 e^2-16 b c d e+16 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{5 d e^4 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1} (c d-b e)}-\frac{16 \sqrt{-b} \sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (2 c d-b e) F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{5 e^4 \sqrt{b x+c x^2} \sqrt{d+e x}}-\frac{2 \left (b x+c x^2\right )^{3/2}}{5 e (d+e x)^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*x + c*x^2)^(3/2)/(d + e*x)^(7/2),x]

[Out]

(-2*(c*d^2*(8*c*d - 7*b*e) + e*(10*c^2*d^2 - 10*b*c*d*e + b^2*e^2)*x)*Sqrt[b*x + c*x^2])/(5*d*e^3*(c*d - b*e)*
(d + e*x)^(3/2)) - (2*(b*x + c*x^2)^(3/2))/(5*e*(d + e*x)^(5/2)) + (2*Sqrt[-b]*Sqrt[c]*(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)
])/(5*d*e^4*(c*d - b*e)*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) - (16*Sqrt[-b]*Sqrt[c]*(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)])/(5*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 810

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
- b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
- c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(m + 2*p + 2))))*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] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 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{\left (b x+c x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx &=-\frac{2 \left (b x+c x^2\right )^{3/2}}{5 e (d+e x)^{5/2}}+\frac{3 \int \frac{(b+2 c x) \sqrt{b x+c x^2}}{(d+e x)^{5/2}} \, dx}{5 e}\\ &=-\frac{2 \left (c d^2 (8 c d-7 b e)+e \left (10 c^2 d^2-10 b c d e+b^2 e^2\right ) x\right ) \sqrt{b x+c x^2}}{5 d e^3 (c d-b e) (d+e x)^{3/2}}-\frac{2 \left (b x+c x^2\right )^{3/2}}{5 e (d+e x)^{5/2}}-\frac{2 \int \frac{-\frac{1}{2} b c d (8 c d-7 b e)-\frac{1}{2} c \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}{5 d e^3 (c d-b e)}\\ &=-\frac{2 \left (c d^2 (8 c d-7 b e)+e \left (10 c^2 d^2-10 b c d e+b^2 e^2\right ) x\right ) \sqrt{b x+c x^2}}{5 d e^3 (c d-b e) (d+e x)^{3/2}}-\frac{2 \left (b x+c x^2\right )^{3/2}}{5 e (d+e x)^{5/2}}-\frac{(8 c (2 c d-b e)) \int \frac{1}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{5 e^4}+\frac{\left (c \left (16 c^2 d^2-16 b c d e+b^2 e^2\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{b x+c x^2}} \, dx}{5 d e^4 (c d-b e)}\\ &=-\frac{2 \left (c d^2 (8 c d-7 b e)+e \left (10 c^2 d^2-10 b c d e+b^2 e^2\right ) x\right ) \sqrt{b x+c x^2}}{5 d e^3 (c d-b e) (d+e x)^{3/2}}-\frac{2 \left (b x+c x^2\right )^{3/2}}{5 e (d+e x)^{5/2}}-\frac{\left (8 c (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}{5 e^4 \sqrt{b x+c x^2}}+\frac{\left (c \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}{5 d e^4 (c d-b e) \sqrt{b x+c x^2}}\\ &=-\frac{2 \left (c d^2 (8 c d-7 b e)+e \left (10 c^2 d^2-10 b c d e+b^2 e^2\right ) x\right ) \sqrt{b x+c x^2}}{5 d e^3 (c d-b e) (d+e x)^{3/2}}-\frac{2 \left (b x+c x^2\right )^{3/2}}{5 e (d+e x)^{5/2}}+\frac{\left (c \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}{5 d e^4 (c d-b e) \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}-\frac{\left (8 c (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}{5 e^4 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ &=-\frac{2 \left (c d^2 (8 c d-7 b e)+e \left (10 c^2 d^2-10 b c d e+b^2 e^2\right ) x\right ) \sqrt{b x+c x^2}}{5 d e^3 (c d-b e) (d+e x)^{3/2}}-\frac{2 \left (b x+c x^2\right )^{3/2}}{5 e (d+e x)^{5/2}}+\frac{2 \sqrt{-b} \sqrt{c} \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 )}{5 d e^4 (c d-b e) \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}-\frac{16 \sqrt{-b} \sqrt{c} (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 )}{5 e^4 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 1.19756, size = 369, normalized size = 1.04 $-\frac{2 (x (b+c x))^{3/2} \left (b e x (b+c x) \left (b^2 e^4 x^2-b c d e \left (7 d^2+16 d e x+11 e^2 x^2\right )+c^2 d^2 \left (8 d^2+18 d e x+11 e^2 x^2\right )\right )-c \sqrt{\frac{b}{c}} (d+e x)^2 \left (-i b e x^{3/2} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (b^2 e^2-9 b c d e+8 c^2 d^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right ),\frac{c d}{b e}\right )+i b e x^{3/2} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (b^2 e^2-16 b c d e+16 c^2 d^2\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right )|\frac{c d}{b e}\right )+\sqrt{\frac{b}{c}} (b+c x) (d+e x) \left (b^2 e^2-16 b c d e+16 c^2 d^2\right )\right )\right )}{5 b d e^4 x^2 (b+c x)^2 (d+e x)^{5/2} (c d-b e)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*x + c*x^2)^(3/2)/(d + e*x)^(7/2),x]

[Out]

(-2*(x*(b + c*x))^(3/2)*(b*e*x*(b + c*x)*(b^2*e^4*x^2 - b*c*d*e*(7*d^2 + 16*d*e*x + 11*e^2*x^2) + c^2*d^2*(8*d
^2 + 18*d*e*x + 11*e^2*x^2)) - Sqrt[b/c]*c*(d + e*x)^2*(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*ArcSinh[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)])))/(5*b*d*e^4*(c*d - b*e)*x^2*(b + c*x
)^2*(d + e*x)^(5/2))

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Maple [B]  time = 0.302, size = 1885, normalized size = 5.3 \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)^(7/2),x)

[Out]

2/5*(x*(c*x+b))^(1/2)*(-11*x^3*b^2*c^2*d*e^4-5*x^3*b*c^3*d^2*e^3-16*x^2*b^2*c^2*d^2*e^3+11*x^2*b*c^3*d^3*e^2-7
*x*b^2*c^2*d^3*e^2+32*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b^2*c^2*d^2*e^3*(-c*x/b)^(1/2)*((
c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)-16*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b*c^3*d
^3*e^2*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)+16*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-
c*d))^(1/2))*x*b^3*c*d^2*e^3*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)-48*EllipticF(((c*x+
b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^2*c^2*d^3*e^2*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(
1/2)+32*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b*c^3*d^4*e*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e
*x+d)*c/(b*e-c*d))^(1/2)-34*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^3*c*d^2*e^3*(-c*x/b)^(1/2)*
((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)+64*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^2*c^2
*d^3*e^2*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)-32*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*
e-c*d))^(1/2))*x*b*c^3*d^4*e*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)+8*EllipticF(((c*x+b
)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b^3*c*d*e^4*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2
)-24*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b^2*c^2*d^2*e^3*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(
-(e*x+d)*c/(b*e-c*d))^(1/2)+8*x*b*c^3*d^4*e+EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^4*d^2*e^3*(-c
*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)-11*x^4*b*c^3*d*e^4+8*EllipticF(((c*x+b)/b)^(1/2),(b
*e/(b*e-c*d))^(1/2))*b^3*c*d^3*e^2*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)-24*EllipticF(
((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^2*d^4*e*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))
^(1/2)-17*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*c*d^3*e^2*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-
(e*x+d)*c/(b*e-c*d))^(1/2)+32*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^2*d^4*e*(-c*x/b)^(1/2)*
((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)+x^4*b^2*c^2*e^5+11*x^4*c^4*d^2*e^3+x^3*b^3*c*e^5+18*x^3*c^4*d^3
*e^2+8*x^2*c^4*d^4*e+2*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^4*d*e^4*(-c*x/b)^(1/2)*((c*x+b)/
b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)+16*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b*c^3*d^3*e^2*
(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)-17*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(
1/2))*x^2*b^3*c*d*e^4*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)+EllipticE(((c*x+b)/b)^(1/2
),(b*e/(b*e-c*d))^(1/2))*x^2*b^4*e^5*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)+16*Elliptic
F(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*d^5*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(
1/2)-16*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*d^5*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d
)*c/(b*e-c*d))^(1/2))/(c*x+b)/x/(b*e-c*d)/c/(e*x+d)^(5/2)/e^4/d

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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{7}{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)^(7/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(3/2)/(e*x + d)^(7/2), 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^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}}, 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)^(7/2),x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x)^(3/2)*sqrt(e*x + d)/(e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x + d^4), 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{7}{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)**(7/2),x)

[Out]

Integral((x*(b + c*x))**(3/2)/(d + e*x)**(7/2), 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{7}{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)^(7/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x)^(3/2)/(e*x + d)^(7/2), x)