3.308 $$\int \frac{(b x+c x^2)^{5/2}}{(d+e x)^3} \, dx$$

Optimal. Leaf size=282 $\frac{5 \sqrt{b x+c x^2} \left (5 b^2 e^2-4 c e x (2 c d-b e)-20 b c d e+16 c^2 d^2\right )}{8 e^5}-\frac{5 (2 c d-b e) \left (b^2 e^2-16 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 \sqrt{c} e^6}+\frac{5 \left (b x+c x^2\right )^{3/2} (-3 b e+8 c d+2 c e x)}{12 e^3 (d+e x)}+\frac{5 \sqrt{d} (4 c d-3 b e) \sqrt{c d-b e} (4 c d-b e) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{8 e^6}-\frac{\left (b x+c x^2\right )^{5/2}}{2 e (d+e x)^2}$

[Out]

(5*(16*c^2*d^2 - 20*b*c*d*e + 5*b^2*e^2 - 4*c*e*(2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(8*e^5) + (5*(8*c*d - 3*b*
e + 2*c*e*x)*(b*x + c*x^2)^(3/2))/(12*e^3*(d + e*x)) - (b*x + c*x^2)^(5/2)/(2*e*(d + e*x)^2) - (5*(2*c*d - b*e
)*(16*c^2*d^2 - 16*b*c*d*e + b^2*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(8*Sqrt[c]*e^6) + (5*Sqrt[d]*(4*
c*d - 3*b*e)*Sqrt[c*d - b*e]*(4*c*d - b*e)*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x
+ c*x^2])])/(8*e^6)

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Rubi [A]  time = 0.32579, antiderivative size = 282, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.333, Rules used = {732, 812, 814, 843, 620, 206, 724} $\frac{5 \sqrt{b x+c x^2} \left (5 b^2 e^2-4 c e x (2 c d-b e)-20 b c d e+16 c^2 d^2\right )}{8 e^5}-\frac{5 (2 c d-b e) \left (b^2 e^2-16 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 \sqrt{c} e^6}+\frac{5 \left (b x+c x^2\right )^{3/2} (-3 b e+8 c d+2 c e x)}{12 e^3 (d+e x)}+\frac{5 \sqrt{d} (4 c d-3 b e) \sqrt{c d-b e} (4 c d-b e) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{8 e^6}-\frac{\left (b x+c x^2\right )^{5/2}}{2 e (d+e x)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(5*(16*c^2*d^2 - 20*b*c*d*e + 5*b^2*e^2 - 4*c*e*(2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(8*e^5) + (5*(8*c*d - 3*b*
e + 2*c*e*x)*(b*x + c*x^2)^(3/2))/(12*e^3*(d + e*x)) - (b*x + c*x^2)^(5/2)/(2*e*(d + e*x)^2) - (5*(2*c*d - b*e
)*(16*c^2*d^2 - 16*b*c*d*e + b^2*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(8*Sqrt[c]*e^6) + (5*Sqrt[d]*(4*
c*d - 3*b*e)*Sqrt[c*d - b*e]*(4*c*d - b*e)*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x
+ c*x^2])])/(8*e^6)

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 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*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] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 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 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 724

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

Rubi steps

\begin{align*} \int \frac{\left (b x+c x^2\right )^{5/2}}{(d+e x)^3} \, dx &=-\frac{\left (b x+c x^2\right )^{5/2}}{2 e (d+e x)^2}+\frac{5 \int \frac{(b+2 c x) \left (b x+c x^2\right )^{3/2}}{(d+e x)^2} \, dx}{4 e}\\ &=\frac{5 (8 c d-3 b e+2 c e x) \left (b x+c x^2\right )^{3/2}}{12 e^3 (d+e x)}-\frac{\left (b x+c x^2\right )^{5/2}}{2 e (d+e x)^2}-\frac{5 \int \frac{(b (8 c d-3 b e)+8 c (2 c d-b e) x) \sqrt{b x+c x^2}}{d+e x} \, dx}{8 e^3}\\ &=\frac{5 \left (16 c^2 d^2-20 b c d e+5 b^2 e^2-4 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{8 e^5}+\frac{5 (8 c d-3 b e+2 c e x) \left (b x+c x^2\right )^{3/2}}{12 e^3 (d+e x)}-\frac{\left (b x+c x^2\right )^{5/2}}{2 e (d+e x)^2}+\frac{5 \int \frac{-2 b c d \left (16 c^2 d^2-20 b c d e+5 b^2 e^2\right )-2 c (2 c d-b e) \left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) x}{(d+e x) \sqrt{b x+c x^2}} \, dx}{32 c e^5}\\ &=\frac{5 \left (16 c^2 d^2-20 b c d e+5 b^2 e^2-4 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{8 e^5}+\frac{5 (8 c d-3 b e+2 c e x) \left (b x+c x^2\right )^{3/2}}{12 e^3 (d+e x)}-\frac{\left (b x+c x^2\right )^{5/2}}{2 e (d+e x)^2}+\frac{(5 d (4 c d-3 b e) (c d-b e) (4 c d-b e)) \int \frac{1}{(d+e x) \sqrt{b x+c x^2}} \, dx}{8 e^6}-\frac{\left (5 (2 c d-b e) \left (16 c^2 d^2-16 b c d e+b^2 e^2\right )\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{16 e^6}\\ &=\frac{5 \left (16 c^2 d^2-20 b c d e+5 b^2 e^2-4 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{8 e^5}+\frac{5 (8 c d-3 b e+2 c e x) \left (b x+c x^2\right )^{3/2}}{12 e^3 (d+e x)}-\frac{\left (b x+c x^2\right )^{5/2}}{2 e (d+e x)^2}-\frac{(5 d (4 c d-3 b e) (c d-b e) (4 c d-b e)) \operatorname{Subst}\left (\int \frac{1}{4 c d^2-4 b d e-x^2} \, dx,x,\frac{-b d-(2 c d-b e) x}{\sqrt{b x+c x^2}}\right )}{4 e^6}-\frac{\left (5 (2 c d-b e) \left (16 c^2 d^2-16 b c d e+b^2 e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{8 e^6}\\ &=\frac{5 \left (16 c^2 d^2-20 b c d e+5 b^2 e^2-4 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{8 e^5}+\frac{5 (8 c d-3 b e+2 c e x) \left (b x+c x^2\right )^{3/2}}{12 e^3 (d+e x)}-\frac{\left (b x+c x^2\right )^{5/2}}{2 e (d+e x)^2}-\frac{5 (2 c d-b e) \left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 \sqrt{c} e^6}+\frac{5 \sqrt{d} (4 c d-3 b e) \sqrt{c d-b e} (4 c d-b e) \tanh ^{-1}\left (\frac{b d+(2 c d-b e) x}{2 \sqrt{d} \sqrt{c d-b e} \sqrt{b x+c x^2}}\right )}{8 e^6}\\ \end{align*}

Mathematica [A]  time = 2.12393, size = 322, normalized size = 1.14 $\frac{\sqrt{x (b+c x)} \left (\frac{e \sqrt{x} \left (3 b^2 e^2 \left (25 d^2+40 d e x+11 e^2 x^2\right )-2 b c e \left (230 d^2 e x+150 d^3+55 d e^2 x^2-13 e^3 x^3\right )+4 c^2 \left (20 d^2 e^2 x^2+90 d^3 e x+60 d^4-5 d e^3 x^3+2 e^4 x^4\right )\right )}{(d+e x)^2}+\frac{30 \sqrt{d} \left (19 b^2 c d e^2-3 b^3 e^3-32 b c^2 d^2 e+16 c^3 d^3\right ) \tan ^{-1}\left (\frac{\sqrt{x} \sqrt{b e-c d}}{\sqrt{d} \sqrt{b+c x}}\right )}{\sqrt{b+c x} \sqrt{b e-c d}}+\frac{15 \left (-18 b^2 c d e^2+b^3 e^3+48 b c^2 d^2 e-32 c^3 d^3\right ) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{b} \sqrt{c} \sqrt{\frac{c x}{b}+1}}\right )}{24 e^6 \sqrt{x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*((e*Sqrt[x]*(3*b^2*e^2*(25*d^2 + 40*d*e*x + 11*e^2*x^2) - 2*b*c*e*(150*d^3 + 230*d^2*e*x +
55*d*e^2*x^2 - 13*e^3*x^3) + 4*c^2*(60*d^4 + 90*d^3*e*x + 20*d^2*e^2*x^2 - 5*d*e^3*x^3 + 2*e^4*x^4)))/(d + e*x
)^2 + (15*(-32*c^3*d^3 + 48*b*c^2*d^2*e - 18*b^2*c*d*e^2 + b^3*e^3)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[
b]*Sqrt[c]*Sqrt[1 + (c*x)/b]) + (30*Sqrt[d]*(16*c^3*d^3 - 32*b*c^2*d^2*e + 19*b^2*c*d*e^2 - 3*b^3*e^3)*ArcTan[
(Sqrt[-(c*d) + b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])/(Sqrt[-(c*d) + b*e]*Sqrt[b + c*x])))/(24*e^6*Sqrt[x])

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Maple [B]  time = 0.23, size = 5534, normalized size = 19.6 \begin{align*} \text{output too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.16874, size = 4356, normalized size = 15.45 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(15*(32*c^3*d^5 - 48*b*c^2*d^4*e + 18*b^2*c*d^3*e^2 - b^3*d^2*e^3 + (32*c^3*d^3*e^2 - 48*b*c^2*d^2*e^3
+ 18*b^2*c*d*e^4 - b^3*e^5)*x^2 + 2*(32*c^3*d^4*e - 48*b*c^2*d^3*e^2 + 18*b^2*c*d^2*e^3 - b^3*d*e^4)*x)*sqrt(c
)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 30*(16*c^3*d^4 - 16*b*c^2*d^3*e + 3*b^2*c*d^2*e^2 + (16*c^3*d
^2*e^2 - 16*b*c^2*d*e^3 + 3*b^2*c*e^4)*x^2 + 2*(16*c^3*d^3*e - 16*b*c^2*d^2*e^2 + 3*b^2*c*d*e^3)*x)*sqrt(c*d^2
- b*d*e)*log((b*d + (2*c*d - b*e)*x + 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d)) - 2*(8*c^3*e^5*x^4
+ 240*c^3*d^4*e - 300*b*c^2*d^3*e^2 + 75*b^2*c*d^2*e^3 - 2*(10*c^3*d*e^4 - 13*b*c^2*e^5)*x^3 + (80*c^3*d^2*e^3
- 110*b*c^2*d*e^4 + 33*b^2*c*e^5)*x^2 + 20*(18*c^3*d^3*e^2 - 23*b*c^2*d^2*e^3 + 6*b^2*c*d*e^4)*x)*sqrt(c*x^2
+ b*x))/(c*e^8*x^2 + 2*c*d*e^7*x + c*d^2*e^6), 1/48*(60*(16*c^3*d^4 - 16*b*c^2*d^3*e + 3*b^2*c*d^2*e^2 + (16*c
^3*d^2*e^2 - 16*b*c^2*d*e^3 + 3*b^2*c*e^4)*x^2 + 2*(16*c^3*d^3*e - 16*b*c^2*d^2*e^2 + 3*b^2*c*d*e^3)*x)*sqrt(-
c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/((c*d - b*e)*x)) - 15*(32*c^3*d^5 - 48*b*c^2*d^4
*e + 18*b^2*c*d^3*e^2 - b^3*d^2*e^3 + (32*c^3*d^3*e^2 - 48*b*c^2*d^2*e^3 + 18*b^2*c*d*e^4 - b^3*e^5)*x^2 + 2*(
32*c^3*d^4*e - 48*b*c^2*d^3*e^2 + 18*b^2*c*d^2*e^3 - b^3*d*e^4)*x)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)
*sqrt(c)) + 2*(8*c^3*e^5*x^4 + 240*c^3*d^4*e - 300*b*c^2*d^3*e^2 + 75*b^2*c*d^2*e^3 - 2*(10*c^3*d*e^4 - 13*b*c
^2*e^5)*x^3 + (80*c^3*d^2*e^3 - 110*b*c^2*d*e^4 + 33*b^2*c*e^5)*x^2 + 20*(18*c^3*d^3*e^2 - 23*b*c^2*d^2*e^3 +
6*b^2*c*d*e^4)*x)*sqrt(c*x^2 + b*x))/(c*e^8*x^2 + 2*c*d*e^7*x + c*d^2*e^6), 1/24*(15*(32*c^3*d^5 - 48*b*c^2*d^
4*e + 18*b^2*c*d^3*e^2 - b^3*d^2*e^3 + (32*c^3*d^3*e^2 - 48*b*c^2*d^2*e^3 + 18*b^2*c*d*e^4 - b^3*e^5)*x^2 + 2*
(32*c^3*d^4*e - 48*b*c^2*d^3*e^2 + 18*b^2*c*d^2*e^3 - b^3*d*e^4)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)
/(c*x)) + 15*(16*c^3*d^4 - 16*b*c^2*d^3*e + 3*b^2*c*d^2*e^2 + (16*c^3*d^2*e^2 - 16*b*c^2*d*e^3 + 3*b^2*c*e^4)*
x^2 + 2*(16*c^3*d^3*e - 16*b*c^2*d^2*e^2 + 3*b^2*c*d*e^3)*x)*sqrt(c*d^2 - b*d*e)*log((b*d + (2*c*d - b*e)*x +
2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d)) + (8*c^3*e^5*x^4 + 240*c^3*d^4*e - 300*b*c^2*d^3*e^2 + 75*
b^2*c*d^2*e^3 - 2*(10*c^3*d*e^4 - 13*b*c^2*e^5)*x^3 + (80*c^3*d^2*e^3 - 110*b*c^2*d*e^4 + 33*b^2*c*e^5)*x^2 +
20*(18*c^3*d^3*e^2 - 23*b*c^2*d^2*e^3 + 6*b^2*c*d*e^4)*x)*sqrt(c*x^2 + b*x))/(c*e^8*x^2 + 2*c*d*e^7*x + c*d^2*
e^6), 1/24*(30*(16*c^3*d^4 - 16*b*c^2*d^3*e + 3*b^2*c*d^2*e^2 + (16*c^3*d^2*e^2 - 16*b*c^2*d*e^3 + 3*b^2*c*e^4
)*x^2 + 2*(16*c^3*d^3*e - 16*b*c^2*d^2*e^2 + 3*b^2*c*d*e^3)*x)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*
e)*sqrt(c*x^2 + b*x)/((c*d - b*e)*x)) + 15*(32*c^3*d^5 - 48*b*c^2*d^4*e + 18*b^2*c*d^3*e^2 - b^3*d^2*e^3 + (32
*c^3*d^3*e^2 - 48*b*c^2*d^2*e^3 + 18*b^2*c*d*e^4 - b^3*e^5)*x^2 + 2*(32*c^3*d^4*e - 48*b*c^2*d^3*e^2 + 18*b^2*
c*d^2*e^3 - b^3*d*e^4)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (8*c^3*e^5*x^4 + 240*c^3*d^4*e -
300*b*c^2*d^3*e^2 + 75*b^2*c*d^2*e^3 - 2*(10*c^3*d*e^4 - 13*b*c^2*e^5)*x^3 + (80*c^3*d^2*e^3 - 110*b*c^2*d*e^
4 + 33*b^2*c*e^5)*x^2 + 20*(18*c^3*d^3*e^2 - 23*b*c^2*d^2*e^3 + 6*b^2*c*d*e^4)*x)*sqrt(c*x^2 + b*x))/(c*e^8*x^
2 + 2*c*d*e^7*x + c*d^2*e^6)]

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

[Out]

Timed out

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Giac [B]  time = 1.66918, size = 988, 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]

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

[Out]

5/4*(16*c^3*d^4 - 32*b*c^2*d^3*e + 19*b^2*c*d^2*e^2 - 3*b^3*d*e^3)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + b*x))*e
+ sqrt(c)*d)/sqrt(-c*d^2 + b*d*e))*e^(-6)/sqrt(-c*d^2 + b*d*e) + 5/16*(32*c^3*d^3 - 48*b*c^2*d^2*e + 18*b^2*c*
d*e^2 - b^3*e^3)*e^(-6)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/sqrt(c) + 1/24*sqrt(c*x^2 + b
*x)*(2*(4*c^2*x*e^(-3) - (18*c^4*d*e^14 - 13*b*c^3*e^15)*e^(-18)/c^2)*x + 3*(48*c^4*d^2*e^13 - 54*b*c^3*d*e^14
+ 11*b^2*c^2*e^15)*e^(-18)/c^2) + 1/4*(40*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*c^3*d^4*e + 72*(sqrt(c)*x - sqrt(
c*x^2 + b*x))^2*c^(7/2)*d^5 - 120*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b*c^(5/2)*d^4*e + 72*(sqrt(c)*x - sqrt(c*x
^2 + b*x))*b*c^3*d^5 - 80*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*b*c^2*d^3*e^2 - 124*(sqrt(c)*x - sqrt(c*x^2 + b*x)
)*b^2*c^2*d^4*e + 18*b^2*c^(5/2)*d^5 + 51*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b^2*c^(3/2)*d^3*e^2 - 27*b^3*c^(3/
2)*d^4*e + 49*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*b^2*c*d^2*e^3 + 59*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b^3*c*d^3*e
^2 - 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b^3*sqrt(c)*d^2*e^3 + 9*b^4*sqrt(c)*d^3*e^2 - 9*(sqrt(c)*x - sqrt(c*x
^2 + b*x))^3*b^3*d*e^4 - 7*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b^4*d^2*e^3)*e^(-6)/((sqrt(c)*x - sqrt(c*x^2 + b*x)
)^2*e + 2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c)*d + b*d)^2