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

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

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Rubi [A]  time = 0.33, antiderivative size = 282, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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 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 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]

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]

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*}

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Mathematica [A]  time = 1.42, size = 308, normalized size = 1.09 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\frac {30 \sqrt {d} \sqrt {c d-b e} \left (3 b^2 e^2-16 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt {c d-b e}}{\sqrt {d} \sqrt {b+c x}}\right )}{\sqrt {b+c x}}+\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 (150 d^3+230 d^2 e x+55 d e^2 x^2-13 e^3 x^3\right )+4 c^2 \left (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\right )\right )}{(d+e x)^2}+\frac {15 \left (b^3 e^3-18 b^2 c d e^2+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}} \end {gather*}

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]*Sqrt[c*d - b*e]*(16*c^2*d^2 - 16*b*c*d*e + 3*b^2*e^2)*ArcTanh[(Sqr
t[c*d - b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])/Sqrt[b + c*x]))/(24*e^6*Sqrt[x])

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IntegrateAlgebraic [B]  time = 118.87, size = 9721, normalized size = 34.47 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Result too large to show

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fricas [B]  time = 0.71, size = 2033, normalized size = 7.21

result too large to display

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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giac [B]  time = 0.50, size = 732, normalized size = 2.60 \begin {gather*} \frac {5 \, {\left (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}\right )} \arctan \left (-\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} + b d e}}\right ) e^{\left (-6\right )}}{4 \, \sqrt {-c d^{2} + b d e}} + \frac {5 \, {\left (32 \, c^{3} d^{3} - 48 \, b c^{2} d^{2} e + 18 \, b^{2} c d e^{2} - b^{3} e^{3}\right )} e^{\left (-6\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{16 \, \sqrt {c}} + \frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, c^{2} x e^{\left (-3\right )} - \frac {{\left (18 \, c^{4} d e^{14} - 13 \, b c^{3} e^{15}\right )} e^{\left (-18\right )}}{c^{2}}\right )} x + \frac {3 \, {\left (48 \, c^{4} d^{2} e^{13} - 54 \, b c^{3} d e^{14} + 11 \, b^{2} c^{2} e^{15}\right )} e^{\left (-18\right )}}{c^{2}}\right )} + \frac {{\left (40 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} c^{3} d^{4} e + 72 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} c^{\frac {7}{2}} d^{5} - 120 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b c^{\frac {5}{2}} d^{4} e + 72 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b c^{3} d^{5} - 80 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} b c^{2} d^{3} e^{2} - 124 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{2} c^{2} d^{4} e + 18 \, b^{2} c^{\frac {5}{2}} d^{5} + 51 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b^{2} c^{\frac {3}{2}} d^{3} e^{2} - 27 \, b^{3} c^{\frac {3}{2}} d^{4} e + 49 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} b^{2} c d^{2} e^{3} + 59 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{3} c d^{3} e^{2} - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b^{3} \sqrt {c} d^{2} e^{3} + 9 \, b^{4} \sqrt {c} d^{3} e^{2} - 9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} b^{3} d e^{4} - 7 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{4} d^{2} e^{3}\right )} e^{\left (-6\right )}}{4 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} e + 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} d + b d\right )}^{2}} \end {gather*}

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

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maple [B]  time = 0.08, size = 5534, normalized size = 19.62 \begin {gather*} \text {output too large to display} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-c*d>0)', see `assume?` for
more details)Is b*e-c*d zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}}{{\left (d+e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x \left (b + c x\right )\right )^{\frac {5}{2}}}{\left (d + e x\right )^{3}}\, dx \end {gather*}

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]

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

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