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

Optimal. Leaf size=314 \[ -\frac {5 \sqrt {b x+c x^2} \left (-b^3 e^3-2 c e x \left (b^2 e^2-16 b c d e+16 c^2 d^2\right )+48 b^2 c d e^2-112 b c^2 d^2 e+64 c^3 d^3\right )}{64 c e^5}+\frac {5 \left (-b^4 e^4-16 b^3 c d e^3+144 b^2 c^2 d^2 e^2-256 b c^3 d^3 e+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{3/2} e^6}-\frac {5 d^{3/2} (c d-b e)^{3/2} (2 c d-b e) \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{2 e^6}-\frac {5 \left (b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{24 e^3}-\frac {\left (b x+c x^2\right )^{5/2}}{e (d+e x)} \]

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Rubi [A]  time = 0.39, antiderivative size = 314, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {732, 814, 843, 620, 206, 724} \begin {gather*} -\frac {5 \sqrt {b x+c x^2} \left (-2 c e x \left (b^2 e^2-16 b c d e+16 c^2 d^2\right )+48 b^2 c d e^2-b^3 e^3-112 b c^2 d^2 e+64 c^3 d^3\right )}{64 c e^5}+\frac {5 \left (144 b^2 c^2 d^2 e^2-16 b^3 c d e^3-b^4 e^4-256 b c^3 d^3 e+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{3/2} e^6}-\frac {5 d^{3/2} (c d-b e)^{3/2} (2 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 )}{2 e^6}-\frac {5 \left (b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{24 e^3}-\frac {\left (b x+c x^2\right )^{5/2}}{e (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(64*c^3*d^3 - 112*b*c^2*d^2*e + 48*b^2*c*d*e^2 - b^3*e^3 - 2*c*e*(16*c^2*d^2 - 16*b*c*d*e + b^2*e^2)*x)*Sq
rt[b*x + c*x^2])/(64*c*e^5) - (5*(8*c*d - 7*b*e - 6*c*e*x)*(b*x + c*x^2)^(3/2))/(24*e^3) - (b*x + c*x^2)^(5/2)
/(e*(d + e*x)) + (5*(128*c^4*d^4 - 256*b*c^3*d^3*e + 144*b^2*c^2*d^2*e^2 - 16*b^3*c*d*e^3 - b^4*e^4)*ArcTanh[(
Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*c^(3/2)*e^6) - (5*d^(3/2)*(c*d - b*e)^(3/2)*(2*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])])/(2*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 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)^2} \, dx &=-\frac {\left (b x+c x^2\right )^{5/2}}{e (d+e x)}+\frac {5 \int \frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{d+e x} \, dx}{2 e}\\ &=-\frac {5 (8 c d-7 b e-6 c e x) \left (b x+c x^2\right )^{3/2}}{24 e^3}-\frac {\left (b x+c x^2\right )^{5/2}}{e (d+e x)}-\frac {5 \int \frac {\left (-b c d (8 c d-7 b e)-c \left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) x\right ) \sqrt {b x+c x^2}}{d+e x} \, dx}{16 c e^3}\\ &=-\frac {5 \left (64 c^3 d^3-112 b c^2 d^2 e+48 b^2 c d e^2-b^3 e^3-2 c e \left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) x\right ) \sqrt {b x+c x^2}}{64 c e^5}-\frac {5 (8 c d-7 b e-6 c e x) \left (b x+c x^2\right )^{3/2}}{24 e^3}-\frac {\left (b x+c x^2\right )^{5/2}}{e (d+e x)}+\frac {5 \int \frac {\frac {1}{2} b c d \left (64 c^3 d^3-112 b c^2 d^2 e+48 b^2 c d e^2-b^3 e^3\right )+\frac {1}{2} c \left (128 c^4 d^4-256 b c^3 d^3 e+144 b^2 c^2 d^2 e^2-16 b^3 c d e^3-b^4 e^4\right ) x}{(d+e x) \sqrt {b x+c x^2}} \, dx}{64 c^2 e^5}\\ &=-\frac {5 \left (64 c^3 d^3-112 b c^2 d^2 e+48 b^2 c d e^2-b^3 e^3-2 c e \left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) x\right ) \sqrt {b x+c x^2}}{64 c e^5}-\frac {5 (8 c d-7 b e-6 c e x) \left (b x+c x^2\right )^{3/2}}{24 e^3}-\frac {\left (b x+c x^2\right )^{5/2}}{e (d+e x)}-\frac {\left (5 d^2 (c d-b e)^2 (2 c d-b e)\right ) \int \frac {1}{(d+e x) \sqrt {b x+c x^2}} \, dx}{2 e^6}+\frac {\left (5 \left (128 c^4 d^4-256 b c^3 d^3 e+144 b^2 c^2 d^2 e^2-16 b^3 c d e^3-b^4 e^4\right )\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{128 c e^6}\\ &=-\frac {5 \left (64 c^3 d^3-112 b c^2 d^2 e+48 b^2 c d e^2-b^3 e^3-2 c e \left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) x\right ) \sqrt {b x+c x^2}}{64 c e^5}-\frac {5 (8 c d-7 b e-6 c e x) \left (b x+c x^2\right )^{3/2}}{24 e^3}-\frac {\left (b x+c x^2\right )^{5/2}}{e (d+e x)}+\frac {\left (5 d^2 (c d-b e)^2 (2 c d-b e)\right ) \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 )}{e^6}+\frac {\left (5 \left (128 c^4 d^4-256 b c^3 d^3 e+144 b^2 c^2 d^2 e^2-16 b^3 c d e^3-b^4 e^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{64 c e^6}\\ &=-\frac {5 \left (64 c^3 d^3-112 b c^2 d^2 e+48 b^2 c d e^2-b^3 e^3-2 c e \left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) x\right ) \sqrt {b x+c x^2}}{64 c e^5}-\frac {5 (8 c d-7 b e-6 c e x) \left (b x+c x^2\right )^{3/2}}{24 e^3}-\frac {\left (b x+c x^2\right )^{5/2}}{e (d+e x)}+\frac {5 \left (128 c^4 d^4-256 b c^3 d^3 e+144 b^2 c^2 d^2 e^2-16 b^3 c d e^3-b^4 e^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{3/2} e^6}-\frac {5 d^{3/2} (c d-b e)^{3/2} (2 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 )}{2 e^6}\\ \end {align*}

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Mathematica [A]  time = 1.27, size = 347, normalized size = 1.11 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (-\frac {960 c^{3/2} d^{3/2} \sqrt {c d-b e} \left (b^2 e^2-3 b c d e+2 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 {\sqrt {c} e \sqrt {x} \left (15 b^3 e^3 (d+e x)+2 b^2 c e^2 \left (-360 d^2-205 d e x+59 e^2 x^2\right )+8 b c^2 e \left (210 d^3+110 d^2 e x-35 d e^2 x^2+17 e^3 x^3\right )-16 c^3 \left (60 d^4+30 d^3 e x-10 d^2 e^2 x^2+5 d e^3 x^3-3 e^4 x^4\right )\right )}{d+e x}-\frac {15 \left (b^4 e^4+16 b^3 c d e^3-144 b^2 c^2 d^2 e^2+256 b c^3 d^3 e-128 c^4 d^4\right ) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {\frac {c x}{b}+1}}\right )}{192 c^{3/2} e^6 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*((Sqrt[c]*e*Sqrt[x]*(15*b^3*e^3*(d + e*x) + 2*b^2*c*e^2*(-360*d^2 - 205*d*e*x + 59*e^2*x^2)
+ 8*b*c^2*e*(210*d^3 + 110*d^2*e*x - 35*d*e^2*x^2 + 17*e^3*x^3) - 16*c^3*(60*d^4 + 30*d^3*e*x - 10*d^2*e^2*x^
2 + 5*d*e^3*x^3 - 3*e^4*x^4)))/(d + e*x) - (15*(-128*c^4*d^4 + 256*b*c^3*d^3*e - 144*b^2*c^2*d^2*e^2 + 16*b^3*
c*d*e^3 + b^4*e^4)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[b]*Sqrt[1 + (c*x)/b]) - (960*c^(3/2)*d^(3/2)*Sqrt
[c*d - b*e]*(2*c^2*d^2 - 3*b*c*d*e + b^2*e^2)*ArcTanh[(Sqrt[c*d - b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])/Sqrt
[b + c*x]))/(192*c^(3/2)*e^6*Sqrt[x])

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IntegrateAlgebraic [A]  time = 2.53, size = 383, normalized size = 1.22 \begin {gather*} -\frac {5 \sqrt {c d-b e} \left (b^2 d^{3/2} e^2-3 b c d^{5/2} e+2 c^2 d^{7/2}\right ) \tanh ^{-1}\left (\frac {-e \sqrt {b x+c x^2}+\sqrt {c} d+\sqrt {c} e x}{\sqrt {d} \sqrt {c d-b e}}\right )}{e^6}+\frac {\sqrt {b x+c x^2} \left (15 b^3 d e^3+15 b^3 e^4 x-720 b^2 c d^2 e^2-410 b^2 c d e^3 x+118 b^2 c e^4 x^2+1680 b c^2 d^3 e+880 b c^2 d^2 e^2 x-280 b c^2 d e^3 x^2+136 b c^2 e^4 x^3-960 c^3 d^4-480 c^3 d^3 e x+160 c^3 d^2 e^2 x^2-80 c^3 d e^3 x^3+48 c^3 e^4 x^4\right )}{192 c e^5 (d+e x)}-\frac {5 \left (-b^4 e^4-16 b^3 c d e^3+144 b^2 c^2 d^2 e^2-256 b c^3 d^3 e+128 c^4 d^4\right ) \log \left (-2 c^{3/2} \sqrt {b x+c x^2}+b c+2 c^2 x\right )}{128 c^{3/2} e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-960*c^3*d^4 + 1680*b*c^2*d^3*e - 720*b^2*c*d^2*e^2 + 15*b^3*d*e^3 - 480*c^3*d^3*e*x + 880
*b*c^2*d^2*e^2*x - 410*b^2*c*d*e^3*x + 15*b^3*e^4*x + 160*c^3*d^2*e^2*x^2 - 280*b*c^2*d*e^3*x^2 + 118*b^2*c*e^
4*x^2 - 80*c^3*d*e^3*x^3 + 136*b*c^2*e^4*x^3 + 48*c^3*e^4*x^4))/(192*c*e^5*(d + e*x)) - (5*Sqrt[c*d - b*e]*(2*
c^2*d^(7/2) - 3*b*c*d^(5/2)*e + b^2*d^(3/2)*e^2)*ArcTanh[(Sqrt[c]*d + Sqrt[c]*e*x - e*Sqrt[b*x + c*x^2])/(Sqrt
[d]*Sqrt[c*d - b*e])])/e^6 - (5*(128*c^4*d^4 - 256*b*c^3*d^3*e + 144*b^2*c^2*d^2*e^2 - 16*b^3*c*d*e^3 - b^4*e^
4)*Log[b*c + 2*c^2*x - 2*c^(3/2)*Sqrt[b*x + c*x^2]])/(128*c^(3/2)*e^6)

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fricas [A]  time = 1.66, size = 1875, normalized size = 5.97

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)^2,x, algorithm="fricas")

[Out]

[-1/384*(15*(128*c^4*d^5 - 256*b*c^3*d^4*e + 144*b^2*c^2*d^3*e^2 - 16*b^3*c*d^2*e^3 - b^4*d*e^4 + (128*c^4*d^4
*e - 256*b*c^3*d^3*e^2 + 144*b^2*c^2*d^2*e^3 - 16*b^3*c*d*e^4 - b^4*e^5)*x)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x
^2 + b*x)*sqrt(c)) - 960*(2*c^4*d^4 - 3*b*c^3*d^3*e + b^2*c^2*d^2*e^2 + (2*c^4*d^3*e - 3*b*c^3*d^2*e^2 + b^2*c
^2*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*(48*c^4*e^5*x^4 - 960*c^4*d^4*e + 1680*b*c^3*d^3*e^2 - 720*b^2*c^2*d^2*e^3 + 15*b^3*c*d*e^4 - 8*(10*c^
4*d*e^4 - 17*b*c^3*e^5)*x^3 + 2*(80*c^4*d^2*e^3 - 140*b*c^3*d*e^4 + 59*b^2*c^2*e^5)*x^2 - 5*(96*c^4*d^3*e^2 -
176*b*c^3*d^2*e^3 + 82*b^2*c^2*d*e^4 - 3*b^3*c*e^5)*x)*sqrt(c*x^2 + b*x))/(c^2*e^7*x + c^2*d*e^6), -1/384*(192
0*(2*c^4*d^4 - 3*b*c^3*d^3*e + b^2*c^2*d^2*e^2 + (2*c^4*d^3*e - 3*b*c^3*d^2*e^2 + b^2*c^2*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*(128*c^4*d^5 - 256*b*c^3*d^4*e
+ 144*b^2*c^2*d^3*e^2 - 16*b^3*c*d^2*e^3 - b^4*d*e^4 + (128*c^4*d^4*e - 256*b*c^3*d^3*e^2 + 144*b^2*c^2*d^2*e
^3 - 16*b^3*c*d*e^4 - b^4*e^5)*x)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(48*c^4*e^5*x^4 - 9
60*c^4*d^4*e + 1680*b*c^3*d^3*e^2 - 720*b^2*c^2*d^2*e^3 + 15*b^3*c*d*e^4 - 8*(10*c^4*d*e^4 - 17*b*c^3*e^5)*x^3
+ 2*(80*c^4*d^2*e^3 - 140*b*c^3*d*e^4 + 59*b^2*c^2*e^5)*x^2 - 5*(96*c^4*d^3*e^2 - 176*b*c^3*d^2*e^3 + 82*b^2*
c^2*d*e^4 - 3*b^3*c*e^5)*x)*sqrt(c*x^2 + b*x))/(c^2*e^7*x + c^2*d*e^6), -1/192*(15*(128*c^4*d^5 - 256*b*c^3*d^
4*e + 144*b^2*c^2*d^3*e^2 - 16*b^3*c*d^2*e^3 - b^4*d*e^4 + (128*c^4*d^4*e - 256*b*c^3*d^3*e^2 + 144*b^2*c^2*d^
2*e^3 - 16*b^3*c*d*e^4 - b^4*e^5)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - 480*(2*c^4*d^4 - 3*b*
c^3*d^3*e + b^2*c^2*d^2*e^2 + (2*c^4*d^3*e - 3*b*c^3*d^2*e^2 + b^2*c^2*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)) - (48*c^4*e^5*x^4 - 960*c^4*d^4*e + 16
80*b*c^3*d^3*e^2 - 720*b^2*c^2*d^2*e^3 + 15*b^3*c*d*e^4 - 8*(10*c^4*d*e^4 - 17*b*c^3*e^5)*x^3 + 2*(80*c^4*d^2*
e^3 - 140*b*c^3*d*e^4 + 59*b^2*c^2*e^5)*x^2 - 5*(96*c^4*d^3*e^2 - 176*b*c^3*d^2*e^3 + 82*b^2*c^2*d*e^4 - 3*b^3
*c*e^5)*x)*sqrt(c*x^2 + b*x))/(c^2*e^7*x + c^2*d*e^6), -1/192*(960*(2*c^4*d^4 - 3*b*c^3*d^3*e + b^2*c^2*d^2*e^
2 + (2*c^4*d^3*e - 3*b*c^3*d^2*e^2 + b^2*c^2*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*(128*c^4*d^5 - 256*b*c^3*d^4*e + 144*b^2*c^2*d^3*e^2 - 16*b^3*c*d^2*e^3 - b
^4*d*e^4 + (128*c^4*d^4*e - 256*b*c^3*d^3*e^2 + 144*b^2*c^2*d^2*e^3 - 16*b^3*c*d*e^4 - b^4*e^5)*x)*sqrt(-c)*ar
ctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (48*c^4*e^5*x^4 - 960*c^4*d^4*e + 1680*b*c^3*d^3*e^2 - 720*b^2*c^2*d^
2*e^3 + 15*b^3*c*d*e^4 - 8*(10*c^4*d*e^4 - 17*b*c^3*e^5)*x^3 + 2*(80*c^4*d^2*e^3 - 140*b*c^3*d*e^4 + 59*b^2*c^
2*e^5)*x^2 - 5*(96*c^4*d^3*e^2 - 176*b*c^3*d^2*e^3 + 82*b^2*c^2*d*e^4 - 3*b^3*c*e^5)*x)*sqrt(c*x^2 + b*x))/(c^
2*e^7*x + c^2*d*e^6)]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.06, size = 2534, normalized size = 8.07

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4/e/(b*e-c*d)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(3/2)*x*b*c-55/4/e^4/(b*e-c*d)*d^3*((x+d/e
)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*b*c^2+25/4/e^3/(b*e-c*d)*d^2*ln(((x+d/e)*c+1/2*(b*e-2*c*d)/
e)/c^(1/2)+((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))*c^(1/2)*b^3-125/8/e^4/(b*e-c*d)*d^3*ln((
(x+d/e)*c+1/2*(b*e-2*c*d)/e)/c^(1/2)+((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))*c^(3/2)*b^2-5/
e^7/(b*e-c*d)*d^6/(-(b*e-c*d)*d/e^2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)^(
1/2)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d/e))*c^4-5/4/e^2/(b*e-c*d)*d*((x+d/e)^2*c-
(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(3/2)*x*c^2-75/128/e^2/(b*e-c*d)*d*ln(((x+d/e)*c+1/2*(b*e-2*c*d)/e)/c^(
1/2)+((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/c^(1/2)*b^4-25/8/e^2/(b*e-c*d)*d*((x+d/e)^2*c-
(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(3/2)*b*c-5/2/e^4/(b*e-c*d)*d^3*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d
)*(x+d/e)/e)^(1/2)*x*c^3+15/e^5/(b*e-c*d)*d^4*ln(((x+d/e)*c+1/2*(b*e-2*c*d)/e)/c^(1/2)+((x+d/e)^2*c-(b*e-c*d)*
d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))*c^(5/2)*b+25/2/e^3/(b*e-c*d)*d^2*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*
(x+d/e)/e)^(1/2)*b^2*c-5/2/e^3/(b*e-c*d)*d^2/(-(b*e-c*d)*d/e^2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/
e)/e+2*(-(b*e-c*d)*d/e^2)^(1/2)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d/e))*b^4+1/e/(b
*e-c*d)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(5/2)*c+35/24/e/(b*e-c*d)*((x+d/e)^2*c-(b*e-c*d)*d
/e^2+(b*e-2*c*d)*(x+d/e)/e)^(3/2)*b^2+1/(b*e-c*d)/d/(x+d/e)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e
)^(7/2)-1/(b*e-c*d)/d*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(5/2)*b-5/e^6/(b*e-c*d)*d^5*ln(((x+d
/e)*c+1/2*(b*e-2*c*d)/e)/c^(1/2)+((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))*c^(7/2)-245/64/e^2
/(b*e-c*d)*d*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*b^3+5/e^5/(b*e-c*d)*d^4*((x+d/e)^2*c-(b
*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*c^3+5/32/e/(b*e-c*d)*b^3*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(
x+d/e)/e)^(1/2)*x+5/64/e/(b*e-c*d)/c*b^4*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)-5/128/e/(b*
e-c*d)/c^(3/2)*ln(((x+d/e)*c+1/2*(b*e-2*c*d)/e)/c^(1/2)+((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1
/2))*b^5+5/3/e^3/(b*e-c*d)*d^2*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(3/2)*c^2-c/(b*e-c*d)/d*((x
+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(5/2)*x-45/2/e^5/(b*e-c*d)*d^4/(-(b*e-c*d)*d/e^2)^(1/2)*ln((-
2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)^(1/2)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x
+d/e)/e)^(1/2))/(x+d/e))*b^2*c^2+35/2/e^6/(b*e-c*d)*d^5/(-(b*e-c*d)*d/e^2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2
*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)^(1/2)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d/e))
*b*c^3-85/32/e^2/(b*e-c*d)*d*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*x*b^2*c+5/e^3/(b*e-c*d)
*d^2*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*x*b*c^2+25/2/e^4/(b*e-c*d)*d^3/(-(b*e-c*d)*d/e^
2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)^(1/2)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+
(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d/e))*b^3*c

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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)^2,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 )}^2} \,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)^2,x)

[Out]

int((b*x + c*x^2)^(5/2)/(d + e*x)^2, 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 )^{2}}\, 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)**2,x)

[Out]

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

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