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

Optimal. Leaf size=360 \[ \frac {\left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{3/2} e^5}-\frac {\sqrt {a+b x+c x^2} \left (-2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (7 b d-4 a e)+4 b c e^2 (12 b d-11 a e)-b^3 e^3+64 c^3 d^3\right )}{32 c e^4}-\frac {(2 c d-b e) \left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e^5}-\frac {\left (a+b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{12 e^2} \]

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Rubi [A]  time = 0.60, antiderivative size = 360, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {814, 843, 621, 206, 724} \begin {gather*} \frac {\left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{3/2} e^5}-\frac {\sqrt {a+b x+c x^2} \left (-2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (7 b d-4 a e)+4 b c e^2 (12 b d-11 a e)-b^3 e^3+64 c^3 d^3\right )}{32 c e^4}-\frac {(2 c d-b e) \left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e^5}-\frac {\left (a+b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{12 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((64*c^3*d^3 - b^3*e^3 + 4*b*c*e^2*(12*b*d - 11*a*e) - 16*c^2*d*e*(7*b*d - 4*a*e) - 2*c*e*(16*c^2*d^2 + b^2*e
^2 - 4*c*e*(4*b*d - 3*a*e))*x)*Sqrt[a + b*x + c*x^2])/(32*c*e^4) - ((8*c*d - 7*b*e - 6*c*e*x)*(a + b*x + c*x^2
)^(3/2))/(12*e^2) + ((128*c^4*d^4 - b^4*e^4 - 8*b^2*c*e^3*(2*b*d - 3*a*e) - 64*c^3*d^2*e*(4*b*d - 3*a*e) + 48*
c^2*e^2*(3*b^2*d^2 - 4*a*b*d*e + a^2*e^2))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(64*c^(3/2)
*e^5) - ((2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2)^(3/2)*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b
*d*e + a*e^2]*Sqrt[a + b*x + c*x^2])])/e^5

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 621

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

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 {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx &=-\frac {(8 c d-7 b e-6 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^2}-\frac {\int \frac {\left (c \left (7 b^2 d e+4 a c d e-8 b \left (c d^2+a e^2\right )\right )-c \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{d+e x} \, dx}{8 c e^2}\\ &=-\frac {\left (64 c^3 d^3-b^3 e^3+4 b c e^2 (12 b d-11 a e)-16 c^2 d e (7 b d-4 a e)-2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{32 c e^4}-\frac {(8 c d-7 b e-6 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^2}+\frac {\int \frac {\frac {1}{2} c \left (d \left (4 b c d-b^2 e-4 a c e\right ) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right )+4 c e (b d-2 a e) \left (7 b^2 d e+4 a c d e-8 b \left (c d^2+a e^2\right )\right )\right )+\frac {1}{2} c \left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\right )\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{32 c^2 e^4}\\ &=-\frac {\left (64 c^3 d^3-b^3 e^3+4 b c e^2 (12 b d-11 a e)-16 c^2 d e (7 b d-4 a e)-2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{32 c e^4}-\frac {(8 c d-7 b e-6 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^2}-\frac {\left ((2 c d-b e) \left (c d^2-b d e+a e^2\right )^2\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{e^5}+\frac {\left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{64 c e^5}\\ &=-\frac {\left (64 c^3 d^3-b^3 e^3+4 b c e^2 (12 b d-11 a e)-16 c^2 d e (7 b d-4 a e)-2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{32 c e^4}-\frac {(8 c d-7 b e-6 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^2}+\frac {\left (2 (2 c d-b e) \left (c d^2-b d e+a e^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac {-b d+2 a e-(2 c d-b e) x}{\sqrt {a+b x+c x^2}}\right )}{e^5}+\frac {\left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{32 c e^5}\\ &=-\frac {\left (64 c^3 d^3-b^3 e^3+4 b c e^2 (12 b d-11 a e)-16 c^2 d e (7 b d-4 a e)-2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{32 c e^4}-\frac {(8 c d-7 b e-6 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^2}+\frac {\left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{3/2} e^5}-\frac {(2 c d-b e) \left (c d^2-b d e+a e^2\right )^{3/2} \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{e^5}\\ \end {align*}

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Mathematica [A]  time = 0.74, size = 342, normalized size = 0.95 \begin {gather*} \frac {3 \left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )+2 \sqrt {c} \left (e \sqrt {a+x (b+c x)} \left (8 c^2 e \left (a e (15 e x-32 d)+b \left (42 d^2-20 d e x+13 e^2 x^2\right )\right )+2 b c e^2 (94 a e-72 b d+31 b e x)+3 b^3 e^3-16 c^3 \left (12 d^3-6 d^2 e x+4 d e^2 x^2-3 e^3 x^3\right )\right )+96 c (2 c d-b e) \left (e (a e-b d)+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac {2 a e-b d+b e x-2 c d x}{2 \sqrt {a+x (b+c x)} \sqrt {e (a e-b d)+c d^2}}\right )\right )}{192 c^{3/2} e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(128*c^4*d^4 - b^4*e^4 - 8*b^2*c*e^3*(2*b*d - 3*a*e) - 64*c^3*d^2*e*(4*b*d - 3*a*e) + 48*c^2*e^2*(3*b^2*d^2
- 4*a*b*d*e + a^2*e^2))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] + 2*Sqrt[c]*(e*Sqrt[a + x*(b +
c*x)]*(3*b^3*e^3 + 2*b*c*e^2*(-72*b*d + 94*a*e + 31*b*e*x) - 16*c^3*(12*d^3 - 6*d^2*e*x + 4*d*e^2*x^2 - 3*e^3
*x^3) + 8*c^2*e*(a*e*(-32*d + 15*e*x) + b*(42*d^2 - 20*d*e*x + 13*e^2*x^2))) + 96*c*(2*c*d - b*e)*(c*d^2 + e*(
-(b*d) + a*e))^(3/2)*ArcTanh[(-(b*d) + 2*a*e - 2*c*d*x + b*e*x)/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + x*(
b + c*x)])]))/(192*c^(3/2)*e^5)

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IntegrateAlgebraic [A]  time = 2.48, size = 500, normalized size = 1.39 \begin {gather*} \frac {\left (-48 a^2 c^2 e^4-24 a b^2 c e^4+192 a b c^2 d e^3-192 a c^3 d^2 e^2+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 {a+b x+c x^2}+b c+2 c^2 x\right )}{64 c^{3/2} e^5}-\frac {2 \left (b^2 d e^2 \sqrt {-a e^2+b d e-c d^2}+2 c^2 d^3 \sqrt {-a e^2+b d e-c d^2}-3 b c d^2 e \sqrt {-a e^2+b d e-c d^2}+2 a c d e^2 \sqrt {-a e^2+b d e-c d^2}-a b e^3 \sqrt {-a e^2+b d e-c d^2}\right ) \tan ^{-1}\left (\frac {-e \sqrt {a+b x+c x^2}+\sqrt {c} d+\sqrt {c} e x}{\sqrt {-a e^2+b d e-c d^2}}\right )}{e^5}+\frac {\sqrt {a+b x+c x^2} \left (188 a b c e^3-256 a c^2 d e^2+120 a c^2 e^3 x+3 b^3 e^3-144 b^2 c d e^2+62 b^2 c e^3 x+336 b c^2 d^2 e-160 b c^2 d e^2 x+104 b c^2 e^3 x^2-192 c^3 d^3+96 c^3 d^2 e x-64 c^3 d e^2 x^2+48 c^3 e^3 x^3\right )}{96 c e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(-192*c^3*d^3 + 336*b*c^2*d^2*e - 144*b^2*c*d*e^2 - 256*a*c^2*d*e^2 + 3*b^3*e^3 + 188*a
*b*c*e^3 + 96*c^3*d^2*e*x - 160*b*c^2*d*e^2*x + 62*b^2*c*e^3*x + 120*a*c^2*e^3*x - 64*c^3*d*e^2*x^2 + 104*b*c^
2*e^3*x^2 + 48*c^3*e^3*x^3))/(96*c*e^4) - (2*(2*c^2*d^3*Sqrt[-(c*d^2) + b*d*e - a*e^2] - 3*b*c*d^2*e*Sqrt[-(c*
d^2) + b*d*e - a*e^2] + b^2*d*e^2*Sqrt[-(c*d^2) + b*d*e - a*e^2] + 2*a*c*d*e^2*Sqrt[-(c*d^2) + b*d*e - a*e^2]
- a*b*e^3*Sqrt[-(c*d^2) + b*d*e - a*e^2])*ArcTan[(Sqrt[c]*d + Sqrt[c]*e*x - e*Sqrt[a + b*x + c*x^2])/Sqrt[-(c*
d^2) + b*d*e - a*e^2]])/e^5 + ((-128*c^4*d^4 + 256*b*c^3*d^3*e - 144*b^2*c^2*d^2*e^2 - 192*a*c^3*d^2*e^2 + 16*
b^3*c*d*e^3 + 192*a*b*c^2*d*e^3 + b^4*e^4 - 24*a*b^2*c*e^4 - 48*a^2*c^2*e^4)*Log[b*c + 2*c^2*x - 2*c^(3/2)*Sqr
t[a + b*x + c*x^2]])/(64*c^(3/2)*e^5)

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

[Out]

Timed out

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r: Bad Argument Type

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maple [B]  time = 0.06, size = 3119, normalized size = 8.66 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6/e^3/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d
^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*a*b*d^2*c-4/e^4*ln(
((x+d/e)*c+1/2*(b*e-2*c*d)/e)/c^(1/2)+((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(3/
2)*d^3*b+3/4*c/e*(c*x^2+b*x+a)^(1/2)*x*a+1/4/e*(c*x^2+b*x+a)^(3/2)*b+1/3/e*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+
(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)*b+2/e^5*ln(((x+d/e)*c+1/2*(b*e-2*c*d)/e)/c^(1/2)+((x+d/e)^2*c+(b*e-2*c*d)*(x+d/
e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(5/2)*d^4+3/4*c^(1/2)/e*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^2
+3/64/c^(3/2)/e*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^4-3/32/c/e*(c*x^2+b*x+a)^(1/2)*b^3-3/16/e*(c*x^2
+b*x+a)^(1/2)*x*b^2+1/2*c/e*(c*x^2+b*x+a)^(3/2)*x-1/16/e/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-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*b^4-2/e^4*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a
*e^2-b*d*e+c*d^2)/e^2)^(1/2)*c^2*d^3+3/8/e*(c*x^2+b*x+a)^(1/2)*b*a+1/8/e/c*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+
(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b^3-2/3/e^2*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)*c
*d-3/2/e^2*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b^2*d+1/e*((x+d/e)^2*c+(b*e-2*c*d
)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*a*b+1/4/e*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^
2)^(1/2)*x*b^2+2/e^6/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a
*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*c^3*d
^5+7/2/e^3*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b*d^2*c+3/4/e/c^(1/2)*ln(((x+d/e)
*c+1/2*(b*e-2*c*d)/e)/c^(1/2)+((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*a*b^2-1/e/((a
*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(
1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*a^2*b+1/e^3*((x+d/e)^2*c+(b*e
-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*c^2*d^2-1/e^3/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c*
d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a
*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*b^3*d^2+3/e^3*ln(((x+d/e)*c+1/2*(b*e-2*c*d)/e)/c^(1/2)+((x+d/e)^2*c+(b*
e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(3/2)*d^2*a+9/4/e^3*ln(((x+d/e)*c+1/2*(b*e-2*c*d)/e)/c^(1
/2)+((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(1/2)*b^2*d^2-1/4/e^2/c^(1/2)*ln(((x+
d/e)*c+1/2*(b*e-2*c*d)/e)/c^(1/2)+((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*b^3*d-3/e
^2*c^(1/2)*ln(((x+d/e)*c+1/2*(b*e-2*c*d)/e)/c^(1/2)+((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2
)^(1/2))*a*b*d+4/e^4/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a
*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*b^2*d
^3*c-1/e^2*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*b*c*d+2/e^2/((a*e^2-b*d*e+c*d^2
)/e^2)^(1/2)*ln(((b*e-2*c*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*
c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*a^2*c*d+4/e^4/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)
*ln(((b*e-2*c*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d
)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*a*c^2*d^3+2/e^2/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-
2*c*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/
e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*a*b^2*d-5/e^5/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c*d)*(x+d/
e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d
*e+c*d^2)/e^2)^(1/2))/(x+d/e))*b*d^4*c^2-3/8/c^(1/2)/e*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^2*a-2/e^2
*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*a*c*d

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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((2*c*x+b)*(c*x^2+b*x+a)^(3/2)/(e*x+d),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-2*c*d>0)', see `assume?` f
or more details)Is b*e-2*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 (b+2\,c\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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