3.2348 \(\int \frac {(a+b x+c x^2)^{3/2}}{(d+e x)^2} \, dx\)
Optimal. Leaf size=227 \[ \frac {3 \left (-4 c e (2 b d-a e)+b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c} e^4}-\frac {3 (2 c d-b e) \sqrt {a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{2 e^4}-\frac {3 \sqrt {a+b x+c x^2} (-3 b e+4 c d-2 c e x)}{4 e^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{e (d+e x)} \]
[Out]
-(c*x^2+b*x+a)^(3/2)/e/(e*x+d)+3/8*(8*c^2*d^2+b^2*e^2-4*c*e*(-a*e+2*b*d))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2
+b*x+a)^(1/2))/e^4/c^(1/2)-3/2*(-b*e+2*c*d)*arctanh(1/2*(b*d-2*a*e+(-b*e+2*c*d)*x)/(a*e^2-b*d*e+c*d^2)^(1/2)/(
c*x^2+b*x+a)^(1/2))*(a*e^2-b*d*e+c*d^2)^(1/2)/e^4-3/4*(-2*c*e*x-3*b*e+4*c*d)*(c*x^2+b*x+a)^(1/2)/e^3
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Rubi [A] time = 0.27, antiderivative size = 227, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used =
{732, 814, 843, 621, 206, 724} \[ \frac {3 \left (-4 c e (2 b d-a e)+b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c} e^4}-\frac {3 (2 c d-b e) \sqrt {a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{2 e^4}-\frac {3 \sqrt {a+b x+c x^2} (-3 b e+4 c d-2 c e x)}{4 e^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{e (d+e x)} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*x + c*x^2)^(3/2)/(d + e*x)^2,x]
[Out]
(-3*(4*c*d - 3*b*e - 2*c*e*x)*Sqrt[a + b*x + c*x^2])/(4*e^3) - (a + b*x + c*x^2)^(3/2)/(e*(d + e*x)) + (3*(8*c
^2*d^2 + b^2*e^2 - 4*c*e*(2*b*d - a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*Sqrt[c]*e^4
) - (3*(2*c*d - b*e)*Sqrt[c*d^2 - b*d*e + a*e^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])])/(2*e^4)
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 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 (a+b x+c x^2\right )^{3/2}}{(d+e x)^2} \, dx &=-\frac {\left (a+b x+c x^2\right )^{3/2}}{e (d+e x)}+\frac {3 \int \frac {(b+2 c x) \sqrt {a+b x+c x^2}}{d+e x} \, dx}{2 e}\\ &=-\frac {3 (4 c d-3 b e-2 c e x) \sqrt {a+b x+c x^2}}{4 e^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{e (d+e x)}-\frac {3 \int \frac {c \left (3 b^2 d e+4 a c d e-4 b \left (c d^2+a e^2\right )\right )-c \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{8 c e^3}\\ &=-\frac {3 (4 c d-3 b e-2 c e x) \sqrt {a+b x+c x^2}}{4 e^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{e (d+e x)}-\frac {\left (3 (2 c d-b e) \left (c d^2-b d e+a e^2\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{2 e^4}+\frac {\left (3 \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{8 e^4}\\ &=-\frac {3 (4 c d-3 b e-2 c e x) \sqrt {a+b x+c x^2}}{4 e^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{e (d+e x)}+\frac {\left (3 (2 c d-b e) \left (c d^2-b d e+a e^2\right )\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^4}+\frac {\left (3 \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\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 )}{4 e^4}\\ &=-\frac {3 (4 c d-3 b e-2 c e x) \sqrt {a+b x+c x^2}}{4 e^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{e (d+e x)}+\frac {3 \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c} e^4}-\frac {3 (2 c d-b e) \sqrt {c d^2-b d e+a e^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 )}{2 e^4}\\ \end {align*}
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Mathematica [A] time = 0.42, size = 214, normalized size = 0.94 \[ \frac {\frac {3 \left (4 c e (a e-2 b d)+b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c}}+12 (2 c d-b e) \sqrt {e (a e-b d)+c d^2} \tanh ^{-1}\left (\frac {2 a e-b d+b e x-2 c d x}{2 \sqrt {a+x (b+c x)} \sqrt {e (a e-b d)+c d^2}}\right )-\frac {8 e^3 (a+x (b+c x))^{3/2}}{d+e x}+6 e \sqrt {a+x (b+c x)} (3 b e-4 c d+2 c e x)}{8 e^4} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*x + c*x^2)^(3/2)/(d + e*x)^2,x]
[Out]
(6*e*(-4*c*d + 3*b*e + 2*c*e*x)*Sqrt[a + x*(b + c*x)] - (8*e^3*(a + x*(b + c*x))^(3/2))/(d + e*x) + (3*(8*c^2*
d^2 + b^2*e^2 + 4*c*e*(-2*b*d + a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/Sqrt[c] + 12*(2*
c*d - b*e)*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*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)])])/(8*e^4)
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fricas [A] time = 39.53, size = 1583, normalized size = 6.97 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*x^2+b*x+a)^(3/2)/(e*x+d)^2,x, algorithm="fricas")
[Out]
[1/16*(3*(8*c^2*d^3 - 8*b*c*d^2*e + (b^2 + 4*a*c)*d*e^2 + (8*c^2*d^2*e - 8*b*c*d*e^2 + (b^2 + 4*a*c)*e^3)*x)*s
qrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 12*(2*c^2*d^2 -
b*c*d*e + (2*c^2*d*e - b*c*e^2)*x)*sqrt(c*d^2 - b*d*e + a*e^2)*log((8*a*b*d*e - 8*a^2*e^2 - (b^2 + 4*a*c)*d^2
- (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*x^2 - 4*sqrt(c*d^2 - b*d*e + a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d
- 2*a*e + (2*c*d - b*e)*x) - 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x)/(e^2*x^2 + 2*d*e*x + d^2)) + 4
*(2*c^2*e^3*x^2 - 12*c^2*d^2*e + 9*b*c*d*e^2 - 4*a*c*e^3 - (6*c^2*d*e^2 - 5*b*c*e^3)*x)*sqrt(c*x^2 + b*x + a))
/(c*e^5*x + c*d*e^4), -1/16*(24*(2*c^2*d^2 - b*c*d*e + (2*c^2*d*e - b*c*e^2)*x)*sqrt(-c*d^2 + b*d*e - a*e^2)*a
rctan(-1/2*sqrt(-c*d^2 + b*d*e - a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x)/(a*c*d^2 - a*b*d
*e + a^2*e^2 + (c^2*d^2 - b*c*d*e + a*c*e^2)*x^2 + (b*c*d^2 - b^2*d*e + a*b*e^2)*x)) - 3*(8*c^2*d^3 - 8*b*c*d^
2*e + (b^2 + 4*a*c)*d*e^2 + (8*c^2*d^2*e - 8*b*c*d*e^2 + (b^2 + 4*a*c)*e^3)*x)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*
x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(2*c^2*e^3*x^2 - 12*c^2*d^2*e + 9*b*c*d*e^2
- 4*a*c*e^3 - (6*c^2*d*e^2 - 5*b*c*e^3)*x)*sqrt(c*x^2 + b*x + a))/(c*e^5*x + c*d*e^4), -1/8*(3*(8*c^2*d^3 - 8
*b*c*d^2*e + (b^2 + 4*a*c)*d*e^2 + (8*c^2*d^2*e - 8*b*c*d*e^2 + (b^2 + 4*a*c)*e^3)*x)*sqrt(-c)*arctan(1/2*sqrt
(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 6*(2*c^2*d^2 - b*c*d*e + (2*c^2*d*e - b*c*e^
2)*x)*sqrt(c*d^2 - b*d*e + a*e^2)*log((8*a*b*d*e - 8*a^2*e^2 - (b^2 + 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e + (b
^2 + 4*a*c)*e^2)*x^2 - 4*sqrt(c*d^2 - b*d*e + a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x) - 2
*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x)/(e^2*x^2 + 2*d*e*x + d^2)) - 2*(2*c^2*e^3*x^2 - 12*c^2*d^2*e
+ 9*b*c*d*e^2 - 4*a*c*e^3 - (6*c^2*d*e^2 - 5*b*c*e^3)*x)*sqrt(c*x^2 + b*x + a))/(c*e^5*x + c*d*e^4), -1/8*(12
*(2*c^2*d^2 - b*c*d*e + (2*c^2*d*e - b*c*e^2)*x)*sqrt(-c*d^2 + b*d*e - a*e^2)*arctan(-1/2*sqrt(-c*d^2 + b*d*e
- a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x)/(a*c*d^2 - a*b*d*e + a^2*e^2 + (c^2*d^2 - b*c*d
*e + a*c*e^2)*x^2 + (b*c*d^2 - b^2*d*e + a*b*e^2)*x)) + 3*(8*c^2*d^3 - 8*b*c*d^2*e + (b^2 + 4*a*c)*d*e^2 + (8*
c^2*d^2*e - 8*b*c*d*e^2 + (b^2 + 4*a*c)*e^3)*x)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)
/(c^2*x^2 + b*c*x + a*c)) - 2*(2*c^2*e^3*x^2 - 12*c^2*d^2*e + 9*b*c*d*e^2 - 4*a*c*e^3 - (6*c^2*d*e^2 - 5*b*c*e
^3)*x)*sqrt(c*x^2 + b*x + a))/(c*e^5*x + c*d*e^4)]
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*x^2+b*x+a)^(3/2)/(e*x+d)^2,x, algorithm="giac")
[Out]
Timed out
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maple [B] time = 0.11, size = 3450, normalized size = 15.20 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c*x^2+b*x+a)^(3/2)/(e*x+d)^2,x)
[Out]
c/(a*e^2-b*d*e+c*d^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)*x+3/2*c^(1/2)/(a*e^2-b
*d*e+c*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-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^
2)^(1/2))*a^2+9/4/(a*e^2-b*d*e+c*d^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-3/
e/(a*e^2-b*d*e+c*d^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-6/e^3/(a*e^2-b*d
*e+c*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-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+9/2/e^2/(a*e^2-b*d*e+c*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-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+3/e^5/(a*e^2-b*d*e+c*d^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))*c^3*d^5+3/2*c/(a*e^2-b*d*e+c*d^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*a+1/(a*e^2-b*d*e+c*d^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)*b-1/(a*e^2-b*d*e+c*d^2)/(x+d/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)^(5/2)-9/e^2/(a*e^2-b*d*e+c*d^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*d^2*c+6/e^3/(a*e^2-b*d*e+c*d^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))*b^2*d^3*c-3/2/e/(a*e^2-b*d*e+c*d^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-9/2/e/(a*e^2-b*d*e+c*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-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(1/2)*d*a*b+3/e/(a*e^2-b*d*e+c*d^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-15/2/e^4/(a*e^2-b*d*e+c*d^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))*b*d^4*c^2+6/e^3/(a*e^2-b*d*e+c*
d^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*c^2*d^3+3/e/(a*e^2-
b*d*e+c*d^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+21/4/
e^2/(a*e^2-b*d*e+c*d^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*d^2*c-3/2/e^2/(a*e
^2-b*d*e+c*d^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))*b^3*d^2-3/
8/e/(a*e^2-b*d*e+c*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-2*c*d)*(x+d/e)/e+(a*e^2-b*d
*e+c*d^2)/e^2)^(1/2))/c^(1/2)*b^3*d+3/8/(a*e^2-b*d*e+c*d^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^2-9/4/e/(a*e^2-b*d*e+c*d^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/(a*e^2-b*d*e+c*d^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/e^4/(a*e^2-b*d*e+c*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-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/e^3/(a*e^2-b*d*e+c*d^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^2*d^3-3/2/(a*e^2-b*d*e+c*d^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*b+3/2/e^2/(a*e^2-b*d*e+c*d^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*c^2*d^2+27/8/e^2/(a*e^2-b*d*e+c*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-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(1/2)*d^2*b^2
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*x^2+b*x+a)^(3/2)/(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(a*e^2-b*d*e>0)', see `assume?`
for more details)Is a*e^2-b*d*e +c*d^2 zero or nonzero?
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{{\left (d+e\,x\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*x + c*x^2)^(3/2)/(d + e*x)^2,x)
[Out]
int((a + b*x + c*x^2)^(3/2)/(d + e*x)^2, x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*x**2+b*x+a)**(3/2)/(e*x+d)**2,x)
[Out]
Integral((a + b*x + c*x**2)**(3/2)/(d + e*x)**2, x)
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