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

Optimal. Leaf size=225 \[ -\frac {3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (-2 a e+x (2 c d-b e)+b d)}{64 (d+e x)^2 \left (a e^2-b d e+c d^2\right )^2}+\frac {3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{128 \left (a e^2-b d e+c d^2\right )^{5/2}}+\frac {\left (a+b x+c x^2\right )^{3/2} (-2 a e+x (2 c d-b e)+b d)}{8 (d+e x)^4 \left (a e^2-b d e+c d^2\right )} \]

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Rubi [A]  time = 0.15, antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {720, 724, 206} \begin {gather*} -\frac {3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (-2 a e+x (2 c d-b e)+b d)}{64 (d+e x)^2 \left (a e^2-b d e+c d^2\right )^2}+\frac {3 \left (b^2-4 a c\right )^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 )}{128 \left (a e^2-b d e+c d^2\right )^{5/2}}+\frac {\left (a+b x+c x^2\right )^{3/2} (-2 a e+x (2 c d-b e)+b d)}{8 (d+e x)^4 \left (a e^2-b d e+c d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b^2 - 4*a*c)*(b*d - 2*a*e + (2*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2])/(64*(c*d^2 - b*d*e + a*e^2)^2*(d + e*
x)^2) + ((b*d - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(3/2))/(8*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^4) + (3
*(b^2 - 4*a*c)^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])
])/(128*(c*d^2 - b*d*e + a*e^2)^(5/2))

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 720

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*
(d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^p)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(p*(b^2 -
4*a*c))/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, b, c, d, e}, 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] && EqQ[m +
2*p + 2, 0] && GtQ[p, 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 (a+b x+c x^2\right )^{3/2}}{(d+e x)^5} \, dx &=\frac {(b d-2 a e+(2 c d-b e) x) \left (a+b x+c x^2\right )^{3/2}}{8 \left (c d^2-b d e+a e^2\right ) (d+e x)^4}-\frac {\left (3 \left (b^2-4 a c\right )\right ) \int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^3} \, dx}{16 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {3 \left (b^2-4 a c\right ) (b d-2 a e+(2 c d-b e) x) \sqrt {a+b x+c x^2}}{64 \left (c d^2-b d e+a e^2\right )^2 (d+e x)^2}+\frac {(b d-2 a e+(2 c d-b e) x) \left (a+b x+c x^2\right )^{3/2}}{8 \left (c d^2-b d e+a e^2\right ) (d+e x)^4}+\frac {\left (3 \left (b^2-4 a c\right )^2\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{128 \left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac {3 \left (b^2-4 a c\right ) (b d-2 a e+(2 c d-b e) x) \sqrt {a+b x+c x^2}}{64 \left (c d^2-b d e+a e^2\right )^2 (d+e x)^2}+\frac {(b d-2 a e+(2 c d-b e) x) \left (a+b x+c x^2\right )^{3/2}}{8 \left (c d^2-b d e+a e^2\right ) (d+e x)^4}-\frac {\left (3 \left (b^2-4 a c\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 )}{64 \left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac {3 \left (b^2-4 a c\right ) (b d-2 a e+(2 c d-b e) x) \sqrt {a+b x+c x^2}}{64 \left (c d^2-b d e+a e^2\right )^2 (d+e x)^2}+\frac {(b d-2 a e+(2 c d-b e) x) \left (a+b x+c x^2\right )^{3/2}}{8 \left (c d^2-b d e+a e^2\right ) (d+e x)^4}+\frac {3 \left (b^2-4 a c\right )^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 )}{128 \left (c d^2-b d e+a e^2\right )^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.66, size = 222, normalized size = 0.99 \begin {gather*} -\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b^2-4 a c\right ) \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 )}{8 \left (e (a e-b d)+c d^2\right )^{3/2}}+\frac {\sqrt {a+x (b+c x)} (-2 a e+b (d-e x)+2 c d x)}{4 (d+e x)^2 \left (e (a e-b d)+c d^2\right )}\right )+\frac {2 (a+x (b+c x))^{3/2} (2 a e-b d+b e x-2 c d x)}{(d+e x)^4}}{16 \left (e (a e-b d)+c d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*((2*(-(b*d) + 2*a*e - 2*c*d*x + b*e*x)*(a + x*(b + c*x))^(3/2))/(d + e*x)^4 + 3*(b^2 - 4*a*c)*((Sqrt[a +
x*(b + c*x)]*(-2*a*e + 2*c*d*x + b*(d - e*x)))/(4*(c*d^2 + e*(-(b*d) + a*e))*(d + e*x)^2) + ((b^2 - 4*a*c)*Ar
cTanh[(-(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*(c*d^2
+ e*(-(b*d) + a*e))^(3/2))))/(c*d^2 + e*(-(b*d) + a*e))

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IntegrateAlgebraic [F]  time = 180.04, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [B]  time = 17.67, size = 2570, normalized size = 11.42

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)^5,x, algorithm="fricas")

[Out]

[1/256*(3*((b^4 - 8*a*b^2*c + 16*a^2*c^2)*e^4*x^4 + 4*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*d*e^3*x^3 + 6*(b^4 - 8*a*
b^2*c + 16*a^2*c^2)*d^2*e^2*x^2 + 4*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^3*e*x + (b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^
4)*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*(40*a^3*b*d*e^4 - 16*a^4*e^5 - (
3*b^3*c - 20*a*b*c^2)*d^5 + (3*b^4 - 22*a*b^2*c - 40*a^2*c^2)*d^4*e - (a*b^3 - 84*a^2*b*c)*d^3*e^2 - 2*(13*a^2
*b^2 + 28*a^3*c)*d^2*e^3 + (16*c^4*d^5 - 40*b*c^3*d^4*e + 2*(13*b^2*c^2 + 28*a*c^3)*d^3*e^2 + (b^3*c - 84*a*b*
c^2)*d^2*e^3 - (3*b^4 - 22*a*b^2*c - 40*a^2*c^2)*d*e^4 + (3*a*b^3 - 20*a^2*b*c)*e^5)*x^3 + (24*b*c^3*d^5 - 4*(
17*b^2*c^2 - 8*a*c^3)*d^4*e + 5*(11*b^3*c + 4*a*b*c^2)*d^3*e^2 - (11*b^4 + 74*a*b^2*c + 8*a^2*c^2)*d^2*e^3 + (
13*a*b^3 + 68*a^2*b*c)*d*e^4 - 2*(a^2*b^2 + 20*a^3*c)*e^5)*x^2 - (24*a^3*b*e^5 - 2*(b^2*c^2 + 20*a*c^3)*d^5 +
(13*b^3*c + 68*a*b*c^2)*d^4*e - (11*b^4 + 74*a*b^2*c + 8*a^2*c^2)*d^3*e^2 + 5*(11*a*b^3 + 4*a^2*b*c)*d^2*e^3 -
4*(17*a^2*b^2 - 8*a^3*c)*d*e^4)*x)*sqrt(c*x^2 + b*x + a))/(c^3*d^10 - 3*b*c^2*d^9*e - 3*a^2*b*d^5*e^5 + a^3*d
^4*e^6 + 3*(b^2*c + a*c^2)*d^8*e^2 - (b^3 + 6*a*b*c)*d^7*e^3 + 3*(a*b^2 + a^2*c)*d^6*e^4 + (c^3*d^6*e^4 - 3*b*
c^2*d^5*e^5 - 3*a^2*b*d*e^9 + a^3*e^10 + 3*(b^2*c + a*c^2)*d^4*e^6 - (b^3 + 6*a*b*c)*d^3*e^7 + 3*(a*b^2 + a^2*
c)*d^2*e^8)*x^4 + 4*(c^3*d^7*e^3 - 3*b*c^2*d^6*e^4 - 3*a^2*b*d^2*e^8 + a^3*d*e^9 + 3*(b^2*c + a*c^2)*d^5*e^5 -
(b^3 + 6*a*b*c)*d^4*e^6 + 3*(a*b^2 + a^2*c)*d^3*e^7)*x^3 + 6*(c^3*d^8*e^2 - 3*b*c^2*d^7*e^3 - 3*a^2*b*d^3*e^7
+ a^3*d^2*e^8 + 3*(b^2*c + a*c^2)*d^6*e^4 - (b^3 + 6*a*b*c)*d^5*e^5 + 3*(a*b^2 + a^2*c)*d^4*e^6)*x^2 + 4*(c^3
*d^9*e - 3*b*c^2*d^8*e^2 - 3*a^2*b*d^4*e^6 + a^3*d^3*e^7 + 3*(b^2*c + a*c^2)*d^7*e^3 - (b^3 + 6*a*b*c)*d^6*e^4
+ 3*(a*b^2 + a^2*c)*d^5*e^5)*x), 1/128*(3*((b^4 - 8*a*b^2*c + 16*a^2*c^2)*e^4*x^4 + 4*(b^4 - 8*a*b^2*c + 16*a
^2*c^2)*d*e^3*x^3 + 6*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^2*e^2*x^2 + 4*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^3*e*x +
(b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^4)*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)) + 2*(40*a^3*b*d*e^4 - 16*a^4*e^5 - (3*b^3*c - 20*a*b*c^2)*d^5 + (3*b^
4 - 22*a*b^2*c - 40*a^2*c^2)*d^4*e - (a*b^3 - 84*a^2*b*c)*d^3*e^2 - 2*(13*a^2*b^2 + 28*a^3*c)*d^2*e^3 + (16*c^
4*d^5 - 40*b*c^3*d^4*e + 2*(13*b^2*c^2 + 28*a*c^3)*d^3*e^2 + (b^3*c - 84*a*b*c^2)*d^2*e^3 - (3*b^4 - 22*a*b^2*
c - 40*a^2*c^2)*d*e^4 + (3*a*b^3 - 20*a^2*b*c)*e^5)*x^3 + (24*b*c^3*d^5 - 4*(17*b^2*c^2 - 8*a*c^3)*d^4*e + 5*(
11*b^3*c + 4*a*b*c^2)*d^3*e^2 - (11*b^4 + 74*a*b^2*c + 8*a^2*c^2)*d^2*e^3 + (13*a*b^3 + 68*a^2*b*c)*d*e^4 - 2*
(a^2*b^2 + 20*a^3*c)*e^5)*x^2 - (24*a^3*b*e^5 - 2*(b^2*c^2 + 20*a*c^3)*d^5 + (13*b^3*c + 68*a*b*c^2)*d^4*e - (
11*b^4 + 74*a*b^2*c + 8*a^2*c^2)*d^3*e^2 + 5*(11*a*b^3 + 4*a^2*b*c)*d^2*e^3 - 4*(17*a^2*b^2 - 8*a^3*c)*d*e^4)*
x)*sqrt(c*x^2 + b*x + a))/(c^3*d^10 - 3*b*c^2*d^9*e - 3*a^2*b*d^5*e^5 + a^3*d^4*e^6 + 3*(b^2*c + a*c^2)*d^8*e^
2 - (b^3 + 6*a*b*c)*d^7*e^3 + 3*(a*b^2 + a^2*c)*d^6*e^4 + (c^3*d^6*e^4 - 3*b*c^2*d^5*e^5 - 3*a^2*b*d*e^9 + a^3
*e^10 + 3*(b^2*c + a*c^2)*d^4*e^6 - (b^3 + 6*a*b*c)*d^3*e^7 + 3*(a*b^2 + a^2*c)*d^2*e^8)*x^4 + 4*(c^3*d^7*e^3
- 3*b*c^2*d^6*e^4 - 3*a^2*b*d^2*e^8 + a^3*d*e^9 + 3*(b^2*c + a*c^2)*d^5*e^5 - (b^3 + 6*a*b*c)*d^4*e^6 + 3*(a*b
^2 + a^2*c)*d^3*e^7)*x^3 + 6*(c^3*d^8*e^2 - 3*b*c^2*d^7*e^3 - 3*a^2*b*d^3*e^7 + a^3*d^2*e^8 + 3*(b^2*c + a*c^2
)*d^6*e^4 - (b^3 + 6*a*b*c)*d^5*e^5 + 3*(a*b^2 + a^2*c)*d^4*e^6)*x^2 + 4*(c^3*d^9*e - 3*b*c^2*d^8*e^2 - 3*a^2*
b*d^4*e^6 + a^3*d^3*e^7 + 3*(b^2*c + a*c^2)*d^7*e^3 - (b^3 + 6*a*b*c)*d^6*e^4 + 3*(a*b^2 + a^2*c)*d^5*e^5)*x)]

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

[Out]

Timed out

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maple [B]  time = 0.07, size = 15932, normalized size = 70.81 \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+a)^(3/2)/(e*x+d)^5,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+a)^(3/2)/(e*x+d)^5,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 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{{\left (d+e\,x\right )}^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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