3.13.11 \(\int \frac {(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^5} \, dx\) [1211]

Optimal. Leaf size=123 \[ -\frac {3 \sqrt {a+b x+c x^2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{8 c d^5 (b+2 c x)^4}+\frac {3 \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{128 c^{5/2} \sqrt {b^2-4 a c} d^5} \]

[Out]

-1/8*(c*x^2+b*x+a)^(3/2)/c/d^5/(2*c*x+b)^4+3/128*arctan(2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2))/c^(5
/2)/d^5/(-4*a*c+b^2)^(1/2)-3/64*(c*x^2+b*x+a)^(1/2)/c^2/d^5/(2*c*x+b)^2

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {698, 702, 211} \begin {gather*} \frac {3 \text {ArcTan}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{128 c^{5/2} d^5 \sqrt {b^2-4 a c}}-\frac {3 \sqrt {a+b x+c x^2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{8 c d^5 (b+2 c x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 698

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[b*(p/(d*e*(m + 1))), 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] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] &&
GtQ[p, 0] && LtQ[m, -1] &&  !(IntegerQ[m/2] && LtQ[m + 2*p + 3, 0]) && IntegerQ[2*p]

Rule 702

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[4*c, Subst[Int[1/(b^2*e
- 4*a*c*e + 4*c*e*x^2), x], x, Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^5} \, dx &=-\frac {\left (a+b x+c x^2\right )^{3/2}}{8 c d^5 (b+2 c x)^4}+\frac {3 \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^3} \, dx}{16 c d^2}\\ &=-\frac {3 \sqrt {a+b x+c x^2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{8 c d^5 (b+2 c x)^4}+\frac {3 \int \frac {1}{(b d+2 c d x) \sqrt {a+b x+c x^2}} \, dx}{128 c^2 d^4}\\ &=-\frac {3 \sqrt {a+b x+c x^2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{8 c d^5 (b+2 c x)^4}+\frac {3 \text {Subst}\left (\int \frac {1}{2 b^2 c d-8 a c^2 d+8 c^2 d x^2} \, dx,x,\sqrt {a+b x+c x^2}\right )}{32 c d^4}\\ &=-\frac {3 \sqrt {a+b x+c x^2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{8 c d^5 (b+2 c x)^4}+\frac {3 \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{128 c^{5/2} \sqrt {b^2-4 a c} d^5}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.40, size = 119, normalized size = 0.97 \begin {gather*} \frac {-\frac {\sqrt {c} \sqrt {a+x (b+c x)} \left (3 b^2+20 b c x+4 c \left (2 a+5 c x^2\right )\right )}{(b+2 c x)^4}-\frac {3 \tan ^{-1}\left (\frac {b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}}{64 c^{5/2} d^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((Sqrt[c]*Sqrt[a + x*(b + c*x)]*(3*b^2 + 20*b*c*x + 4*c*(2*a + 5*c*x^2)))/(b + 2*c*x)^4) - (3*ArcTan[(b + 2*
c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)])/Sqrt[b^2 - 4*a*c]])/Sqrt[b^2 - 4*a*c])/(64*c^(5/2)*d^5)

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(341\) vs. \(2(103)=206\).
time = 0.69, size = 342, normalized size = 2.78

method	result	size
default	\(\frac {-\frac {c \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{4}}+\frac {c^{2} \left (-\frac {2 c \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{2}}+\frac {6 c^{2} \left (\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{3}+\frac {\left (4 a c -b^{2}\right ) \left (\frac {\sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{2 c \sqrt {\frac {4 a c -b^{2}}{c}}}\right )}{4 c}\right )}{4 a c -b^{2}}\right )}{4 a c -b^{2}}}{32 d^{5} c^{5}}\)	\(342\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/d^5/c^5*(-1/(4*a*c-b^2)*c/(x+1/2*b/c)^4*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)+c^2/(4*a*c-b^2)*(-2/(4*
a*c-b^2)*c/(x+1/2*b/c)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)+6*c^2/(4*a*c-b^2)*(1/3*((x+1/2*b/c)^2*c+1/4
*(4*a*c-b^2)/c)^(3/2)+1/4*(4*a*c-b^2)/c*(1/2*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)-1/2*(4*a*c-b^2)/c/((4*a*c
-b^2)/c)^(1/2)*ln((1/2*(4*a*c-b^2)/c+1/2*((4*a*c-b^2)/c)^(1/2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2))/(x+1/2
*b/c))))))

________________________________________________________________________________________

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)/(2*c*d*x+b*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(4*a*c-b^2>0)', see `assume?` f
or more deta

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (103) = 206\).
time = 6.85, size = 622, normalized size = 5.06 \begin {gather*} \left [-\frac {3 \, {\left (16 \, c^{4} x^{4} + 32 \, b c^{3} x^{3} + 24 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x + b^{4}\right )} \sqrt {-b^{2} c + 4 \, a c^{2}} \log \left (-\frac {4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c - 4 \, \sqrt {-b^{2} c + 4 \, a c^{2}} \sqrt {c x^{2} + b x + a}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) + 4 \, {\left (3 \, b^{4} c - 4 \, a b^{2} c^{2} - 32 \, a^{2} c^{3} + 20 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} + 20 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{256 \, {\left (16 \, {\left (b^{2} c^{7} - 4 \, a c^{8}\right )} d^{5} x^{4} + 32 \, {\left (b^{3} c^{6} - 4 \, a b c^{7}\right )} d^{5} x^{3} + 24 \, {\left (b^{4} c^{5} - 4 \, a b^{2} c^{6}\right )} d^{5} x^{2} + 8 \, {\left (b^{5} c^{4} - 4 \, a b^{3} c^{5}\right )} d^{5} x + {\left (b^{6} c^{3} - 4 \, a b^{4} c^{4}\right )} d^{5}\right )}}, -\frac {3 \, {\left (16 \, c^{4} x^{4} + 32 \, b c^{3} x^{3} + 24 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x + b^{4}\right )} \sqrt {b^{2} c - 4 \, a c^{2}} \arctan \left (\frac {\sqrt {b^{2} c - 4 \, a c^{2}} \sqrt {c x^{2} + b x + a}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (3 \, b^{4} c - 4 \, a b^{2} c^{2} - 32 \, a^{2} c^{3} + 20 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} + 20 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{128 \, {\left (16 \, {\left (b^{2} c^{7} - 4 \, a c^{8}\right )} d^{5} x^{4} + 32 \, {\left (b^{3} c^{6} - 4 \, a b c^{7}\right )} d^{5} x^{3} + 24 \, {\left (b^{4} c^{5} - 4 \, a b^{2} c^{6}\right )} d^{5} x^{2} + 8 \, {\left (b^{5} c^{4} - 4 \, a b^{3} c^{5}\right )} d^{5} x + {\left (b^{6} c^{3} - 4 \, a b^{4} c^{4}\right )} d^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^5,x, algorithm="fricas")

[Out]

[-1/256*(3*(16*c^4*x^4 + 32*b*c^3*x^3 + 24*b^2*c^2*x^2 + 8*b^3*c*x + b^4)*sqrt(-b^2*c + 4*a*c^2)*log(-(4*c^2*x
^2 + 4*b*c*x - b^2 + 8*a*c - 4*sqrt(-b^2*c + 4*a*c^2)*sqrt(c*x^2 + b*x + a))/(4*c^2*x^2 + 4*b*c*x + b^2)) + 4*
(3*b^4*c - 4*a*b^2*c^2 - 32*a^2*c^3 + 20*(b^2*c^3 - 4*a*c^4)*x^2 + 20*(b^3*c^2 - 4*a*b*c^3)*x)*sqrt(c*x^2 + b*
x + a))/(16*(b^2*c^7 - 4*a*c^8)*d^5*x^4 + 32*(b^3*c^6 - 4*a*b*c^7)*d^5*x^3 + 24*(b^4*c^5 - 4*a*b^2*c^6)*d^5*x^
2 + 8*(b^5*c^4 - 4*a*b^3*c^5)*d^5*x + (b^6*c^3 - 4*a*b^4*c^4)*d^5), -1/128*(3*(16*c^4*x^4 + 32*b*c^3*x^3 + 24*
b^2*c^2*x^2 + 8*b^3*c*x + b^4)*sqrt(b^2*c - 4*a*c^2)*arctan(1/2*sqrt(b^2*c - 4*a*c^2)*sqrt(c*x^2 + b*x + a)/(c
^2*x^2 + b*c*x + a*c)) + 2*(3*b^4*c - 4*a*b^2*c^2 - 32*a^2*c^3 + 20*(b^2*c^3 - 4*a*c^4)*x^2 + 20*(b^3*c^2 - 4*
a*b*c^3)*x)*sqrt(c*x^2 + b*x + a))/(16*(b^2*c^7 - 4*a*c^8)*d^5*x^4 + 32*(b^3*c^6 - 4*a*b*c^7)*d^5*x^3 + 24*(b^
4*c^5 - 4*a*b^2*c^6)*d^5*x^2 + 8*(b^5*c^4 - 4*a*b^3*c^5)*d^5*x + (b^6*c^3 - 4*a*b^4*c^4)*d^5)]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx + \int \frac {b x \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx + \int \frac {c x^{2} \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx}{d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a*sqrt(a + b*x + c*x**2)/(b**5 + 10*b**4*c*x + 40*b**3*c**2*x**2 + 80*b**2*c**3*x**3 + 80*b*c**4*x**
4 + 32*c**5*x**5), x) + Integral(b*x*sqrt(a + b*x + c*x**2)/(b**5 + 10*b**4*c*x + 40*b**3*c**2*x**2 + 80*b**2*
c**3*x**3 + 80*b*c**4*x**4 + 32*c**5*x**5), x) + Integral(c*x**2*sqrt(a + b*x + c*x**2)/(b**5 + 10*b**4*c*x +
40*b**3*c**2*x**2 + 80*b**2*c**3*x**3 + 80*b*c**4*x**4 + 32*c**5*x**5), x))/d**5

________________________________________________________________________________________

Giac [F(-1)] Timed out
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)/(2*c*d*x+b*d)^5,x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{{\left (b\,d+2\,c\,d\,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)/(b*d + 2*c*d*x)^5,x)

[Out]

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

________________________________________________________________________________________