\(\int \frac {\sqrt {b d+2 c d x}}{(a+b x+c x^2)^3} \, dx\) [1315]

Optimal result

Integrand size = 26, antiderivative size = 192 \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {5 c (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac {5 c^2 \sqrt {d} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{9/4}}-\frac {5 c^2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{9/4}} \]

[Out]

-1/2*(2*c*d*x+b*d)^(3/2)/(-4*a*c+b^2)/d/(c*x^2+b*x+a)^2+5/2*c*(2*c*d*x+b*d)^(3/2)/(-4*a*c+b^2)^2/d/(c*x^2+b*x+
a)+5*c^2*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))*d^(1/2)/(-4*a*c+b^2)^(9/4)-5*c^2*arctanh((d*(2
*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))*d^(1/2)/(-4*a*c+b^2)^(9/4)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {701, 708, 335, 304, 209, 212} \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^3} \, dx=\frac {5 c^2 \sqrt {d} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{9/4}}-\frac {5 c^2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{9/4}}+\frac {5 c (b d+2 c d x)^{3/2}}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {(b d+2 c d x)^{3/2}}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2} \]

[In]

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

[Out]

-1/2*(b*d + 2*c*d*x)^(3/2)/((b^2 - 4*a*c)*d*(a + b*x + c*x^2)^2) + (5*c*(b*d + 2*c*d*x)^(3/2))/(2*(b^2 - 4*a*c
)^2*d*(a + b*x + c*x^2)) + (5*c^2*Sqrt[d]*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2 - 4*
a*c)^(9/4) - (5*c^2*Sqrt[d]*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2 - 4*a*c)^(9/4)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 701

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*c*(d + e*x)^(m + 1
)*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)*(b^2 - 4*a*c))), x] - Dist[2*c*e*((m + 2*p + 3)/(e*(p + 1)*(b^2 - 4*a*
c))), Int[(d + e*x)^m*(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]
&& EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && RationalQ[m] && IntegerQ[2*p]

Rule 708

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}-\frac {(5 c) \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^2} \, dx}{2 \left (b^2-4 a c\right )} \\ & = -\frac {(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {5 c (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac {\left (5 c^2\right ) \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx}{2 \left (b^2-4 a c\right )^2} \\ & = -\frac {(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {5 c (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac {(5 c) \text {Subst}\left (\int \frac {\sqrt {x}}{a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}} \, dx,x,b d+2 c d x\right )}{4 \left (b^2-4 a c\right )^2 d} \\ & = -\frac {(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {5 c (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac {(5 c) \text {Subst}\left (\int \frac {x^2}{a-\frac {b^2}{4 c}+\frac {x^4}{4 c d^2}} \, dx,x,\sqrt {d (b+2 c x)}\right )}{2 \left (b^2-4 a c\right )^2 d} \\ & = -\frac {(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {5 c (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}-\frac {\left (5 c^2 d\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c} d-x^2} \, dx,x,\sqrt {d (b+2 c x)}\right )}{\left (b^2-4 a c\right )^2}+\frac {\left (5 c^2 d\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c} d+x^2} \, dx,x,\sqrt {d (b+2 c x)}\right )}{\left (b^2-4 a c\right )^2} \\ & = -\frac {(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {5 c (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac {5 c^2 \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{9/4}}-\frac {5 c^2 \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{9/4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.93 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.39 \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^3} \, dx=\left (\frac {1}{2}+\frac {i}{2}\right ) c^2 \sqrt {d (b+2 c x)} \left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) (b+2 c x) \left (b^2-5 b c x-c \left (9 a+5 c x^2\right )\right )}{c^2 \left (b^2-4 a c\right )^2 (a+x (b+c x))^2}-\frac {5 \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{9/4} \sqrt {b+2 c x}}+\frac {5 \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{9/4} \sqrt {b+2 c x}}-\frac {5 \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{9/4} \sqrt {b+2 c x}}\right ) \]

[In]

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

[Out]

(1/2 + I/2)*c^2*Sqrt[d*(b + 2*c*x)]*(((-1/2 + I/2)*(b + 2*c*x)*(b^2 - 5*b*c*x - c*(9*a + 5*c*x^2)))/(c^2*(b^2
- 4*a*c)^2*(a + x*(b + c*x))^2) - (5*ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/((b^2 - 4*a*c)
^(9/4)*Sqrt[b + 2*c*x]) + (5*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/((b^2 - 4*a*c)^(9/4)*S
qrt[b + 2*c*x]) - (5*ArcTanh[((1 + I)*(b^2 - 4*a*c)^(1/4)*Sqrt[b + 2*c*x])/(Sqrt[b^2 - 4*a*c] + I*(b + 2*c*x))
])/((b^2 - 4*a*c)^(9/4)*Sqrt[b + 2*c*x]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(372\) vs. \(2(164)=328\).

Time = 2.86 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.94

method	result	size
pseudoelliptic	\(\frac {\left (d \left (2 c x +b \right )\right )^{\frac {3}{2}}}{8 \left (a +x \left (c x +b \right )\right )^{2} \left (-\frac {b^{2}}{4}+a c \right ) d}+\frac {5 c \left (d \left (2 c x +b \right )\right )^{\frac {3}{2}}}{32 \left (c \,x^{2}+b x +a \right ) \left (-\frac {b^{2}}{4}+a c \right )^{2} d}+\frac {5 d^{5} c^{2} \sqrt {2}\, \ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right )}{4 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {9}{4}}}+\frac {5 d^{5} c^{2} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )}{2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {9}{4}}}-\frac {5 d^{5} c^{2} \sqrt {2}\, \arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )}{2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {9}{4}}}\)	\(373\)
derivativedivides	\(64 c^{2} d^{5} \left (\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (\left (2 c d x +b d \right )^{2}+4 a c \,d^{2}-b^{2} d^{2}\right )^{2}}+\frac {\frac {5 \left (2 c d x +b d \right )^{\frac {3}{2}}}{32 \left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (\left (2 c d x +b d \right )^{2}+4 a c \,d^{2}-b^{2} d^{2}\right )}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{256 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {5}{4}}}}{4 a c \,d^{2}-b^{2} d^{2}}\right )\)	\(377\)
default	\(64 c^{2} d^{5} \left (\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (\left (2 c d x +b d \right )^{2}+4 a c \,d^{2}-b^{2} d^{2}\right )^{2}}+\frac {\frac {5 \left (2 c d x +b d \right )^{\frac {3}{2}}}{32 \left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (\left (2 c d x +b d \right )^{2}+4 a c \,d^{2}-b^{2} d^{2}\right )}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{256 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {5}{4}}}}{4 a c \,d^{2}-b^{2} d^{2}}\right )\)	\(377\)

[In]

int((2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/8*(d*(2*c*x+b))^(3/2)/(a+x*(c*x+b))^2/(-1/4*b^2+a*c)/d+5/32*c*(d*(2*c*x+b))^(3/2)/(c*x^2+b*x+a)/(-1/4*b^2+a*
c)^2/d+5/4*d^5*c^2/(d^2*(4*a*c-b^2))^(9/4)*2^(1/2)*ln(((d^2*(4*a*c-b^2))^(1/2)-(d^2*(4*a*c-b^2))^(1/4)*(d*(2*c
*x+b))^(1/2)*2^(1/2)+d*(2*c*x+b))/((d^2*(4*a*c-b^2))^(1/4)*(d*(2*c*x+b))^(1/2)*2^(1/2)+(d^2*(4*a*c-b^2))^(1/2)
+d*(2*c*x+b)))+5/2*d^5*c^2/(d^2*(4*a*c-b^2))^(9/4)*2^(1/2)*arctan((2^(1/2)*(d*(2*c*x+b))^(1/2)+(d^2*(4*a*c-b^2
))^(1/4))/(d^2*(4*a*c-b^2))^(1/4))-5/2*d^5*c^2/(d^2*(4*a*c-b^2))^(9/4)*2^(1/2)*arctan((-2^(1/2)*(d*(2*c*x+b))^
(1/2)+(d^2*(4*a*c-b^2))^(1/4))/(d^2*(4*a*c-b^2))^(1/4))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 1980, normalized size of antiderivative = 10.31 \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^3} \, dx=\text {Too large to display} \]

[In]

integrate((2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

-1/2*(5*(c^8*d^2/(b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*
b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9))^(1/4)*(a^2*b^4 - 8*a
^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3
+ (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)*log(125*sqrt(2*c*d*x + b*d)*c
^6*d + 125*(b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 +
28672*a^6*b^2*c^6 - 16384*a^7*c^7)*(c^8*d^2/(b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256
*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9
*c^9))^(3/4)) - 5*(c^8*d^2/(b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 1
29024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9))^(1/4)*(a^2
*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b
*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)*log(125*sqrt(2*c*d*
x + b*d)*c^6*d - 125*(b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*
b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*(c^8*d^2/(b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c
^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 -
262144*a^9*c^9))^(3/4)) + 5*(c^8*d^2/(b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^
10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9))^
(1/4)*(I*a^2*b^4 - 8*I*a^3*b^2*c + 16*I*a^4*c^2 + I*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 + 2*I*(b^5*c - 8*
a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + I*(b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*I*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2
)*x)*log(125*sqrt(2*c*d*x + b*d)*c^6*d - 125*(I*b^14 - 28*I*a*b^12*c + 336*I*a^2*b^10*c^2 - 2240*I*a^3*b^8*c^3
+ 8960*I*a^4*b^6*c^4 - 21504*I*a^5*b^4*c^5 + 28672*I*a^6*b^2*c^6 - 16384*I*a^7*c^7)*(c^8*d^2/(b^18 - 36*a*b^1
6*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 58
9824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9))^(3/4)) + 5*(c^8*d^2/(b^18 - 36*a*b^16*c + 576*a^2*b^1
4*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7
+ 589824*a^8*b^2*c^8 - 262144*a^9*c^9))^(1/4)*(-I*a^2*b^4 + 8*I*a^3*b^2*c - 16*I*a^4*c^2 - I*(b^4*c^2 - 8*a*b^
2*c^3 + 16*a^2*c^4)*x^4 - 2*I*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 - I*(b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2
- 2*I*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)*log(125*sqrt(2*c*d*x + b*d)*c^6*d - 125*(-I*b^14 + 28*I*a*b^12*c
- 336*I*a^2*b^10*c^2 + 2240*I*a^3*b^8*c^3 - 8960*I*a^4*b^6*c^4 + 21504*I*a^5*b^4*c^5 - 28672*I*a^6*b^2*c^6 +
16384*I*a^7*c^7)*(c^8*d^2/(b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 12
9024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9))^(3/4)) - (1
0*c^3*x^3 + 15*b*c^2*x^2 - b^3 + 9*a*b*c + 3*(b^2*c + 6*a*c^2)*x)*sqrt(2*c*d*x + b*d))/(a^2*b^4 - 8*a^3*b^2*c
+ 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 -
6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^3} \, dx=\text {Timed out} \]

[In]

integrate((2*c*d*x+b*d)**(1/2)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^3,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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 645 vs. \(2 (164) = 328\).

Time = 0.30 (sec) , antiderivative size = 645, normalized size of antiderivative = 3.36 \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{\sqrt {2} b^{6} d - 12 \, \sqrt {2} a b^{4} c d + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d - 64 \, \sqrt {2} a^{3} c^{3} d} - \frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{\sqrt {2} b^{6} d - 12 \, \sqrt {2} a b^{4} c d + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d - 64 \, \sqrt {2} a^{3} c^{3} d} + \frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{2 \, {\left (\sqrt {2} b^{6} d - 12 \, \sqrt {2} a b^{4} c d + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d - 64 \, \sqrt {2} a^{3} c^{3} d\right )}} - \frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{2 \, {\left (\sqrt {2} b^{6} d - 12 \, \sqrt {2} a b^{4} c d + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d - 64 \, \sqrt {2} a^{3} c^{3} d\right )}} - \frac {2 \, {\left (9 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} c^{2} d^{3} - 36 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a c^{3} d^{3} - 5 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} c^{2} d\right )}}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}^{2}} \]

[In]

integrate((2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

-5*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) + 2*sqrt(2*c*d*x
+ b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4))/(sqrt(2)*b^6*d - 12*sqrt(2)*a*b^4*c*d + 48*sqrt(2)*a^2*b^2*c^2*d - 64*sq
rt(2)*a^3*c^3*d) - 5*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4
) - 2*sqrt(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4))/(sqrt(2)*b^6*d - 12*sqrt(2)*a*b^4*c*d + 48*sqrt(2)*a^
2*b^2*c^2*d - 64*sqrt(2)*a^3*c^3*d) + 5/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*log(2*c*d*x + b*d + sqrt(2)*(-b^2*d
^2 + 4*a*c*d^2)^(1/4)*sqrt(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2))/(sqrt(2)*b^6*d - 12*sqrt(2)*a*b^4*c*d
+ 48*sqrt(2)*a^2*b^2*c^2*d - 64*sqrt(2)*a^3*c^3*d) - 5/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*log(2*c*d*x + b*d -
sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqrt(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2))/(sqrt(2)*b^6*d - 12*sqr
t(2)*a*b^4*c*d + 48*sqrt(2)*a^2*b^2*c^2*d - 64*sqrt(2)*a^3*c^3*d) - 2*(9*(2*c*d*x + b*d)^(3/2)*b^2*c^2*d^3 - 3
6*(2*c*d*x + b*d)^(3/2)*a*c^3*d^3 - 5*(2*c*d*x + b*d)^(7/2)*c^2*d)/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*(b^2*d^2 -
4*a*c*d^2 - (2*c*d*x + b*d)^2)^2)

Mupad [B] (verification not implemented)

Time = 9.57 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.67 \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^3} \, dx=\frac {\frac {10\,c^2\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{7/2}}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {18\,c^2\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{4\,a\,c-b^2}}{{\left (b\,d+2\,c\,d\,x\right )}^4-{\left (b\,d+2\,c\,d\,x\right )}^2\,\left (2\,b^2\,d^2-8\,a\,c\,d^2\right )+b^4\,d^4+16\,a^2\,c^2\,d^4-8\,a\,b^2\,c\,d^4}+\frac {5\,c^2\,\sqrt {d}\,\mathrm {atan}\left (\frac {b^4\,\sqrt {b\,d+2\,c\,d\,x}+16\,a^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}-8\,a\,b^2\,c\,\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{9/4}}+\frac {c^2\,\sqrt {d}\,\mathrm {atan}\left (\frac {b^4\,\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}+a^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}\,16{}\mathrm {i}-a\,b^2\,c\,\sqrt {b\,d+2\,c\,d\,x}\,8{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )\,5{}\mathrm {i}}{{\left (b^2-4\,a\,c\right )}^{9/4}} \]

[In]

int((b*d + 2*c*d*x)^(1/2)/(a + b*x + c*x^2)^3,x)

[Out]

((10*c^2*d*(b*d + 2*c*d*x)^(7/2))/(b^4 + 16*a^2*c^2 - 8*a*b^2*c) + (18*c^2*d^3*(b*d + 2*c*d*x)^(3/2))/(4*a*c -
b^2))/((b*d + 2*c*d*x)^4 - (b*d + 2*c*d*x)^2*(2*b^2*d^2 - 8*a*c*d^2) + b^4*d^4 + 16*a^2*c^2*d^4 - 8*a*b^2*c*d
^4) + (5*c^2*d^(1/2)*atan((b^4*(b*d + 2*c*d*x)^(1/2) + 16*a^2*c^2*(b*d + 2*c*d*x)^(1/2) - 8*a*b^2*c*(b*d + 2*c
*d*x)^(1/2))/(d^(1/2)*(b^2 - 4*a*c)^(9/4))))/(b^2 - 4*a*c)^(9/4) + (c^2*d^(1/2)*atan((b^4*(b*d + 2*c*d*x)^(1/2
)*1i + a^2*c^2*(b*d + 2*c*d*x)^(1/2)*16i - a*b^2*c*(b*d + 2*c*d*x)^(1/2)*8i)/(d^(1/2)*(b^2 - 4*a*c)^(9/4)))*5i
)/(b^2 - 4*a*c)^(9/4)