\(\int \frac {(b d+2 c d x)^{11/2}}{(a+b x+c x^2)^2} \, dx\) [1297]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [B] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F(-2)]
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 26, antiderivative size = 179 \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx=36 c \left (b^2-4 a c\right ) d^5 \sqrt {b d+2 c d x}+\frac {36}{5} c d^3 (b d+2 c d x)^{5/2}-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}-18 c \left (b^2-4 a c\right )^{5/4} d^{11/2} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-18 c \left (b^2-4 a c\right )^{5/4} d^{11/2} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ) \]

[Out]

36/5*c*d^3*(2*c*d*x+b*d)^(5/2)-d*(2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)-18*c*(-4*a*c+b^2)^(5/4)*d^(11/2)*arctan((d*
(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))-18*c*(-4*a*c+b^2)^(5/4)*d^(11/2)*arctanh((d*(2*c*x+b))^(1/2)/(-4*
a*c+b^2)^(1/4)/d^(1/2))+36*c*(-4*a*c+b^2)*d^5*(2*c*d*x+b*d)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {700, 706, 708, 335, 218, 212, 209} \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx=-18 c d^{11/2} \left (b^2-4 a c\right )^{5/4} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-18 c d^{11/2} \left (b^2-4 a c\right )^{5/4} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )+36 c d^5 \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}+\frac {36}{5} c d^3 (b d+2 c d x)^{5/2} \]

[In]

Int[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^2,x]

[Out]

36*c*(b^2 - 4*a*c)*d^5*Sqrt[b*d + 2*c*d*x] + (36*c*d^3*(b*d + 2*c*d*x)^(5/2))/5 - (d*(b*d + 2*c*d*x)^(9/2))/(a
 + b*x + c*x^2) - 18*c*(b^2 - 4*a*c)^(5/4)*d^(11/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]
- 18*c*(b^2 - 4*a*c)^(5/4)*d^(11/2)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]

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 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[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 700

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

Rule 706

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

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 {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}+\left (9 c d^2\right ) \int \frac {(b d+2 c d x)^{7/2}}{a+b x+c x^2} \, dx \\ & = \frac {36}{5} c d^3 (b d+2 c d x)^{5/2}-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}+\left (9 c \left (b^2-4 a c\right ) d^4\right ) \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx \\ & = 36 c \left (b^2-4 a c\right ) d^5 \sqrt {b d+2 c d x}+\frac {36}{5} c d^3 (b d+2 c d x)^{5/2}-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}+\left (9 c \left (b^2-4 a c\right )^2 d^6\right ) \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx \\ & = 36 c \left (b^2-4 a c\right ) d^5 \sqrt {b d+2 c d x}+\frac {36}{5} c d^3 (b d+2 c d x)^{5/2}-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}+\frac {1}{2} \left (9 \left (b^2-4 a c\right )^2 d^5\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )} \, dx,x,b d+2 c d x\right ) \\ & = 36 c \left (b^2-4 a c\right ) d^5 \sqrt {b d+2 c d x}+\frac {36}{5} c d^3 (b d+2 c d x)^{5/2}-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}+\left (9 \left (b^2-4 a c\right )^2 d^5\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b^2}{4 c}+\frac {x^4}{4 c d^2}} \, dx,x,\sqrt {d (b+2 c x)}\right ) \\ & = 36 c \left (b^2-4 a c\right ) d^5 \sqrt {b d+2 c d x}+\frac {36}{5} c d^3 (b d+2 c d x)^{5/2}-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}-\left (18 c \left (b^2-4 a c\right )^{3/2} d^6\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 (18 c \left (b^2-4 a c\right )^{3/2} d^6\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 ) \\ & = 36 c \left (b^2-4 a c\right ) d^5 \sqrt {b d+2 c d x}+\frac {36}{5} c d^3 (b d+2 c d x)^{5/2}-\frac {d (b d+2 c d x)^{9/2}}{a+b x+c x^2}-18 c \left (b^2-4 a c\right )^{5/4} d^{11/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-18 c \left (b^2-4 a c\right )^{5/4} d^{11/2} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.79 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.65 \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx=\left (\frac {1}{5}+\frac {i}{5}\right ) c (d (b+2 c x))^{11/2} \left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (45 b^4-360 a b^2 c+720 a^2 c^2-36 b^2 (b+2 c x)^2+144 a c (b+2 c x)^2-4 (b+2 c x)^4\right )}{c (b+2 c x)^5 (a+x (b+c x))}-\frac {45 i \left (b^2-4 a c\right )^{5/4} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{11/2}}+\frac {45 i \left (b^2-4 a c\right )^{5/4} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{11/2}}+\frac {45 i \left (b^2-4 a c\right )^{5/4} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{(b+2 c x)^{11/2}}\right ) \]

[In]

Integrate[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^2,x]

[Out]

(1/5 + I/5)*c*(d*(b + 2*c*x))^(11/2)*(((-1/2 + I/2)*(45*b^4 - 360*a*b^2*c + 720*a^2*c^2 - 36*b^2*(b + 2*c*x)^2
 + 144*a*c*(b + 2*c*x)^2 - 4*(b + 2*c*x)^4))/(c*(b + 2*c*x)^5*(a + x*(b + c*x))) - ((45*I)*(b^2 - 4*a*c)^(5/4)
*ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b + 2*c*x)^(11/2) + ((45*I)*(b^2 - 4*a*c)^(5/4)*A
rcTan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b + 2*c*x)^(11/2) + ((45*I)*(b^2 - 4*a*c)^(5/4)*Arc
Tanh[((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*c*x)^(11/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(372\) vs. \(2(151)=302\).

Time = 2.95 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.08

method result size
derivativedivides \(16 c \,d^{3} \left (-8 a c \,d^{2} \sqrt {2 c d x +b d}+2 b^{2} d^{2} \sqrt {2 c d x +b d}+\frac {\left (2 c d x +b d \right )^{\frac {5}{2}}}{5}+d^{4} \left (\frac {\left (-a^{2} c^{2}+\frac {1}{2} a \,b^{2} c -\frac {1}{16} b^{4}\right ) \sqrt {2 c d x +b d}}{a c \,d^{2}-\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {9 \left (4 a^{2} c^{2}-2 a \,b^{2} c +\frac {1}{4} b^{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\right )\) \(373\)
default \(16 c \,d^{3} \left (-8 a c \,d^{2} \sqrt {2 c d x +b d}+2 b^{2} d^{2} \sqrt {2 c d x +b d}+\frac {\left (2 c d x +b d \right )^{\frac {5}{2}}}{5}+d^{4} \left (\frac {\left (-a^{2} c^{2}+\frac {1}{2} a \,b^{2} c -\frac {1}{16} b^{4}\right ) \sqrt {2 c d x +b d}}{a c \,d^{2}-\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {9 \left (4 a^{2} c^{2}-2 a \,b^{2} c +\frac {1}{4} b^{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\right )\) \(373\)
pseudoelliptic \(-\frac {144 \left (-\frac {\left (d \left (2 c x +b \right )\right )^{\frac {5}{2}} \left (c \,x^{2}+b x +a \right ) c \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}}}{45}+\frac {d^{2} \left (4 a c -b^{2}\right ) \left (2 \left (\frac {8 c^{2} x^{2}}{9}+\left (\frac {8 b x}{9}+a \right ) c -\frac {b^{2}}{36}\right ) \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} \sqrt {d \left (2 c x +b \right )}-c \,d^{2} \sqrt {2}\, \left (c \,x^{2}+b x +a \right ) \left (-\frac {b^{2}}{4}+a c \right ) \left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+\ln \left (\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}{\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}\right )-2 \arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right )\right )}{8}\right ) d^{3}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} \left (c \,x^{2}+b x +a \right )}\) \(387\)

[In]

int((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

16*c*d^3*(-8*a*c*d^2*(2*c*d*x+b*d)^(1/2)+2*b^2*d^2*(2*c*d*x+b*d)^(1/2)+1/5*(2*c*d*x+b*d)^(5/2)+d^4*((-a^2*c^2+
1/2*a*b^2*c-1/16*b^4)*(2*c*d*x+b*d)^(1/2)/(a*c*d^2-1/4*b^2*d^2+1/4*(2*c*d*x+b*d)^2)+9/8*(4*a^2*c^2-2*a*b^2*c+1
/4*b^4)/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*(ln((2*c*d*x+b*d+(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)*2^(1/
2)+(4*a*c*d^2-b^2*d^2)^(1/2))/(2*c*d*x+b*d-(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)*2^(1/2)+(4*a*c*d^2-b^
2*d^2)^(1/2)))+2*arctan(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)-2*arctan(-2^(1/2)/(4*a*c*d^2-
b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1))))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 816, normalized size of antiderivative = 4.56 \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {45 \, \left ({\left (b^{10} c^{4} - 20 \, a b^{8} c^{5} + 160 \, a^{2} b^{6} c^{6} - 640 \, a^{3} b^{4} c^{7} + 1280 \, a^{4} b^{2} c^{8} - 1024 \, a^{5} c^{9}\right )} d^{22}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (-9 \, {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {2 \, c d x + b d} d^{5} + 9 \, \left ({\left (b^{10} c^{4} - 20 \, a b^{8} c^{5} + 160 \, a^{2} b^{6} c^{6} - 640 \, a^{3} b^{4} c^{7} + 1280 \, a^{4} b^{2} c^{8} - 1024 \, a^{5} c^{9}\right )} d^{22}\right )^{\frac {1}{4}}\right ) - 45 \, \left ({\left (b^{10} c^{4} - 20 \, a b^{8} c^{5} + 160 \, a^{2} b^{6} c^{6} - 640 \, a^{3} b^{4} c^{7} + 1280 \, a^{4} b^{2} c^{8} - 1024 \, a^{5} c^{9}\right )} d^{22}\right )^{\frac {1}{4}} {\left (-i \, c x^{2} - i \, b x - i \, a\right )} \log \left (-9 \, {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {2 \, c d x + b d} d^{5} + 9 i \, \left ({\left (b^{10} c^{4} - 20 \, a b^{8} c^{5} + 160 \, a^{2} b^{6} c^{6} - 640 \, a^{3} b^{4} c^{7} + 1280 \, a^{4} b^{2} c^{8} - 1024 \, a^{5} c^{9}\right )} d^{22}\right )^{\frac {1}{4}}\right ) - 45 \, \left ({\left (b^{10} c^{4} - 20 \, a b^{8} c^{5} + 160 \, a^{2} b^{6} c^{6} - 640 \, a^{3} b^{4} c^{7} + 1280 \, a^{4} b^{2} c^{8} - 1024 \, a^{5} c^{9}\right )} d^{22}\right )^{\frac {1}{4}} {\left (i \, c x^{2} + i \, b x + i \, a\right )} \log \left (-9 \, {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {2 \, c d x + b d} d^{5} - 9 i \, \left ({\left (b^{10} c^{4} - 20 \, a b^{8} c^{5} + 160 \, a^{2} b^{6} c^{6} - 640 \, a^{3} b^{4} c^{7} + 1280 \, a^{4} b^{2} c^{8} - 1024 \, a^{5} c^{9}\right )} d^{22}\right )^{\frac {1}{4}}\right ) - 45 \, \left ({\left (b^{10} c^{4} - 20 \, a b^{8} c^{5} + 160 \, a^{2} b^{6} c^{6} - 640 \, a^{3} b^{4} c^{7} + 1280 \, a^{4} b^{2} c^{8} - 1024 \, a^{5} c^{9}\right )} d^{22}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (-9 \, {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {2 \, c d x + b d} d^{5} - 9 \, \left ({\left (b^{10} c^{4} - 20 \, a b^{8} c^{5} + 160 \, a^{2} b^{6} c^{6} - 640 \, a^{3} b^{4} c^{7} + 1280 \, a^{4} b^{2} c^{8} - 1024 \, a^{5} c^{9}\right )} d^{22}\right )^{\frac {1}{4}}\right ) + {\left (64 \, c^{4} d^{5} x^{4} + 128 \, b c^{3} d^{5} x^{3} + 48 \, {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} d^{5} x^{2} + 16 \, {\left (11 \, b^{3} c - 36 \, a b c^{2}\right )} d^{5} x - {\left (5 \, b^{4} - 216 \, a b^{2} c + 720 \, a^{2} c^{2}\right )} d^{5}\right )} \sqrt {2 \, c d x + b d}}{5 \, {\left (c x^{2} + b x + a\right )}} \]

[In]

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

1/5*(45*((b^10*c^4 - 20*a*b^8*c^5 + 160*a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22)
^(1/4)*(c*x^2 + b*x + a)*log(-9*(b^2*c - 4*a*c^2)*sqrt(2*c*d*x + b*d)*d^5 + 9*((b^10*c^4 - 20*a*b^8*c^5 + 160*
a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22)^(1/4)) - 45*((b^10*c^4 - 20*a*b^8*c^5 +
 160*a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22)^(1/4)*(-I*c*x^2 - I*b*x - I*a)*log
(-9*(b^2*c - 4*a*c^2)*sqrt(2*c*d*x + b*d)*d^5 + 9*I*((b^10*c^4 - 20*a*b^8*c^5 + 160*a^2*b^6*c^6 - 640*a^3*b^4*
c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22)^(1/4)) - 45*((b^10*c^4 - 20*a*b^8*c^5 + 160*a^2*b^6*c^6 - 640*a^3
*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22)^(1/4)*(I*c*x^2 + I*b*x + I*a)*log(-9*(b^2*c - 4*a*c^2)*sqrt(
2*c*d*x + b*d)*d^5 - 9*I*((b^10*c^4 - 20*a*b^8*c^5 + 160*a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 10
24*a^5*c^9)*d^22)^(1/4)) - 45*((b^10*c^4 - 20*a*b^8*c^5 + 160*a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8
 - 1024*a^5*c^9)*d^22)^(1/4)*(c*x^2 + b*x + a)*log(-9*(b^2*c - 4*a*c^2)*sqrt(2*c*d*x + b*d)*d^5 - 9*((b^10*c^4
 - 20*a*b^8*c^5 + 160*a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22)^(1/4)) + (64*c^4*
d^5*x^4 + 128*b*c^3*d^5*x^3 + 48*(5*b^2*c^2 - 12*a*c^3)*d^5*x^2 + 16*(11*b^3*c - 36*a*b*c^2)*d^5*x - (5*b^4 -
216*a*b^2*c + 720*a^2*c^2)*d^5)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)

Sympy [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \]

[In]

integrate((2*c*d*x+b*d)**(11/2)/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^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(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. 646 vs. \(2 (151) = 302\).

Time = 0.31 (sec) , antiderivative size = 646, normalized size of antiderivative = 3.61 \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx=32 \, \sqrt {2 \, c d x + b d} b^{2} c d^{5} - 128 \, \sqrt {2 \, c d x + b d} a c^{2} d^{5} + \frac {16}{5} \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} c d^{3} - 9 \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} b^{2} c d^{5} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a c^{2} d^{5}\right )} \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 ) - 9 \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} b^{2} c d^{5} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a c^{2} d^{5}\right )} \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 ) - \frac {9}{2} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} b^{2} c d^{5} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a c^{2} d^{5}\right )} \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 ) + \frac {9}{2} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} b^{2} c d^{5} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a c^{2} d^{5}\right )} \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 ) + \frac {4 \, {\left (\sqrt {2 \, c d x + b d} b^{4} c d^{7} - 8 \, \sqrt {2 \, c d x + b d} a b^{2} c^{2} d^{7} + 16 \, \sqrt {2 \, c d x + b d} a^{2} c^{3} d^{7}\right )}}{b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}} \]

[In]

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

32*sqrt(2*c*d*x + b*d)*b^2*c*d^5 - 128*sqrt(2*c*d*x + b*d)*a*c^2*d^5 + 16/5*(2*c*d*x + b*d)^(5/2)*c*d^3 - 9*(s
qrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*b^2*c*d^5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a*c^2*d^5)*arctan(1/2*s
qrt(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)) - 9*(sqrt(
2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*b^2*c*d^5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a*c^2*d^5)*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)) - 9/2*(sqrt(2
)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*b^2*c*d^5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a*c^2*d^5)*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)) + 9/2*(sqrt(2)*(-b^2
*d^2 + 4*a*c*d^2)^(1/4)*b^2*c*d^5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a*c^2*d^5)*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)) + 4*(sqrt(2*c*d*x + b*d)*b^
4*c*d^7 - 8*sqrt(2*c*d*x + b*d)*a*b^2*c^2*d^7 + 16*sqrt(2*c*d*x + b*d)*a^2*c^3*d^7)/(b^2*d^2 - 4*a*c*d^2 - (2*
c*d*x + b*d)^2)

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 834, normalized size of antiderivative = 4.66 \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {16\,c\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{5/2}}{5}-\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (64\,a^2\,c^3\,d^7-32\,a\,b^2\,c^2\,d^7+4\,b^4\,c\,d^7\right )}{{\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2}-18\,c\,d^{11/2}\,\mathrm {atan}\left (\frac {9\,c\,d^{11/2}\,{\left (b^2-4\,a\,c\right )}^{5/4}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (1327104\,a^4\,c^6\,d^{14}-1327104\,a^3\,b^2\,c^5\,d^{14}+497664\,a^2\,b^4\,c^4\,d^{14}-82944\,a\,b^6\,c^3\,d^{14}+5184\,b^8\,c^2\,d^{14}\right )-c\,d^{11/2}\,{\left (b^2-4\,a\,c\right )}^{5/4}\,\left (-36864\,a^3\,c^4\,d^9+27648\,a^2\,b^2\,c^3\,d^9-6912\,a\,b^4\,c^2\,d^9+576\,b^6\,c\,d^9\right )\,9{}\mathrm {i}\right )+9\,c\,d^{11/2}\,{\left (b^2-4\,a\,c\right )}^{5/4}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (1327104\,a^4\,c^6\,d^{14}-1327104\,a^3\,b^2\,c^5\,d^{14}+497664\,a^2\,b^4\,c^4\,d^{14}-82944\,a\,b^6\,c^3\,d^{14}+5184\,b^8\,c^2\,d^{14}\right )+c\,d^{11/2}\,{\left (b^2-4\,a\,c\right )}^{5/4}\,\left (-36864\,a^3\,c^4\,d^9+27648\,a^2\,b^2\,c^3\,d^9-6912\,a\,b^4\,c^2\,d^9+576\,b^6\,c\,d^9\right )\,9{}\mathrm {i}\right )}{c\,d^{11/2}\,{\left (b^2-4\,a\,c\right )}^{5/4}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (1327104\,a^4\,c^6\,d^{14}-1327104\,a^3\,b^2\,c^5\,d^{14}+497664\,a^2\,b^4\,c^4\,d^{14}-82944\,a\,b^6\,c^3\,d^{14}+5184\,b^8\,c^2\,d^{14}\right )-c\,d^{11/2}\,{\left (b^2-4\,a\,c\right )}^{5/4}\,\left (-36864\,a^3\,c^4\,d^9+27648\,a^2\,b^2\,c^3\,d^9-6912\,a\,b^4\,c^2\,d^9+576\,b^6\,c\,d^9\right )\,9{}\mathrm {i}\right )\,9{}\mathrm {i}-c\,d^{11/2}\,{\left (b^2-4\,a\,c\right )}^{5/4}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (1327104\,a^4\,c^6\,d^{14}-1327104\,a^3\,b^2\,c^5\,d^{14}+497664\,a^2\,b^4\,c^4\,d^{14}-82944\,a\,b^6\,c^3\,d^{14}+5184\,b^8\,c^2\,d^{14}\right )+c\,d^{11/2}\,{\left (b^2-4\,a\,c\right )}^{5/4}\,\left (-36864\,a^3\,c^4\,d^9+27648\,a^2\,b^2\,c^3\,d^9-6912\,a\,b^4\,c^2\,d^9+576\,b^6\,c\,d^9\right )\,9{}\mathrm {i}\right )\,9{}\mathrm {i}}\right )\,{\left (b^2-4\,a\,c\right )}^{5/4}-32\,c\,d^5\,\sqrt {b\,d+2\,c\,d\,x}\,\left (4\,a\,c-b^2\right )+c\,d^{11/2}\,\mathrm {atan}\left (\frac {b^2\,\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}-a\,c\,\sqrt {b\,d+2\,c\,d\,x}\,4{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{5/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{5/4}\,18{}\mathrm {i} \]

[In]

int((b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^2,x)

[Out]

(16*c*d^3*(b*d + 2*c*d*x)^(5/2))/5 - ((b*d + 2*c*d*x)^(1/2)*(4*b^4*c*d^7 + 64*a^2*c^3*d^7 - 32*a*b^2*c^2*d^7))
/((b*d + 2*c*d*x)^2 - b^2*d^2 + 4*a*c*d^2) - 18*c*d^(11/2)*atan((9*c*d^(11/2)*(b^2 - 4*a*c)^(5/4)*((b*d + 2*c*
d*x)^(1/2)*(1327104*a^4*c^6*d^14 + 5184*b^8*c^2*d^14 - 82944*a*b^6*c^3*d^14 + 497664*a^2*b^4*c^4*d^14 - 132710
4*a^3*b^2*c^5*d^14) - c*d^(11/2)*(b^2 - 4*a*c)^(5/4)*(576*b^6*c*d^9 - 36864*a^3*c^4*d^9 - 6912*a*b^4*c^2*d^9 +
 27648*a^2*b^2*c^3*d^9)*9i) + 9*c*d^(11/2)*(b^2 - 4*a*c)^(5/4)*((b*d + 2*c*d*x)^(1/2)*(1327104*a^4*c^6*d^14 +
5184*b^8*c^2*d^14 - 82944*a*b^6*c^3*d^14 + 497664*a^2*b^4*c^4*d^14 - 1327104*a^3*b^2*c^5*d^14) + c*d^(11/2)*(b
^2 - 4*a*c)^(5/4)*(576*b^6*c*d^9 - 36864*a^3*c^4*d^9 - 6912*a*b^4*c^2*d^9 + 27648*a^2*b^2*c^3*d^9)*9i))/(c*d^(
11/2)*(b^2 - 4*a*c)^(5/4)*((b*d + 2*c*d*x)^(1/2)*(1327104*a^4*c^6*d^14 + 5184*b^8*c^2*d^14 - 82944*a*b^6*c^3*d
^14 + 497664*a^2*b^4*c^4*d^14 - 1327104*a^3*b^2*c^5*d^14) - c*d^(11/2)*(b^2 - 4*a*c)^(5/4)*(576*b^6*c*d^9 - 36
864*a^3*c^4*d^9 - 6912*a*b^4*c^2*d^9 + 27648*a^2*b^2*c^3*d^9)*9i)*9i - c*d^(11/2)*(b^2 - 4*a*c)^(5/4)*((b*d +
2*c*d*x)^(1/2)*(1327104*a^4*c^6*d^14 + 5184*b^8*c^2*d^14 - 82944*a*b^6*c^3*d^14 + 497664*a^2*b^4*c^4*d^14 - 13
27104*a^3*b^2*c^5*d^14) + c*d^(11/2)*(b^2 - 4*a*c)^(5/4)*(576*b^6*c*d^9 - 36864*a^3*c^4*d^9 - 6912*a*b^4*c^2*d
^9 + 27648*a^2*b^2*c^3*d^9)*9i)*9i))*(b^2 - 4*a*c)^(5/4) - 32*c*d^5*(b*d + 2*c*d*x)^(1/2)*(4*a*c - b^2) + c*d^
(11/2)*atan((b^2*(b*d + 2*c*d*x)^(1/2)*1i - a*c*(b*d + 2*c*d*x)^(1/2)*4i)/(d^(1/2)*(b^2 - 4*a*c)^(5/4)))*(b^2
- 4*a*c)^(5/4)*18i