\(\int \frac {(b d+2 c d x)^{13/2}}{(a+b x+c x^2)^2} \, dx\) [1296]

Optimal result

Integrand size = 26, antiderivative size = 181 \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {44}{3} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{3/2}+\frac {44}{7} c d^3 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}+22 c \left (b^2-4 a c\right )^{7/4} d^{13/2} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-22 c \left (b^2-4 a c\right )^{7/4} d^{13/2} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ) \]

[Out]

44/3*c*(-4*a*c+b^2)*d^5*(2*c*d*x+b*d)^(3/2)+44/7*c*d^3*(2*c*d*x+b*d)^(7/2)-d*(2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a
)+22*c*(-4*a*c+b^2)^(7/4)*d^(13/2)*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))-22*c*(-4*a*c+b^2)^(7
/4)*d^(13/2)*arctanh((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 181, 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, 304, 209, 212} \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx=22 c d^{13/2} \left (b^2-4 a c\right )^{7/4} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-22 c d^{13/2} \left (b^2-4 a c\right )^{7/4} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )+\frac {44}{3} c d^5 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}+\frac {44}{7} c d^3 (b d+2 c d x)^{7/2} \]

[In]

Int[(b*d + 2*c*d*x)^(13/2)/(a + b*x + c*x^2)^2,x]

[Out]

(44*c*(b^2 - 4*a*c)*d^5*(b*d + 2*c*d*x)^(3/2))/3 + (44*c*d^3*(b*d + 2*c*d*x)^(7/2))/7 - (d*(b*d + 2*c*d*x)^(11
/2))/(a + b*x + c*x^2) + 22*c*(b^2 - 4*a*c)^(7/4)*d^(13/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqr
t[d])] - 22*c*(b^2 - 4*a*c)^(7/4)*d^(13/2)*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 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 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)^{11/2}}{a+b x+c x^2}+\left (11 c d^2\right ) \int \frac {(b d+2 c d x)^{9/2}}{a+b x+c x^2} \, dx \\ & = \frac {44}{7} c d^3 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}+\left (11 c \left (b^2-4 a c\right ) d^4\right ) \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx \\ & = \frac {44}{3} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{3/2}+\frac {44}{7} c d^3 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}+\left (11 c \left (b^2-4 a c\right )^2 d^6\right ) \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx \\ & = \frac {44}{3} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{3/2}+\frac {44}{7} c d^3 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}+\frac {1}{2} \left (11 \left (b^2-4 a c\right )^2 d^5\right ) \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 ) \\ & = \frac {44}{3} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{3/2}+\frac {44}{7} c d^3 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}+\left (11 \left (b^2-4 a c\right )^2 d^5\right ) \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 ) \\ & = \frac {44}{3} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{3/2}+\frac {44}{7} c d^3 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}-\left (22 c \left (b^2-4 a c\right )^2 d^7\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 (22 c \left (b^2-4 a c\right )^2 d^7\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 ) \\ & = \frac {44}{3} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{3/2}+\frac {44}{7} c d^3 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}+22 c \left (b^2-4 a c\right )^{7/4} d^{13/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-22 c \left (b^2-4 a c\right )^{7/4} d^{13/2} \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 = 0.81 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.60 \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx=\left (\frac {1}{21}+\frac {i}{21}\right ) c (d (b+2 c x))^{13/2} \left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (77 b^4-616 a b^2 c+1232 a^2 c^2-44 b^2 (b+2 c x)^2+176 a c (b+2 c x)^2-12 (b+2 c x)^4\right )}{c (b+2 c x)^5 (a+x (b+c x))}-\frac {231 \left (b^2-4 a c\right )^{7/4} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{13/2}}+\frac {231 \left (b^2-4 a c\right )^{7/4} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{13/2}}-\frac {231 \left (b^2-4 a c\right )^{7/4} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{(b+2 c x)^{13/2}}\right ) \]

[In]

Integrate[(b*d + 2*c*d*x)^(13/2)/(a + b*x + c*x^2)^2,x]

[Out]

(1/21 + I/21)*c*(d*(b + 2*c*x))^(13/2)*(((-1/2 + I/2)*(77*b^4 - 616*a*b^2*c + 1232*a^2*c^2 - 44*b^2*(b + 2*c*x
)^2 + 176*a*c*(b + 2*c*x)^2 - 12*(b + 2*c*x)^4))/(c*(b + 2*c*x)^5*(a + x*(b + c*x))) - (231*(b^2 - 4*a*c)^(7/4
)*ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b + 2*c*x)^(13/2) + (231*(b^2 - 4*a*c)^(7/4)*Arc
Tan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b + 2*c*x)^(13/2) - (231*(b^2 - 4*a*c)^(7/4)*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*c*x)^(13/2))

Maple [B] (verified)

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

Time = 3.16 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.06

method	result	size
derivativedivides	\(16 c \,d^{3} \left (-\frac {8 a c \,d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}{3}+\frac {2 b^{2} d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}{3}+\frac {\left (2 c d x +b d \right )^{\frac {7}{2}}}{7}+d^{4} \left (\frac {\left (-a^{2} c^{2}+\frac {1}{2} a \,b^{2} c -\frac {1}{16} b^{4}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{a c \,d^{2}-\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {\left (44 a^{2} c^{2}-22 a \,b^{2} c +\frac {11}{4} b^{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\right )\)	\(373\)
default	\(16 c \,d^{3} \left (-\frac {8 a c \,d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}{3}+\frac {2 b^{2} d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}{3}+\frac {\left (2 c d x +b d \right )^{\frac {7}{2}}}{7}+d^{4} \left (\frac {\left (-a^{2} c^{2}+\frac {1}{2} a \,b^{2} c -\frac {1}{16} b^{4}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{a c \,d^{2}-\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {\left (44 a^{2} c^{2}-22 a \,b^{2} c +\frac {11}{4} b^{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\right )\)	\(373\)
pseudoelliptic	\(\frac {88 d^{3} \left (-\frac {2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (\frac {8 c^{2} x^{2}}{11}+\left (\frac {8 b x}{11}+a \right ) c -\frac {3 b^{2}}{44}\right ) d^{2} \left (-\frac {b^{2}}{4}+a c \right ) \left (d \left (2 c x +b \right )\right )^{\frac {3}{2}}}{3}+\left (\frac {2 \left (d \left (2 c x +b \right )\right )^{\frac {7}{2}} \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{77}+\frac {\sqrt {2}\, d^{4} \left (4 a c -b^{2}\right )^{2} \left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+\ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right )-2 \arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right )}{16}\right ) \left (c \,x^{2}+b x +a \right ) c \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (c \,x^{2}+b x +a \right )}\)	\(377\)

[In]

int((2*c*d*x+b*d)^(13/2)/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

16*c*d^3*(-8/3*a*c*d^2*(2*c*d*x+b*d)^(3/2)+2/3*b^2*d^2*(2*c*d*x+b*d)^(3/2)+1/7*(2*c*d*x+b*d)^(7/2)+d^4*((-a^2*
c^2+1/2*a*b^2*c-1/16*b^4)*(2*c*d*x+b*d)^(3/2)/(a*c*d^2-1/4*b^2*d^2+1/4*(2*c*d*x+b*d)^2)+1/8*(44*a^2*c^2-22*a*b
^2*c+11/4*b^4)/(4*a*c*d^2-b^2*d^2)^(1/4)*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 = 1211, normalized size of antiderivative = 6.69 \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \]

[In]

integrate((2*c*d*x+b*d)^(13/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

1/21*(231*((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 21504*a^5*b^4*
c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c^11)*d^26)^(1/4)*(c*x^2 + b*x + a)*log(-1331*(b^10*c^3 - 20*a*b^8*c^4 +
160*a^2*b^6*c^5 - 640*a^3*b^4*c^6 + 1280*a^4*b^2*c^7 - 1024*a^5*c^8)*sqrt(2*c*d*x + b*d)*d^19 + 1331*((b^14*c^
4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2
*c^10 - 16384*a^7*c^11)*d^26)^(3/4)) - 231*((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 +
8960*a^4*b^6*c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c^11)*d^26)^(1/4)*(I*c*x^2 + I*b*x + I*a
)*log(-1331*(b^10*c^3 - 20*a*b^8*c^4 + 160*a^2*b^6*c^5 - 640*a^3*b^4*c^6 + 1280*a^4*b^2*c^7 - 1024*a^5*c^8)*sq
rt(2*c*d*x + b*d)*d^19 + 1331*I*((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^
6*c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c^11)*d^26)^(3/4)) - 231*((b^14*c^4 - 28*a*b^12*c^5
+ 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7
*c^11)*d^26)^(1/4)*(-I*c*x^2 - I*b*x - I*a)*log(-1331*(b^10*c^3 - 20*a*b^8*c^4 + 160*a^2*b^6*c^5 - 640*a^3*b^4
*c^6 + 1280*a^4*b^2*c^7 - 1024*a^5*c^8)*sqrt(2*c*d*x + b*d)*d^19 - 1331*I*((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2
*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c^11)*d^2
6)^(3/4)) - 231*((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 21504*a^
5*b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c^11)*d^26)^(1/4)*(c*x^2 + b*x + a)*log(-1331*(b^10*c^3 - 20*a*b^8*
c^4 + 160*a^2*b^6*c^5 - 640*a^3*b^4*c^6 + 1280*a^4*b^2*c^7 - 1024*a^5*c^8)*sqrt(2*c*d*x + b*d)*d^19 - 1331*((b
^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 21504*a^5*b^4*c^9 + 28672*a
^6*b^2*c^10 - 16384*a^7*c^11)*d^26)^(3/4)) + (384*c^5*d^6*x^5 + 960*b*c^4*d^6*x^4 + 32*(41*b^2*c^3 - 44*a*c^4)
*d^6*x^3 + 48*(21*b^3*c^2 - 44*a*b*c^3)*d^6*x^2 + 2*(115*b^4*c + 88*a*b^2*c^2 - 1232*a^2*c^3)*d^6*x - (21*b^5
- 440*a*b^3*c + 1232*a^2*b*c^2)*d^6)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)

Sympy [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \]

[In]

integrate((2*c*d*x+b*d)**(13/2)/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((2*c*d*x+b*d)^(13/2)/(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 = 3.57 \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {32}{3} \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} c d^{5} - \frac {128}{3} \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a c^{2} d^{5} + \frac {16}{7} \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} c d^{3} - 11 \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} b^{2} c d^{5} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{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 ) - 11 \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} b^{2} c d^{5} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{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 {11}{2} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} b^{2} c d^{5} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{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 {11}{2} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} b^{2} c d^{5} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{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 ({\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{4} c d^{7} - 8 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a b^{2} c^{2} d^{7} + 16 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} 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)^(13/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

32/3*(2*c*d*x + b*d)^(3/2)*b^2*c*d^5 - 128/3*(2*c*d*x + b*d)^(3/2)*a*c^2*d^5 + 16/7*(2*c*d*x + b*d)^(7/2)*c*d^
3 - 11*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*b^2*c*d^5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*a*c^2*d^5)*arc
tan(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)) -
11*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*b^2*c*d^5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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)) + 1
1/2*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*b^2*c*d^5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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)) - 11/2*(s
qrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*b^2*c*d^5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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*((2*c*d*x +
b*d)^(3/2)*b^4*c*d^7 - 8*(2*c*d*x + b*d)^(3/2)*a*b^2*c^2*d^7 + 16*(2*c*d*x + b*d)^(3/2)*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 = 9.33 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.38 \[ \int \frac {(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {16\,c\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{7/2}}{7}-\frac {{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (64\,a^2\,c^3\,d^7-32\,a\,b^2\,c^2\,d^7+4\,b^4\,c\,d^7\right )}{{\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2}+22\,c\,d^{13/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{7/4}}{\sqrt {d}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )\,{\left (b^2-4\,a\,c\right )}^{7/4}-\frac {32\,c\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (4\,a\,c-b^2\right )}{3}+c\,d^{13/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{7/4}\,1{}\mathrm {i}}{\sqrt {d}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )\,{\left (b^2-4\,a\,c\right )}^{7/4}\,22{}\mathrm {i} \]

[In]

int((b*d + 2*c*d*x)^(13/2)/(a + b*x + c*x^2)^2,x)

[Out]

(16*c*d^3*(b*d + 2*c*d*x)^(7/2))/7 - ((b*d + 2*c*d*x)^(3/2)*(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) + 22*c*d^(13/2)*atan(((b*d + 2*c*d*x)^(1/2)*(b^2 - 4*a*c)^(7/4))/(d
^(1/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)))*(b^2 - 4*a*c)^(7/4) + c*d^(13/2)*atan(((b*d + 2*c*d*x)^(1/2)*(b^2 - 4*
a*c)^(7/4)*1i)/(d^(1/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)))*(b^2 - 4*a*c)^(7/4)*22i - (32*c*d^5*(b*d + 2*c*d*x)^(
3/2)*(4*a*c - b^2))/3