\(\int \frac {(b d+2 c d x)^{9/2}}{(a+b x+c x^2)^2} \, dx\) [1298]

Optimal result

Integrand size = 26, antiderivative size = 152 \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+14 c \left (b^2-4 a c\right )^{3/4} d^{9/2} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-14 c \left (b^2-4 a c\right )^{3/4} d^{9/2} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ) \]

[Out]

28/3*c*d^3*(2*c*d*x+b*d)^(3/2)-d*(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)+14*c*(-4*a*c+b^2)^(3/4)*d^(9/2)*arctan((d*(
2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))-14*c*(-4*a*c+b^2)^(3/4)*d^(9/2)*arctanh((d*(2*c*x+b))^(1/2)/(-4*a*
c+b^2)^(1/4)/d^(1/2))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {700, 706, 708, 335, 304, 209, 212} \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx=14 c d^{9/2} \left (b^2-4 a c\right )^{3/4} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-14 c d^{9/2} \left (b^2-4 a c\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+\frac {28}{3} c d^3 (b d+2 c d x)^{3/2} \]

[In]

Int[(b*d + 2*c*d*x)^(9/2)/(a + b*x + c*x^2)^2,x]

[Out]

(28*c*d^3*(b*d + 2*c*d*x)^(3/2))/3 - (d*(b*d + 2*c*d*x)^(7/2))/(a + b*x + c*x^2) + 14*c*(b^2 - 4*a*c)^(3/4)*d^
(9/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])] - 14*c*(b^2 - 4*a*c)^(3/4)*d^(9/2)*ArcTanh[Sqr
t[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)^{7/2}}{a+b x+c x^2}+\left (7 c d^2\right ) \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx \\ & = \frac {28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+\left (7 c \left (b^2-4 a c\right ) d^4\right ) \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx \\ & = \frac {28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+\frac {1}{2} \left (7 \left (b^2-4 a c\right ) d^3\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 {28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+\left (7 \left (b^2-4 a c\right ) d^3\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 {28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}-\left (14 c \left (b^2-4 a c\right ) d^5\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 (14 c \left (b^2-4 a c\right ) d^5\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 {28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+14 c \left (b^2-4 a c\right )^{3/4} d^{9/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-14 c \left (b^2-4 a c\right )^{3/4} d^{9/2} \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 = 0.68 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.67 \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx=\left (\frac {1}{3}+\frac {i}{3}\right ) c (d (b+2 c x))^{9/2} \left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-7 b^2+28 a c+4 (b+2 c x)^2\right )}{c (b+2 c x)^3 (a+x (b+c x))}-\frac {21 \left (b^2-4 a c\right )^{3/4} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{9/2}}+\frac {21 \left (b^2-4 a c\right )^{3/4} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{9/2}}-\frac {21 \left (b^2-4 a c\right )^{3/4} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{(b+2 c x)^{9/2}}\right ) \]

[In]

Integrate[(b*d + 2*c*d*x)^(9/2)/(a + b*x + c*x^2)^2,x]

[Out]

(1/3 + I/3)*c*(d*(b + 2*c*x))^(9/2)*(((1/2 - I/2)*(-7*b^2 + 28*a*c + 4*(b + 2*c*x)^2))/(c*(b + 2*c*x)^3*(a + x
*(b + c*x))) - (21*(b^2 - 4*a*c)^(3/4)*ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b + 2*c*x)^
(9/2) + (21*(b^2 - 4*a*c)^(3/4)*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b + 2*c*x)^(9/2) -
(21*(b^2 - 4*a*c)^(3/4)*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)^(9/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(314\) vs. \(2(126)=252\).

Time = 2.68 (sec) , antiderivative size = 315, normalized size of antiderivative = 2.07

method	result	size
derivativedivides	\(16 c \,d^{3} \left (\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{3}-d^{2} \left (\frac {\left (-\frac {a c}{4}+\frac {b^{2}}{16}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{a c \,d^{2}-\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {\left (7 a c -\frac {7 b^{2}}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\right )\)	\(315\)
default	\(16 c \,d^{3} \left (\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{3}-d^{2} \left (\frac {\left (-\frac {a c}{4}+\frac {b^{2}}{16}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{a c \,d^{2}-\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {\left (7 a c -\frac {7 b^{2}}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\right )\)	\(315\)
pseudoelliptic	\(\frac {56 \left (\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (\frac {4 c^{2} x^{2}}{7}+\left (\frac {4 b x}{7}+a \right ) c -\frac {3 b^{2}}{28}\right ) \left (d \left (2 c x +b \right )\right )^{\frac {3}{2}}-\frac {3 \sqrt {2}\, c \,d^{2} \left (c \,x^{2}+b x +a \right ) \left (-\frac {b^{2}}{4}+a c \right ) \left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+\ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right )-2 \arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right )}{2}\right ) d^{3}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (6 c \,x^{2}+6 b x +6 a \right )}\)	\(335\)

[In]

int((2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

16*c*d^3*(1/3*(2*c*d*x+b*d)^(3/2)-d^2*((-1/4*a*c+1/16*b^2)*(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*(7*a*c-7/4*b^2)/(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*arct
an(-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 = 659, normalized size of antiderivative = 4.34 \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {21 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (343 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} \sqrt {2 \, c d x + b d} d^{13} + 343 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {3}{4}}\right ) + 21 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {1}{4}} {\left (-i \, c x^{2} - i \, b x - i \, a\right )} \log \left (343 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} \sqrt {2 \, c d x + b d} d^{13} + 343 i \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {3}{4}}\right ) + 21 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {1}{4}} {\left (i \, c x^{2} + i \, b x + i \, a\right )} \log \left (343 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} \sqrt {2 \, c d x + b d} d^{13} - 343 i \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {3}{4}}\right ) - 21 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (343 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} \sqrt {2 \, c d x + b d} d^{13} - 343 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {3}{4}}\right ) - {\left (32 \, c^{3} d^{4} x^{3} + 48 \, b c^{2} d^{4} x^{2} + 2 \, {\left (5 \, b^{2} c + 28 \, a c^{2}\right )} d^{4} x - {\left (3 \, b^{3} - 28 \, a b c\right )} d^{4}\right )} \sqrt {2 \, c d x + b d}}{3 \, {\left (c x^{2} + b x + a\right )}} \]

[In]

integrate((2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

-1/3*(21*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(1/4)*(c*x^2 + b*x + a)*log(343*(b^4*c^
3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(2*c*d*x + b*d)*d^13 + 343*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^
3*c^7)*d^18)^(3/4)) + 21*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(1/4)*(-I*c*x^2 - I*b*x
- I*a)*log(343*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(2*c*d*x + b*d)*d^13 + 343*I*((b^6*c^4 - 12*a*b^4*c^5
+ 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(3/4)) + 21*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18
)^(1/4)*(I*c*x^2 + I*b*x + I*a)*log(343*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(2*c*d*x + b*d)*d^13 - 343*I*
((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(3/4)) - 21*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^
2*c^6 - 64*a^3*c^7)*d^18)^(1/4)*(c*x^2 + b*x + a)*log(343*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(2*c*d*x +
b*d)*d^13 - 343*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(3/4)) - (32*c^3*d^4*x^3 + 48*b*
c^2*d^4*x^2 + 2*(5*b^2*c + 28*a*c^2)*d^4*x - (3*b^3 - 28*a*b*c)*d^4)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)

Sympy [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \]

[In]

integrate((2*c*d*x+b*d)**(9/2)/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((2*c*d*x+b*d)^(9/2)/(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. 441 vs. \(2 (126) = 252\).

Time = 0.32 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.90 \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx=-7 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - 7 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) + \frac {7}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d^{3} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) - \frac {7}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d^{3} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + \frac {16}{3} \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c d^{3} + \frac {4 \, {\left ({\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} c d^{5} - 4 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a c^{2} d^{5}\right )}}{b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}} \]

[In]

integrate((2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

-7*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) + 2*sqr
t(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4)) - 7*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d^3*arctan(-1/2*sqr
t(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)) + 7/2*sqrt(2
)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d^3*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)) - 7/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d^3*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)) + 16/3*(2*c*d*x + b*d)^(3/2)*c
*d^3 + 4*((2*c*d*x + b*d)^(3/2)*b^2*c*d^5 - 4*(2*c*d*x + b*d)^(3/2)*a*c^2*d^5)/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x
+ b*d)^2)

Mupad [B] (verification not implemented)

Time = 9.30 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.09 \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {16\,c\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{3}+\frac {{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (16\,a\,c^2\,d^5-4\,b^2\,c\,d^5\right )}{{\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2}+14\,c\,d^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}+c\,d^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}\,14{}\mathrm {i} \]

[In]

int((b*d + 2*c*d*x)^(9/2)/(a + b*x + c*x^2)^2,x)

[Out]

(16*c*d^3*(b*d + 2*c*d*x)^(3/2))/3 + ((b*d + 2*c*d*x)^(3/2)*(16*a*c^2*d^5 - 4*b^2*c*d^5))/((b*d + 2*c*d*x)^2 -
b^2*d^2 + 4*a*c*d^2) + 14*c*d^(9/2)*atan((b*d + 2*c*d*x)^(1/2)/(d^(1/2)*(b^2 - 4*a*c)^(1/4)))*(b^2 - 4*a*c)^(
3/4) + c*d^(9/2)*atan(((b*d + 2*c*d*x)^(1/2)*1i)/(d^(1/2)*(b^2 - 4*a*c)^(1/4)))*(b^2 - 4*a*c)^(3/4)*14i