\(\int \frac {(b d+2 c d x)^{5/2}}{(a+b x+c x^2)^3} \, dx\) [1313]

Optimal result

Integrand size = 26, antiderivative size = 178 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {3 c d (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {3 c^2 d^{5/2} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{5/4}}+\frac {3 c^2 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{5/4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 178, 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, 701, 708, 335, 304, 209, 212} \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {3 c^2 d^{5/2} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/4}}+\frac {3 c^2 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/4}}-\frac {3 c d (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2} \]

[In]

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

[Out]

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

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 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 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 {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}+\frac {1}{2} \left (3 c d^2\right ) \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^2} \, dx \\ & = -\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {3 c d (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {\left (3 c^2 d^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 )} \\ & = -\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {3 c d (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {(3 c d) \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 )} \\ & = -\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {3 c d (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {(3 c d) \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 )} \\ & = -\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {3 c d (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {\left (3 c^2 d^3\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 )}{b^2-4 a c}-\frac {\left (3 c^2 d^3\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 )}{b^2-4 a c} \\ & = -\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {3 c d (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {3 c^2 d^{5/2} \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 )^{5/4}}+\frac {3 c^2 d^{5/2} \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 )^{5/4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.51 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.50 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx=\left (\frac {1}{2}+\frac {i}{2}\right ) c^2 (d (b+2 c x))^{5/2} \left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (b^2+3 b c x+c \left (-a+3 c x^2\right )\right )}{c^2 \left (b^2-4 a c\right ) (b+2 c x) (a+x (b+c x))^2}+\frac {3 \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/4} (b+2 c x)^{5/2}}-\frac {3 \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/4} (b+2 c x)^{5/2}}+\frac {3 \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{5/4} (b+2 c x)^{5/2}}\right ) \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(323\) vs. \(2(150)=300\).

Time = 2.79 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.82

method	result	size
derivativedivides	\(64 c^{2} d^{5} \left (\frac {\frac {3 \left (2 c d x +b d \right )^{\frac {7}{2}}}{32 d^{2} \left (4 a c -b^{2}\right )}-\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{32}}{\left (\left (2 c d x +b d \right )^{2}+4 a c \,d^{2}-b^{2} d^{2}\right )^{2}}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{256 d^{2} \left (4 a c -b^{2}\right ) \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\)	\(324\)
default	\(64 c^{2} d^{5} \left (\frac {\frac {3 \left (2 c d x +b d \right )^{\frac {7}{2}}}{32 d^{2} \left (4 a c -b^{2}\right )}-\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{32}}{\left (\left (2 c d x +b d \right )^{2}+4 a c \,d^{2}-b^{2} d^{2}\right )^{2}}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{256 d^{2} \left (4 a c -b^{2}\right ) \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\)	\(324\)
pseudoelliptic	\(-\frac {3 d \left (-\frac {\sqrt {2}\, c^{2} d^{2} \left (c \,x^{2}+b x +a \right )^{2} \ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right )}{2}+\arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right ) \sqrt {2}\, c^{2} d^{2} \left (c \,x^{2}+b x +a \right )^{2}-\arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right ) \sqrt {2}\, c^{2} d^{2} \left (c \,x^{2}+b x +a \right )^{2}+\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (-3 c^{2} x^{2}+\left (-3 b x +a \right ) c -b^{2}\right ) \left (d \left (2 c x +b \right )\right )^{\frac {3}{2}}}{3}\right )}{8 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (c \,x^{2}+b x +a \right )^{2} \left (-\frac {b^{2}}{4}+a c \right )}\)	\(376\)

[In]

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

[Out]

64*c^2*d^5*(16*(3/512/d^2/(4*a*c-b^2)*(2*c*d*x+b*d)^(7/2)-1/512*(2*c*d*x+b*d)^(3/2))/((2*c*d*x+b*d)^2+4*a*c*d^
2-b^2*d^2)^2+3/256/d^2/(4*a*c-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*
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 = 0.56 (sec) , antiderivative size = 1296, normalized size of antiderivative = 7.28 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/2*(3*(c^8*d^10/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(1
/4)*((b^2*c^2 - 4*a*c^3)*x^4 + a^2*b^2 - 4*a^3*c + 2*(b^3*c - 4*a*b*c^2)*x^3 + (b^4 - 2*a*b^2*c - 8*a^2*c^2)*x
^2 + 2*(a*b^3 - 4*a^2*b*c)*x)*log(27*sqrt(2*c*d*x + b*d)*c^6*d^7 + 27*(c^8*d^10/(b^10 - 20*a*b^8*c + 160*a^2*b
^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(3/4)*(b^8 - 16*a*b^6*c + 96*a^2*b^4*c^2 - 256*a^
3*b^2*c^3 + 256*a^4*c^4)) - 3*(c^8*d^10/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*
c^4 - 1024*a^5*c^5))^(1/4)*((b^2*c^2 - 4*a*c^3)*x^4 + a^2*b^2 - 4*a^3*c + 2*(b^3*c - 4*a*b*c^2)*x^3 + (b^4 - 2
*a*b^2*c - 8*a^2*c^2)*x^2 + 2*(a*b^3 - 4*a^2*b*c)*x)*log(27*sqrt(2*c*d*x + b*d)*c^6*d^7 - 27*(c^8*d^10/(b^10 -
20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(3/4)*(b^8 - 16*a*b^6*c +
96*a^2*b^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^4)) + 3*(c^8*d^10/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*
b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(1/4)*(I*(b^2*c^2 - 4*a*c^3)*x^4 + I*a^2*b^2 - 4*I*a^3*c + 2*I*(b^
3*c - 4*a*b*c^2)*x^3 + I*(b^4 - 2*a*b^2*c - 8*a^2*c^2)*x^2 + 2*I*(a*b^3 - 4*a^2*b*c)*x)*log(27*sqrt(2*c*d*x +
b*d)*c^6*d^7 - 27*(c^8*d^10/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a
^5*c^5))^(3/4)*(I*b^8 - 16*I*a*b^6*c + 96*I*a^2*b^4*c^2 - 256*I*a^3*b^2*c^3 + 256*I*a^4*c^4)) + 3*(c^8*d^10/(b
^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(1/4)*(-I*(b^2*c^2 -
4*a*c^3)*x^4 - I*a^2*b^2 + 4*I*a^3*c - 2*I*(b^3*c - 4*a*b*c^2)*x^3 - I*(b^4 - 2*a*b^2*c - 8*a^2*c^2)*x^2 - 2*I
*(a*b^3 - 4*a^2*b*c)*x)*log(27*sqrt(2*c*d*x + b*d)*c^6*d^7 - 27*(c^8*d^10/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2
- 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(3/4)*(-I*b^8 + 16*I*a*b^6*c - 96*I*a^2*b^4*c^2 + 256*I
*a^3*b^2*c^3 - 256*I*a^4*c^4)) - (6*c^3*d^2*x^3 + 9*b*c^2*d^2*x^2 + (5*b^2*c - 2*a*c^2)*d^2*x + (b^3 - a*b*c)*
d^2)*sqrt(2*c*d*x + b*d))/((b^2*c^2 - 4*a*c^3)*x^4 + a^2*b^2 - 4*a^3*c + 2*(b^3*c - 4*a*b*c^2)*x^3 + (b^4 - 2*
a*b^2*c - 8*a^2*c^2)*x^2 + 2*(a*b^3 - 4*a^2*b*c)*x)

Sympy [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((2*c*d*x+b*d)^(5/2)/(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. 567 vs. \(2 (150) = 300\).

Time = 0.32 (sec) , antiderivative size = 567, normalized size of antiderivative = 3.19 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx=\frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} d \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{\sqrt {2} b^{4} - 8 \, \sqrt {2} a b^{2} c + 16 \, \sqrt {2} a^{2} c^{2}} + \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} d \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{\sqrt {2} b^{4} - 8 \, \sqrt {2} a b^{2} c + 16 \, \sqrt {2} a^{2} c^{2}} - \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} d \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{2 \, {\left (\sqrt {2} b^{4} - 8 \, \sqrt {2} a b^{2} c + 16 \, \sqrt {2} a^{2} c^{2}\right )}} + \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} d \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{2 \, {\left (\sqrt {2} b^{4} - 8 \, \sqrt {2} a b^{2} c + 16 \, \sqrt {2} a^{2} c^{2}\right )}} - \frac {2 \, {\left ({\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} c^{2} d^{5} - 4 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a c^{3} d^{5} + 3 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} c^{2} d^{3}\right )}}{{\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}^{2} {\left (b^{2} - 4 \, a c\right )}} \]

[In]

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

[Out]

3*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*d*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^4 - 8*sqrt(2)*a*b^2*c + 16*sqrt(2)*a^2*c^2) + 3*(-b^2*d^2 +
4*a*c*d^2)^(3/4)*c^2*d*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^4 - 8*sqrt(2)*a*b^2*c + 16*sqrt(2)*a^2*c^2) - 3/2*(-b^2*d^2 + 4*a*c*d^2)^
(3/4)*c^2*d*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^4 - 8*sqrt(2)*a*b^2*c + 16*sqrt(2)*a^2*c^2) + 3/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*d*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^4 - 8*sqrt(2)*a*b^2*c + 16*sqrt(2)*a^2*c^2) - 2*((2*c*d*x + b*d)^(3/2)*b^2*c^2*d^5 - 4*(2*c*d*x + b*d)^(3/2
)*a*c^3*d^5 + 3*(2*c*d*x + b*d)^(7/2)*c^2*d^3)/((b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)^2*(b^2 - 4*a*c))

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.41 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx=\frac {3\,c^2\,d^{5/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{5/4}}-\frac {3\,c^2\,d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{5/4}}-\frac {2\,c^2\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}-\frac {6\,c^2\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{7/2}}{4\,a\,c-b^2}}{{\left (b\,d+2\,c\,d\,x\right )}^4-{\left (b\,d+2\,c\,d\,x\right )}^2\,\left (2\,b^2\,d^2-8\,a\,c\,d^2\right )+b^4\,d^4+16\,a^2\,c^2\,d^4-8\,a\,b^2\,c\,d^4} \]

[In]

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

[Out]

(3*c^2*d^(5/2)*atanh(((b*d + 2*c*d*x)^(1/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/(d^(1/2)*(b^2 - 4*a*c)^(9/4))))/(b
^2 - 4*a*c)^(5/4) - (3*c^2*d^(5/2)*atan(((b*d + 2*c*d*x)^(1/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/(d^(1/2)*(b^2 -
4*a*c)^(9/4))))/(b^2 - 4*a*c)^(5/4) - (2*c^2*d^5*(b*d + 2*c*d*x)^(3/2) - (6*c^2*d^3*(b*d + 2*c*d*x)^(7/2))/(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)