\(\int \frac {(b d+2 c d x)^{5/2}}{(a+b x+c x^2)^2} \, dx\) [1300]

Optimal result

Integrand size = 26, antiderivative size = 131 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {d (b d+2 c d x)^{3/2}}{a+b x+c x^2}+\frac {6 c d^{5/2} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac {6 c d^{5/2} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\sqrt [4]{b^2-4 a c}} \]

[Out]

-d*(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)+6*c*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)^(1/4)-6*c*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)^(1/4)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {700, 708, 335, 304, 209, 212} \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {6 c d^{5/2} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac {6 c d^{5/2} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac {d (b d+2 c d x)^{3/2}}{a+b x+c x^2} \]

[In]

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

[Out]

-((d*(b*d + 2*c*d*x)^(3/2))/(a + b*x + c*x^2)) + (6*c*d^(5/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*
Sqrt[d])])/(b^2 - 4*a*c)^(1/4) - (6*c*d^(5/2)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2
- 4*a*c)^(1/4)

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.66 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.76 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=(1+i) c (d (b+2 c x))^{5/2} \left (-\frac {\frac {1}{2}-\frac {i}{2}}{c (b+2 c x) (a+x (b+c x))}-\frac {3 \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c} (b+2 c x)^{5/2}}+\frac {3 \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c} (b+2 c x)^{5/2}}-\frac {3 \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{\sqrt [4]{b^2-4 a c} (b+2 c x)^{5/2}}\right ) \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(276\) vs. \(2(109)=218\).

Time = 2.91 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.11

method	result	size
derivativedivides	\(16 c \,d^{3} \left (-\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{16 \left (a c \,d^{2}-\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}\right )}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\)	\(277\)
default	\(16 c \,d^{3} \left (-\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{16 \left (a c \,d^{2}-\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}\right )}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\)	\(277\)
pseudoelliptic	\(\frac {\left (-2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \left (2 c x +b \right )+3 \sqrt {2}\, c d \left (c \,x^{2}+b x +a \right ) \left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+\ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right )-2 \arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right )\right ) d^{2}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (2 c \,x^{2}+2 b x +2 a \right )}\)	\(308\)

[In]

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

[Out]

16*c*d^3*(-1/16*(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)+3/32/(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 = 0.27 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.87 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {3 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (27 \, \sqrt {2 \, c d x + b d} c^{3} d^{7} + 27 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {3}{4}} {\left (b^{2} - 4 \, a c\right )}\right ) - 3 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (27 \, \sqrt {2 \, c d x + b d} c^{3} d^{7} - 27 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {3}{4}} {\left (b^{2} - 4 \, a c\right )}\right ) - 3 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} {\left (i \, c x^{2} + i \, b x + i \, a\right )} \log \left (27 \, \sqrt {2 \, c d x + b d} c^{3} d^{7} + 27 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {3}{4}} {\left (i \, b^{2} - 4 i \, a c\right )}\right ) - 3 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} {\left (-i \, c x^{2} - i \, b x - i \, a\right )} \log \left (27 \, \sqrt {2 \, c d x + b d} c^{3} d^{7} + 27 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {3}{4}} {\left (-i \, b^{2} + 4 i \, a c\right )}\right ) + {\left (2 \, c d^{2} x + b d^{2}\right )} \sqrt {2 \, c d x + b d}}{c x^{2} + b x + a} \]

[In]

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

[Out]

-(3*(c^4*d^10/(b^2 - 4*a*c))^(1/4)*(c*x^2 + b*x + a)*log(27*sqrt(2*c*d*x + b*d)*c^3*d^7 + 27*(c^4*d^10/(b^2 -
4*a*c))^(3/4)*(b^2 - 4*a*c)) - 3*(c^4*d^10/(b^2 - 4*a*c))^(1/4)*(c*x^2 + b*x + a)*log(27*sqrt(2*c*d*x + b*d)*c
^3*d^7 - 27*(c^4*d^10/(b^2 - 4*a*c))^(3/4)*(b^2 - 4*a*c)) - 3*(c^4*d^10/(b^2 - 4*a*c))^(1/4)*(I*c*x^2 + I*b*x
+ I*a)*log(27*sqrt(2*c*d*x + b*d)*c^3*d^7 + 27*(c^4*d^10/(b^2 - 4*a*c))^(3/4)*(I*b^2 - 4*I*a*c)) - 3*(c^4*d^10
/(b^2 - 4*a*c))^(1/4)*(-I*c*x^2 - I*b*x - I*a)*log(27*sqrt(2*c*d*x + b*d)*c^3*d^7 + 27*(c^4*d^10/(b^2 - 4*a*c)
)^(3/4)*(-I*b^2 + 4*I*a*c)) + (2*c*d^2*x + b*d^2)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

integrate((2*c*d*x+b*d)^(5/2)/(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. 439 vs. \(2 (109) = 218\).

Time = 0.30 (sec) , antiderivative size = 439, normalized size of antiderivative = 3.35 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {4 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c d^{3}}{b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}} - \frac {3 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{2} - 4 \, a c} - \frac {3 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{2} - 4 \, a c} + \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{2} - 4 \, \sqrt {2} a c} - \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{2} - 4 \, \sqrt {2} a c} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx=\frac {6\,c\,d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{1/4}}-\frac {4\,c\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{{\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2}-\frac {6\,c\,d^{5/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{1/4}} \]

[In]

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

[Out]

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